TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.3

Question 1.
Find the sum of the following APs :

i) 2, 7, 12, ………., to 10 terms.
Solution:
Given A.P. : 2, 7, 12, ……… to 10 term
a = 2; d = a2 – a1 = 7 – 2 = 5; n = 10
sn = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s10 = – \(\frac{10}{2}\)[2 × 2 +(10 – 1)5]
= 5[4 + 9 × 5]
= 5 [4 + 45]
= 5 × 49 = 245

ii) – 37, – 33, – 29, …… to 12 terms.
Solution:
Given A.P :
– 37, – 33, – 29, …., to 12 terms
a = – 37; d = a2 – a1 = (- 33) – (- 37)
= – 33 + 37 = 4 and n = 12
Sn = \(\frac{n}{2}\)[2a+ (n – 1)d]
s12 = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6[-74 + 11 × 4]
= 6[-74 + 44] = 6 × (- 30) = -180

iii) 0.6, 1.7, 2.8, …, to 100 terms.
Solution:
Given A.P : 0.6, 1.7, 2.8, …., S100
a = 0.6; d = a2 – a1 = 1.7 – 0.6 = 1.1 and n = 100
s = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s100 = \(\frac{100}{2}\)[2 × 0.6 + (100 – 1) × 1.1]
= 50[1.2 + 99 × 1.1]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505.

iv) \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\) ….. to 11 terms.
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 1

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 2.
Find the sums given below :

i) 7 + 10\(\frac{1}{2}\) + 14 + ….. + 84
Solution:
Given A.P. : 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
∴ a = 7, d = a2 – a1 = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\)
and the last term l = an
But, an = a +(n – 1) d
∴ 84 = 7 + (n – 1) × 3\(\frac{1}{2}\)
⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\)
⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22
⇒ n = 22 + 1 = 23
Now, Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 7; l = 84
S23 = \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = \(\frac{2093}{2}\) = 1046\(\frac{1}{2}\)

ii) 34 + 32 + 30 + ……. + 10
Solution:
Given A.P : 34 + 32 + 30 + …….. + 10
Here a = 34; d = a2 – a1 = 32 – 34 = -2
and the last term ‘l‘ = an = 10
But, an = a + (n – 1) d
⇒ 10 = 34 + (n – 1)(-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) ∴ n = 13
Also, Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 34, l = 10
S13 = \(\frac{13}{2}\)(34 + 10)
\(\frac{13}{2}\) × 44 = 13 × 22 = 286

iii) -5 + (-8) + (-11) + …. + (-230).
Solution:
Given AP:
-5 + (-8) + (-11) + …….. + (-230)
Here first term, a = -5;
Last term, l = an = -230:
d = a2 – a1 = (-8) – (-5)
= -8 + 5 = -3
But, an = a + (n – 1) d
∴ (-230) = (-5) + (n – 1) × (-3)
⇒ -230 + 5 = -3n + 3
⇒ -3n + 3 = -225
⇒ -3n = -225 – 3 ⇒ 3n = 228
∴ n = \(\frac{228}{3}\) = 76
Now Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S76 = \(\frac{76}{2}\)[(-5) + (-230)]
= 38 × (-235) = -8930

Question 3.
In an AP:

i) Given a = 5, d = 3, an = 50, find n and Sn.
Solution:
Given:
a = 5; d = 3 ; an = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1)3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S16 = \(\frac{16}{2}\)[5 + 50] = 8 × 55 = 440.

ii) Given a = 7, a13 = 35, find d and S13.
Solution:
Given : a = 7
a13 = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ d = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S13 = \(\frac{13}{2}\)[7 + 35]
= \(\frac{13}{2}\) × 42 = 13 × 21 = 273

iii) Given a12 = 37, d = 3, find a and S12.
Solution:
Given : a12 = a + 11d = 37
d = 3
So, a12 = a + 11 × 3 = 37
a + 33 = 37
a = 37 – 33 = 4
Now, Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S12 = \(\frac{12}{2}\)[4 + 37] = 6 × 41 = 246.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

iv) Given a3 = 15, S10 = 125, find d and a10.
Solution:
Given : a3 = a + 2d = 15
⇒ a = 15 – 2d —- (1)
S10 = 125 but take S10 as 175
i.e., S10 = 175
We know that,
Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
⇒ S10 = \(\frac{10}{2}\)[a + l]
⇒ 175 = \(\frac{10}{2}\)[2a + 9d]
⇒ 175 = 5[2a + 9d]
⇒ 35 = 2(15 – 2d) + 9d [∵ a = 15 – 2d]
⇒ 35 = 30 – 4d + 9d
⇒ 35 – 30 = 5d ⇒ d = \(\frac{5}{5}\) = 1
Substituting d = 1 in equation (1) we get
a = 15 – 2 × 1 = 15 – 2 = 13
Now, an = a + (n – 1) d
a10 = a + 9d = 13 + 9 × 1
= 13 + 9 = 22
∴ a10 = 22 ; d = 1.

v) Given a = 2, d = 8, Sn = 0, find n and an.
Solution:
Given : a = 2; d = 8 and Sn = 90
But Sn = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1) d]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[2 × 2 + (n – 1) × 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[4 + 8n – 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[8n – 4]
⇒ 90 = \(\frac{4 n}{2}\)[2n – 1]
⇒ 90 = 2n[2n – 1]
⇒ 4n2 – 2n – 90 = 0
⇒ 2(2n2 – n – 45) = 0
⇒ 2n2 – n – 45 = 0
⇒ 2n2 – 10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (n – 5) (2n + 9) = 0
⇒ n – 5 = 0 (or) 2n + 9 = 0
⇒ n = 5 (or) n = \(\frac{-9}{2}\) (discarded)
∴ n = 5
Now an = a5 = a + 4d = 2 + 4 × 8
= 2 + 32 = 34.

vi) Given an = 4, d = 2, Sn = -14, find n and a.
Solution:
Given an = a + (n – 1) d = 4 —– (1)
d = 2; Sn = -14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2 ⇒ a = 6 – 2n
Sn = \(\frac{\mathrm{n}}{2}\) [a + an]
-14 = \(\frac{\mathrm{n}}{2}\) [(6 – 2n) + 4] [∵ a = 6 – 2n]
-14 × 2 = n(10 – 2n)
⇒ 10n – 2n2 = -28
⇒ 2n2 – 10n – 28 = 0
⇒ n2 – 5n – 14 = 0
⇒ n2 – 7n + 2n – 14 = 0
⇒ n(n – 7) + 2(n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 (or) n = – 2
∴ n = 7
Now a = 6 – 2n = 6 – 2 × 7
= 6 – 14 = -8
∴ a = – 8; n = 7.

vii) Given l = 28, S = 144 and there are total 9 terms. Find a.
Solution:
Given :
l = a9 = a + 8d = 28 and S9 = 144
But, Sn = \(\frac{9}{2}\)(a + l)
144 = \(\frac{9}{2}\)[a + 28]
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 32 – 28 = 4.

Question 4.
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?
Solution:
Given A.P in which a = 17
Last term = l = 350
Common difference, d = 9
We know that, an = a + (n – 1) d
350 = 17 + (n – 1)9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, Sn = \(\frac{n}{2}\)(a + l)
S38 = \(\frac{38}{2}\)[17 + 350]
= 19 × 367 = 6973
∴ n = 38
Sn = 6973.

Question 5.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given A.P in which
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 2
Substituting d = 4 in equation (1), we get a + 4 = 14
⇒ a = 14 – 4 = 10
Now, Sn = \(\frac{n}{2}\)[a + l] (or) \(\frac{n}{2}\) [2a (n – 1) d]
∴ S51 = \(\frac{51}{2}\)[2 × 10 + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 50 × 4]
= \(\frac{51}{2}\) × (20 + 200)
= \(\frac{51}{2}\) × 220
= 51 × 110 = 5610.

Question 6.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. (T.S. & A.P. Mar. 15, 16)
Solution:
Given :
A.P such that S7 = 49
S17 = 289
We know that,
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
S7 = 49 = \(\frac{7}{2}\)[2a + (7 – 1) d]
⇒ \(\frac{49}{7}\) = \(\frac{1}{2}\)[2a + 6d]
⇒ 7 = a + 3d —– (1)
Also, S17 = 289 = \(\frac{17}{2}\) [2a + (17 – 1)d]
\(\frac{289}{17}\) = \(\frac{1}{2}\)(2a + 16d)
⇒ 17 = a + 8d —— (2)
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 3
⇒ d = \(\frac{10}{5}\)
Substituting d = 2 in equation (1), we get
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 2
Now, Sn = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
Sn = \(\frac{\mathrm{n}}{2}\) [2 × 1 + (n -1)2]
= \(\frac{\mathrm{n}}{2}\) [2 + 2n – 2)d] = \(\frac{\mathrm{n} .2 \mathrm{n}}{2}\)
∴ Sum of first n terms Sn = n2.
Shortcut : S7 = 49 = 72
S17 = 289 = 172
∴ Sn = n2

Question 7.
Show that a1, a2, ……. an, …. form an AP where an is defined as below :
i) an = 3 + 4n
ii) an = 9 – 5n. Also find the sum of the first 15 terms in each case.
Solution:
i) Given an = 3 + 4n
Then a1 = 3 + 4 × 1 = 3 + 4 = 7
a2 = 3 + 4 × 2 = 3 + 8 = 11
a3 = 3 + 4 × 3 = 3 + 12 = 15
a4 = 3 + 4 × 4 = 3 + 16 = 19
Now the pattern is 7, 11, 15, ……..
where a = a1 = 7; a2 = 11; a3 = 15, …. and a2 – a1 = 11 – 7 = 4;
a3 – a2 = 15 -11 = 4; Here d = 4. Hence a1, a2, …., an … forms an A.P.
s15 = \(\frac{15}{2}\) [2 × 7 + (15 – 1)4]
= \(\frac{15}{2}\) [14 + 14.4] = \(\frac{15}{2}\) [70]
= 15 × 35 = 525.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

ii) an = 9 – 5n
Solution:
Given : an = 9 – 5n
a1 = 9 – 5 × 1 = 9 – 5 = 4
a2 = 9 – 5 × 2 = 9 – 10 = -1
a3 = 9 – 5 × 3 = 9 – 15 = -6
a4 = 9 – 5 × 4 = 9 – 20 = -11
………………………………..
Also a2 – a1 = -1 – 4 = – 5
a3 – a2 = – 6 – (-1) = – 6 + 1 = -5
a4 – a3 = – 11 – (- 6)
= -11 + 6
= -5
∴ d = a2 – a1 = a3 – a2
= a4 – a3 = ……. = -5
Thus the difference between any two succes-sive terms is constant.
Hence {an} forms A.P.
S15 = \(\frac{15}{2}\)[2 × 4 + (15 – 1) × (-5)]
= \(\frac{15}{2}\) [8 + 14 (-5)] = \(\frac{15}{2}\) × -62
= 15 × – 31 = – 465.

Question 8.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (remember the first term is S1) ? What is the sum of first two terms ? What is the second term ? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given an A.P in which S<sub?n = 4n – n2
Taking n = 1 we get
S1 = 4 × 1 – 12 = 4 – 1 = 3
n = 2; S2 = a1 + a2 = 4 × 2 – 22
= 8 – 4 = 4
n = 3; S3 = a1 + a2 + a3
= 4 × 3 – 32 = 12 – 9 = 3
n = 4; S4 = a1 + a2 + a3 + a4
= 4 × 4 – 42 = 16 – 16 = 0
Hence, S1 = a1 = 3
∴ a2 = S2 – S1 = 4 – 3 = 1
a3 = S3 – S2 = 3 – 4 = -1
a4 = S4 – S3 = 0 – 3 = -3
So, d = a2 – a1 = 1 – 3 = -2
Now, a10 = a + 9d [∴ an = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
an = 3 + (n – 1) × (-2)
= 3 – 2n + 2 = 5 – 2n.

Question 9.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The given numbers are the first 40 positive multiples of 6
⇒ 6 × 1, 6 × 2, 6 × 3, ……., 6 × 40
⇒ 6, 12, 18, …… 240
Sn = \(\frac{\mathrm{n}}{2}\) [a + l]
= \(\frac{40}{2}\) [6 + 240]
= 20 × 246 = 4920
∴ S40 = 4920.

Question 10.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Given :
Total l Sum of all cash prizes = ₹ 700
Each prize differs by ₹ 20
Let the prizes (in ascending order) be
x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120
∴ Sum of the prizes = S7 = \(\frac{\mathrm{n}}{2}\) [a + l]
⇒ 700 = \(\frac{7}{2}\) [x + x + 120]
⇒ 700 × \(\frac{2}{7}\) = 2x + 120
⇒ 100 = x + 60
⇒ x = 100 – 60 = 40
∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Question 11.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g : a section of Class I will plant 1 tree, a section of Class 11 will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Given : Classes : From I to XII
Section : 3 in each class.
∴ Trees planted by each class = 3 × class number
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 4
∴ Total trees planted = 3 + 6 + 9 + 12 + ………+ 36; n = 12
∴ Sn = \(\frac{\mathrm{n}}{2}\) [a + l]
S12 = \(\frac{12}{2}\) [3 + 36]
= 6 × 39 = 234
∴ Total plants = 234

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 12.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles ? (Take π = \(\frac{22}{7}\))
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 6
[Hint: Length of successive semi-circles is l1, l2, l3, l4,……… with centres at A, B, A, B,…., respectively.]
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 5
Given : l1; l2, l3, l4, …….l13 are the semi-circles
with centres alternately at A and B; with radii
r1 = 0.5 cm [1 × 0.5]
r2 = 1.0 cm [2 × 0.5]
r3 = 1.5 cm [3 × 0.5]
r4 = 2.0 cm [4 × 0.5]
………………………..
r13 = 13 × 0.5 = 6.5
[∴ Radii are in A.P. as a1 = 0.5 and d = 0.5]
Now, the total length of the spiral = l1 + l2 +….. + l13 [… 13 given]
But circumference of a semi-circle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ……… + π × 6.5
= π × \(\frac{1}{2}\)[1 + 2 + 3 + ……. + 13]
Sum of the first n – natural numbers is \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)]

Question 13.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row ?
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 7
Solution:
Given : Total logs = 200
Number of logs stacked in the first row = 20
Number of logs stacked in the second row = 19
Number of logs stacked in the third row = 18
………………………………
The number series is 20, 19, 18, …. is an A.P.
where a = 20 and d = a2 – a1 = 19 – 20 = -1
Also, Sn = 200
∴ Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
200 = \(\frac{n}{2}\) [2 × 20 + (n – 1) × (-1)]
200 = \(\frac{n}{2}\) [40 – n + 1]
200 = \(\frac{n}{2}\) [41 – n]
400 = 41n – n2
⇒ n2 – 41n + 400 = 0
⇒ n2 – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n = 25 (or) 16
There can’t be 25 rows as we are starting with 20 logs in the first row.
∴ Number of rows must be 16.
∴ n = 16 and t16 = a + (n – 1) d = 20 + (16 – 1) (-1) = 20 – 15 = 5

Question 14.
In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 8
A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run ?
[Hint : To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Given : Balls are placed at an equal distance of 3 m from one another.
Distance of first ball from the bucket = 5 m
Distance of second ball from the bucket = 5 + 3 = 8m (5 + 1 × 3)
Distance of third ball from the bucket = 8 + 3 = 11 m (5 + 2 × 3)
Distance of fourth ball from the bucket = 11 + 3 = 14m (5 + 3 × 3)
…………………………………………………..
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32m.
1st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2nd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3rd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
……………………………………………………
10th ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 32 = 64 m
Total distance = 10 m + 16 m + 22 m +………….+ 64 m.
Clearly, this is an A.P in which a = 10; d = a2 – a1 = 16 – 10 = 6 and n = 10.
Sn = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
= \(\frac{10}{2}\) [2 × 10 + (10 – 1) × 6]
= 5[20 + 54] = 5 × 74 = 370 m
∴ Total distance = 370 m.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 1.
Find the condition for the points (a, 0), (h, k), and (0, b), where ab ≠ 0, to be collinear. [Mar. ’10]
Solution:
Let A(a, 0), B(h, k), C (0, b) be the given points.
Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{h}-\mathrm{a}}\)
Slope of \(\overline{\mathrm{BC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{b}-\mathrm{k}}{0-\mathrm{h}}=\frac{\mathrm{b}-\mathrm{k}}{-\mathrm{h}}\)
Since points A, B, C are collinear, then
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{BC}}\)
\(\frac{k}{h-a}=\frac{b-k}{-h}\)
⇒ -hk = (h – a) (b – k)
⇒ -hk = bh – hk – ab + ak
⇒ bh + ak = ab
⇒ \(\frac{b h}{a b}+\frac{a k}{a b}=1\)
\(\frac{h}{a}+\frac{k}{b}=1\), which is the required condition.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. [Mar. ’18 (AP & TS); May ’12; B.P.]
Solution:
Let A = (2, 5), B = (x, 3) are the given points.
Given, Slope of \(\overline{\mathrm{AB}}\) = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q2

Question 3.
Find the value of y, if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4) and (0, 6). [Mar. ’17 (TS), ’14, ’08]
Solution:
Let A = (3, y), B = (2, 7), C = (-1, 4), and D = (0, 6) are the given points.
Given,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q3
Since, the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are parallel then the
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{CD}}\)
⇒ y – 7 = 2
⇒ y = 9

Question 4.
Find the equation of the straight line which make 150° with the X-axis in the positive direction and which pass through the point (-2, -1). [May ’04]
Solution:
Given that inclination of a straight line is θ = 150°
The slope of a line is, m = tan θ
= tan (150°)
= tan (90° + 60°)
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
Let the given point A(x1, y1) is (-2, -1).
∴ The equation of the straight line passing through A(-2, -1) and having slope \(\frac{-1}{\sqrt{3}}\) is y – y1 = m(x – x1)
⇒ y + 1 = latex]\frac{-1}{\sqrt{3}}[/latex] (x + 2)
⇒ √3(y + 1) = -1(x + 2)
⇒ √3y + √3 = -x – 2
⇒ x + √3y + √3 = -2
⇒ x + √3y + √3 + 2 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 5.
Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. [May ’05]
Solution:
Let l1, l2 are the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes
Case I: Inclination of a straight line l1 is θ = 45°
Slope of a line l1 is, m = tan θ = tan 45° = 1
Let the given point O = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q5
∴ The equation of a straight line l1 passing through O(0, 0) and having slope ‘1’ is y – y1 = m(x – x1)
y – 0 = 1(x – 0)
⇒ y = x
⇒ x – y = 0
Case II: Inclination of a straight line l2 is θ = 135°
The slope of a line l2 is, m = tan θ
= tan 135°
= tan (90° + 45°)
= -tan 45°
= -1
Let the given point O = (0, 0)
∴ The equation of a straight line l2 passing through O(0, 0) and having slope ‘-1’ is y – y1 = m(x – x1)
⇒ y – 0 = -1(x – 0)
⇒ y = -x
⇒ x + y = 0
∴ Required equations of the straight lines are x – y = 0; x + y = 0

Question 6.
Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. [Mar. ’15 (TS) ’13 (Old), ’07, ’00; May ’10, ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
Given that, the straight line making equal intercepts on the co-ordinate axis, then a = b
From (1),
\(\frac{x}{a}+\frac{y}{a}=1\)
x + y = a ……….(2)
Since equation (2) passes through the point (-4, 5) then,
-4 + 5 = a
∴ a = 1
Substitute the value of ’a’ in equation (2)
∴ x + y = 1

Question 7.
Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero. [Mar. ’15 (AP), ’13; May ’15 (TS), ’02]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the straight line-making intercepts whose sum is ‘0’
i.e., a + b = 0
b = -a
From (1)
\(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = -a …….(2)
Since equation (2) passes through the point (-2, 4) then,
-2 – 4 = a
∴ a = -6
Substitute the value of ‘a’ in equation (2)
x – y = -6
x – y + 6 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 8.
Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3. [Mar. ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the ratio of intercepts = 2 : 3
X-intercept = 2a
Y-intercept = 3a
From (1),
\(\frac{x}{2 a}+\frac{y}{3 a}=1\)
\(\frac{3 x+2 y}{6 a}=1\)
3x + 2y = 6a ……..(2)
Since equation (2) passes through point (3, -4) then,
3(3) + 2(-4)= 6a
⇒ 9 – 8 = 6a
⇒ 6a = 1
⇒ a = \(\frac{1}{6}\)
Substitute the value of ‘a’ in equation (2)
3x + 2y = 6(\(\frac{1}{6}\))
∴ 3x + 2y = 1

Question 9.
Find the equation of the straight line passing through the points \(\left(a t_1^2, 2 at_1\right)\) and \(\left(a t_2^2, 2 at_2\right)\). [Mar. ’14. ’04; May ’15 (AP), ’00]
Solution:
Let A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) are the given points
The equation of the straight line passing through the points A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) is
(y – y1) (x2 – x1) = (x – x1) (y2 – y1)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q9

Question 10.
Find the equation of the straight line passing through A(-1, 3) and (i) parallel (ii) perpendicular to the straight line passing through B(2, -5) and C(4, 6). [May ’12; Mar. ’11]
Solution:
A(-1, 3), B(2, -5), C(4, 6) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q10
(i) Slope of the parallel line = m = \(\frac{11}{2}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{11}{2}\) is y – y1 = m(x – x1)
y – 3 = \(\frac{11}{2}\)(x + 1)
⇒ 2y – 6 = 11x + 11
⇒ 11x – 2y + 17 = 0
(ii) Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{11 / 2}=\frac{-2}{11}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{-2}{11}\) is y – y1 = \(\frac{-1}{m}\) (x – x1)
y – 3 = \(\frac{-2}{11}\) (x + 1)
⇒ 11y – 33 = -2x – 2
⇒ 2x + 11y – 31 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 11.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equation of the altitude through B. [May ’13 (Old)]
Solution:
Slope of \(\overline{\mathrm{AC}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q11
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is
y – y1 = m(x – x1)
y – 9 = \(\frac{-12}{5}\) (x + 4)
5y – 45 = -12x – 48
12x + 5y + 3 = 0

Question 12.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight line. [May ’90]
Solution:
Let a, b be the intercepts of a line.
∴ The line cuts the X-axis at A(a, 0), Y-axis at B(0, b)
Midpoint of \(\overline{\mathrm{AB}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12.1

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X-axis measured in the counterclockwise direction. [May ’94]
Solution:
Given, the equation of the straight line is y = -√3x + 3
Comparing this equation with y = mx + c, We get
m = -√3 (∵ m = tan θ)
⇒ tan θ = -√3
⇒ tan θ = tan \(\frac{2 \pi}{3}\)
⇒ θ = \(\frac{2 \pi}{3}\)
∴ The angle made by the straight line is θ = \(\frac{2 \pi}{3}\)

Question 14.
Transform the equation √3x + y = 4 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’16 (TS)]
Solution:
Given, the equation of the straight line is √3x + y = 4
(a) Slope-intercept form:
√3x + y = 4
y = -√3x + 4 which is of the form y = mx + c
where, slope (m) = -√3, y-intercept (c) = 4
(b) Intercept form:
Given, the equation of the straight line is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14
which is in the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = \(\frac{4}{\sqrt{3}}\), y-intercept (b) = 4
(c) Normal form:
Given the equation of the straight line is √3x + y = 4
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.2
which is in the form x cos α + y sin α = p
∴ α = \(\frac{\pi}{6}\), p = 2

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 15.
Transform the equation x + y + 1 = 0 into normal form. [Mar. ’17 (AP), ’08; May ’10; B.P.; Mar. ’18 (TS)]
Solution:
Given, equation of the straight line is x + y + 1 = 0
x + y = -1
– x – y = 1
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q15

Question 16.
Transform the equation 4x – 3y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(a) Slope-intercept form:
Given equation of the line is 4x – 3y + 12 = 0
3y = 4x + 12
y = \(\frac{4 x+12}{3}=\left(\frac{4}{3}\right) x+4\)
which is in the form of y = mx + c
∴ Slope = \(\frac{4}{3}\), y-intercept = 4
(b) Intercept form:
Given equation is 4x – 3y + 12 = 0
\(\frac{x}{-3}+\frac{y}{4}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept = -3, y-intercept = 4
(c) Normal form:
Given the equation of the straight line is 4x – 3y = -12
-4x + 3y = 12
On dividing both sides by \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q16

Question 17.
Transform the equation x + y – 2 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [Mar. ’12]
Solution:
(a) Slope-intercept form:
Given equation of the straight line is, x + y – 2 = 0
y = -x + 2
which is in the form of y = mx + c
∴ Slope, m = -1, y-intercept, c = 2
(b) Intercept form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
\(\frac{x+y}{2}\) = 1
\(\frac{x}{2}+\frac{y}{2}\) = 1
which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = 2, y-intercept (b) = 2
(c) Normal form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17.1

Question 18.
Transform the equation √3x + y + 10 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’04]
Solution:
(a) Slope-intercept form:
Given the equation of the straight line is, √3x + y + 10 = 0
y = -√3x – 10 = -√3x + (-10)
which is in the form of y = mx + c
∴ Slope, m = -√3, y-intercept, c = -10
(b) Intercept form:
Given the equation of the straight line is √3x + y + 10 = 0
√3x + y = -10 × 1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18
(c) Normal form:
Given equation of the straight line is, √3x + y + 10 = 0
√3x + y = -10
-√3x – y = 10
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.2

Question 19.
If the area of the triangle is formed by the straight lines, x = 0, y = 0, and 3x + 4y = a [a > 0] is ‘6’. Find the value of ‘a’. [May ’11, Mar. ’09, ’07]
Solution:
Given equations of the straight lines are (a > 0) 3x + 4y = a, x = 0 and y = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -a
The area of the triangle formed by this line and the co-ordinate axis is equal to \(\frac{c^2}{2|a b|}\)
Given that area of the triangle = 6
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q19

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 20.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent. [Mar. ’17 (TS), ’13; May ’15 (TS)]
Solution:
Given, the equation of the straight lines
x + p = 0 ……..(1)
y + 2 = 0 ………(2)
3x + 2y + 5 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q20
∴ Point of intersection of the lines (2) & (3) is (\(\frac{-1}{3}\), -2)
since given lines are concurrent, then, the point of intersection (\(\frac{-1}{3}\), -2) lies on (1)
x + p = 0
⇒ \(\frac{-1}{3}\) + p = 0
⇒ p = \(\frac{1}{3}\)

Question 21.
Find the ratio in which the straight line 2x + 3y = 5 divides the line joining the points (0, 0) and (-2, 1). [Mar. ’14]
Solution:
Given the equation of the straight line is L = 2x + 3y – 5 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = 5
Let the given points are A(x1, y1) = (0, 0) and B(x2, y2) = (-2, 1)
Required ratio = \(\frac{-\left(a x_1+b {y}_1+c\right)}{a x_2+b y_2+c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q21

Question 22.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0. [Mar. ’19 (AP); May ’13]
Solution:
Given, equations of the straight lines are 3x + 4y – 3 = 0, 6x + 8y – 1 = 0
6x + 8y – 6 = 0 …..(1)
6x + 8y – 1 = 0 …..(2)
Comparing (1) with ax + by + c1 = 0, we get
a = 6, b = 8, c1 = -6
Comparing (2) with ax + by + c2 = 0, we get
a = 6, b = 8, c2 = -1
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q22

Question 23.
Find the equation of ‘k’, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°. [Mar. ’12, ’08, ’82; May ’11, ’02]
Solution:
Given, the equations of the straight lines are
4x – y + 7 = 0 ……..(1)
kx – 5y – 9 = 0 ……..(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = 4, b1 = -1, c1 = 7
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = k, b2 = -5, c2 = -9
Given that, θ = 45°
If ‘θ’ is the angle between the given lines then,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23.1
Squaring on both sides
⇒ 17(k2 + 25) = 2(4k + 5)2
⇒ 17k2 + 425 = 2(16k2 + 25 + 40k)
⇒ 17k2 + 425 = 32k2 + 50 + 80k
⇒ 15k2 + 80k – 375 = 0
⇒ 3k2 + 16k – 375 = 0
⇒ 3k2 + 25k – 9k – 75 = 0
⇒ k(3k + 25) – 3(3k + 25) = 0
⇒ (3k + 25) (k – 3) = 0
⇒ 3k + 25 = 0; k – 3 = 0
⇒ 3k = -25; k = 3
⇒ k = 3 or \(\frac{-25}{3}\)

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 24.
Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and pass through the point (5, 4). [Mar. ’13, ’03]
Solution:
Given, the equation of the straight line is 2x + 3y + 7 = 0
Given points (5, 4)
The equation of the straight line parallel to 2x + 3y + 7 = 0 is 2x + 3y + k = 0
Since equation (1) passes through the point (5, 4) then,
2(5) + 3(4) + k = 0
⇒ 10 + 12 + k = 0
⇒ 22 + k = 0
⇒ k = -22
∴ The required equation of the straight line is 2x + 3y – 22 = 0

Question 25.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8k – 1)y – 6 = 0 are perpendicular. [Mar. ’10]
Solution:
Given, the equations of the straight lines are
y – 3kx + 4 = 0 ………(1)
(2k – 1)x – (8k – 1)y – 6 = 0 ……….(2)
Slope of the line (1) is m1 = \(\frac{-(-3 k)}{1}\) = 3k
Slope of the line (2) is m2 = \(\frac{-(2 k-1)}{-(8 k-1)}=\frac{(2 k-1)}{(8 k-1)}\)
Since the given lines are perpendicular then m1 × m2 = -1
\(3 \mathrm{k}\left(\frac{2 \mathrm{k}-1}{8 \mathrm{k}-1}\right)\) = -1
⇒ 3k(2k – 1) = -1(8k – 1)
⇒ 6k2 – 3k = -8k + 1
⇒ 6k2 + 5k – 1 = 0
⇒ 6k2 + 6k – k – 1 = 0
⇒ 6k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(6k – 1) = 0
⇒ k + 1 = 0 (or) 6k – 1 = 0
⇒ k = -1 (or) k = \(\frac{1}{6}\)

Question 26.
Find the perpendicular distance from the point (3, 4) to the straight line 3x – 4y + 10 = 0. [Mar. ’16 (AP); May. ’15 (AP)]
Solution:
Given the equation of the straight line is 3x – 4y + 10 = 0
Comparing with ax + by + c = 0, we get
a = 3; b = -4, c = 40
Let the given point p(x1, y1) = (3, 4)
The perpendicular distance from the point (3, 4) to the line 3x – 4y + 10 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q26

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 27.
Find the slopes of the lines x + y = 0 and x – y = 0. [Mar. ’17 (AP)]
Solution:
Slope of x + y = 0 is \(\frac{-a}{b}=\frac{-1}{1}\) = -1
Slope of x – y = 0 is \(\frac{-a}{b}=\frac{-1}{-1}\) = 1

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 1.
Transform the equation \(\frac{x}{a}+\frac{y}{b}=1\) into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). [May ’08, ’04, ’02, ’97, ’95, ’90; Mar. ’07, ’02, ’00]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q1

Question 2.
Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3, 2). [Mar. ’16 (AP); ’15 (AP); B.P.]
Solution:
Let the given point A(x1, y1) = (3, 2)
Given the equation of the straight line is 3x – 4y – 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q2
Distance |r| = 5
Required points = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 + r sin θ)
= (3 ± 5 . \(\frac{4}{5}\), 2 ± 5 . \(\frac{3}{5}\))
= (3 ± 4, 2 ± 3)
= (3 + 4, 2 + 3); (3 – 4, 2 – 3)
= (7, 5), (-1, 1)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 3.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent. [May ’07, ’95; Mar. ’05, ’80; Mar. ’18 (TS)]
Solution:
Given, the equations of the straight lines are,
2x – 3y + k = 0 ……..(1)
3x – 4y – 13 = 0 ……..(2)
8x – 11y – 33 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3.1
∴ The point of intersection (11, 5) lies on (1)
2x – 3y + k = 0
⇒ 2(11) – 3(5) + k = 0
⇒ 22 – 15 + k = 0
⇒ 7 + k = 0
⇒ k = -7

Question 4.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a3 + b3 + c3 = 3abc. [Mar. ’19 (AP); Mar. ’08; May. ’00]
Solution:
Given, the equations of the straight lines are,
ax + by + c = 0 ……(1)
bx + cy + a = 0 ………(2)
cx + ay + b = 0 ……..(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4
\(\frac{y}{b c-a^2}=\frac{1}{a c-b^2}\) ⇒ y = \(\frac{b c-a^2}{a c-b^2}\)
∴ Point of intersection of the straight lines (1) & (2) is \(\left(\frac{a b-c^2}{a c-b^2}, \frac{b c-a^2}{a c-b^2}\right)\)
Since the given lines are concurrent, then the point of intersection lies on line (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4.1
⇒ c(ab – c2) + a(bc – a2) + b(ac – b2) = 0
⇒ abc – c3 + abc – a3 + abc – b3 = 0
⇒ 3abc – a3 – b3 – c3 = 0
⇒ a3 + b3 + c3 = 3abc

Question 5.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{a}+\frac{y}{b}=1\) meets the coordinate axes at A and B. Show that the locus of the midpoint of \(\overline{\mathbf{A B}}\) is 2(a + b) xy = ab(x + y). [May ’05]
Solution:
Given equations of the straight lines are,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5
The point of intersection of lines (1) & (2) is
C = \(\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)\)
The equation of the straight line \(\overline{\mathbf{A B}}\) is the intercept from \(\frac{x}{p}+\frac{y}{q}=1\) ……(3)
The straight line (3) meets the X-axis at A(p, 0), Y-axis at B(0, q).
Let Q(x1, y1) be any point on the locus.
Since Q(x1, y1) is the midpoint of \(\overline{\mathbf{A B}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.1
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.2
then, \(y_1\left(\frac{a b}{a+b}\right)+\left(\frac{a b}{a+b}\right) x_1=2 x_1 y_1\)
\(\frac{a b y_1+a b x_1}{a+b}=2 x_1 y_1\)
ab(x1 + y1) = (a + b) 2x1y1
∴ The locus of the midpoint ‘Q’ of AB is 2(a + b) xy = ab(x + y)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 6.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when (p, q) bisects \(\overline{\mathbf{A B}}\). [May ’90]
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, C(p, q) bisects \(\overline{\mathbf{A B}}\), then C is the midpoint of \(\overline{\mathbf{A B}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6.1

Question 7.
A triangle of area 24 sq. units is formed by a straight line and the coordinate axes in the first quadrant, find the equation of the straight line if it passes through (3, 4). [May ’07]
Solution:
Equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q7
Since equation (1) passes through the point (3, 4) then,
\(\frac{3}{a}+\frac{4}{b}=1\)
\(\frac{3 b+4 a}{a b}\) = 1
3b + 4a = ab
4a = ab – 3b
4a = b(a – 3)
b = \(\frac{4 a}{a-3}\) ……(2)
Given that, area of ΔOAB = 24 sq.units
\(\frac{1}{2}\)ab = 24
ab = 48
\(a\left(\frac{4 a}{a-3}\right)=48\)
a2 = 12(a – 3)
a = 12a – 36
a2 – 12a + 36 = 0
(a – 6)2 = 0
a = 6
from (2), b = \(\frac{4(6)}{6-3}\) = 8
The equation of the straight line is, from (1)
\(\frac{x}{6}+\frac{y}{8}\) = 1
\(\frac{4 x+3 y}{24}\) = 1
4x + 3y = 24

Question 8.
If 3a + 2b + 4c = 0, then show that the equation ax + by + c = 0, represents a family of concurrent straight lines and find the point of concurrency. [May ’10]
Solution:
Given that,
3a + 2b + 4c = 0
4c = -3a – 2b
c = \(\frac{-3 a-2 b}{4}\)
Now, ax + by + c = 0
ax + by + \(\left(\frac{-3 a-2 b}{4}\right)\) = 0
\(\frac{4 a x+4 b y-3 a-2 b}{4}\) = 0
4ax + 4by – 3a – 2b = 0
a(4x – 3) + b(4y – 2) = 0
(4x – 3) + \(\frac{b}{a}\) (4y – 2) = 0
This is of the form L1 + λL2 = 0
Here, ax + by + c = 0, represents a set of lines passing through the point of intersection of the lines
L1 = 4x – 3 = 0 …….(1)
L2 = 4y – 2 = 0 ………(2)
Solving (1) & (2)
From (1), 4x – 3 = 0
4x = 3
x = \(\frac{3}{4}\)
from (2), 4y – 2 = 0
4y = 2
y = \(\frac{1}{2}\)
∴ The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)
∴ ax + by + c = 0 represents a set of concurrent lines.
The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 9.
Find the point on the straight line 3x + y + 4 = 0, which is equidistant from the points (-5, 6) and (3, 2). [Mar. ’13; Nov. ’98]
Solution:
Given, equation of the straight line is 3x + y + 4 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9
Let the given points are A(-5, 6) & B(3, 2)
Let P(x, y) be a point on the straight line 3x + y + 4 = 0,
Given that, PA = PB
\(\sqrt{(x+5)^2+(y-6)^2}=\sqrt{(x-3)^2+(y-2)^2}\)
Squaring on both sides
(x + 5)2 + (y – 6)2 = (x – 3)2 + (y – 2)2
x2 + 25 + 10x + y2 + 36 – 12y = x2 + 9 – 6x + y2 – 4y + 4
10x – 12y + 61 = -6x – 4y + 13
16x – 8y + 48 = 0
2x – y + 6 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9.1

Question 10.
A straight line through Q(√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line √3x – 4y + 8 = 0 at P, find the distance PQ. [Mar. ’19 (TS); Mar. ’04]
Solution:
Given equation of the straight line is √3x – 4y + 8 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10
Given point Q(x1, y1) = (√3, 2)
Inclination of a straight line, θ = \(\frac{\pi}{6}\) = 30°
Slope of a straight line, m = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope \(\frac{1}{\sqrt{3}}\) and passing through the
point Q(√3, 2) is, y – y1 = m(x – x1)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – √3)
√3y – 2√3 = x – √3
x – √3y + √3 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10.1

Question 11.
The line \(\frac{x}{a}-\frac{y}{b}=1\) meets the X-axis at P. Find the equation of the line perpendicular to the line at P. [May ’03]
Solution:
Given, the equation of the straight line is \(\frac{x}{a}-\frac{y}{b}=1\) …..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q11
Since line (1) meets the X-axis at P.
Then y-coordinate = 0
\(\frac{x}{a}-\frac{0}{b}=1\)
x = a
∴ The coordinates of P = (a, 0)
The slope of the line (1) is m = \(\frac{\frac{-1}{a}}{\frac{-1}{b}}=\frac{b}{a}\)
Slope of the perpendicular line = \(\frac{-1}{b/a}=\frac{-a}{b}\)
∴ The equation of the line passing through P(a, 0) and having slope \(\frac{-a}{b}\) is,
y – y1 = \(\frac{-1}{m}\) (x – x1)
y – 0 = \(\frac{-a}{b}\) (x – a)
y = \(\frac{-a}{b}\) (x – a)
by = -ax + a2
ax + by – a2 = 0
which is the required equation of a straight line.

Question 12.
Find the equation of the line perpendicular to the line 3x + 4y + 6 = 0 and make an intercept -4 on the X-axis. [Mar. ’10]
Solution:
Given, equation of the straight line is 3x + 4y + 6 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q12
Slope of the line (1) is m = \(\frac{-3}{4}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\left(\frac{-3}{4}\right)}=\frac{4}{3}\)
Given that, the required a straight line making an intercept -4 on X-axis. Then P = (-4, 0).
Equation of the straight line passing through P(-4, 0) and having slope \(\frac{4}{3}\) is
(y – y1) = \(\frac{-1}{m}\) (x – x1)
y – 0 = \(\frac{4}{3}\) (x + 4)
3y = 4x + 16
4x – 3y + 16 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 13.
Find the equation of the straight line making non-zero equal intercepts on the coordinate axes and passing through the point of intersection of the lines 2x – 5y + 1 = 0 and x – 3y – 4 = 0. [Mar. ’06, ’00]
Solution:
Given the equation of the lines are
2x – 5y + 1 = 0 …….(1)
x – 3y – 4 = 0 …….(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q13
\(\frac{x}{23}=\frac{y}{9}=\frac{1}{-1}\)
\(\frac{x}{23}\) = -1; \(\frac{y}{9}\) = -1
x = -23; y = -9
∴ Point of intersection of lines (1) & (2) is, P = (-23, -9)
The equation of the straight line in the intercept form is, \(\frac{x}{a}+\frac{y}{b}=1\) = 1 ……..(3)
The straight line (3) makes equal intercepts on the coordinate axes
from (3),
\(\frac{x}{a}+\frac{y}{a}\)
x + y = a ……..(4)
Since equation (4) passes through the point P(-23, -9) then,
-23 – 9 = a
a = -32
∴ The equation of the straight line is x + y = -32
x + y + 32 = 0

Question 14.
Find the length of the perpendicular drawn from the point of intersection of the lines 3x + 2y + 4 = 0 and 2x + 5y – 1 = 0 to the straight line 7x + 24y – 15 = 0. [May ’01; Mar. ’91]
Solution:
Given, the equation of the straight lines are
3x + 2y + 4 = 0 ……(1)
2x + 5y – 1 = 0 ……..(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14
∴ The point of intersection of lines (1) & (2) is, P = (-2, 1)
Given the equation of the straight line is 7x + 24y – 15 = 0
Comparing with ax + by + c = 0 then a = 7, b = 24, c = -15
Point of intersection P(x1, y1) = (-2, 1)
The perpendicular distance from P(-2, 1) to the straight line
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14.1

Question 15.
If θ is the angle between the lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{b}+\frac{y}{a}=1\), find the value of sin θ, where a > b. [May ’09]
Solution:
Given, the equation of the straight lines are
\(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{b x+a y}{a b}\) = 1
bx + ay = ab
bx + ay – ab = 0 ……..(1)
\(\frac{x}{b}+\frac{y}{a}=1\)
ax + by = ab
ax + by – ab = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = b, b1 = a, c1 = -ab
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = a, b2 = b, c2 = -ab
If ‘θ’ is the angle between lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q15

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 16.
Find the equations of the straight lines passing through (1, 3) and (i) parallel to (ii) perpendicular to the line passing through the points (3, -5) and (-6, 1). [May ’15 (AP)]
Solution:
The slope of the line passing through the points (3, -5) and (-6, 1) is
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1+5}{-6-3}=\frac{-6}{9}=\frac{-2}{3}\)
(i) Equation of the line passing through (1, 3) and parallel to the line passing through the points (3, -5) and (-6, 1) is y – y1 = m(x – x1)
⇒ y – 3 = \(\frac{-2}{3}\) (x – 1)
⇒ 3y – 9 = -2x + 2
⇒ 2x + 3y – 11 = 0
(ii) Equation of the line passing through (1, 3) and perpendicular to the line passing through the points (3, -5) and (-6, 1) is y – y1 = \(\frac{-1}{m}\) (x – x1)
⇒ y – 3 = \(\frac{1}{2}\) (x – 1)
⇒ 2y – 6 = 3x – 3
⇒ 3x – 2y + 3 = 0

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Students can practice TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 1.
Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.
Solution:
Number of distinct line segments that can be formed out of n-points = \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
Given : No. of line segments \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) = 10
⇒ n2 – n = 20
⇒ n2 – n – 20 = 0
⇒ n2 – 5n + 4n – 20 = 0
⇒ n(n- 5) + 4(n – 5) = 0
⇒ (n – 5) (n + 4) = 0
⇒ n – 5 = 0 (or) n + 4 = 0
⇒ n = 5 (or) – 4
∴ n = 5 [n – can’t be negative]

Question 2.
A two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number.
Solution:
Let the digit in the units place = x
Let the digit in the tens place = y
∴ The number = 10y + x
By interchanging the digits the number becomes 10x + y
By problem (10x + y) – (10y + x) = 18
⇒ 9x + 9y = 18
⇒ 9(x – y) = 18
⇒ x – y = \(\frac{18}{9}\) = 2
⇒ y = x – 2
(i.e.) digit in the tens place = x – 2
digit in the units place = x
Product of the digits = (x – 2) x
By problem x2 – 2x = 8
x2 – 2x – 8 = 0
⇒ x2 – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 (or) x + 2 = 0
⇒ x = 4 (or) x = -2
x = 4 [ ∵ x can’t be negative]
∴ The number is 24.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 3.
A piece of wire 8m in length, cut into two pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m2 ?
[Hint: x + y = 8, (\(\frac{x}{4}\))2 + (\(\frac{y}{4}\))2 = 2
⇒ (\(\frac{x}{4}\))2 + \(\left(\frac{8-x}{4}\right)\) = 2
Solution:
Let the length of the first piece = x m
Then length of the second piece = (8 – x) m
∴ Side of the 1st square = \(\frac{x}{4}\) m and
Side of the second square = \(\frac{8-x}{4}\) m
Sum of the areas = 2 m2
\(\left(\frac{x}{4}\right)^2\) + \(\left(\frac{8-x}{4}\right)^2\) = 2
⇒ x2 + 64 + x2 – 16x = 16 × 2 = 32
⇒ 2x2 – 16x + 64 = 32
⇒ 2x2 – 16x + 32 = 0
⇒ 2(x2 – 8x + 16) = 0
⇒ x2 – 4x – 4x + 16 = 0
⇒ x(x – 4) – 4(x – 4) = 0
⇒ (x – 4) (x – 4) = 0
∴ x = 4
The cut should be made at the centre making two equal pieces of length 4m, 4m.

Question 4.
Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself ?
Solution:
Let the time taken by Vinay to complete the job = x days
Then the time taken by Praveen to complete the job = x + 5 days
Both worked for 6 days to complete a job.
∴ Total work done by them is
\(\frac{6}{x}\) + \(\frac{6}{x+5}\) = 1
⇒ 6(2x + 5) = x2 + 5x
⇒ x2 – 7x – 30 = 0
⇒ x2 – 10x + 3x – 30 = 0
⇒ x(x – 10) + 3(x – 10) = 0
⇒ x – 10 = 0 (or) x + 3 = 0
⇒ x = 10(or) x = -3
x = 10 (∵ x can’t be negative)
∴ Time taken by Vinay = x = 10 days
Time taken by Praveen = x + 5 = 15 days.

Question 5.
Show that the sum of roots of a quadratic equation is \(\frac{-\mathbf{b}}{\mathbf{a}}\).
Solution:
Let the Q.E. = ax2 + bx + c = 0 (a ≠ 0)
⇒ ax2 + bx = -c
⇒ x2 + \(\frac{\mathrm{b}}{\mathrm{a}}\)x = \(\frac{-\mathrm{c}}{\mathrm{a}}\)
⇒ x2 + 2.\(x \cdot \frac{b}{2 a}\) = \(-\frac{c}{a}\)
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 6
sum of the roots
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 7
∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 6.
Show that the product of the roots of a Q.E is \(\frac{\mathbf{c}}{\mathrm{a}}\).
Solution:
Let the Q.E. = ax2 + bx + c = 0(a ≠ 0)
⇒ ax2 + bx = -c
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 8
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 9
Product of the roots
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 10

Question 7.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\), find the fraction.
Solution:
Let the numerator = x
then denominator = 2x + 1
Then the fraction = \(\frac{x}{2 x+1}\)
Its reciprocal = \(\frac{2 x+1}{x}\)
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 11
⇒ 105x2 + 84x + 21 = 116x2 + 58x
⇒ 11x2 – 26x – 21 = 0
⇒ 11x2 – 33x + 7x – 21 = 0
⇒ 11x (x – 3) + 7(x – 3) = 0
⇒ (x – 3)(11x + 7) = 0
⇒ x – 3 = 0 (or) 11x + 3 = 0
⇒ x = 3 (or) \(\frac{-3}{11}\)
∴x = 3
Numerator = 3;
Denominator = 2 × 3 + 1 = 7
Fraction = \(\frac{3}{7}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Exercise 7(c)

I. Find the values of the following integrals.

Question 1.
\(\int_0^{\frac{\pi}{2}} \sin ^{10} x d x\) (May ’06; Mar. ’03)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q1

Question 2.
\(\int_0^{\frac{\pi}{2}} \cos ^{11} x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 3.
\(\int_0^{\frac{\pi}{2}} \cos ^7 x \sin ^2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q3

Question 4.
\(\int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^4 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q4

Question 5.
\(\int_0^\pi \sin ^3 x \cos ^6 x d x\)
Solution:
We have \(\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\)
if f(2a – x) = f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q5

Question 6.
\(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\)
Solution:
Take f(x) = sin2x cos4x
Then f(π – x) = sin2(π – x) cos4(π – x)
= sin2x cos4x
= f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6.1

Question 7.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta \cos ^7 \theta d \theta\)
Solution:
f(θ) = sin2θ cos7θ
f(-θ) = sin2(-θ) cos7(-θ)
= sin2θ cos7θ
= f(θ); and f is even
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q7

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 8.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta\)
Solution:
Let f(θ) = sin3θ cos3θ
∴ f(-θ) = sin3(-θ) cos3(-θ)
= -sin3θ cos3θ
= -f(θ)
∴ f is an odd function.
Hence \(\int_{-a}^a f(x) d x=0\) when f is odd.
∴ \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta=0\)

Question 9.
\(\int_0^a x\left(a^2-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Take x = a sin θ then dx = a cos θ dθ
Upper limit when x = a is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9
Let cos θ = t then -sin θ dθ = dt
Upper limit when θ = \(\frac{\pi}{2}\) is t = 0
Lower limit when θ = 0 is t = 1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9.1

Question 10.
\(\int_0^2 x^{\frac{3}{2}} \sqrt{2-x} d x\)
Solution:
Take x = 2 sin2θ, then dx = 4 sin θ cos θ dθ
Upper limit when x = 2 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10.1

II. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^5(1-x)^{\frac{5}{2}} d x\)
Solution:
Let x = sin2θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 2.
\(\int_0^4\left(16-x^2\right)^{\frac{5}{2}} d x\)
Solution:
Let x = 4 sin θ then dx = 4 cos θ dθ
Upper limit when x = 4 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q2

Question 3.
\(\int_{-3}^3\left(9-x^2\right)^{\frac{3}{2}} x d x\)
Solution:
Let x = 3 sin θ then dx = 3 cos θ dθ
Upper limit when x = 3 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = -3 is θ = \(-\frac{\pi}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q3

Question 4.
\(\int_0^5 x^3\left(25-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Let x = 5 sin θ then
Upper limit when x = 5 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4.1

Question 5.
\(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x\)
Solution:
Take f(x) = sin8x cos7x
then f(-x) = sin8(-x) cos7(-x)
= sin8x cos7x
= f(x)
f is an even function of x.
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=2 \int_0^\pi \sin ^8 x \cos ^7 x d x\)
Now f(x) = sin8x cos7x
and f(π – x) = sin8(π – x) cos7(π – x)
= -sin8x cos7x
= -f(x)
Hence \(\int_0^\pi \sin ^8 x \cos ^7 x d x=0\)
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=0\)
[By the result that f = [0, 2a] → R is integrable on [0, a] and if f(2a – x) = -f(x) ∀ x ∈ [a, 2a] then \(\int_0^{2 a} f(x) d x=0\)]

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 6.
\(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)
Solution:
Let x = 3 cos2θ + 7 sin2θ then
dx = -6 cos θ sin θ + 14 sin θ cos θ = 8 cos θ sin θ
Upper limit when x = 7 is
7 = 3 cos2θ + 7 sin2θ
⇒ 7(1 – sin2θ) = 3 cos2θ
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
The lower limit when x = 3 is
3 = 3 cos2θ + 7 sin2θ
⇒ 3 sin2θ = 7 sin2θ
⇒ 4 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6.1

Question 7.
\(\int_2^6 \sqrt{(6-x)(x-2)} d x\)
Solution:
Put x = 2 cos2θ + 6 sin2θ
then dx = (-4 cos θ sin θ + 12 sin θ cos θ) dθ = 8 sin θ cos θ dθ
Upper limit when x = 6 is 6 = 2 cos2θ + 6 sin2θ
⇒ 6 cos2θ = 2 cos2θ
⇒ 4 cos2θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 2 is 2 = 2 cos2θ + 6 sin2θ
⇒ 2 sin2θ = 6 sin2θ
⇒ 4 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7.1

Question 8.
\(\int_0^{\frac{\pi}{2}} \tan ^5 x \cos ^8 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q8

III. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^{7 / 2}(1-x)^{5 / 2} d x\)
Solution:
Let x = sin2θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 2.
\(\int_0^\pi(1+\cos x)^3 d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2.1

Question 3.
\(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)
Solution:
Take x = 4 cos2θ + 9 sin2θ then
dx = (-8 cos θ sin θ + 18 sin θ cos θ) dθ = 10 cos θ sin θ
Upper limit when x = 9 is 9 = 4 cos2θ + 9 sin2θ
⇒ 9(1 – sin2θ) = 4 cos2θ
⇒ 5 cos2θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 4 is 4 = 4 cos2θ + 9 sin2θ
⇒ 4(1 – cos2θ) = 9 sin2θ
⇒ 5 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q3

Question 4.
\(\int_0^5 x^2(\sqrt{5-x})^7 d x\)
Solution:
Let x = 5 sin2θ then dx = 10 sin θ cos θ dθ
Upper limit when x = 5 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 5.
\(\int_0^{2 \pi}(1+\cos x)^5(1-\cos x)^3 d \dot{x}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.2

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Try This

Question 1.
Check whether the following equations are quadratic or not.
i) x2 – 6x – 4 = 0
ii) x3 – 6x2 + 2x – 1 = 0
iii) 7x = 2x2
iv) x2 + \(\frac{1}{\mathrm{x}^2}\) = 2
v) (2x + 1) (3x + 1) = 6(x – 1)(x – 2)
vi) 3y2 = 192 (Page No. 102)
Solution:
i) x2 – 6x – 4 = 0
Yes, It’s a quadratic equation.

ii) x3 – 6x2 + 2x – 1 = 0
No, It is not a quadratic equation.
[∵ degree is 3]

iii) 7x = 2x2
Yes, It’s a quadratic equation.

iv) x2 + \(\frac{1}{x^2}\) = 2
No, It is not a quadratic equation.
[∵ degree is 4]

v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2)
No, It is not a quadratic equation.
[coefficient of x2 on both sides is same i.e., 6]

vi) 3y2 = 192
Yes, It’s a quadratic equation.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Try This

Question 1.
Verify whether 1 and \(\frac{3}{2}\) are the roots of the equation 2x2 – 5x + 3 = 0. (Page No. 107)
Solution:
Let the given
Q.E. be p(x) = 2x2 – 5x + 3
Now p(1) = 2(1)2 – 5(1) + 3
= 2 – 5 + 3 = 0
∴ 1 is a root of 2x2 – 5x + 3 = 0
also p(\(\frac{3}{2}\)) = 2(\(\frac{3}{2}\))2 – 5(\(\frac{3}{2}\)) + 3
= 2 × \(\frac{9}{4}\) – \(\frac{15}{2}\) + 3
= \(\frac{9}{2}\) + 3 – \(\frac{15}{2}\) = \(\frac{9+6-15}{2}\) = 0
∴ \(\frac{3}{2}\) is also a root of 2x2 – 5x + 3 = 0

Do This

Question 1.
Solve the equations by completing the square. (Page No. 113)

i) x2 – 10x + 9 = 0
Solution:
Given x2 – 10x + 9 = 0
⇒ x2 – 10x = – 9
⇒ x2 – 2. x. 5 = – 9
⇒ x2 – 2.x. 5 + 52 = -9 + 52
⇒ (x – 5)2 = 16
∴ x – 5 = ± 4
∴ x – 5 = 4 (or) x – 5 = – 4
⇒ x = 9 (or) x = 1
⇒ x = 9 (or) 1

ii) x2 – 5x + 5 = 0
Solution:
Given : x2 – 5x + 5 = 0
⇒ x2 – 5x = -5
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 1

iii) x2 + 7x – 6 = 0
Solution:
x2 + 7x – 6 = 0
x2 + 7x = 6
x2 + 2.x.\(\frac{7}{2}\) = 6
x2 + 2.x.\(\frac{7}{2}\) + (\(\frac{7}{2}\))2 = 6 + (\(\frac{7}{2}\))2
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 2

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Think – Discuss

Question 1.
We have three methods to solve a quadratic equation. Among these three, which method would you like to use ? Why ? (Page No. 115)
Solution:
If the Q.E. has district and real roots, we use factorisation. If Q.E. has no real roots, we use quadratic formula.

Try This

Question 1.
Explain the benefits of evaluating the discriminate of a quadratic equation before attempting to solve it. What does its value signifies ? (Page No. 122)
Solution:
The discriminant b2 – 4ac of a Q.E. ax2 + bx + c = 0 give the clear idea about the nature of Q the roots of the Q.E. If the discriminant D = b2 – 4ac > 0, the Q.E. has distinct and real roots.

If b2 – 4ac = 0, the Q.E. had equal roots. If b2 – 4ac < 0, the Q.E. has no real roots. By the 3 value of the discriminant, we can state the nature of the roots of a Q.E. without actually finding them.

Question 2.
Write three quadratic equations one having two distinct real solutions, one having no real solution and one having exactly one real solution. (Page No. 122)
Solution:
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 3

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P :
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 1
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 2

Question 2.
Find the
i) 30th term of the A.P. : 10, 7, 4, …… (A.P. June 15)
ii) 11th term of the A.P. : -3, \(\frac{-1}{2}\) ,2, …….
Solution:
i) Given A.P. = 10, 7, 4, ….
a1 = 10; d = a2 – a1 = 7 – 10 = – 3
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29 × (- 3)
= 10 – 87
= -77

ii) Given A.P. = -3, \(\frac{-1}{2}\), 2,……..
a1 = -3; d = a2 – a1 = \(\frac{-1}{2}\) – (-3)
= \(\frac{-1}{2}\) + 3 = = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
an = a + (n – 1)d
a11 = -3 + (11 – 1) × (\(\frac{5}{2}\))
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25 = 22

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 3.
Find the respective terms for the following APs.

i) a1 = 2; a3 = 26, find a2.
Solution:
Given : a1 = a = 2 —— (1)
a3 = a + 2d = 26 —— (2)
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a2 = a + d = 2 + 12 = 14

ii) a2 = 13; a4 = 3, find a1, a3.
Solution:
Given : a2 = a + d = 13 —— (1)
a4 = a + 3d = 3 —- (2)
Solving equations (1) and (2);
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 3
Substituting, d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18
i.e., a1 = 18
a3 = a + 2d = 18 + 2(-5)
= 18 – 10 = 8

iii) a1 = 5; a4 = 9\(\frac{1}{2}\), a2, a3.
Solution:
Given : a1 = a = 5 —– (1)
a4 = a + 3d = 9\(\frac{1}{2}\) —– (2)
Solving equations (1) and (2);
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 4

iv) a1 = – 4; a6 = 6, find a2, a3, a4, a5.
Solution:
Given : a1 = a = – 4 —– (1)
a6 = a + 5d = 6 —– (2)
Solving equations (1) and (2);
(-4) + 5d = 6 ⇒ 5d = 6 + 4
⇒ 5d = 10 ⇒ d = \(\frac{10}{5}\) = 2
Now, a2 = a + d = -4 + 2 = -2;
a3 = a + 2d = -4 + 2 × 2
= -4 + 4 = 0;
a4 = a + 3d = (-4) + 3 × 2
= -4 + 6 = 2;
a5 = a + 4d = -4 + 4 × 2
= -4 + 8 = 4

v) a2 = 38; a6 = -22, find a1, a3, a4, a5.
Solution:
Given : a2 = a + d = 38 —– (1)
a6 = a + 5d = -22 —– (2)
Subtracting (2) from (1) we get
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 5
Now substituting, d = -15 in equation (1), we get
a + (-15) = 38 ⇒ a = 38 + 15 = 53 Thus,
a1 = a = 53;
a3 = a + 2d = 53 + 2 × (-15) = 53 – 30 = 23;
a4 = a + 3d = 53 + 3 × (-15) = 53 – 45 = 8;
a5 = a + 4d = 53 + 4 × (-15) = 53 – 60 = -7

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 4.
Which term of the AP :
3, 8, 13, 18, ……. is 78 ?
Solution:
Given : 3, 8, 13, 18,
Here a = 3; d = a2 – a1 = 8 – 3 = 5
Let ’78’ be the nth term of the given A.P.
∴ an = a + (n – 1) d
78 = 3 + (n – 1)5
78 = 3 + 5n – 5
⇒ 5n = 78 + 2 ⇒ n = \(\frac{80}{5}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs :
i) 7, 13, 19, ……., 205 (A.P. Mar. ’16)
Solution:
Given : A.P : 7, 13, 19, ………
Here a1 = a = 7; d = a2 – a1 = 13 – 7 = 6
Let 205 be the nth term of the given A.P.
Then, an= a + (n – 1) d
205 = 7 + (n – 1) 6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, ………, -47.
Solution:
Given A.P = 18, 15\(\frac{1}{2}\), 13, ….
Here a1 = a = 18
d = a2 – a1 = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the nth term of the given A.P.
an = a + (n – 1) × d
⇒ -47 = \(\frac{18 \times 2+(n-1)(-5)}{2}\)
⇒ – 94 = 36 – 5n + 5
⇒ 5n = 94 + 41 ⇒ n = \(\frac{135}{5}\) = 27
i.e., 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP : 11, 8, 5, 2, ….
Solution:
Given A.P. = 11, 8, 5, 2,
Here a = a2 – a1 = 8 – 11 = -3
If possible, take -150 as the n,sup>th term of the given A.P.
an = a + (n – 1) d
-150 = 11 + (n – 1) × (- 3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = -150
⇒ 3n = 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:
Given : An A.P. whose
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 6
Substituting d = 7 in the equation (1) we get,
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = – 32
Now, the 31st term = a + 30d
= (- 32) + 30 × 7
= – 32 + 210 = 178

Question 8.
If the 3rd and the 9th terms of an A.P are 4 and – 8 respectively, which term of this A.P is zero ?
Solution:
Given : An A.P whose
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 7
Substituting d = – 2 in equation (1) we get a + 2 × (- 2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let nth term of the given A.P be equal to zero.
an = a + (n – 1) d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5th term of the given A.P is zero.

Question 9.
The 17th term of an A.P exceeds its 10th term by 7. Find the common difference.
Solution:
Given an A.P. in which a17 = a10 + 7
⇒ a17 – a10 = 7
We know that an = a + (n – 1) d
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 8

Question 10.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Solution:
Let the first A.P be : a, a + d, a + 2d,
Second A.P be : b, b + d, b + 2d, b + 3d,….
Also, general term, an = a + (n – 1) d
Given that, a100 – b100 = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000 terms,
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 9
∴ The difference between their 1000th terms is (a – b) = 100.

Note : If the common difference for any two A.Ps are equal then difference nth terms of two A.Ps is same for all natural values of n.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 11.
How many three-digit numbers are divisible by 7 ?
Solution:
The list of three digit numbers divisible by 7 , is,
105, 112, 119, ………, 994
Here a = 105; d = 7; an = 994
an = a + (n – 1) d
994 = 105 + (n – 1) 7
(n – 1)7 = 889
n -1 = \(\frac{889}{7}\)
n – 1 = 127
∴ n = 127 + 1 = 128
There are 128 three digit numbers which are divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Given numbers : 10 to 250
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 10
∴ Multiples of 4 between 10 and 250 are
First term : 10 + (4 – 2) = 12
Last term : 250 – 2 = 248
∴ 12, 16, 20, 24, ………, 248
a = a1 = 12; d = 4: an = 248
an = a + (n – 1) d
248 = 12 + (n – 1) × 4
⇒ (n – 1) 4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the nth terms of two APs : 63, 65, 67, … and 3, 10, 17, …. equal ?
Solution:
Given : The first A.P is 63, 65, 67, ….
where a = 63, d = a2 – a1
⇒ d = 65 – 63 = 2
and the second A.P is 3, 10, 17, ….
where a = 3, d = a2 – a1 = 10 – 3 = 7
Suppose the nth terms of the two A.Ps are equal, where an = a + (n – 1) d
⇒ 63 + (n -1) 2 = 3 + (n – 1) 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 +4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Given : An A.P in which
a3 = a + 2d = 16 —– (1)
and a7 = a5 + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A. P is a, a + d, a + 2d, a + 3d, …
⇒ 4, 4 + 6, 4 + 12, 4 + 18, …
⇒ A.P: 4, 10, 16, 22,….

Question 15.
Find the 20th term from the end of the AP : 3, 8, 13,…….., 253.
Solution:
Given : An A.P : 3, 8, 13, …., 253
Here a = a1 = 3
d = a2 – a1 = 8 – 3 = 5
an = 253, where 253 is the last term
an = a + (n – 1) d
∴ 253 = 3 + (n – 1) 5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a32 = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155
= 158

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 16.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the (A.P. Mar. ’15)
Solution:
Given an A.P in which
a4 + a8 = 24 ⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 —– (1)
and a6 + a10 = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 —– (2)
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 11
Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a , a + d, a + 2d, ……..
i.e., – 13, (- 13 + 5), (- 13 + 2 × 5)
⇒ -13, -8, -3, ………

Question 17.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach Rs. 7000?
Solution:
Given : Salary of Subba Rao in 1995 = ₹ 5000
Annual increment = ₹ 200
i.e., His salary increases by Rs. 200 every year.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 12
Clearly 5000, 5200, 5400, ……… forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached ₹ 7000
after x – years,
i.e., an = 7000
But, an = a + (n – 1) d
7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ n – 1 = \(\frac{2000}{200}\)
= 10
⇒ n = 10 + 1 = 11
∴ In 11th year his salary reached ₹ 7000.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Telangana TSBIE TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities Textbook Questions and Answers.

TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities

Long Answer Questions

Question 1.
What is meant by industry? Explain various types of industries with suitable examples.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort. E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale. E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products. E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses. E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community. E.g: Banking, trans-port, insurance etc.

Question 2.
What is commerce? Describe its branches.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers. It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Branches of commerce:
The activities of commerce may be classified into following two broad categories.

  • Trade
  • Aids to trade

I) Trade: Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade may be classified into a) Home trade, b) Foreign trade. Home trade may again sub-divided into (i) Wholesale trade and (ii) Retail trade.

a) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’ Home trade is carried on within the boundaries of a nation. Home trade again is of two types: wholesale trade and retail trade.

  • Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in whole sale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.
  • Retail trade: It involves buying and selling of goods in small quantities. Traders engaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

b) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘International trade’.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the U.K.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as re-export trade.

II) Aids to trade: Trade or exchange of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities, which directly or indirectly facilitate smooth exchange of goods and services.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Aids to trade includes transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in below:

1) Transport: There will be a vast distance between centers of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport.

2) Communication: Communication means transmitting or exchange of information froth one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modern means of communication like telephone, email, video conference etc play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production and consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So, it creates time utility by supplying the goods at right time to consumer.

4) Insurance: Insurance reduces the problem of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a larger number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Advertising fills the knowledge gap. Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspapers, Magazines, TV, Internet etc.

Question 3.
Discuss the significance of commerce in the present scenario.
Answer:
Commerce can also be defined as The sum total of those processes. Which are engaged in the removal of hindrances of person, place and time in the exchange of commodities’.

Importance of commerce:
The importance of commerce is explained below:
1) Commerce tries to satisfy increasing human wants: Human wants and desires are never ending. Commerce has made distribution and movement of goods possible from one part of the world to the other. Today we can buy anything produced anywhere in the world. Hence commerce facilitates the people to satisfy their needs, desires and wants by distribution and exchange of the goods and services.

2) Commerce helps us to increase our standard of living: Standard of living refers to quality of life enjoyed by the members of a society. When a man consumes more products his standard of living improves. Commerce helps us to get what we want at the right time, right place and at the right price and thus helps in improving our standard of living.

3) Commerce links producers and consumers: Commerce makes a link between producers and consumers through retailers and wholesalers and also through the aids to trade. Thus it creates and facilitates the contact between the centres of production and consumption.

4) Commerce generates employment opportunities: The growth of commerce and trade cause the growth of agencies of trade such as banking, transport, warehousing, insurance, advertising etc. Thus, development of commerce generates more and more employment opportunities.

5) Commerce increases national income and wealth: When production increases, national income also increases. It also helps to earn foreign exchange by way of exports and duties levied on imports.

6) Commerce helps in expansion of aids-to-trade: With the growth in trade and commerce there is a growing need for expansion and modernization of aids to trade. Aids to trade such as banking communication, advertising and publicity, transport, insurance etc. are expanded and modernised for the smooth conduct of commerce.

7) Commerce encourages international trade: With the help of transport and communication development, countries can exchange their surplus. Commodities and earn foreign exchange. Thus, the commerce ensures faster economic growth of the country.

8) Commerce benefit underdeveloped countries: Underdeveloped countries can import skilled labour and technical knowhow from developed countries, while the advanced countries can import raw materials from underdeveloped countries. This helps in laying down the seeds of industrialisation in the underdeveloped countries.

9) Commerce helps during emergencies: During emergencies like floods, earthquakes and wars, commerce helps in reaching the essential requirements like food stuff, medicines and relief measures to the affected area.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 4.
Define trade and explain various types of aids to trade.
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Aids to trade:
Trade or exchange / distribution of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities which directly or indirectly facilitate smooth exchange of goods and services.

Aids to trade includes Transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in following points.

1) Transport: There will be a vast distance between centres of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport, (i) Land (ii) Water (iii) Air.

2) Communication: Communication means transmitting or exchange of information from one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modem means of communication like telephone, email, teleconference, video-conference etc. play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production & consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So it creates time utility by supplying the goods at right time to consumers.

4) Insurance: Insurance reduces the problems of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a large number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspaper, Magazines, TV; Internet etc.

Question 5.
Explain inter-relationship between industry, commerce and trade.
Answer:
Business is divided into two categories: Industry and commerce. Commerce is again sub-divided into trade and aids to trade practically all of them are closely related to each other. They are inseparable. All of them are parts of the whole business system. Industry and commerce are closely related to each other. Industry cannot exist without commerce and commerce cannot exist without industry because every producer has to find his market for his product to sell. But the producer has no direct connection with the buyer or consumers. Hence, industry needs commerce.

BUSINESS = INDUSTRY + COMMERCE
Commerce is concerned with the sale, transfer or exchange of goods and services. Hence commerce needs industry for the production of goods and services. Commerce makes the necessary arrangement for linking between producers and ultimate consumers. It includes all those activities that are involved in buying, selling, transporting, banking, warehousing of goods, and insurance of safeguarding the goods.

COMMERCE = TRADE + AIDS TO TRADE
Thus industry, commerce, and trade are closely related to one another and are inter-dependent as shown in the figure below. In conclusion, we can say that industry, trade and commerce are inter-related with each other. Industry is concered with production of goods and services and commerce arranges its sales; but the actual operation of sales is in the hands of trade. So they cannot work independently.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 6.
Describe the various Hindrances of Commerce.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the place of production to the ultimate consumers. Whereas trade involves buying and selling of goods, commerce has a wider meaning. Commerce include trade and aids to trade. The aids to trade include transport, banking, insurance, warehousing, advertisement and salesmanship.

Hindrances of trade: In the course of exchange of goods various problems are encountered. The hindrances in the way of smooth trade may be place, person, finance, time, knowledge and risk.

1) Hindrances of place: Generally, all the goods are not consumed at the same place where are produced. The goods are to be taken from a place where there is less demand, to the place where there is more demand. The activity of movement of the goods is called transportation. Thus transport eliminates the hindrances of place.

2) Hindrances of persons: In the present day world the consumers are in millions and it is not possible for the producers to know the consumers who are in need of goods produced by them. A chain of middlemen like wholesalers, retailers, dealers etc. Purchase goods from the producers and take them to the customers. Thus, middlemen remove the hindrances of persons.

3) Hindrances of finance: There is always time lag between the production and sale of goods. It takes time to collect money and hence need finance for trade. Commerce makes exchange of goods and services possible by removing these hindrances through the agency of banks.

4) Hindrances of time: As the goods are produced in anticipation of demand, there is a need to store the until they are required for consumption. Warehousing eliminates the hindrances of time and provides time utility to goods.

5) Hindrances of knowledge: The consumers may not be aware of the availability of various goods in the market. The absence of information is another hindrance. This is eliminated through advertising. Advertisement is done through T.V., radio, news papers, magazines, wall posters, hoardings etc.

6) Hindrances of risk: There are risks involved in production, transporting goods from one place to another, warehousing. The businessmen would like to cover these risks. Insurance companies undertake to compensate the loss suffered due to such risks. So, insurances eliminates hindrances of risks.

Question 7.
Distinguish between trade, commerce and industry.
Answer:
Differences between trade, commerce and industry.

Basis of DifferenceTradeCommerceIndustry
1. MeaningIt is related to the purchase and sale of goods.It is related to the activities which deals with taking of goods from producers to consumers.All those activities which deal with conversion of raw material into finished goods.
2. CapitalMore capital is required than commerce.It requires less capital.Capital needs are high.
3. ScopeIt deals with purchase and sale of goods.It includes trade and aids to trade.Primary manufacturing, processing etc., industries are covered under industry.
4. RiskIt involves greater risk.The risk involved is comparatively less.It involves greater risk compared to any other activity.
5. UtilityTrade creates possession utility.Commerce creates time and place utility.Industry creates form utility.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Short Answer Questions

Question 1.
Define Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Question 2.
What is Trade?
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade is classified into two types: (I) Home trade (II) Foreign trade

I) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’. Home trade is carried on within the boundaries of a nation.
Home trade is again divided into two types:

  • Wholesale trade and
  • Retail trade

(i) Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in wholesale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.

(ii) Retail trade: It involves buying and selling of goods in small quantities. Traders en-gaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

(II) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘international trade’. Foreign trade is again may be classified into three categories as mentioned below.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the united kingdom.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as ‘re-export’ trade.

Question 3.
State the types of foreign trade.
Answer:
When trade takes place between two countries it is called foreign trade or external trade or international trade. Two countries are involved in foreign trade. External trade generally requires permission from the respective countries. The hindrances of place, time, risk, exchange are overcome with the help of various agencies. Foreign trade may be classified into export trade, import trade and entrepot trade.

1) Export trade: This trade refers to sale of goods to foreign countries.
E.g.: India exports tea to U.K.

2) Import trade: The purchase of goods from other countries is known as import trade.
Ex: India buys petrol from Iran.

3) Entrepot trade: When goods are imported (purchased) from one country with a view to exporting (selling) them to other country, it is called entrepot trade or re-export trade.
Ex: India imported petrol from Iran and export the same to Nepal.

Question 4.
Explain the classification of industries.
Answer:
Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort.
E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale.
E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products.
E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses.
E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and Steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community.
E.g: Banking, Trans-port, Insurance etc.

Question 5.
Define ‘Entrepot trade’ with example.
Answer:
When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
E.g.:

  • India importing wheat from U.S. and exporting the same to Srilanka.
  • India imports petrol from Iran and export the same to Nepal.

Very Short Answer Questions

Question 1.
Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods from raw material to the finished product.
E.g.: Weaving woollen yam into cloth. Thus industry creates form utility in goods.

Question 2.
Commerce.
Answer:

  • Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers.
  • It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Question 3.
Trade.
Answer:

  • All the activities engaged in buying and selling of goods and services a re called trade.
  • Therefore trade includes sale, transfer or exchange of goods and services with the intention of making profit. The object of trade is to make goods available to those who need them and willing to pay for them.
  • Trade is the final stage of business activities and involves transfer of ownership.

Question 4.
Home Trade.
Answer:

  • The trade is carried on within the boundaries of the nation is called Home Trade. Home trade is also called as internal trade or domestic trade.
  • Home trade refers to a trade where buying and selling of goods takes place between the persons who belong to the same country. So, home For movement of goods internal transport system is used.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 5.
Entrepot trade.
Answer:

  • When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
  • E.g.: India importing wheat from U.S. and exporting the same to Srilanka.

Question 6.
Transportation.
Answer:
There is a vast distance between centres of production and centres of consumption. Goods are to be moved from the places of production to the place where they are demanded.

  • The activity which is concerned with movement of goods is called transportation. Trans-port create place utility.
  • There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of transport.

Question 7.
Warehousing.
Answer:

  • There is time gap between production and consumption. Hence, it became necessary to make arrangements for storage or warehousing.
  • It is one of the aid to trade, which facilitates to store the goods until they get demand or consumed.
  • Some goods are to be stored in warehouses till they are demanded. Warehousing creates time utility.

Question 8.
Genetic Industries.
Answer:

  • Genetic industry is related to the reproducing and multiplying certain species of animals and plants with the object of earning profit from their sale.
  • E.g.: Nurseries, cattle breeding, poultry farm, fish hatcheries etc.

Question 9.
Extractive industries.
Answer:

  • These industries are engaged in raising some form of wealth from the soil, climate, air, water or from beneath the surface of the earth. Generally the products of extractive industry comes in raw form and they are used by manufacturing and construction industries for producing finished products.
  • E.g.: Mining, coal, mineral, iron ore oil, extraction of timber and rubber from forests.

Question 10.
Banking.
Answer:

  • Banking solve the problem of finance.
  • Producers and traders require money for carrying on production and trade. Banks are the institutions which supply funds for industries and trade.
  • They pool savings from the public and make them available to industry. So, banking is an important function of commerce.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?

i) The taxi fare after each km when the fare is ₹ 20 for the first km and rises by ₹ 8 for each additional km.
Solution:
Fare for the first km = ₹ 20 = a
Fare for each km after the first = ₹ 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

ii) The amount of air present in a cylinder when a vacuum pump removes 1/4th of the air remaining in the cylinder at a time.
Solution:
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes \(\frac{1}{4}\)th of the volume
i.e., \(\frac{1}{4}\) × 1024 = 256
Remaining air present in the cylinder = 768
At second time it removes \(\frac{1}{4}\)th of 768
i.e., \(\frac{1}{4}\)th × 768 = 192
Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes \(\frac{1}{4}\)th of 576
i.e., \(\frac{1}{4}\) × 576 = 144
Remaining air in the cylinder
= 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1st, 2nd, 3rd …. times is 1024, 768, 576, 432,…….
Here, a2 – a1 = 768 – 1024 = -256
a3 – a2 = 576 – 768 = – 192
a4 – a3 = 432 – 576 = – 144
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.R

iii) The cost of digging a well, after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
Solution:
Cost for digging the first metre = ₹ 150
Cost for digging subsequent metres = ₹ 50 each i.e.,
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 1
The list is 150, 200, 250, 300, 350, …..
Here d = a2 – a1 = a3 – a2
= a4 – a3 = ……… = 50
∴ The given situation represents an A.P

iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution:
Amount deposited initially = P = ₹ 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ A = P(1 + \(\frac{\mathrm{R}}{100}\))n
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 2
The terms 10800, 11664, 12597.12, ….
a2 – a2 = 800
a3 – a2 = 864

a4 – a3 = 933.12
…………………….

Here a = 10,000
But a2 – a1 ≠ a3
– a2 ≠ a4 – a3

∴ The given situation doesn’t represent an A.P.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

Question 2.
Write first four of the AP, when first term a and the common difference d are given as follows :
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = -3
iv) a = -1, d = 1/2
v) a = -1.25, d = -0.25
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 3

Question 3.
For the following A.Ps, write the first term and the common difference :

i) 3, 1, -1, – 3, …..
ii) -5,-1, 3, 7, …….
iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\),………
iv) 0.6, 1.7, 2.8, 3.9,…….. (A.P. Mar ’15)
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 4

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

i) 2, 4, 8, 16, ……..
ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\),………..
iii) – 1.2, -3.2, -5.2, -7.2, …….
iv) -10, -6, -2, 2,……….
v) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),……….
vi) 0.2, 0.22, 0.222, 0.2222,……….
vii) 0, -4, -8, -12,……….
viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\),…………
ix) 1, 3, 9, 27,……………….
x) a, 2a, 3a, 4a,………
xi) a, a2, a3, a4,………..
xii) \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\),…………
xiii) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\),…………
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 5
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 6
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 7

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Telangana TSBIE TS Inter 1st Year Commerce Study Material 1st Lesson Introduction to Business Textbook Questions and Answers.

TS Inter 1st Year Commerce Study Material 1st Lesson Introduction to Business

Long Answer Questions

Question 1.
Define Business. Explain its characteristic features.
Answer:
Meaning:

  • Business means the state of being busy.
  • Business is an economic activity involving production, exchange, distribution and sale of goods and services with an objective of making profits and maximization of wealth.
  • The primary intention of business is making profits.

Definitions:

  • L.H. Haney: Defined business as “A human activity directed towards producing or acquiring wealth through buying and selling of goods”.
  • Stephen son: Defines business as, “The regular production or purchase and sale of goods undertaken with an objective of earning profits and acquiring wealth through the satisfaction of human wants”.
  • According to keith and carlo: “Business is a sum of all activities involved in the production and distribution of goods and services for private profits”.

Business may be defined as an economic activity which involves regular transfer or exchange of goods or services for a price.

Characteristics (or) Features:
Following are the main characteristic features of business:
i) Economic Activity: Business is an economic activity. It is performed with the main motive of earning money or profit. It does not include the activities undertaken out of love, affection, charity and religious commitments etc.

ii) Deals with goods and services: Every business enterprise produces or buys goods and services with the intention of selling them to others, to earn profit. Goods deals in business, may be consumer goods or capital goods.

  • Consumer goods are meant for direct use by the ultimate consumers, e.g. cloths, note books, tea, bread, etc.
  • Capital goods also known as producer goods, which are not intended for direct consumption but used for production of other goods, Equipment, Tools, Raw mate-rials, Machinery etc.

Services like transport, warehousing, banking, insurance, etc. may be considered as intangible and invisible goods. Services facilitate buying and selling of goods by over coming various hindrances in trade.

iii) Creation of utilities: Business makes goods more useful to satisfy human wants. It adds, time, place, form and possession utilities to various types of goods.
For example: It carries goods from place of production to the place of consumption (place utility). It makes goods available for use in future through storage (time utility).

iv) Continuity in dealings: Dealings in goods and services become business only if undertaken on a regular basis. A single isolated transaction of purchase and sale does not constitute business.
Therefore, regulartity of dealings is an essential feature of business.

v) Sale, transfer or exchange: All business activities involve transfer or exchange of goods and services for some consideration. The consideration called price, which is expressed in terms of money.
For example: If a person cooks and serves food to his family, it is not business. But when he cooks food and sells it to others for a price, it becomes business.

vi) Profit motive: The primary objective of business is to earn profits. Profits are essential for the survival as well as growth of any business. Profits must be earned through legal and fair means. Business should never exploit any part of the society to make money.

vii) Risk and uncertainity: Risk means fear of loss. It implies the uncertainly of profit or the possibility of loss. Business enterprises function in uncertain and uncontrollable environment.

Changes in customers tastes and fashions, demand, competition, government policies etc., create risk, flood, fire, earthquake, strike by employees, theft etc., also cause loss. A businessman can reduce risks through correct forecasting and insurance.

viii) Business is also a social institution: Business is also a social institution because it helps to improve the living standards of people through effective utilisation of scarce resources of the society.

ix) Art as well as science: Business can be pressumed as an art as well as science. It is art because it requires personal skills and experience. It is also a science because it is based m certain principles and laws.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 2.
Describe various objectives of a business.
Answer:
“An objective is a goal or motive directed towards achievement of predetermined purpose or aim. Hence, every business enterprise has certain objectives. These objectives are broadly classified into four categories viz, economic objectives, social objectives, human objectives and national objectives.

I. Economic Objectives:
Business is basically an economic activity. Therefore, its primary objectives are economic in nature. The main economic objectives of business are as follows.
i) Earning profits:

  • Any business enterprise is established for earning some profits. It is the main in-centive and hope to start business.
  • Just as a person cannot live without food, a business firm cannot survive without profits.
  • Profit is also necessary for the expansion and growth of business.
  • Profit also serves as the barometer of economic stability, efficiency and progress of a business enterprise.
  • Therefore, profit is essential for the survival of every business.

ii) Creating customers:

  • A business man can earn profits only when there are enough customers to buy and pay for his goods and services.
  • In the words of peter F. Drucker, “There is only one valid definition of business purpose; to create a customer.
  • The customer is the foundation of business and keeps it in existence.
  • In order to earn profits, business must supply better quality goods and services at reasonable prices. Therefore, creation and satisfaction of customers is an important economic objective of business.

iii) Innovation:

  • Application and adoption of new methods, devices and technologies is called innovation. It also refers to creation of new things resulting from the study and experimentation, research and development.
  • It comprises all efforts made in perfecting the product, minimising the costs and maximizing benefits to customers. Business firms invest money, time and efforts in Research and Development (R&D) to introduce innovation.

iv) Optimum Utilization of Resources:

  • The best usage of resources like material, machine, men and money is called optimum utilization of resources. It is also one of the important economic objectives.
  • Waste control and reprocessing mechanism, providing proper training to its workers and effective spending of money may prompt the business concern to reach the objective of optimum utilization of resources.

II. Social Objectives:
Business does not exist in a vaccum. It is part of society. It cannot survive and grow without the support of society. Business must therefore discharge social responsibilities in addition to earning profits.

According to Henry Ford, ‘The primary aim of business should be service and subsidiary aim should be earning of profits.”

The social objectives of business are as follows:
i) Supplying desired goods and Services at reasonable prices: Supply of goods and services to consumers at reasonable price with the accepted quality is the responsibility of any business entity. It is also the social obligation of business to avoid malpractices like, adulteration, smuggling, black marketing and misleading advertising.

ii) Fair Remuneration to employees: Business must give fair remuneration and compensation to employees for their work. In addition to wages and salary a reasonable part of profits should be distributed among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees. It is the obligation of business to provide healthy and safe work environment for employees.

iii) Employment creation: In a country like India unemployment Has become a serious problem and no government can offer jobs to all. Therefore, provision of adequate and full employment opportunities is a significant service to society.

iv) Promotion of social welfare: Business should provide support to social, cultural and religious, organisation. Biteiness enterprise can build schools, college, Libraria, dharmashalas, Hospitals, sports bodies, and research institutions. They can help non-government organisations (NGOs) like CRY (Child Relief and You), Help Age, and other which render services to weaker sections of society.

v) Payment of dues and Taxes to government: Every business enterprise should pay taxes and dues (income tax and GST) to the government honestly and at the right time.

III. Human objectives: Human objectives are concerned with the well being of labour, satisfaction of customers and shareholders. The human objective are as follows.
i) Welfare of workers or staff: The workers should be provided win physical comfort, appreciation and dignity of labour and conditions which will motivate the workers to give their best. Adequate provisions should be made for their health, safety and social security.

ii) Development of human resources: Human resources are the most valuable asset of business and their development will help in the growth of the business. This can be done by training the employees and conducting workshops on skill development and attitude.

iii) Labour participation: The workers should be allowed to take part in decision making process and helps them in their development.

iv) Labour management co-operation: The employees should be looked upon as human beings. The employees help in increasing probability and should be rewarded for their hardwork.

IV. National objectives: The following are the national objectives of the business.
i) Optimum utilisation of resources: Juditions allocation and optimum utilisation of scarce resources is essential for rapid economic growth of the country.

ii) National Self-reliance: It is the duty of the business to help the government, in increasing exports and in reducing dependence on imports. It will help nation to become self-reliant.

iii) Development of small scale industries and MSMEs : Small scale industries are necessary for generating employment opportunities and for providing inputs to large scale industries.

iv) Development of backward areas: In order to achieve balanced regional development, industries should be started at backward regions which helps to raise the standard of living in backward areas.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 3.
Discuss the social responsibility of business.
Answer:
Meaning: Every business operates within a society. It uses the resources of the society and depends on the society for its functioning. This creates an obligation on the part of the business to look after the welfare of the society. Therefore, all the activities of the business should be that they will protect and contribute to the interests of the society. Social responsibility of business refers to all such duties and obligations of business directed towards the welfare of the society.

Responsibility towards different interested groups: The business is associated with owners, employees, suppliers, customers, government and society. They are called as interested groups because, every activity of the business will affect the interests directly or indirectly. Various responsibilities of business towards different groups given below:

1) Responsibility towards owners: Owners are the persons who own the business. They contribute the capital and bears the business risks. The responsibilities of business towards them are:

  • Running the business efficiently.
  • Proper utilisation of capital and other resources.
  • Growth and appreciation of capital
  • Regular and fair returns on capital invested by way of dividends.

2) Responsibility towards employees: The future of the business depends on the efforts and efficiency of the employees. So, it is the responsibility of the take care of their interest. The responsibilities are:

  • Regular and fair payment of their wages and salaries.
  • Better working conditions and welfare amenities.
  • To improve the skill and efficiency by providing proper training.
  • Job security and social security like group insurance, pension, retirement benefits etc.

3) Responsibility towards suppliers: Suppliers are those persons who supply raw ma-terials and other items to the business. The responsibilites towards them are:

  • Timely payment of dues.
  • Dealing on fair terms and conditions.
  • Availing reasonable credit period.

4) Responsibility towards customers: In order to survive, the business responsibility is to provide following facilities of customers:

  • Providing qualitative goods and services.
  • Charging reasonable prices.
  • Giving delivery of goods within stipulated time.
  • After sale service.
  • Avoiding unfair means like under weighing the product, adulteration etc.

5) Responsibility towards government: Business is governed by the rules and regulations framed by the government. The responsibilities towards government are:

  • Payment of fees, duties and taxes honestly and regularly.
  • Setting up units as per guidelines of the government.
  • Following the pollution control norms.
  • Not to indulge in unlawful activities.

6. Responsibility towards society: Business, being a part of the society has to maintain relationship with other members of the society. The responsibility of business towards society is:

  • To help weaker and backward sections of the society.
  • To generate employment.
  • To protect environment.
  • To conserve natural resources and wild life.
  • To promote sports, social and cultural values.

Question 4.
Classify and describe each type of Economic activities.
Answer:
All human activities are directed towards satisfying human wants. Depending upon the nature of wants, human activities may be categorised as economic and non-economic. Economic activities are undertaken to create utilities. Non – economic activities do not have economic values and these are primarily tend to satisfy social, religious, cultural or sentimental requirements of human beings.

Classification of economic activities: Economic activities are broadly classified into three. They are:

  • Business
  • Profession
  • Employment

1) Business: The business is an activity which is primarily pursued with the object of earning profit. A business activity involves production, exchange of goods and services to earn profits or to earn a living. The word business literally means a state of being busy. Every person is engaged in some kind of occupation, a farmer works in the field, a worker works in the factory, a clerk does his work in the office, a teacher teaches in the class, a salesman is busy in selling the goods and an entrepreneur is busy in running his factory. The primary aim of all these persons is to earn their live¬lihood while doing some work.

2) Profession: Profession is an occupation involving the provision of personal services of a specialised and expert nature. The services are based on professional service of specialised nature. The service is based on professional education, knowledge, train¬ing etc. The specialised service is provided for a professional fees charged from the clients. For instance, a doctor helps his patients through his expert knowledge of the science of medicine and charges a fees for the service. Minimum educational qualifications are prescribed for entry into a profession and every profession requires a high degree of formal education and specialised training in a particular field. For example, a chartered accountant needs to be a member of the Institute of Chartered Accountants of India (ICA1). A person entering law profession has to acquire LL.B degree in order to become a lawyer.

3) Employment or Service: Employment or service involves working under a contract of employment for or under someone known as employer in return for a wage or salary. The person engaged under employment works as per the directions of the employer. There is an employer – employee relationship. A professional may also work under the contract of employment. A Chartered Accountant may be employed by the company The service may be of government department or in a private organisation.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Short Answer Questions

Question 1.
What are the Business objectives?
Answer:
Business objectives are classified as economic, Social, human and national objectives.
1. Economic objectives:

  • Earning profit
  • Creation of customers
  • Innovation
  • Optimum utilization of resources.

2. Social objectives:

  • Supplying desired goods and services at reasonable prices.
  • Fair remuneration to employees
  • Creation of employment opportunities
  • Promotion of social welfare
  • Payment of dues and taxes to Government.

3. Human objectives:

  • Welfare of workers or staff
  • Development of human resources
  • Labour participation in management
  • Labour management co-operation.

4. National objectives:

  • Optimum utilisation of natural resources
  • Self-reliance.
  • Development of small scale industries and MSMES
  • Development of backward regions.

Question 2.
What are the economic activities.
Answer:
Economic activities are broadly classified into business, profession and employment.

Business: The word business literally means a state of being busy. Every person is engaged in some kind of occupation or other work. The business is an activity which is primarily pursued with the object of earning profit. So, a business activity involves production, exchange of goods and services to earn profits or to earn a living.

Profession: Profession is an occupation involving the provision of personal services of a specialised and expert nature. The service is based on professional education, knowledge, training etc. The specialised service is provided for a professional fees charged from the clients. For example, a doctor helps his patients through his expert knowledge of science of medicine and charges a fees for the service.

Employment: Employment involves working under a contract of employment for or under someone known as employer in return for a salary. The person engaged under employment works as per the directions of the employer.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 3.
What are the social objectives?
Answer:
Business does not exist in a vaccum. It is a part of society. It cannot survive and grow without the support of society. Business must therefore discharge social responsibilities in addition to earning profits.

According to Henry Ford, “The primary aim of business should be service and subsidiary aim should be earning of profit”.

The social objectives of business are as follows:
i) Supplying desired goods and services at reasonable prices: Supply of goods and services to consumers at reasonable price with the accepted quality is the responsibility of any business entity. It is also the social obligation of business to avoid malpractices like, adulteration, smuggling, black marketing and misleading advertising.

ii) Fair Remuneration to employees: Business must be given fair remuneration and compensation to employees for their work. In addition to wages and salary a reasonable part of profits should be distributed among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees. It is the obligation of business to provide healthy and safe work environment for employees.

iii) Employment Generation: In a country like India unemployment has become a serious problem and no government can offer jobs to all. Therefore, provision of adequate and full employment opportunities is a significant service to society.

iv) Promotion of social welfare: Business should provide support to social, cultural and religious organisations. Business enterprise can build schools, colleges, libraries, dharamashalas, hospitals, sports bodies and research institutions. They can help NGO’s like CRY (Child Relief and You). Help Age, and others which render services to weaker sections of society.

v) Payment of dues and taxes to government: Every business enterprise should pay taxes and dues (income tax and GST) to the government honestly and at the right time.

Question 4.
What are the National objectives?
Answer:
National objectives of business are as follows:
i) Optimum utilisation of resources: Business should use the nation’s resources in the best possible manner. Judicious allocation and optimum utilisation of scarce resources is essential for rapid and balanced economic growth of the country.

ii) National self-reliance: It is the duty of business to help the government in increasing exports and in reducing dependance on imports. This will help a country to achieve economic independence.

iii) Development of MSMEs: Big business firms are expected to encourage growth of micro, small and medium enterprises (MSMEs) which are necessary for generating employment. MSMEs can be developed as ancillaries which provide inputs to large scale industries.

iv) Development of backward classes: Business is expected to give preference to the industrialisation of backward regions of the country. It will also help to raise standard of living in backward areas. Balanced regional development is necessary for peace and progress in the country. Government offers special incentives to the businessmen who setup factories in notified backward areas.

Question 5.
What is the role of profit in business?.
Answer:
Earning profit is necessary not only to pay adequate returns to the investors, but also to maintain the stability of the business. The role of profit in business is given below:

  • Profits ensure adequate funds for future expansion and innovation, there by increasing the wealth of country.
  • Profits are also helpful to attract new capital from outside sources like banks, Financial Institutions and Investors.
  • Profits are serves as the barometer of economic stability, efficiency and progress of a business enterprise.
  • Profits enable a businessman to stay in business by maintaining intact the wealth producing capacity of its resources.
  • Profits ensure continuous flow of capital for the modernisation of business.

Just as a person cannot line without food, a businesss firm cannot survive without profit. Therefore, profits are essential for every business organisation.

Question 6.
Explain the History of commerce.
Answer:
Commerce is a wider system which is the custome of a gradual evolution spread over a long period of human history and it passes through the following different stages.
i) Household economy: This is the first stage of economic development. At this stage, division of labour was unknown concept at the family level. There was no commercial inter relationship between families. While men engaged in jobs like hunting, fishing, making weapons for hunting etc, women were engaged in fruit gathering, cultivation of lands etc. Therefore, commerce was unknown at this stage.

ii) Primitive Barter Economy: Gradually the need of the families increased and families started specializing in different occupations and the need for exchange of goods, etc., between different families. Thus, commerce originated with introduction of barter systesms. In barter system, the goods are exchanged only. Hence barter system was considered as a popular element of commerce.

iii) The raise of trade: In course of time, the needs of people increased. In the begining, goods were exchanged at particular fixed places, but gradually trade appeared on the scene. Home trade began to develop and assume importance and money appeared as an instrument and medium of exchange. Further, the systems of weights and measurement also came into existence.

iv) Town Economy: At this stage, trade began to be undertaken for catering to the needs of local markets which gradually developed into large town. Further, traders were divided into wholesale and retail merchants. Division of labour become significant and prices for goods began to be fixed regularly and traders started using the credit system in their transactions.

v) International Trade: At this stage, goods were produced not only for selling in the local markets but also in foreign markets. This expansion of trade was due to the industrial revolution which made large scale manufacturing of goods possible. Commercial banks, insurance companies, transport companies, warehousing companies etc began to setup.

vi) E-commerce: It is an innovative business idea in the modem economy e-commerce means electronic commerce. It is the process of buy and sell the goods and services through an electronic medium electric commerce emerged in the early 1990s, and its use has increased at a rapid rate.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 7.
What are the economic objectives of the business?
Answer:
Economic objectives:
Business is basically an economic activity. Therefore, its primary objectives are economic in nature. The main economic objectives of business are as follows:

i) Earning profits: Any business enterprise is established for earning some profits. It is the main incentive and hope to start business. Just as a person cannot live without food, a business firm cannot survive without profit. Profit also serves as the barometer of economic activity, efficiency and progress of a business enterprise. Therefore profit is essential for the survival of every business.

ii) Creating customers: A business man can earn profits only when there are enough customers to buy and pay for his goods and services. In the words of Peter F. Drucker, ‘There is only one valid definition of business purpose, to create a customer. The customer is the foundation of business and keeps it in existence. In order to earn profit, business must supply better quality, goods and services at reasonable prices. Therefore, creation and satisfaction of customers is an important economic objective of business.

iii) Innovation: Application and Adoption of new methods, devices and technologies is called Innovation. It also refers to creation of new things resulting from the study and experimentation, research and development. It comprises all efforts made in prefecting the product, minimising the costs and maximizing benefits to customers. Business firms invest money, time and efforts in Research and Development (R and D) to introduce innovation.

iv) Optimum Utilization of Resources: The best usage of resources like material, machine men and money is called optimum utilization of resources. It is also one of the important economic objectives waste control and reprocessing mechanism, providing proper training to its workers and effective spending of money may prompt the business concern to reach the objective of optimum utilization of resources.

Question 8.
What is profession and explain features and objectives of profession.
Answer:
Profession: A profession is an occupation or vocation intended to render personal service of a specialised and expert nature. Profession is based on educational qualification, knowledge and competencies. For providing this service, the service provider may be a doctor, an engineer, lawyer, or a chartered accountant will charge fee on their clients which will be their primary earning.

Following are the salient features of profession:

  • A person enters into any particular profession should have Specialized qualification, knowledge and training in that respective profession.
  • Person enters into the specialized profession must be the member of respective professional body.
    Ex: Doctors in IMA (Indian Medical Association), Lawyers in Indian Bar Council and chartered Accountant in ICA (Institute of Charatered Accountants of India).
  • Person choosing the profession must follows the ethics.
  • Professional fee from the clients for providing their specialized services.
  • Rendering service legibly is the key factor in any profession.
  • Service motive must be an integral part of any profession.
  • Some professionals instead of work on their own may also work in organizations as employees and consultants.

Following are the important objectives of any profession:

  • To secure respected position in the society.
  • To spread the efficiency, skill and knowledge possessed, for the needy people.
  • To build career and register growth and development in their respective field.
  • To render service.
  • To enjoy the freedom of work at the cost low risk.

Question 9.
What is employment and explain the features and objectives of employment.
Answer:
Employment: An employment is a contract of service. A person who works under the contract for a salary or wage is called an employee and the person who has given the job to the employee is called employer. An employee works under an agreement as per the rules of service and performs tasks assigned to him by the employer.

The relationship between two parties will be employer – employee. Working in Banks, offices, hotels, schools and companies are the examples of employment.

Following are the important features of employment:

  • Employment contract commences when employee signs the agreement and joins organisation.
  • It involves rendering service for periodical remuneration called wage or salary.
  • Employment is the contractual relationship between employee and employer.
  • Employment will have very low risk and uncertainty, when compared with other economic activities.
  • Employee need not invest any capital.
  • Employees need to follow service rules and regulations framed by the employer strictly.
  • Certain employments required necessary qualifications too.
  • There are certain terms and conditions, of work like hours of work, duration of work, leave facility, salary / wages and place of worts in employment, which must be followed by the both employer and employee.

The important object of employment can be prevented as under:

  • The basic objective of employment is to earn for their livelihood and support family.
  • To get secured and assured wages and salaries.
  • To free from risk and uncertainty.
  • To keep away from disputes, conflicts and devastations.
  • To focus on skill development and career paths for job seekers.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Very Short Answer Questions

Question 1.
Business.
Answer:

  • Business means the state of being busy.
  • Business is an economic activity involving production, exchange, distribution and sale of goods and services with an objective of making profits and maximization of wealth. The primary intension of business is making profits.

Definition: L.H.Maney defined business as “a human activity directed towards producing or acquiring wealth through buying and selling of goods”.

Question 2.
Human activities.
Answer:

  • The activities performed by the human being in their daily life are called human activities. These are undertaken by them to fulfill their needs, desires, wants and luxuries.
  • Human activities are divided into two types:
    • Economic activities
    • Non-economic activities.

Question 3.
Profession.
Answer:

  • A profession is an occupation or vocation intended to render personal service of a specialized and expert nature. It is part of economic activities undertaken by the human being.
  • Always profession is based on educational qualification, knowledge and competencies. A fee is charged for providing professional service.

Question 4.
Employment.
Answer:

  • Employment is the relationship between two parties. One party render this service on contractual basis for the remuneration wage or salary.
  • Hence the relationship between these two parties will be employer – employee.

Question 5.
Risk and uncertainity.
Answer:

  • Risk means fear or loss. It implies the uncertainity of profit or the possibility of loss. Business enterprises function is uncertain and uncontrollable environment.
  • Changes in customers tastes and fashions, demand, competition, Government policies etc., create risk. Flood, fire, earthquake, strike by employees, theft etc., also cause loss. A businessman can reduce risks through correct forecasting and insurance.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.2

Question 1.
Find the roots of the following quadratic equations by factorisation.

i) x2 – 3x – 10 = 0
Solution:
Given : x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
⇒ x = 5 or – 2
are the roots of the given Q.E.

ii) 2x2 + x – 6 = 0 (A.P.Mar. 16)
Solution:
Given : 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or 2x = 3
⇒ x = -2 or \(\frac{3}{2}\)
are the roots of the given Q.E.

iii) \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0
Solution:
Given : \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x2 + 5x + 2x + 5\(\sqrt{2}\) = 0
⇒ x(\(\sqrt{2}\)x + 5) + \(\sqrt{2}\)(\(\sqrt{2}\)x + 5) = 0
⇒ (\(\sqrt{2}\)x + 5)(x + \(\sqrt{2}\)) = 0
⇒ \(\sqrt{2}\)x + 5 = 0 or x + \(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x = – 5 or x = –\(\sqrt{2}\)
⇒ x = \(\frac{-5}{\sqrt{2}}\) or –\(\sqrt{2}\) are the roots of the given Q.E.

iv) 2x2 – x + \(\frac{1}{8}\) = 0
Solution:
Given : 2x2 – x + \(\frac{1}{8}\) = 0
⇒ \(\frac{16 x^2-8 x+1}{8}\) = 0
⇒ 16x2 – 8x + 1 = 0
⇒ 16x2 – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) – 1(4x – 1) = 0
⇒ (4x – 1) (4x- 1) = 0
⇒ 4x – 1 = 0
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\), \(\frac{1}{4}\)
are the roots of given Q.E.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

v) 100x2 – 20x + 1 = 0
Solution:
Given : 100x2 – 20x + 1 = 0
⇒ 100x2 – 10x – 10x + 1 = 0
⇒ 10x (10x – 1) – 1(10x – 1) = 0
⇒ (10x – 1) (10x – 1) = 0
⇒ 10x – 1 = 0
⇒ 10x = 1
x = \(\frac{1}{10}\), \(\frac{1}{10}\)
are the roots of the given Q.E.

vi) x(x + 4) = 12
Solution:
Given : x(x + 4) = 12
⇒ x2 + 4x = 12
⇒ x2 + 4x – 12 = 0
⇒ x2 + 6x – 2x – 12 = 0
⇒ x(x + 6) – 2(x + 6) = 0
⇒ (x + 6) (x – 2) = 0
⇒ x + 6 = 0 or x – 2 = 0
⇒ x = -6 or x = 2
⇒ x = -6 or 2
are the roots of the given Q.E.

vii) 3x2 – 5x + 2 = 0
Solution:
Given : 3x2 – 5x + 2 = 0
⇒ 3x2 – 3x – 2x + 2 = 0
⇒ 3x(x – 1) – 2(x – 1) = 0
⇒ (x – 1) (3x – 2) = 0
⇒ x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or x = \(\frac{2}{3}\)
⇒ x = 1 or \(\frac{2}{3}\)
are the roots of the given Q.E.

viii) x – \(\frac{3}{x}\) = 2
Solution:
Given : x – \(\frac{3}{x}\) = 2
⇒ \(\frac{x^2-3}{x}\) = 2
⇒ x2 – 3 = 2x
⇒ x2 – 2x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ x = 3 or x = -1
⇒ x = 3 or – 1
are the roots of the given Q.E.

ix) 3(x – 4)2 – 5(x – 4) = 12
Solution:
Take (x -4) = a, then the given Q.E.
⇒ reduces to 3a2 – 5a = 12
⇒ 3a2 – 5a – 12 = 0
⇒ 3a2 – 9a + 4a – 12 = 0
⇒ 3a(a – 3) + 4(a – 3) = 0
⇒ (a – 3) (3a + 4) = 0 = 0
⇒ a – 3 = 0 or 3a + 4 = 0
⇒ a = 3 or a = \(\frac{-4}{3}\)
but a = x – 4
∴ x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)
⇒ x = 7 or x = 4 – \(\frac{4}{3}\)
⇒ x = 7 or \(\frac{8}{3}\)
are the roots of the given Q.E.

Question 2.
Find two numbers whose sum is 27 and product is 182. (A.P.Mar. 15)
Solution:
Let a number be x.
Then the other number = 27 – x
Product of the numbers = x(27 – x)
= 27x – x2
By problem 27x – x2 = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 or x – 14 = 0
⇒ x = 13 or 14.
The numbers are 13; 27 – 13 = 14 or 14 and 27 – 14 = 13.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 3.
Find two consecutive positive integers, sum of whose squares is 613.
Solution:
Let a positive integer be x.
Then the second integer = x + 1
Sum of the squares of the above integers = x2 + (x + 1)2
= x2 + x2 + 2x + 1
= 2x2 + 2x + 1
By problem 2x2 + 2x + 1 = 613
⇒ 2x2 + 2x – 612 = 0
⇒ x2 + x – 306 = 0
⇒ x2 + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17) (x + 18) = 0
⇒ x – 17 = 0 (or) x + 18 = 0
⇒ x = 17 (or) -18
Then the numbers are x = (17; 17 + 1) or x = (-18; -18 + 1)
i.e., 17, 18 or -17, -18.

Question 4.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Solution:
Let the base of the right triangle = x cm
Then its altitude = x – 7 cm
By Pythagoras Theorem
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2 3
(base)2 + (height)2 = (hypotenuse)2
⇒ x2 + (x – 7)2 = 132
⇒ x2 + x2 – 14x + 49 = 169
⇒ 2x2 – 14x + 49 – 169 = 0
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
⇒ x = 12 (or) x = – 5
But x can’t be negative.
∴ x = 12
x – 7 = 12 – 7 = 5
The two sides are 12 cm and 5 cm.

Question 5.
A cottage industry produces a certain number of pottery articles in a day. It was observed in a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. if the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced be x.
Then the cost of each article = 2x + 3
Total cost of the articles produced = x[2x + 3] = 2x2 + 3x
By problem 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ 2x + 15 = 0 (or) x – 6 = 0
⇒ x = \(\frac{-15}{2}\) or x = 6
But x can’t be negative, x = 6
∴ x = 6
2x + 3 = 2 × 6 + 3 = 15
∴ Number of articles produced = 6
Cost of each article = ₹ 15.

Question 6.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Solution:
Let the length of the rectangle = x
Given perimeter = 2(l + b) = 28
⇒ l + b = \(\frac{28}{2}\) = 14
∴ Breadth of the rectangle = 14 – x
Area = Length × Breadth = x(14 – x)
= 14x – x2
By problem, 14x – x2 = 40
⇒ x2 – 14x + 40 = 0
⇒ x2 – 10x – 4x + 40 = 0
⇒ x(x – 10) – 4(x – 10) = 0
⇒ (x – 10) (x – 4) = 0
⇒ x – 10 = 0 (or) x – 4 = 0
⇒ x = 10 (or) 4
∴ Length = 10 m or 4 m
Then breadth 14 – 10 = 4 m (or) = 14 – 4 = 10 m

Question 7.
The base of a triangel is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Solution:
Let the altitude of the triangle h = x cm
Then its base ‘b’ = x + 4
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)(x + 4)(x) = \(\frac{x^2+4 x}{2}\)
By problem \(\frac{x^2+4 x}{2}\) = 48
⇒ x2 + 4x = 2 × 48
⇒ x2 + 4x – 96 = 0
⇒ x2 + 12x – 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12) (x – 8) = 0
⇒ x + 12 = 0 (or) x – 8 = 0
⇒ x = -12 (or) x = 8
But x can’t be negative.
∴ x = 8 and x + 4 = 8 + 4 = 12
Hence altitude = 8 cm and base = 12 cm.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 8.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Solution:
Let the speed of the slower train = x kmph
Then speed of the faster train = x + 5 kmph
Distance = Speed × Time
Distance travelled by the first train = 2(x + 5) = 2x + 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)2 = (side)2 + (side)2
⇒ (2x)2 + (2x + 10)2 = 502
⇒ 4x2 + (4x2 + 40x + 100) = 2500
⇒ 4x2 + 4x2 + 40x + 100 = 2500
⇒ 8x2 + 40x – 2400 = 0
⇒ x2 + 5x – 300 = 0
⇒ x2 + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) x – 20
⇒ x = 15 (or) – 20
But x can’t be negative.
∴ Speed of the slower train x = 15 kmph.
Speed of the faster train x + 5 = 15 + 5 = 20 kmph.

Question 9.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was ₹ 1600, how many boys are there in the class ?
Solution:
Let the number of boys in the class = x
Then number of girls in the class = 60 – x [∵ total students = 60]
Money contributed by the boys = x(60 – x) = 60x – x2 [∵ given]
Money contributed by the girls = (60 – x) x = 60x – x2
Money contributed by the class = 120x – 2x2
By problem 120x – 2x2 = 1600
⇒ x2 – 60x + 800 = 0
⇒ x2 – 40x – 20x + 800 = 0
⇒ x(x – 40) – 20(x – 40) = 0
⇒ (x – 40) (x – 20) = 0
⇒ x = 40 (or) 20
∴ Boys = 40 or 20
Girls = 20 or 40.

Question 10.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed ?
Solution:
Let the speed of the boat in still water be x kmph.
Speed of the current = 3 kmph
Then speed of the boat in upstream = (x – 3)kmph
Speed of the boat in down stream = (x + 3) kmph
By problem total time taken = 6hrs
∴ \(\frac{24}{x-3}\) + \(\frac{24}{x+3}\) = 6
⇒ 24[\(\frac{1}{x-3}\) + \(\frac{1}{x+3}\)] = 6
⇒ 24\(\left[\frac{x+3+x-3}{(x+3)(x-3)}\right]\) = 6
⇒ 24(2x) = 6(x2 – 9)
⇒ 8x = x2 – 9
⇒ x2 – 8x – 9 = 0
⇒ x2 – 9x + x – 9 = 0
⇒ x(x – 9) + 1(x – 9) = 0
⇒ (x – 9) (x + 1) = 0
⇒ x – 9 = 0 or x + 1 = 0
x can’t be negative.
∴ x = 9
i.e., speed of the boal in still water = 9 kmph.