Mahesh

Students can practice TS SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion Ex 11.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Exercise 11.2

Question 1.
Among the following, which ratios are in the simplest form ? If not, convert them into the simplest form.
(i) 2 : 3
(ii) 16 : 20
(iii) 5 : 6
(iv) 20 : 60
(v) 8 : 15
(vi) 19 : 2
Answer:
Simplest form:
(i) 2 : 3,
(iii) 5 : 6,
(v) 8 : 15,
(vi) 19 : 2

Convert form:
(ii) \(\frac{16}{4}: \frac{20}{4}\) (i.e) 4 : 5
(iv) \(\frac{20}{20}: \frac{60}{20}\) (i.e) 1 : 3

Question 2.
A bag contains 20 kg of rice and another bag contains 60 kg of wheat. Find the ratio of the amount of rice to that of wheat. What is the ratio of rice to the total weight ?
Sol.
Quantity of rice = 20 kg
Quantity of wheat = 60 kg
Ratio of the amount of rice to that of wheat
= 20 : 60
= 1 : 3
Total weight of bag containing rice and wheat = 20 + 60 = 80 kg
Ratio of rice to the total weight
= 20 : 80
= 1 : 4

Question 3.
There are 32 students in a class out of which 12 are girls. Find :
(i) The ratio of number of boys to number of girls.
Answer:
Total number of students in the class = 32
Number of girls in the class = 12
Number of boys in the class
= 32 – 12 = 20
The ratio of number of boys to number of girls = 20 : 12
= \(\frac{20}{4}: \frac{12}{4}\) = 5 : 3

(ii) The ratio of number of boys to total number of students.
Answer:
The ratio of number of boys to total number of students = 20 : 32
= \(\frac{20}{4}: \frac{32}{4}\) = 5 : 8

(iii) The ratio of number of girls to total number of students.
Answer:
The ratio of number of girls to total number of students = 12:32
= \(\frac{12}{4}: \frac{32}{4}\) = 3 : 8

Question 4.
Draw a four sided closed figure and divide it into some number of equal parts. Shade the figure with any colour so that the ratio of shaded parts to unshaded parts 1 : 3. Draw two more different figures and do the same.
Answer:

Number of shaded parts = 3
Number of unshaded parts = 9
Ratio of shaded parts to unshaded parts = 3 : 9 = \(\frac{3}{3}: \frac{9}{3}\) = 1:3
Number of shaded parts = 2
Number of unshaded parts = 6

Ratio of shaded parts to unshaded parts = 2 : 6 = \(\frac{2}{2}: \frac{6}{2}\) = 1 : 3
Number of shaded parts = 4
Number of unshaded parts = 12
Ratio of shaded parts to un shaded parts = 4 : 12
= \(\frac{4}{4}: \frac{12}{4}\) = 1 : 3

Question 5.
Imran bought 2 litres of oil and Vijay bought 500 ml of oil. Find the ratio of quantities of oil bought by Imran to oil bought by Vijay.
Answer:
Quantity of oil bought by Imran
= 2 litres = 2 × 1000 ml = 2000 ml
Quantity of oil bought by Vijay = 500 ml
The ratio of quantities of oil bought by Imran to oil bought by Vijay
= 2000 : 500
= \(\frac{2000}{500}: \frac{500}{500}\) = 4 : 1

Question 6.
Weight of Abraham is 20 kg and his father’s weight is 60 kg. Find the ratio of weight of Abraham and his father. Express it in the simplest form.
Answer:
Weight of Abraham = 20 kg
Weight of Abraham’s father = 60 kg
Ratio of weight of Abraham and his father = 20 : 60
= \(\frac{20}{20}: \frac{60}{20}\) = 1 : 3
∴ The simplest form is 1:3.

Question 7.
Ramu spent \(\frac{2}{5}\) th of his money on a story book. Find the ratio of money spent to the money with him at the beginning.
Answer:
Let the money with Ramu be Rs. 1.
Then money spent by Ramu on a story book = Rs. 1 × \(\frac{2}{5}\) = Rs. \(\frac{2}{5}\)
The ratio of money spent to the money with him at the beginning = \(\frac{2}{5}\) : 1
= \(\frac{2}{5}\) × 5 : 1 × 5 = 2 : 5

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.3

Question 1.
State which of the following are equations.
(i) x – 3 = 7
(ii) l + 5 > 9
(iii) p – 4 < 10
(iv) 5 + m = – 6
(v) 2s – 2 = 12
(vi) 3x + 5 > 13
(vii) 3x < 15
(viii) 2x – 5 = 3
(ix) 7y + 1 < 22
(x) – 3z + 6 = 12
(xi) 2x – 3y = 3
(xii) z = 4
Answer:
The following are the equations.
(i) x- 3 = 7
(iv) 5 + m = – 6
(v) 2s – 2 = 12
(viii) 2x – 5 = 3
(x) – 3z + 6 = 12
(xi) 2x – 3y = 3
(xii) z = 4

Question 2.
Write LHS and RHS of the following equations.
(i) x – 5 = 6
(ii) 4y = 12
(iii) 2z + 3 = 7
(iv) 3p = 24
(v) 4 = x – 2
(vi) 2a – 3 = – 5
Answer:
The value of expression to the left of the sign ‘=’ is called Left Hand Side (LHS) and that of expression to the right of the sign ‘=’ is called Right Hand Side
(RHS)
(i) x – 5 (LHS) 6(RHS)
(ii) 4y (LHS) 12 (RHS)
(iii) 2z + 3 (LHS) 7 (RHS)
(iv) 3p (LHS) 24 (RHS)
(v) 4 (LHS) x – 2 (RHS)
(vi) 2a – 3 (LHS) -5 (RHS)

Question 3.
Solve the following equations by Trial & Error method.
(i) x + 3 = 5
Answer:

We find that for x = 2, both LHS and RHS are equal. Therefore x = 2 is the solution of the equation.

(ii) y – 2 = 7
Answer:

We find that for y = 9, both LHS and RHS are equal. Therefore y = 9 is the solution of the equation.

(iii) a – 2 = 6
Answer:

We find that for a = 8, both LHS and RHS are equal. Therefore a = 8 is the solution of the equation.

(iv) 5y = 15
Answer:

We find that for y = 3, both LHS and RHS are equal. Therefore y = 3 is the solution of the equation.

(v) 6n = 30
Answer:

We find that for n = 5, both LHS and RHS are equal. Therefore n = 5 is the solution of the equation.

(vi) 3z = 27
Answer:

We find that for z = 9, both LHS and RHS are equal. Therefore z = 9 is the solution of the equation.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchsticks patterns.
(i) A pattern of letter ‘T’
Answer:
The rule which gives the number of sticks required to make the pattern T is 2n.

(ii) A pattern of letter ‘E’
Answer:
The rule which gives the number of sticks required to make the pattern E is 5n.

(iii) A pattern of letter ‘Z’
Answer:
The rule which gives the number of sticks required to make the pattern Z is 3n.

Question 2.
Make a rule between the number of blades required and the number of fans (say n) in a hall ?

Answer:
The number of blades that a fan has = 3
Number of fans = n (say)
∴ The number of blades for ‘n’ fans = 3 × n = 3n
The required rule is 3n.

Question 3.
Find a rule for the following patterns between number of shapes formed and number of matchsticks required.
(a)

Answer:
The first shape has 2 matchsticks, the second shape 4 and the third shape 6.
∴ The required rule 2s.
(‘s’ stands for number of shapes.)

(b)

Answer:
The first shape has 3 matchsticks, the second shape 6 and the third shape 9.
∴ The required rule is 3s.
(‘s’ stands for number of shapes.)

Question 4.
The cost of one pen is ₹ 7 then what is the rule for the cost of ‘n’ pens.
Answer:
Cost of one pen is Rs. 7.
Cost of ‘n’ pens = 7 × n = 7n
The rule for cost of n’ pens is 7n.

Question 5.
The cost of one bag is ₹ 90, what is the rule for the cost of ‘m’ bags ?
Answer:
The cost of one bag = Rs. 90.
Cost of ‘m’ bags = 90 × m = 90m
The rule for the cost of ‘m’ bags is 90m.

Question 6.
The rule for purchase of books is-that the cost of q books is ₹ 23; then find the price of one book ?
Answer:
Cost of ‘q’ books = Rs. 23
∴ Cost of one book = \(\frac{23}{q}\)
The rule for the price of one book is \(\frac{23}{q}\).

Question 7.
John says that he has two books less than Gayathri. Write the relationship using letter x.
Answer:
Let the number of books that Gayathri has be x.
Then the number of books that John has = (x – 2)
(∵ John says that he has two books less than Gayathri)
The required relationship is (x – 2).

Question 8.
Rekha has 3 books more than twice the books with Suresh. Write the relationship using letter y.
Answer:
Let the number of books with Suresh be ‘y’.
Twice the number of books with Suresh = 2 × y – 2y.
As per the problem, the number of ‘ books that Rekha has = (2y + 3)
The required relationship is 2y + 3.

Question 9.
A teacher distributes 6 pencils per student. Can you find how many pencils are needed for the given number of students (use V for the number of students).
Answer:
Let the number of students be ’z’.
The teacher gives 6 pencils to each student.
The number of pencils that are given to ‘z’ students = 6 × z = 6z
The rule that is required is 6z.

Question 10.
Complete each table to generate the given functional relationship.

Answer:

Question 11.
Observe the following pattern.

Count the number of line segments in each shape.
(i) How many line segments will the 9th shape contain ?
(ii) Write the rule for the above pattern.
Answer:
(i) Shape – 1 contains 3 line segments.
Shape – 2 contains 5 line segments.
Shape – 3 contains 7 line segments.
Shape – 4 contains 9 line segments.
Shape – 5 contains 11 line segments.
Shape – 6 contains 13 line segments.
Shape – 7 contains 15 line segments.
Shape – 8 contains 17 line segments.
Shape – 9 contains 19 line segments.
The number of line segments that 9 such shapes contains is 99.
(∵ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)

(ii) The rule for the above pattern is 3 + 2(n – 1).

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise InText Questions

Think, Discuss and Write

Question 1.
How do we compare them ?
To compare them, we trace the line segments \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) on a tracing paper such that they are roughly aligned in the same direction. Do their end points coincide ?
Answer:
We can now say \(\overline{\mathrm{AB}}\) is longer than \(\overline{\mathrm{CD}}\). In the same way we can compare \(\overline{\mathrm{PQ}}\) with \(\overline{\mathrm{RS}}\). We can see \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{RS}}\) are of equal length.
(Or)
We can compare the lengths of \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{RS}}\) by a using a scale or a divider.
Now, we can say \(\overline{\mathrm{PQ}}\) > \(\overline{\mathrm{RS}}\)
∴ \(\overline{\mathrm{PQ}}\) = 3cm; \(\overline{\mathrm{RS}}\) = 25cm
∴ \(\overline{\mathrm{PQ}}\) > \(\overline{\mathrm{RS}}\)

Question 2.
What other errors can you find while measuring the length of line segment ?
For example, to find the length of a pencil, the eye Wrong Right Wrong should be correctly positioned as shown in the figure i.e., just vertically above the mark for both points. Other wise there may be an error due to angular viewing.

Answer:
(i) Eye error
(ii) Parallax error.

Try These

Question 1.
Take a post card and measure the length and breadth with ruler and divider. Do all post cards have the same dimensions ?
Answer:
By using ruler and a divider we can say that the dimensions of a post card are Length = 12 cm, Breadth = 7 cm
Same type of post cards are all having the same dimensions.

Question 2.
Select any three objects eraser, small pencil, etc. Trace their length on a paper. Measure the length of these line segments.
Answer:

The length of an eraser is 3 cm.
The length of a small pencil is 10 cms.

Try These

Question 1.
Use the right angle tester made of straw’s and identify the following angles.

Answer:
(i) Obtuse angle
(ii) Right angle
(iii) Acute angle
(iv) Obtuse angle

Question 2.
List out five daily life situations where you observe acute angles and obtuse angles
Answer:
Acute angles and obtuse angles are formed in following cases

1. The angle between the slant slopes of buffoon cap.
2. At the time of 2’o clock p.m. in a clock.
3. At the time of 10’o clock.
4. When the wrist is bonded.
5. When the knee is bonded.

Question 3.
Draw some angles of your choice. Test them by the “angle tester” and write which are acute and which are obtuse and which are right angles.
Answer:

Think, Discuss and Write

In the adjacent figure ∠AOB and ∠AOC are given. Which angle is clock-wise and which angle is anti clock-wise. Think and discuss with your friends.

Answer:
∠AOB is formed in a clock-wise direction.
∠AOC is formed in an anti clock-wise direction.

Try These

Question 1.
Which angle is greater ? Discuss with your friends.

Verify by measuring the angles using protractor. Is your estimation is correct ? Give reasons.
Answer:
The measure of ∠1 = 30° and that of ∠2 = 30°
∴ Both the angles ∠1 and ∠2 are of equal measure.

Question 2.
Which are acute angles ? Find and write their measures.

Answer:
(i) The measure of angle in figure (i) is 30°. So it is an acute angle,
(ii) The measure of angle in figure (ii) is 50°. So it is an acute angle.
(iii) The measure of angle in figure (iii) is 130°. It is not an acute angle. Because it is an obtuse angle.
(iv) The measure of angle in figure (iv) is 90°. So it is a right angle.
(v) The measure of angle in figure (v) is 80°. So it is an acute angle.

Question 3.
Which are obtuse angles?

Answer:
(i) The measure of angle in figure (i) is 110°. So it is an obtuse angle.
(ii) The measure of angle in figure (ii) is 90°. So it is a right angle.
(iii) The measure of angle in figure (iii) is 130°. So it is an obtuse angle.
(iv) The measure of angle in figure (iv) is 50°. So it is an acute angle.
(v) The measure of angle in figure (v) is 325°. So it is a reflex angle.

Question 4.
Draw any two acute and two obtuse angles of your choice.
Answer:
Acute angles:

AOB and POQ are acute angles.

Obtuse angles:

XOY and MON are obtuse angles.

Question 5.
Classify the following angles into acute, right, obtuse and straight angles.
40°, 140°, 90°, 210°, 44°, 215°, 345°, 125°
10°, 120°, 89°, 270°, 30°, 115°, 180°
Answer:
Acute angles : 40°, 44°, 10°, 89°, 30°
Right angle : 90°
Obtuse angles : 140°, 125°, 120°, 115°
Straight angle : 180°
Reflex angles : 210°, 215°, 345°, 270°

Think, Discuss and Write

Question 1.
If l ⊥ m, then can we say that m ⊥ l?
Answer:
If l ⊥ m then m ⊥ l.
Since both l, m are perpendicular to each other.

Question 2.
How many perpendicular lines can be drawn to a given line ?
Answer:
We can draw only one line which is perpendicular to the given line.

Question 3.
Which letters in English alphabet possess perpendicularity ?
Answer:
L, T possess the perpendicularity.

Try These

Question 1.
Draw two lines on a paper as shown in the figure.
Do they intersect each other ?
Can you call them parallel lines ? Give reason.

Answer:
The above lines are not intersecting. If, they are extended on either sides they will intersect at a point. They are called intersecting lines i.e., they are not parallel lines.

Question 2.
Make a pair of parallel lines. What is the angle formed between them ? Think, discuss with your friends and teacher.
Answer:
The angle between the parallel lines is 0° (zero). Since they do not intersect.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.1

Question 1.
Give any five examples of line segments observed in your classroom.
Eg: edge of black board.
Answer:
Edge of blackboard, edge of a book, edge of threshold, edge of window, height of the room, edge of table, edge of bench.

Question 2.
Why is it better to use a divider than a ruler, while comparing two line segments ?
Answer:
While finding the length of any object, the eye should be correctly positioned (i.e.) just vertically above the marks concerned. Otherwise there may be an error due to angular viewing. To avoid this problem a better way is to use a divider. In the same way, it would be better to use a divider while comparing two line segments.

Question 3.
Measure all the line segments in the figure given below and arrange them in the ascending order of their lengths.

Line segments \(\overline{\mathbf{A B}}, \overline{\mathbf{A C}}, \overline{\mathbf{A D}}, \overline{\mathbf{A E}}, \overline{\mathbf{B C}}, \overline{\mathbf{B D}}, \overline{\mathbf{B E}}, \overline{\mathbf{C D}}, \overline{\mathbf{C E}}, \overline{\mathbf{D E}}\)
Answer:
\(\overline{\mathrm{A}} \overline{\mathrm{B}}\) = 1 cm; \(\overline{\mathrm{A}} \overline{\mathrm{C}}\) = 2 cm; \(\overline{\mathrm{A}} \overline{\mathrm{D}}\) = 3 cm; \(\overline{\mathrm{A}} \overline{\mathrm{E}}\) = 4 cm
\(\overline{\mathrm{B}} \overline{\mathrm{C}}\) = 1 cm; \(\overline{\mathrm{B}} \overline{\mathrm{D}}\) = 2.0 cm; \(\overline{\mathrm{B}} \overline{\mathrm{E}}\) = 3.0 cm
\(\overline{\mathrm{C}} \overline{\mathrm{D}}\) = 1 cm; \(\overline{\mathrm{C}} \overline{\mathrm{E}}\) = 2.0 cm; \(\overline{\mathrm{D}} \overline{\mathrm{E}}\) = 1.0cm
Ascending order: \(\overline{\mathrm{AE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BE}}, \overline{\mathrm{CE}}, \overline{\mathrm{BD}}, \overline{\mathrm{AC}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}\)

Question 4.
Mid point of \(\overline{\mathrm{AB}}\) is located by Swetha and Reshrna like this.

Which one do you feel correct ? Measure the lengths of \(\overline{\mathbf{A C}}, \overline{\mathbf{C B}}\) and verify.
Answer:
By observation, we can say that the mid point of \(\overline{\mathrm{AB}}\) located by Reshrna is correct. (∵ AC = CB)
AC = 1 cm; CB = 2 cm (Swetha)
AC = 1.5 cm; CB = 1.5 cm (Reshrna)

Question 5.
Each of these figures given along side has many line segments. For the almirah we have shown one line segment along the longer edge. Identify and mark all such line segments in these figures.

Answer:
(i) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}\) and \(\overline{\mathrm{AC}}\) are the line segments.

(ii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}, \overline{\mathrm{AE}}, \overline{\mathrm{BF}}, \overline{\mathrm{CG}}, \overline{\mathrm{EF}}\) and \(\overline{\mathrm{FG}}\) are the line segments.

(iii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{BE}}, \overline{\mathrm{AF}}, \overline{\mathrm{FE}},\) and \(\overline{\mathrm{ED}}\) are the line segments.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.2

Question 1.
Write the expressions for the following statements.
(i) q is multiplied by 5.
Answer:
q × 5 = 5q

(ii) y is divided by 4.
Answer:
\(\frac{y}{4}\)

(iii) One fourth of the product of numbers p and q.
Answer:
The product of p and q = p × q = pq
One fourth of the product = \(\frac{\mathrm{pq} \times 1}{4}\) = \(\frac{\mathrm{pq}}{4}\)

(iv) 5 is added to the three times z.
Answer:
3 times z = 3 × z = 3z
If 5 is added to 3z, then we get 3z + 5

(v) 9 times ‘n’ is added to ’10’.
Answer:
9 times ‘n’ = 9 × n = 9n;
If 10 is added 9n, then it becomes 9n + 10

(vi) 16 is subtracted from two times ‘y’.
Answer:
2 times ‘y’ = 2 × y = 2y;
If 16 is subtracted from 2y, it becomes (2y – 16)

(vii) ‘y’ is multiplied by 10 and then x is added to the product.
Answer:
y × 10 + x = 10y + x

Question 2.
Write two statements each for the following expressions.
(i) y – 11
Answer:
(a) In a class of ‘y’ students, 11 students failed in the examination. Find the number of passed students.
(b) The sum of two numbers is y. If one number is 11, then what is the other number ?

(ii) 10a
Answer:
(a) The cost of one pen is Rs. 10. What is the cost of ‘a’ such pens ?
(b) What is the value of 10 times ‘a’ ?

(iii) \(\frac{x}{5}\)
Answer:
(a) The cost of 5 books is .Rs. ’x’.
Then what is the cost of one book?
(b) What is the value of one fifth of x?

(iv) 3m + 11
Answer:
(a) Add 11 to 3 times of m.
(b) The length of a room is 11 more than 3 times of its breadth (say m).

(v) 2y – 5
Answer:
(a) Subtract 5 from 2 times of ‘y’.
(b) The age of A is y and the age of B is 5 less than 2 times of the age of A. What is the age of B ?

Question 3.
Peter has ‘p’ number of balls. Number of balls with David is 3 times the balls with Peter. Write this as an expression.
Answer:
Number of balls that Peter has = p
Number of balls with David is 3 times the balls with Peter.
∴ Number of balls with David is 3p.

Question 4.
Sita has 3 more notebooks than Githa. Find the number of books that Sita has ? Use any letter for the number of books that Githa has.
Answer:
Let the number of books that Githa be x.
Sita has 3 more books than Githa.
∴ The number of books that Sita has = x + 3

Question 5.
Cadets are marching in a parade. There are 5 cadets in each row. What is the rule for the number of cadets, for a given number of rows ? Use ‘n’ for the number of rows.
Answer:
The number of cadets in each row is 5. Let the number of rows be ‘n’.
Total number of cadets in ‘n’ rows = 5 × n = 5n
∴ The required rule is 5n.

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.4

Question 1.
State which of the following sets are empty and which are not ?
a) The set of straight lines passing through a point.
b) Set of odd natural numbers divisible by 2.
c) {x : x is a natural number, x < 5 and x > 7}
d) {x : x is a common point to any two parallel lines}
e) Set of even prime numbers.
Answer:
a) The number of straight lines passing through a point is infinite. So, the given set is non-empty.
b) We know the odd natural numbers are 1, 3, 5, 7,…. are not divisible by 2. Hence the given set is empty.
c) There is no natural number satisfying the given condition. Hence the given set is empty.
d) There is no common point to any two parallel lines because they do not meet when produced on either side. Hence the given set is empty.
e) 2 is the only even prime number. Therefore the set contains one element. Hence the given set is non-empty.

Question 2.
Which of the following sets are finite or infinite ?
a) The set of months in a year.
b) {1, 2, 3,…….., 99, 100}
c) The set of prime numbers less than 99.
Answer:
a) There are 12 months in a year. The set of months in a year contains 12 elements. Hence the set is finite.
b) Obviously, the given set contains 100 elements. Hence the set is finite.
c) We can count the prime numbers less than 99. Hence the set is finite.

Question 3.
State whether each of the following sets is finite or infinite.
a) The set of letters in the english alphabet.
b) The set of lines which are parallel to the x – axis.
c) The set of numbers which are multiples of 5.
d) The set of circles passing through the origin (0, 0).
Answer:
a) The set of letters in the english alphabet contains 26 elements. Hence, the set is finite.
b) We cannot count the number of parallel lines drawn to the x – axis. Hence the set is infinite.
c) The set of numbers which are multiples of 5 is {5, 10, 15, 20, 25, …}. Hence the set contains infinite number of elements. Hence, the set is infinite.
d) The number of circles that can be drawn through the origin (0, 0) is countless. Hence the set is infinite.

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise InText Questions

Do This

Question 1.
Are 953, 9534, 900, 452 divisible by 2? Also check by actual division.
Answer:
(i) The given number is 953.
This number is divisible by 2 because it has not any one of the digits 0, 2, 4, 6 or 8 in its units place.

(ii) The given number is 9534.
This number is divisible by 2 because it has 4 in its units place.

(iii) The given number is 900.
This number is divisible by 2 because it has ‘0’ in its units place.

(iv) The given number is 452.
This number is divisible by 2 because it has 2 in its units place.

Do This

Check whether the following numbers are divisible by 3 ?
(i) 45986
Answer:
The given number is 45986.
Sum of the digits = 4 + 5 + 9 + 8 + 6 = 32
32 is not a multiple of 3.
So the given number is not divisible by 3.

(ii) 36129
Answer:
The given number is 36129.
Sum of the digits = 3 + 6 + 1 + 2 + 9 = 21
21 is a multiple of 3.
So the given number is divisible by 3.

(iii) 7874
Answer:
The given number is 7874.
Sum of the digits = 7 + 8 + 7 + 4 = 26
26 is not a multiple of 3.
So the given number is not divisible by 3.

Try These

Question 1.
Is 7224 divisible by 6 ? Why ?
Answer:
The given number has 4 in its units place.
So it is divisible by 2.
The sum of the digits of the given number is 7 + 2 + 2 + 4 = 15.
It is a multiple of 3.
So the given number is divisible by 3.
If a number is divisible by both 2 and 3 then it is divisible by 6 also.

Question 2.
Give two examples of 4 digit numbers which are divisible by 6.
Answer:
9648 and 3756.

Question 3.
Can you give an example of a number which is divisible by 6 but not by 2 and 3, why ?
Answer:
A number is divisible by 6 only when it is divisible by 2 and 3.
So, it is not possible to give an example for such number.

Do This

Question 1.
Test whether 9846 is divisible by 9 ?
Answer:
Number = 9846
Sum of the digits = 9 + 8 + 4 + 6 = 27 27
\(\frac{27}{9}\) = 3 9
∴ 9846 is divisible 9.

Question 2.
Without acutal division, find whether 8998794 is divisible by 9 ?
Answer:
Number = 8998794
Sum of the digits = 8 + 9 + 9 + 8 + 7 + 9 + 4 = 54
\(\frac{54}{9}\) = 6
∴ 8998794 is divisible by 9.

Question 3.
Check whether 786 is divisible by both 3 and 9 ?
Answer:
Number = 786
Sum of the digits = 7 + 8 + 6 = 21
\(\frac{21}{3}\) = 7
So 786 is divisible by 9.
But 21 is not divisible by 9.
So 786 is not divisible by 9.

Do This

Question 1.
Find the factors of 80.
Answer:
Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.

Question 2.
Do all the factors of a given number divide the number exactly ? Find the factors of 28 and verify by division.
Answer:
Yes, all the factors of a given number divide the number exactly.
Factors of 28 are 1, 2, 4, 7, 14, 28.

Verification:
\(\frac{28}{28}\) = 1, \(\frac{28}{14}\) = 2, \(\frac{28}{7}\) = 4, \(\frac{28}{4}\) = 7, \(\frac{28}{2}\) = 14

Question 3.
3 is a factor of 15 and 24. Is 3 a factor of their difference also ?
Answer:
Yes. (The difference between 15 and 24 is 9 and 9 is a multiple of 3.)

Try These

Question 1.
What is the smallest prime number?
Answer:
The smallest prime number is 2.

Question 2.
What is the smallest composite number?
Answer:
The smallest composite number is 4.

Question 3.
What is the smallest odd composite number?
Sol.
The smallest odd composite number is 6.

Question 4.
Give 5 odd and 5 even composite numbers.
Answer:
The odd composite numbers are
9, 15, 21, 25, 27 etc.
The even composite numbers are
4, 6, 8, 10, 12 etc.

Question 5.
Is 1 prime or composite and why?
Answer:
The number 1 has only one factor i.e. (itself). So, 1 is neither prime nor composite.

Question 6.
Can you guess a prime number which when on reversing Its digits, gives another prime number?
(Hint : Take a 2 digit prime number)
Answer:
13 is a prime number. On reversing its digits it becomes 31, which is also a prime number.

Question 7.
You know 311 is a prime number. Can you find the other two prime numbers just by rearranging the digits?
Answer:
Given prime number is 311.
The other two prime numbers just by rearranging the digits are 113 and 131.

Do This

From the following numbers identify different pairs of co-primes. 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Answer:
(2, 3); (2, 5); (2, 7); (3, 5); (3, 7); (5, 7)

Do This

Question 1.
Write the prime factors of 28 and 36 through division method.
Answer:

∴ Prime factors of 36 is 2 × 2 × 3 × 3

Question 2.
Write the prime factors of 42 by factor tree method.
Answer:

∴ Prime factors of 42 is 2 × 3 × 7

Do This

Find the HCF of 12, 16 and 28 by prime factorization method.
Answer:

The common factors of 12, 16 and 28 is 2 × 2 = 4
Hence, HCF of 12, 16 and 28 is 4.

Do This

Find the HCF of 28, 35 and 49 by division method^
Answer:
First find the HCF of any two numbers.
Let us find the HCF of 28 and 35

Last divisor is 7
∴ HCF of 28 and 35 is 7.
Then find the HCF of the third numbern and the HCF of first two numbers.
Let us find the HCF of 49 and 7.

HCF of 49 and 7 is 7.
∴ The HCF of 28, 35 and 49 is 7.

Think, Discuss and Write

What is the HCF of any two
(i) Consecutive numbers ?
Answer:
The HCF of any two consecutive numbers is 1.

(ii) Consecutive even numbers ?
Answer:
2

(iii) Consecutive odd numbers ?
Answer:
1 (one)

What do you observe ? Discuss with your friends.
Answer:
It was observed that the HCF of two consecutive numbers and consecutive odd numbers is same, i.e., 1.

Try This

Question 1.
find the LCM of
(i) 3, 4
(ii) 10, 11
(iii) 5, 6, 7
(iv) 10, 30
(v) 4, 12, 24
(vi) 3, 12
What do you observe ?
Answer:
(i) LCM of 3 and 4 = 3 × 4 = 12
(ii) LCM of 10 and 11 = 10 × 11 = 110
(iii) LCM of 5, 6 and 7 = 5 × 6 × 7 = 210
(iv) LCM of 10 and 30 = 10 × 1 × 3 = 30
(v) LCM of 4, 12 and 24 = 4 × 3 × 2 = 24
(vi) LCM of 3 and 12 = 3 × 4 = 12
It is observed that the LCM of two numbers will be their product, if the given numbers have no common factor except 1.

Think, Discuss and Write

When will the LCM of two or more numbers be their own product ?
Answer:
If the numbers are co-primes or relatively prime numbers then the LCM of two or more numbers be their own product.

Think, Discuss and Write

Question 1.
What is the LCM and HCF of twin prime numbers?
Answer:
Let the twin primes may be (3, 5)
LCM of 3, 5 is their product 3 × 5 = 15
HCF of 3, 5 is 1.
(for any type of twin prime)

Question 2.
Interpret relationship between LCM and HCF of any two numbers?
Answer:
Consider the two numbers be 14 and 21.
Now find LCM of 14 and 21.

∴ LCM of 14 and 21 = 7 × 2 × 3 = 42
Now find HCF of 14 and 21.

∴ HCF of 14 and 21 is 7.
Relation between LCM and HCF of 14 and 21:
42 × 7 = 14 × 21 = 294
Product of LCM and HCF of two numbers = Product of two numbers.

Do This

Divisibility Rule for 4:

Question 1.
Is 100000 is divisible by 4? Why?
Answer:
100000 = 1000 × 100
The given number is a multiple of 100.
We know, 100 is divisible by 4.
∴ The given number (i.e., 100000) is divisible by 4.

Question 2.
Give an example of a 2 digit number that is divisible by 2 but not divisible by 4?
Answer:
22, 26, 30, 34, 38 98 .
All the above two digit numbers are divisible by 2 but not divisible by 4.

Do This

Question 1.
Is 76104 divisIble by 8?
Answer:
The number formed by the last three digits is 104. It is divisible by 8.
Hence, the given number is divisible by 8.

Question 2.
Write the numbers that are divisible by 8 and lie between 100 and 200.
Answer:
104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192 are ail divisible by 8.
They lie between 100 and 200.

Page No. 92 (45)

Divisibility Rule for 11:

Using the division rule of ’11’. Fill the following table.

Answer:

1221 is a Palindrome number, which on reversing their digits gives the same number.
Thus, every Palindrome number with even number of digits, is always divisible by 11.

Write a Palindrome number of 6 digits and verify whether it is divisible by 11 or not.
Answer:
Palindrome number which on reversing their digits gives the same number.
Every Palindrome number with even number is always divisible by 11.

∴ The 6 digited Palindrome number is 123321. It is divisible by 11.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.3

Question 1.
Which of the following are models for parallel lines, perpendicular lines and which are neither of them.
(i) The vertical window bars.
(ii) Railway lines (Track)
(iii) The adjacent edges of door.
(iv) The letter ‘V’ in English alphabet.
(v) The opposite edges of Blackboard.
Answer:
(i) For parallel lines —— The vertical window bars.
(ii) For parallel lines —— Railway lines (Track)
(iii) For perpendicular lines —— The adjacent edges of door.
(iv) Neither parallel lines nor perpendicular lines —— The letter ‘V’ in English alphabet.
(v) For parallel lines The opposite edges of Blackboard.

Question 2.
Trace the copy of set squares (Geometry box) on a paper and mark the perpendicular edges.
Answer:

Question 3.
ABCD is a rectangle. \(\overline{\mathbf{A C}}\) and \(\overline{\mathbf{B D}}\) are diagonals. Write the pairs of parallel lines, perpendicular lines from the figure in symbolic form. Also write pairs of intersecting lines.

Answer:
(a) Parallel lines : AB || CD, AD || BC.
(b) Perpendicular lines: AD ⊥ DB, AB ⊥ BC, BC ⊥ CD, CD ⊥ DA
(c) Pair of intersecting lines: AC, BD.

Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials Ex 3.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.1

Question 1.
If p(x) = 5x⁷ – 6x⁵ + 7x – 6, find
(i) Coefficient of x⁵
(ii) degree of p(x)
(iii) constant term.
a) If P(x) = 5x⁷ – 6x⁵ + 7x – 6
Solution:
i) coefficient of x⁵ is -6
ii) degree of p(x) = highest degree of x = 7
iii) constant term is -6

Question 2.
State which of the following statements are true and which are false ? Give reasons for your choice.

i) The degree of the polynomial
\(\sqrt{2}\)x² – 3x + 1 is \(\sqrt{2}\).
Solution:
The given statement is false because \(\sqrt{2}\) is the coefficient of x² but not its degree. The degree of the polynomial is 2.

ii) The coefficient of x² in the polynomial p(x) = 3x³ – 4x² + 5x + 7 is 2.
Solution:
The given statement is false because the coeffi-cient of x² in the polynomial is -4 but not 2.

iii) The degree of a constant term is zero.
Solution:
The given statement is true. Because 3x⁰ = 3 the degree is 0

iv) \(\frac{1}{x^2-5 x+6}\) is a quadratic polynomial.
Solution:
The given statement is false because the variable ‘x’ appears in the denominator.

v) The degree of a polynomial is one more than the number of terms ¡n it.
Solution:
The given statement is false. There is a no relationship between the degree of the polynomial and the number of terms in it.

Question 3.
If p(t) = t³ – 1, find the values of p(1), p(-1), p(0), p(2). p(-2) (A.P.Mar. ’15)
Solution:
Given that p(t) = t³ – 1
∴ p(1) = (1)³ – 1 = 1 – 1 = 0
p(-1) = (-1) – 1 = -1 – 1 = -2
p(0) = (0) – 1 = 0 – 1 = -1
p(2) = (2) – 1 = 8 – 1 = 7
= (-2) – 1 = -8 – 1 = -9

Question 4.
Check whether -2 and 2 are the zeroes of the polynomial x⁴ – 16.
Solution:
p(x) = x⁴ – 16
p(-2) = (-2)⁴ – 16 = 16 – 16 = 0
p(2) = (2⁴) – 16 = 16 – 16 = 0
Yes, -2 and 2 are zeroes of the polynomial x⁴ – 16.

Question 5.
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x² – x – 6
Solution:
p(x) = x² – x – 6
= (3)² – 3 – 6 = 9 – 3 – 6 = 9 – 9 = 0
p(-2) = (-2)² – (-2) – 6 = 4 + 2 – 6 = 6 – 6 = 0
Yes, 3 and -2 are zeroes of the polynomial
p(x) = x² – x – 6