Srinivas

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

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Students must practice this TS Intermediate Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

I.
Question 1.
Find the algebraic equation whose roots are 3 times the roots of x3 + 2x2 – 4x + 1 = 0.
Solution:
Given equation is x3 – 2x2 – 4x + 1 = 0 ……………(1)
Let f(x) = x3 + 2x2 – 4x + 1
The equation whose roots are 3 times the roots of f(x) = 0 is given by f(\(\frac{x}{3}\)) = 0.
i.e., \(\left(\frac{x}{3}\right)^3+2\left(\frac{x}{3}\right)^2-4\left(\frac{x}{3}\right)\) + 1 = 0
⇒ \(\frac{x^3}{27}+\frac{2 x^2}{9}-\frac{4 x}{3}\) + 1 = 0
⇒ x3 + 6x2 – 36x + 27 = 0.

Question 2.
Find the algebraic equation whose roots are 2 times the roots of x5 – 2×4 + 3×3 – 2×2 + 4x + 3 = 0.
Solution:
Given equation is x5 – 2x4 + 3x3 – 2x2 + 4x + 3 = 0 ……..(1)
Let f(x) = x5 – 2x4 + 3x3 – 2x2 + 4x + 3
The equation whose roots are 2 times the roots of f(x) = 0 is given by f(\(\frac{x}{2}\)) = 0
i.e., \(\left(\frac{x}{2}\right)^5-2\left(\frac{x}{2}\right)^4+3\left(\frac{x}{2}\right)^3-2\left(\frac{x}{2}\right)^2+4\left(\frac{x}{2}\right)+3\) = 0
⇒ x5 – 4x4 + 12x3 – 16x2 + 64x + 96 = 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 3.
Find the transformed equation whose roots are the negatives of the roots of x4 + 5×3 + lix + = 0.
Solution:
Given equation is x4 + 5x3 + 11x + 3 = 0
Let f(x) = x4 + 5x3 + 11x + 3
The transformed equation whose roots are the negatives of the roots of f(x) = 0 is
f(- x) = 0.
i.e., (- x)4 + 5(- x)3 + 11 (- x) + 3 = 0
⇒ x4 – 5x3 – 11x + 3 = 0.

Question 4.
Find the transformed equation whose roots are the negatives of the root of x7 + 3x5 + x3 – x2 + 7x + 2 = 0.
Solution:
Given equation is
x7 + 3x5 + x3 – x2 + 7x + 2 = 0
Let f(x) = x7 + 3x5 + x3 – x2 + 7x + 2
The transformed equation whose roots are the negatives of the roots of f(x) = 0 is
f(- x) = 0.
i.e., (- x)2 + 3(- x)5 + (- x)3 – (- x)2 + 7(- x) + 2 = 0
x7 + 3x5 + x3 + x2 + 7x – 2 = 0.

Question 5.
Find the polynomial equation whose roots are the reciprocals of the roots of x4 – 3x3 + 7x2 + 5x – 2 = 0.
Solution:
Given equation is x4 – 3x3 + 7x2 + 5x – 2 = 0 ………….(1)
Let f(x) = x4 – 3x3 + 7x2 + 5x – 2
The polynomial equation whose roots are the reciprocals of the roots of (1) is given by
f(\(\frac{1}{x}\)) = 0
i.e., \(\left(\frac{1}{x}\right)^4-3\left(\frac{1}{x}\right)^3+7\left(\frac{1}{x}\right)^2+5\left(\frac{1}{x}\right)\) – 2 = 0
⇒ 1 – 3x + 7x2 + 5x3 – 2x4 = 0
⇒ 2x4 – 5x3 – 7x2 + 3x – 1 = 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 6.
Find the polynomial equation whose roots are the reciprocals of the roots of x5 + 11x4 + x3 + 4x2 – 13x + 6 = 0.
Solution:
Given equation is x5 + 11x4 + x3 + 4x2 – 13x + 6 = 0 …………(1)
Let f(x) = x5 + 11x4 + x3 + 4x2 – 13x + 6
The polynomial equation whose roots are the reciprocals of the roots of f(x) = 0 is
f(\(\frac{1}{x}\)) = 0
i.e., \(\left(\frac{1}{x}\right)^5+11\left(\frac{1}{x}\right)^4+\left(\frac{1}{x}\right)^3\) + 6 = 0
⇒ 6x5 – 13x4 + 4x3 + x2 + 11x + 1 = 0.

II.
Question 1.
Find the polynomial equation whose roots are the squares of the roots of x4 + x3 + 2x2 + x + 1 = 0.
Solution:
Given equation is x4 + x3 + 2x2 + x + 1 = 0
Let f(x) = x4 + x3 + 2x2 + x + 1
The polynomial equation whose roots are squares of the roots of f(x) = 0 is f (√x) = 0.
i.e.. (4√x)4 + (√x)3 + 2(√x)2 + √x + 1 = 0
⇒ x2 + 2x + 1 = √x (x + 1)
⇒ (x + 1)2 = √x (x + 1)
⇒ (x + 1)4 = x (x + 1)2
⇒ x4 + 4x3 + 6x2 + 4x + 1 = x(x2 +2x + 1)
⇒ x4 + 3x3 + 4x2 + 3x + 1 = 0 is the required equation.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 2.
Form the polynomial equation whose roots are the squares of the roots of x3 + 3x2 – 7x + 6 = 0.
Solution:
Given equation is x3 + 3x2 – 7x + 6 = 0 …… (1)
Let f(x) = x3 + 3x2 – 7x + 6
The polynomial equation whose roots are the squares of the roots of f(x) = 0 is f(√x) = 0.
i.e., (√x)3 + 3(√x)2 – 7(√x) + 6 = 0
⇒ x√x – 7√x = – 3x – 6
⇒ √x (x – 7) = – (3x + 6)
⇒ x (x2 + 49 – 14x) = 9x2 + 36 + 36x
⇒ x3 – 23x2 + 13x – 36 = 0 is the required equation.

Question 3.
Form the polynomial equation whose roots are the cubes of the roots of x3 + 3x2 + 2 = 0.
Solution:
Given equation is x3 + 3x2 + 2 = 0 …………. (1)
Let f(x) = x3 + 3x2 + 2
The polynomial equation whose roots are the cubes of the roots of f(x) = 0 is f(\(\sqrt[3]{x}\)) = 0.
i.e., \((\sqrt[3]{x})^3\) + 3 (\((\sqrt[3]{x})^2\)) + 2 = 0
⇒ x + 2 = – 3x2/3
⇒ (x + 2)3 = – 27 (x2)
⇒ x3 + 6x2 + 12x + 8 = – 27x2
⇒ x3 + 33x2 + 12x + 8 = 0 is the required equation.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

III.
Question 1.
Find the polynomial equation whose roots are the translates of those of the equation – 5x3 + 7x2 – 17x + 11 = 0 by – 2.
Solution:
Given equation is
x4 – 5x3 + 7x2 – 17x + 11 = 0 ………(1)
Let f(x) = x4 – 5x3 + 7x2 – 17x + 11
The polynomial equation, whose roots are
the translates of those of the f(x) = 0 by – 2 is f(x + 2) = 0.
Suppose that
f(x + 2) = A0x4 + A1x3 + A2x2+ A3x + A4
By synthetic division, the coefficients A0, A1, A2, A3 and A4 are obtained as follows.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) 1

∴ The roots ol the equation x4 – 3x3 + x2 – 17x + 19 = 0 are the translates of the roots of the given equation by – 2.

Question 2.
Find the polynomial equation whose roots are the translates of those of the equation x5 – 4x4 + 3x2 – 4x + 6 = 0 by – 3.
Solution:
Given equation is x5 – 4x4 + 3x2 – 4x 4 6 = 0 ………….(1)
Let f(x) = x5 – 4x4 + 3x2 – 4x + 6
The polynomial equation, whose roots are the translates of those of the f(x) = 0 by – 3 is f(x + 3) = 0.
Suppose that f(x + 3) = A0x5 + A1x4 + A2x3 + A3x2 + A4x + A5
By synthetic division, the coefficients A0, A1, A2, A3, A4 and A5 are obtained as follows.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) 2

∴ The roots of the equation x5 + 11x4 + 42x3 + 57x2 – 13x – 60 = 0 are the translates of the roots of the given equation by – 3.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 3.
Find the polynomial equation whose roots are the translates of the roots of the equation x4 – x3 – 10x2 + 4x + 24 = 0 by 2.
Solution:
Given equation is x4 – x3 – 10x2 + 4x + 24 = 0
Let f(x) = x4 – x3 – 10x2 + 4x + 24
The polynomial equation whose roots are trans-lates of those of the f(x) = 0 by 2 is f(x – 2) = 0.
Suppose that f(x – 2) = A0x4 + A1x3 + A2x2 + A3x + A4
By synthetic division, the coefficients A0, A1, A2, A3, A4 are obtained as follows – 2.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) 3

∴ The roots of the equation x4 – 9x3 + 20x2 = 0 are the translates of the roots of the given equati on by 2.

Question 4.
Find the polynomial equation whose roots are the translates of the roots of the equation 3x5 – 5x3 + 7 = 0 by 4.
Solution:
Given equation is 3x5 – 5x3 + 7 = 0
Let f(x) = 3x5 – 5x3 + 7
The polynomial equation whose roots are the translates of those of the f(x) = 0 by 4 is f(x – 4) = 0.
Suppose that
f(x – 4) = A0x5 + A1x4 + A2x3 + A3x2 + A4x + A5
By synthetic division, the coefficients A0, A1, A2, A3, A4, A5 are obtained as follows.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) 4

∴ The roots of the equation 3x5 – 60x4 + 475x2 – 1860x3 + 3600x – 2745 = 0 are the translates of the roots of the given equation by 4.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 5.
Transform e jach of the following equations into ones i n which the coefficients of the second hig >hest power of x is zero and also find their transformed equations.
i) x3 – 6x2 + 10x – 3 = 0
ii) x4 + 4x3 + 2x2 – 4x – 2 = 0
iii) x3 – 6x2 + 4x – 7 = 0
iv) x3 + 6x2+ 4x + 4 = 0
Solution:
i) Given equation is x3 – 6x + 10x – 3 = 0
Let f(x) = x3 – 6x2 + 10x – 3
We have to find ’h’ so that the coefficient of the Second highest power of x in f(x + h) is zero.
i.e., Coefficient of x2 in f(x + h) is zero,
f (x + h) = (x + h)3 – 6 (x + h)2 + 10 (x + h) – 3 Coefficients of x2 in f(x + h) is 3h – 6.
We choose ‘h‘ such that 3h – 6 = 0 i.e., h = 2
∴ f (x + 2) = (x + 2)3 – 6 (x + 2)2 + 10 (x + 2) – 3
= x3 + 6x2 + 12x + 8 – 6 (x2 + x + 4) + 10x + 20 – 3
= x3 – 2x + 1
∴ x3 – 2x + 1 = 0 is the required equation.

ii) Given equation is
x4 + 4x3 + 2x2 – 4x – 2 = 0
Let f(x) = x4 + 4x3 + 2x2 – 4x – 2
We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.
i.e., coefficient of x2 in f(x + h) is zero,
f (x + h) = (x + h)4 + 4 (x + h)3 + 2 (x + h)2 , – 4 (x + h) – 2
Coefficient of x2 in f(x + h) is 4h + 4.
We choose ‘h‘ such that 4h + 4 = 0 i.e., h = – 1
∴ f(x – 1) = (x – 1)4 + 4 (x – 1)3 + 2 (x – 1)2 – 4 (x – 1) – 2
= (x4 – 4x3 + 6x2 – 4x + 1) + 4 (x3 – 3x2 + 3x – 1) + 2 (x2 – 2x + 1) – 4 (x – 1) – 2
= x4 – 4x2 + 1
∴ x4 – 4x2 + 1 = 0 is the required equation.

iii) Given equation is x3 – 6x2 + 4x – 7 = 0
Let f(x) = x3 – 6x2 + 4x – 7
We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.
i.e., coefficient of x2 in f(x + h) is zero.
f(x + h) = (x + h)3 – 6 (x + h)2 + 4 (x + h) – 7
Coefficient of x2 in f(x + h) is 3h – 6
We choose ‘h’ such that 3h – 6 = 0 i.e., h = 2
∴ f(x + 2) = (x + 2)3 – 6 (x + 2)2 + 4 (x + 2) – 7
= (x3 + 6x2 + 12x + 8) – 6 (x2 + 4x + 4) + 4 (x + 2) – 7
= x3 – 8x – 15
∴ x3 – 8x – 15 = 0 is the required equation.

iv) Given equation is x3 + 6x2 + 4x + 4 = 0
Let f(x) = x3 + 6x2 + 4x + 4
We have to find ’h’ so that the coefficient of the second highest power of x in f(x + h) is zero, i.e., coefficient of x2 in f(x + h) is zero.
f(x + h) = (x + h)3 + 6(x + h)2 + 4(x + h) + 4
Coefficient of x2 in 1(x + h) is 3h + 6.
We have to choose ‘h’ such that 3h + 6 = 0 i.e., h = – 2
∴ f(x – 2)= (x – 2)3 + 6(x – 2)2 + 4(x – 2) + 4
= (x3 – 6x2 + 12x – 8) + 6 (x2 – 4x + 4) + 4 (x – 2) + 4
= x3 – 8x + 12
∴ x3 – 8x + 12 = is the required equation.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 6.
Transform each of the following equations into ones in which the coefficients of the third highest power of x is zero.
i) x4 + 2x<sup3 – 12x2 + 2x – 1 = 0
ii) x3 + 2x2 + x + 1 = 0
Solution:
i) Given equation is
x4 + 2x3 – 12x2 + 2x – 1 = 0
Let f(x) = x4 + 2x3 – 12x2 + 2x – 1
We have to find h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.
i.e., Coefficient of x2 in [(x + h) is zero.
f(x + h) = (x + h)4 + 2 (x + h)3 – 12 (x + h)2 + 2 (x + h) – 1
Coefficient of x3 in 1(x + h) is 6h2 + 6h – 12
We have to choose ‘h’ such that
6h2 + 6h – 12 = 0
⇒ (h + 2) (h – 1) = 0
⇒ h = – 2 or 1.

Case – (I):
When h = – 2
f(x – 2) = (x – 2)4 + 2 (x – 2)3 – 12(x – 2)2 + 2 (x – 2) – 1
= x4 – 8x3 + 24x2 – 32x + 16 + 2(x3 – 6x2 + 12x – 8) – 12(x2 – 4x + 4) + 2(x – 2) – 1
= x4 – 6x3 + 42x – 53
∴ Tranformed equation is x4 – 6x3 + 42x – 53 = 0.

Case-(ii):
When h = 1
f(x + 1) = (x + 1)4 + 2(x + 1)3 – 12(x + 1)2 + 2 (x + 1) – 1
= (x4 + 4x3 + 6x2 + 4x + 1) + 2(x3 + 3x2 + 3x + 1) – 12 (x2 + 2x + 1) + 2(x + 1) – 1
= x4 + 6x3 – 12x – 8
∴ Tranformed equation is x4 + 6x3 – 12x – 8 = 0
∴ Required equation is x4 – 6x3 + 4x – 53 = 0
or x4 + 6x3 – 12x – 8 = 0.

ii) Given equation is x3 + 2x2 + x + 1 = 0
Let f(x) = x3 + 2x2 + x + 1
We have o find ‘h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.
i.e., Coefficient of x in f(x + h) is zero.
f(x + h) = (x + h)3 + 2(x + h)2 + (x + h) + 1
Coefficient of ‘x3’ in f(x + h) is 3h2 + 4h + 1
We have to Choose ‘h’ such that 3h2 + 4h + 1 = 0
i.e., h = – 1 or h = – \(\frac{1}{3}\).

Case – (I):
When h = – 1
f(x – 1) = (x – 1)3 + 2(x – 1)2 +(x – 1) + 1
= (x3 – 3x2 + 3x – 1)2 + 2 (x2 – 2x + 1) + x – 1 + 1
= x3 – x2 + 1
∴ Transformed euluation is x3 – x2 + 1 = 0.

Case – (ii):
When h = – \(\frac{1}{3}\)
\(f\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)^3+2\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\)
= \(\left(x^3-x^2+\frac{x}{3}-\frac{1}{27}\right)+2\left(x^2-\frac{2}{3} x+\frac{1}{9}\right)\) + x – \(\frac{1}{3}\) + 1
= x3 + x2 + \(\frac{23}{27}\)
∴ Transformed equation is x3 + x2 + \(\frac{23}{27}\)
⇒ 27x3 + 27x2 + 23 = 0
∴ Required equation is x3 – x2 + 1 = 0 or 27x3 + 27x2 + 23 = 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

Question 7.
Solve the following equations.
i) x4 – 10x3 + 26x2 – 10x + 1 = 0.
ii) 2x5 + x4 – 12x3 – 12x2 + x + 2 = 0
Solution:
i) Given equatIon is
x4 – 10x3 + 26x2 – 10x + 1 = 0 (1)
This is even degree reciprocal equation of class one.
On dividing (1) by x2,
x2 – 10x + 26 – \(\frac{10}{x}+\frac{1}{x^2}\) = 0
⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 26 = 0
⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 24 = 0
[Put x + \(\frac{1}{x}\) = y)
⇒ y2 – 10y + 24 = 0
⇒ (y – 4) (y – 6) = 0
⇒ y = 4 or y = 6.

Case – (i):
When y = 4
⇒ x + \(\frac{1}{x}\) = 4
⇒ x2 – 4x + 1 = 0
⇒ x = 2 ± √3.

Case – (ii):
When y = 6
⇒ x + \(\frac{1}{x}\) = 4
⇒ x2 – 6x + 1 = 0
⇒ x = 3 ± 2√2
∴ The roots are 2 ± √3, 3 ± 2√2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

ii) Given equation is
2x5 + x4 – 12x3 – 12x2 + x + 2 = 0 ………………. (1)
Let f(x) = 2x5 + x4 – 12x3 – 12x2 + x + 2
(1) is an odd degree reciprocal equation of class one.
∴ x = – 1 is a root of (1)
x + 1 is a factor of f(x).
We divide f(x) with x + 1.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) 5

∴ f(x) = (x + 1) (2x4 – x3 – 11x2 – x + 2)
Let g(x) = 2x4 – x3 – 11x2 – x + 2
g(x) = 0 is an even degree reciprocal equation 0f class one.
Dividing g(x) = 0 by x2
We get 2x2 – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0
⇒ \(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)\) – 11 = 0
⇒ \(2\left(x+\frac{1}{x}\right)^2-\left(x+\frac{1}{x}\right)\) – 15 = 0
(Put x + \(\frac{1}{x}\) = y)
⇒ 2y2 – y – 15 = 0
⇒ (2y + 5) (y – 3) = 0
∴ y = 3 or y = – 1.

Case – (i) :
When y = 3
x + \(\frac{1}{x}\) = 3
⇒ x2 – 3x + 1 = 0
⇒ x = \(\frac{3 \pm \sqrt{5}}{2}\)

Case – (ii):
When y = \(\frac{-5}{2}\)
x + \(\frac{1}{x}\) = \(\frac{-5}{2}\)
⇒ 2x2 + 2 = – 5x
⇒ 2x2 + 5x + 2 = 0
⇒ (2x + 1) (x + 2) = 0
∴ x = – \(\frac{-1}{2}\) or x = – 2
∴ The roots are – 1, 2, \(\frac{3 \pm \sqrt{5}}{2}\).

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

by

Students must practice this TS Intermediate Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

I.
Question 1.
Solve x3 – 3x2 – 16x + 48 = 0, given that the sum of two roots is zero.
Solution:
Let α, β, γ be the roots of
x3 – 3x2 – 16x + 48 = 0 ………….(1)
Given that the sum of two roots is zero.
Let α + β = 0 …………(2)
But from (1) we have α + β + γ = 3
⇒ γ = 3
Hence αβγ = – 48
⇒ αβ = – 16
We know that,
(α + β)2 – (α – β)2 = 4αβ
⇒ (α – β)2 = 64
⇒ α – β = ± 8.

i) When α – β = 8
⇒ 2α = 8 (∵ from (2))
⇒ α = 4
∴ β = – 4.

ii) When α – β = – 8
⇒ 2α = – 8 (∵ from (2))
⇒ α = – 4
∴ β = 4.
The roots of given equation are 4, – 4 and 3.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 2.
Find the condition that x3 – px2 + qx – r = 0 may have the sum of two of its roots zero.
Solution:
Given equation is x3 – px2 + qx – r = 0
Let α, β, γ be the roots.
∴ α + β + γ = p …………..(1)
Given sum of two of its roots is zero.
∴ (1) ⇒ α + 0 = p
i. e„ α = p
Substituting in given equation, we get p3 – p3 + pq – r = 0
⇒ pq – r = 0
⇒ Pq = r.

Question 3.
Given that the roots of x3 + 3px2 + 3qx + r = 0 are in
i) A.P., show that 2p3 – 3qp + r = 0
ii) G.P., show that p3r = q3
iii) H.P., show that 2q3 = r (3pq – r).
Solution:
Given cubic equation is
x3 + 3px2 + 3qx + r = 0 ……………..(1)
Let α, β, γ be its roots.

i) When the roots are in A.P. :
i.e., 2β = α + γ
from (1) α + β + γ = – 3p
⇒ (α + γ) + P = – 3p
⇒ 2β + β = – 3p
⇒ 3β = – 3p
⇒ β = – P
Substituting in given equation, we get
– p3 + 3p3 – 3pq + r = 0
⇒ 2p3 – 3pq + r = 0.

ii) When the roots are in G.P. :
∴ β2 = αγ
from (1) αβγ = – r
⇒ β3 = – r
⇒ β = – r1/3
Substituting in (1), we get
(- r1/3)3 + 3pr2/3 – 3qr1/3 + r = 0
⇒ 3pr2/3 = 3qr1/3
⇒ P3r = q3.

iii) When the roots are in H.P. :
β = \(\frac{2 \alpha \gamma}{\alpha+\gamma}\)
⇒ β2 = \(\frac{2 \alpha \beta \gamma}{(\alpha+\beta+\gamma)-\beta}\)
⇒ β2 = \(\frac{-2 r}{-3 p-\beta}\)
⇒ β2 + 3pβ2 = 2r (∵ β is a root of (1))
⇒ – 3qβ – r = 2r
β = – \(\frac{r}{q}\)
Substituting in (1) we get
\(\frac{-r^3}{q^3}+\frac{3 p r^2}{q^2}-\frac{3 q r}{q}\) + r = 0
⇒ 2q3 = r(3pq – r).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 4.
Find the condition that x3 – px2 + qx – r = 0 may have the roots in G.P.
Solution:
Let α, β, γ be the roots of
x3 – px2 + qx – r = 0
If α, β, γ are in G.P., then
β2 = αγ
(1) ⇒ αβγ = r
⇒ β3 = r
β = r1/3
Substituting in (1), we get
(r1/3)3 – p(r1/3)2 + q(r1/3) – r = 0
⇒ pr2/3 = qr1/3
⇒ p3r2 = q3r
⇒ q3 = p3r.

II.
Question 1.
Solve 9x3 – 15x2 7x – 1 = 0, given that two of its roots are equal.
Solution:
Given cubic equation is
9x3 – 15x2 – 7x – 1 = 0 ……………. (1)
Suppose α, β, γ are the roots of (1)
∴ α + β + γ = \(\frac{15}{9}=\frac{5}{3}\)
αβ + βγ + γα = \(\frac{7}{9}\)
αβγ = \(\frac{1}{9}\)
According to the problem, α = β (∵ two of its roots are equal)
∴ 2α + γ = \(\frac{5}{3}\)
⇒ γ = \(\frac{5}{3}\) – 2α
Also, α2 + 2αγ = \(\frac{7}{9}\)
⇒ α2 + 2α (\(\frac{5}{3}\) – 2α) = \(\frac{7}{9}\)
⇒ 27α2 – 30α + 7 = 0
⇒ (3α – 1) (9α – 7) = 0
∴ α = \(\frac{1}{3}\) or α = \(\frac{7}{9}\)

Case (i) :
when α = \(\frac{1}{3}\)
γ = \(\frac{5}{3}\) – 2α
= \(\frac{5}{3}\) – \(\frac{2}{3}\) = 1
∴ The roots are \(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Case – (ii):
When α = \(\frac{7}{9}\)
γ = \(\frac{5}{3}\) – 2α
= \(\frac{5}{3}-\frac{14}{9}\) = \(\frac{1}{9}\)
Which is impossible as
αβγ = \(\frac{7}{9} \cdot \frac{7}{9} \cdot \frac{1}{9}\) ≠ \(\frac{1}{9}\)
∴ The roots are \(\frac{1}{3}\), \(\frac{1}{3}\), 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 2.
Given that one root of 2x3 + 3x2 – 8x + 3 = 0 is double the other root, find the roots of equation.
Solution:
Given cubic equation is
2x3 + 3x2 – 8x + 3= 0 ……………..(1)
Suppose α, β, γ are the roots of (1).
∴ α + β + γ = \(\frac{-3}{2}\)
αβ + βγ + γα = \(\frac{-8}{2}\) = – 4 …………….(2)
αβγ = \(\frac{-3}{2}\)
Given one root is double the other.
3α + γ = \(\frac{-3}{2}\)
⇒ γ = \(\frac{-3}{2}\) – 3α
Also from (2):
2α2 – 3α (\(\frac{3}{2}\) + 3α) = – 4
14α2 + 9α – 8 = 0
(2α – 1) (7α + 8) = 0
α = \(\frac{1}{2}\) or α = \(\frac{-8}{7}\).

Case (i):
When α = \(\frac{1}{2}\)
β = 2α = 2 (\(\frac{1}{2}\)) = 1
γ = \(\frac{-3}{2}\) – 3α
= \(\frac{-3}{2} \frac{-3}{2}\) = – 3.
∴ α = \(\frac{1}{2}\), β = 1 and γ = – 3
satisfies αβγ = \(\frac{-3}{2}\)
∴ The roots are \(\frac{1}{2}\), 1, – 3.

Case (ii):
When α = \(\frac{-8}{7}\)
β = 2α = \(\frac{-16}{7}\)
γ = \(\frac{-3}{2}\) – 3α
= \(\frac{-3}{2}+\frac{48}{7}=\frac{75}{14}\)
But α = \(\frac{-8}{7}\), β = \(\frac{-16}{7}\) and γ = \(\frac{75}{14}\) do not satisfy αβγ = \(\frac{-3}{2}\).
Hence the roots of given equation are \(\frac{1}{2}\), 1, – 3.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 3.
Solve x3 – 9x2 + 14x + 24 = 0, given that two of the roots are in the ratio 3 : 2.
Solution:
Given cubic equation is
x3 – 9x2 + 14x + 24 = 0 ……….(1)
Let α, β, γ be the roots of (1)
∴ α + β + γ = 9, αβ + βγ + γα = 14, αβγ = – 24 ……………..(2)
Given two roots are in the ratio 3 : 2,
let α : β = 3 : 2
⇒ β = \(\frac{2 \alpha}{3}\)
Now from (2) \(\frac{5 \alpha}{3}\) + γ = 9
⇒ γ = 9 – \(\frac{5 \alpha}{3}\)
Also, \(\frac{2}{3}\) α2 + (9 – \(\frac{5 \alpha}{3}\)) \(\frac{5 \alpha}{3}\) = 14
⇒ 2α2 + \(\frac{5 \alpha(27-5 \alpha)}{3}\) = 42
⇒ 19α2 – 135α + 126 = 0
⇒ (19α – 21) (α – 6) = 0
⇒ α = \(\frac{21}{19}\) or α = 6.

Case (i):
When α = \(\frac{21}{19}\)
β = \(\frac{2}{3}(\alpha)=\frac{2}{3}\left(\frac{21}{19}\right)=\frac{14}{19}\)
γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}\left(\frac{21}{19}\right)=\frac{136}{19}\)
These values do not satisfy αβγ = – 24.

Case – (ii) :
When α = 6
β = \(\frac{2}{3}(\alpha)=\frac{2}{3}(6)\) = 4
γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}(6)\) = – 1
These values satisfy αβγ = – 24.
∴ The roots of given equation are 6, 4, – 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 4.
Solve the following equations, given that the roots of each are in A.P.
i) 8x3 – 36x2 – 18x + 81 = 0
ii) x3 – 3x2 – 6x + 8 = 0
Solution:
i) Given cubic equation is
8x3 – 36x2 – 18x + 81 = 0 …………….(1)
Given the roots are in A.P.
∴ α – d, α, α + d be the roots.
∴ Sum of the roots 3α = \(\frac{36}{8}\)
⇒ α = \(\frac{3}{2}\)
∴ x – \(\frac{3}{2}\) is a factor of 8x3 – 36x2 – 18x + 81
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 1

8x3 – 36x2 – 18x + 81 = (x – \(\frac{3}{2}\)) (8x2 – 24x – 54)
∴ Equation (1)
⇒ (x – \(\frac{3}{2}\)) (8x2 – 24x – 54) = 0
⇒ (x – \(\frac{3}{2}\)) (2x + 3) (2x – 9) = 0
⇒ x = – \(\frac{3}{2}\) or x = \(\frac{3}{2}\) or x = \(\frac{9}{2}\)
∴ The roots are \(\frac{-3}{2}\), \(\frac{3}{2}\), \(\frac{3}{2}\).

ii) Given roots of cubic equation
x3 – 3x2 – 6x + 8 = 0 ……………(1) are in G.P.
Let α – d, α, α + d be the roots.
∴ Sum of the roots 3α = 3
⇒ α = 1
∴ (x – 1) is a factor of x3 – 3x2 – 6x + 8.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 2

x3 – 3x2 – 6x + 8 = (x – 1) (x2 – 2x – 8)
∴ Equation (1)
⇒ (x – 1) (x2 – 2x – 8) = 0
⇒ (x – 1) (x – 4) (x + 2) = 0
∴ x = 1 or x = 4 or x = – 2.
∴ The roots are – 2, 1, 4.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 5.
Solve the following equations, given that the roots of each are in GP.
i) 3x3 – 26x2 + 52x – 24= 0
ii) 54x3 – 39x2 – 26x + 16 = 0
Solution:
i) Given roots of cubic equation
3x3 – 26x2 + 52x – 24 = 0 ……………. (1) are in G.P.
Let \(\frac{\alpha}{r}\), α, αr be the roots.
∴ Product of the roots α3 = \(\frac{24}{3}\) = 8
⇒ α = 2
∴ (x – 2) is a factor of 3x3 – 26x3 + 52x – 24
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 3

3x3 – 26x2 + 52x – 24 = (x – 2) (3x2 – 20x + 12)
∴ Equation (1) ⇒ (x – 2) (3x2 – 20x + 12) = 0
⇒ (x – 2) (3x – 2) (x – 6) = 0
∴ x = 2 or x = \(\frac{2}{3}\) or x = 6
∴ The roots are \(\frac{2}{3}\), 2, 6.

ii) Given roots of cubic equation.
54x3 – 39x2 – 26x + 16 = 0 (1) are in GP.
Let \(\frac{\alpha}{r}\), α, αr be the roots.
∴ Product of the roots α3 = \(\frac{-16}{54}\)
⇒ α3 = \(\frac{-2}{3}\)
∴ (x + \(\frac{2}{3}\)) is a factor 54x3 – 39x2 – 26x + 16
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 4

54x3 – 39x2 – 26x + 16 = (x + \(\frac{2}{3}\)) (54x2 – 75x + 24)
∴ Equation (1),
(x + \(\frac{2}{3}\)) (54x2 – 75x + 24) = 0
(x + \(\frac{2}{3}\)) (18x2 – 25x + 8) = 0
(x + \(\frac{2}{3}\)) (9x – 8)(2x – 1) = 0
∴ x = – \(\frac{2}{3}\) or x = \(\frac{8}{9}\) or x = \(\frac{1}{2}\)
∴ The roots are \(\frac{8}{9}\), \(\frac{-2}{3}\), \(\frac{1}{2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 6.
Solve the following equations, given that the roots of each are In H.P.
i) 6x3 – 11x2 + 6x – 1 = 0
ii) 15x3 – 23x2 – 9x – 1 = 0
Solution:
i) Given cubic equation is
6x3 – 11x2 + 6x – 1 = 0 …………..(1)
Put y = \(\frac{1}{x}\)
∴ (1) ⇒ \(\frac{6}{y^3}-\frac{11}{y^2}+\frac{6}{y}\) – 1 = 0
⇒ y3 – 6y2 + 11y – 6 = 0 ………… (2)
Given roots of (1) are in H.P.
⇒ Roots of (2) are in AP.
Let a – d, a, a + d be the roots of (2),
∴ Sum of the roots, 3a = 6
⇒ α = 2
∴ (x – 2) is a factor of y3 – 6y2 + 11y – 6
By synthetic division

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 5

∴ y3 – 6y2 + 11y – 6 = (y – 2) (y2 – 4y 3)
∴ Equation (2) = (y – 2) (y2 – 4y + 3) = 0
⇒ (y – 2) (y – 3) (y – 1) = 0
∴ y = 1 or y = 2 ory = – 3
The roots of (2) are 1, 2, 3.
Hence the roots of (1) are 1, \(\frac{1}{2}\), \(\frac{1}{3}\).

ii) Given cubic equation is
15x3 – 23x2 + 9x – 1 = 0 …………….(1)
put y = \(\frac{1}{x}\)
∴ (1) ⇒ y3 – 9y + 23y2 – 15 = 0 ………..(2)
Given roots of (1) are in 1-LP.
⇒ Roots of (2) are in A.P.
Let a – d, a, a + d be the roots of (2),
∴ Sum of the roots, 3α = 9
⇒ α = 3
∴ (y – 3) is a factor of y3 – 9y + 23y2 – 15.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 6

∴ y3 – 9y + 23y2 – 15 = (y – 3) (y2 – 6y + 5)
∴ Equation (2) = (y – 3) (y2 – 6y + 5) = 0
⇒ (y – 3) (y – 1) (y – 5) = 0
∴ y = 1 or y = 3 or y = 5
∴ The roots of (2) are 1, 3, 5.
Hence the roots of (2) are 1, \(\frac{1}{3}\), \(\frac{1}{5}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 7.
Solve the following equations, given that they have multiple roots.
i) x4 – 6x3 + 13x2 – 24x + 36 = 0
ii) 3x4 + 16x3 + 24x2 – 16 = 0
Solution:
i) Given equation,
x4 – 6x3 + 13x2 – 24x + 36 = 0 …………..(1)
Let f(x) = x4 – 6x3 + 13x2 – 24x + 36
f’(x) = 4x3 – 18x2 + 26x – 24
= 2 (2x3 – 9x2 + 13x – 12)
f’(3) = 2(54 – 81 + 39 – 12)
⇒ f'(3) = o
Now
f(3) = 81 – 162 + 117 – 72 + 36
= f(3) = 0
∴ (x – 3) is a factor of f(x) and f’(x).
∴ 3 is the repeated root of f(x) = 0.
Now we divide f(x) by (x – 3) by using synthetic division.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 7

∴ f(x) = (x – 3) (x – 3) (x2 + 4)
∴ Equation (1)
⇒ f(x) = 0
⇒ (x – 3) (x – 3) (x2 + 4) = 0
∴ x = 3 or x2 + 4 = 0
⇒ x = ±2i
∴ The roots of given equation are 3, 3, ± 2i.

ii) Given equation is
3x4 + 16x3 + 24x2 – 16 = 0 …………..(1)
Let f(x) = 3x4 + 16x3+ 24x2 – 16
⇒ f'(x) = 12x3 + 48x2 + 48x
= 12 (x3 + 4x2 + 4x)
= 12x (x + 2)2
⇒ f’ (- 2) = 0
Also f(- 2) = 3(16) + 16(- 8) + 24(4) – 16 = 0
∴ (x + 2) is a factor of f(x) and f'(x).
∴ – 2 is a repeated root of f(x) = 0.
Now we divide f(x) by (x + 2) using synthetic division.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 8

∴ f(x) = (x + 2) (x + 2) (3x2 + 4x – 4)
Equation (1)
⇒ f(x) = 0
⇒ (x + 2) (x + 2) (3x2 + 4x – 4) = 0
⇒ (x + 2) (x + 2) (3x – 2) (x + 2) = 0
⇒ x = – 2 or x = \(\frac{2}{3}\)
∴ The roots of given equation are – 2, – 2, – 2, \(\frac{2}{3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

III.
Question 1.
Solve x4 + x3 – 16x2 – 4x + 48 = 0, given that the product of two of the roots is 6.
Solution:
Given equation is
x4 + x3 – 16x2 – 4x + 48 = 0 ………….(1)
Let α, β, γ, δ be the roots
∴ x4 + x3 – 16x2 – 4x + 48 = (x + α) (x – β) (x – γ) (x – δ) ……….(2)
∴ Sum of the roots α + β + γ + δ = – 1
and product of roots ⇒ αβγδ = 48 …………..(3)
Given product of two roots = 6
Let αβ = 6
∴ γδ = \(\frac{48}{\alpha \beta}\)
γδ = 8
Let α + β = a and γ + δ = b
Now (2)
⇒ x4 + x3 – 16x2 – 4x + 48 = (x2 – (α + β) x + αβ) (x2 – (γ + δ) x + γδ)
⇒ x4 + x3 – 16x2 – 4x + 48 = (x2 – ax + 6) (x2 – bx + 8)
Comparing like terms.
we get, a + b = – 1 and
8a – 6b = 5 ……………..(4)
⇒ 4a + 3b = 2 (5)
(5) ⇒ 4a + 3 (- 1 – a) = 2 (∵ from (4))
⇒ a = 5
∴ b = – 6
∴ x4 + x3 – 16x2 – 4x + 48 = (x2 – 5x + 6) (x2 + 6x + 8)
= (x – 2) (x – 3) (x + 2) (x + 4)
∴ Equation (1),
⇒ (x – 2) (x – 3) (x + 2) (x + 4) = 0
∴ x = – 4; x = – 2 or x = 2 or x = 3
∴ The roots of the given equation are 2, 3, – 4, – 2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 2.
Solve 8x4 – 2x3 – 27x2 + 6x + 9 = 0 given that two roots have the same absolute value, but are opposite in sign.
Solution:
Given equation is
8x4 – 2x3 – 27x2 + 6x + 9 = 0
⇒ x4 – \(\frac{1}{4} x^3-\frac{27}{8} x^2+\frac{3}{4} x+\frac{9}{8}\) = 0 ……………(1)
Let α, β, γ, δ be the roots of (1)
∴ Sum of the roots α + β + γ + δ = \(\frac{1}{4}\)
and product of roots αβγδ = \(\frac{9}{8}\)
But given two roots have same absolute value but are opposite sign.
Let α = – β
⇒ α + β = 0
∴ γ + δ = \(\frac{-1}{4}\)
Let αβ = a and γδ = b
Now(x – α) (x – β) = x2 – (α + β)x + αβ
⇒ (x – α) (x – β) = x2 + a …………(2)
Also (x – γ) (x – δ) = x2 – (γ + δ)x + γδ
= (x – γ) (x – δ) = x2 – \(\frac{1}{4}\) x + b ………….(3)
From (1), (2) and (3)
x4 – \(\frac{1}{4} x^3-\frac{27}{8} x^2+\frac{3}{4} x+\frac{9}{8}\) = (x2 + a) (x2 – \(\frac{1}{4}\) x + b)
Comparing like terms,
\(\frac{3}{4}=\frac{-a}{4}\) and ab = \(\frac{9}{8}\)
a = – 3
∴ b = \(\frac{9}{8(-3)}\)
b = \(\frac{-3}{8}\)
∴ (2) ⇒ (x – α) (x – β) = x2 – 3
& (3) ⇒ (x – γ) (x – δ) = (x2 – \(\frac{1}{4}\) x + \(\frac{3}{8}\))
⇒ \(\frac{1}{8}\) (8x2 – 2x – 3)
⇒ (x – γ) (x – δ) = \(\frac{1}{8}\) (2x + 1) (4x – 3)
(x – γ) (x – δ) = (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\))
∴ Equation (1)
(x2 – 3) (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\)) = 0
⇒ x = ± √3 or x = – \(\frac{1}{2}\) or x = \(\frac{3}{4}\)
∴ The roots of given equation are – √3, √3, – \(\frac{1}{2}\), \(\frac{3}{4}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 3.
Solve 18x3 + 81x2 + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.
Solution:
Given equation is
18x3 + 81x2 + 121x + 60 = 0 ……………(1)
Let α, β, γ, δ be the roots
∴ Sum of roots, α + β + γ = \(\frac{-81}{18}=\frac{-9}{2}\)
αβ + βγ + γδ = \(\frac{121}{18}\)
and product of roots αβγ = \(\)
given one root is equal to halt of the sum of the remaining roots.
∴ Let α = \(\frac{\beta+\gamma}{2}\)
∴ α + 2α = \(\frac{-9}{2}\)
⇒ 3α = \(\frac{-9}{2}\)
⇒ α = \(\frac{-9}{2}\)
∴ x + \(\frac{3}{2}\) is a factor of 18x3 + 81x2 + 121x + 60.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 9

∴ 18x3 + 81x2 + 121x + 60 = (x + \(\frac{3}{2}\)) (18x2 + 54x + 40)
= (x + \(\frac{3}{2}\)) (9x2 + 27x + 60)
∴ 18x3 + 81x2 + 121x + 60 = 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5)
∴ Equation (1),
⇒ 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5) = 0
∴ x = \(\frac{-3}{2}\) or x = \(\frac{-4}{3}\) or x = \(\frac{-5}{3}\).
∴ The roots of given equation are \(\frac{-3}{2}\), \(\frac{-4}{3}\), \(\frac{-5}{3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 4.
Find the condition In order that the equation ax4 + 4bx3 + 6cx2 + 4dx + e = 0 may have two pairs of equal roots.
Solution:
Given equation is
ax4 + 4bx3 + 6cx2 + 4dx + e = 0 ………………..(1)
Given (1) has two pairs of equal roots.
∴ Let α, α, β, β be the root of (1).
(1) ⇒ x4 + \(\frac{4 b}{a} x^3+\frac{6 c}{a} x^2+\frac{4 d}{a} x+\frac{e}{a}\) = 0

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 10

3abc = 2b3 + a2d and ad2 = eb2.
∴ The required conditions are 2b3 + a2d = 3abc and ad2 = eb2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 5.
i) Show that x5 – 5x3 + 5x2 – 1 = 0 has three equal roots and and this root.
ii) Find the repeated roots of x5 – 3x4 – 5x3 + 27x2 – 32x + 12 = 0.
Solution:
i) Given equation is x5 – 5x3 + 5x2 – 1 = 0
Let f(x) = x5 – 5x3 + 5x2 – 1
f’(x) = 5x4 – 15x2 + 10x
f”(x) = 20x3 – 30x + 10
f”(1) = 20 – 30 + 10 = 0
Similarly, f’(1) = 0 and f(1) = 0
∴ (x – 1) is a factor of f”(x), f’(x) & f(x).
Thus f(x) = 0 has three equal roots and it is ‘1’.

ii) Given equation is
x5 – 3x4 – 5x3 + 27x2 – 32x + 12 = 0 …………(1)
Let f(x) = x5 – 3x4 – 5x3 + 27x2 – 32x + 12
f’(x) = 5x4 – 12x3 – 15x2 + 54x – 32
f’(1) = 5 – 12 – 15 + 54 – 32 = 0
Similarly f'(1) = 0 and f(1) = 0
∴ (x – 1) is a factor of f”(x), f'(x) & f(x).
Thus f(x) = 0 has three equal roots and it is ‘1’.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 11

∴ f(x) = (x – 1)2 (x3 – x2 – 8x + 12)
Let g(x) = x3 – x2 – 8x + 12
g’(x) = 3x2 – 2x – 8
g’(2) = 3(4) – 2(2) – 8 = 0
and g(2) = 23 – 22 – 8(2) + 12 = 0.
∴ (x – 2) is a factor of g(x) and g’(x).
∴ 2 is a multiple root of g(x) = 0.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 12

∴ g(x) = (x – 2)2 (x + 3)
∴ f(x) = (x – 1)2 (x – 2)2 (x + 3)
The roots of given equation are 1, 1, 2, 2, 3.
Hence repeated roots are 1 and 2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

Question 6.
Solve the equation 8x3 – 20x2 + 6x + 9 = 0 given that the equation has multiple roots.
Solution:
Given equation is 8x3 – 20x2 + 6x + 9 = 0 …………..(1)
Let f(x) = 8x3 – 20x2 + 6x + 9
f’(x) = 24x2 – 40x + 6
= 2 (12x2 – 20x + 3)
= 2 (2x – 3) (6x – 1)
\(f\left(\frac{3}{2}\right)=8\left(\frac{27}{8}\right)-20\left(\frac{9}{4}\right)+6\left(\frac{3}{2}\right)+9\)
= 27 – 45 + 9 + 9 = 0
∴ f(\(\frac{3}{2}\)) = 0
∴ f(x) and f'(x) has a common factor ‘2x – 3’.
∴ \(\frac{3}{2}\) is a multiple root of f(x) = 0.
By synthetic division,

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) 13

∴ 8x3 – 20x2 + 6x + 9 = 0
(x – \(\frac{3}{2}\))2 (8x + 4) = 0
⇒ x = \(\frac{3}{2}\) or x = \(\frac{-1}{2}\)
∴ The roots of given equation are \(\frac{-1}{2}\), \(\frac{3}{2}\), \(\frac{3}{2}\).

TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

by

Students must practice this TS Intermediate Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Exercise 3(b)

I.
Question 1.
If the quadratic equations ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0, (b ≠ c) have a common root then show that a + 4b + 4c = 0.
Solution:
ax2 + 2bx + c = 0
ax2 + 2cx + c = 0
Let a be common root.
aα2 + 2bα + c = 0
aα2 + 2cα + b = 0
2α (b – c) + c – b = 0
(2α – 1) (b – c) = 0
α = \(\frac{1}{2}\)
\(\frac{a}{4}+\frac{2 b}{2}\) + c = 0
a + 4b + 4c = 0.

Question 2.
If x2 – 6x + 5 = 0 and x2 – 12x + p = 0 have a common root, then find p.
Solution:
x2 – 6x + 5 = 0
(x – 5) (x – 1) = 0
x = 1, 5
Now x2 – 12x + p = 0, can have 1 or 5 as root then 1 – 12 + p = 0
p = 11 (or)
25 – 60 + p = 0
p = 35.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 3.
If x2 – 6x + 5 = 0 and x2 – 3ax + 35 = 0 have a common root, then find a.
Solution:
(x2 – 6x + 5) = 0
(x – 5) (x – 1) = 0
x = 5, 1
Now, 5, 1 satisfy
x2 – 3ax + 35 = 0
25 – 15a + 35 = 0
60 = 15a
a = 4
1 – 3a + 35 = 0
3a = 36
a = 12.

Question 4.
If the equations x2 + ax + b = 0 and x2 + cx + d = 0 have a common root and the first equation have equal roots, then prove that 2 (b + d) = ac.
Solution:
x2 + ax + b = 0
x2 + cx + d = 0
x2 + ax + b = 0 have equal roots.
⇒ a2 – 4b = 0
a2 = 4b
x2 + ax + b = 0
x2 + cx + d = 0
α (a – c) + b – d = 0
as 2α = – a
α = \({-a}{2}\)
\(\frac{a^2}{4}+c\left(\frac{-a}{2}\right)\) + d = 0
a2 – 2ac + 4d = 0
4b + 4d = 2ac
or (b + d) = ac.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 5.
Discuss the signs of the following quadratic expressions when x is real.
i) x2 – 5x + 4
ii) x2 – x + 3
Solution:
i) f(x) = x2 – 5x + 4
f(x) = (x – 4) (x – 1)
y = \(\left(x-\frac{5}{2}\right)^2+4-\frac{25}{4}\)
y = \(\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\)
y + \(\frac{9}{4}\) = \(\left(x-\frac{5}{2}\right)^2\)

TS Inter 2nd Year Maths 2A Solutions Chapter 3 De Moivre’s Theorem Ex 3(b) 1

f(x) < 0
1 < x < 4 f(x) > 0
x > 4 and x < 1
(- ∞ < x < 1) ∪ (4 < x < ∞).

ii) x2 – x + 3
f(x) = x2 – x + 3
= \(\left(x-\frac{1}{2}\right)^2+3-\frac{1}{4}\)
y = \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)
y – \(\frac{11}{4}\) = \(\left(x-\frac{1}{2}\right)^2\)
Now discriminant = 1 – 12 < 0

TS Inter 2nd Year Maths 2A Solutions Chapter 3 De Moivre’s Theorem Ex 3(b) 2

f(x) > 0 ∀ x ∈ Real numbers.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 6.
For what values of x the following expressions are positive?
i) x2 – 5x + 6
ii) 3x2 + 4x + 4
iii) 4x – 5x2 + 2
iv) x2 – 5x + 14
Solution:
i) f(x) = x2 – 5x + 6
f(x) > 0
x2 – 5x + 6 > 0
(x – 3) (x – 1)> 0
x > 3 or x < 1
(- ∞ < x < 1) ∪ (3 < x < ∞)

ii) 3x2 + 4x + 4
f(x) = 3x2 + 4x + 4
f(x) > 0
3x2 + 4x + 4> 0
x2 + \(\frac{4}{3}\)x + \(\frac{4}{3}\) > 0
\(\left(x+\frac{2}{3}\right)^2+\frac{4}{3}-\frac{4}{9}\) > 0
\(\left(x+\frac{2}{3}\right)^2+\frac{8}{9}\) > 0 ∀ x ∈ Real numbers.

iii) f(x) = 4x – 5x2 + 2
= [5x2 – 4x – 2]

TS Inter 2nd Year Maths 2A Solutions Chapter 3 De Moivre’s Theorem Ex 3(b) 3

iv) f(x) = x2 – 5x + 14
f(x) > 0
\(\left(x-\frac{5}{2}\right)^2+14-\frac{25}{4}\) > 0
\(\left(x-\frac{5}{2}\right)^2+\frac{31}{4}\) > 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 7.
For what values of x, the following expres¬sions are negative ?
i) x2 – 7x + 10
ii) 15 + 4x – 3x2
iii) 2x2 + 5x – 3
iv) x2 – 5x – 6
Solution:
i) f(x) = x2 – 7x + 10
f(x) < 0
(x – 5) (x – 2) < 0
2 < x < 5.

ii) f(x) = 15 + 4x – 3x2
f(x) < 0
15 + 4x – 3x2 < 0
3x2 – 9x + 5x – 15 > 0
3x (x – 3) + 5 (x – 3) > 0
(x – 3) (3x + 5) > 0
x > 3, x < \(\frac{-5}{3}\)
(- ∞ < x < \(\frac{-5}{3}\)) ∪ (3 < x < ∞).

iii) if(x) = 2x2 + 5x – 3
f(x) < 0
2x2 + 5x – 3 < 0
2x2 + 6x – x – 3 < 0
2x (x + 3) – 1 (x + 3) < 0
(2x – 1) (x + 3) < 0
– 3 < x < \(\frac{1}{2}\).

iv) f(x) = x2 – 5x – 6
f(x) < 0
x2 – 5x – 6 < 0
(x – 6) (x + 1) < 0
– 1 < x < 6.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 8.
Find the changes in the sign of the following expressions and find their extreme values.
i) x2 – 5x + 6
ii) 15 + 4x – 3x2
Solution:
i) f(x) = x2 – 5x + 6
= (x – 3) (x – 2)
f(x) < 0
2 < x < 3 and f(x) > 0
(- ∞ < x < 2) (3 < x < ∞)
y = \(\left(x-\frac{5}{2}\right)^2+\frac{6-25}{4}\)
ymin. = \(\frac{-1}{4}\)

ii) f(x) = 15 + 4x – 3x2
f(x) <0
15 + 4x – 3x2 < 0
3x2 – 4x – 15 < 0
3x2 – 9x + 5x – 15 < 0
3x (x – 3) 5 (x – 3) < 0
(x – 3) (3x + 5) < 0
\(\frac{-3}{5}\) < x < 3 f(x) > 0
(- ∞ < x < \(\frac{-3}{5}\)) ∪ (3 < x < ∞)
y = 15 + 4x – 3x2

TS Inter 2nd Year Maths 2A Solutions Chapter 3 De Moivre’s Theorem Ex 3(b) 4

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 9.
Find the maximum or minimum of the following expressions as x varies over R.
i) x2 – x + 7
ii) 12x – x2 – 32
iii) 2x+ 5 – 3x2
iv) ax2 + bx + a (a, b ∈ R and a ≠ 0)
Solution:
i) y = x2 – x + 7
y = (x – \(\frac{1}{2}\))2 – \(\frac{1}{4}\) + 7
y = (x – \(\frac{1}{2}\))2 + \(\frac{27}{4}\)
ymin = \(\frac{27}{4}\)

ii) y = 12x – x2 – 32
y = – [x2 – 12x + 32]
= – [(x – 6)2 – 36 +32]
y = – [x – 6]2 + 4
ymax = 4

iii) f(x) = 2x + 5 – 3x2
y = – 3 [x2 – \(\frac{2}{3}\) x – \(\frac{5}{3}\)]
= – 3 \(\left[\left(x-\frac{1}{3}\right)^2-\frac{1}{9}-\frac{15}{9}\right]\)
= – 3 \(\left(x-\frac{1}{3}\right)^2+\frac{16}{3}\)
ymax = \(\frac{16}{3}\)

iv) f(x) = ax2 + bx + a

TS Inter 2nd Year Maths 2A Solutions Chapter 3 De Moivre’s Theorem Ex 3(b) 5

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

II.
Determine the range of the following expressions.
Question 1.
i) \(\frac{x^2+x+1}{x^2-x+1}\)
ii) \(\frac{x+2}{2 x^2+3 x+6}\)
iii) \(\frac{(x-1)(x+2)}{x+3}\)
iv) \(\frac{2 x^2-6 x+5}{x^2-3 x+2}\)
Solution:
\(\frac{x^2+x+1}{x^2-x+1}\)
yx2 – yx + y = x2 + x + 1
x2 (y2 – 1) + x (- y – 1) + y – 1 = 0
Discriminant ≥ 0
(- y- 1)2 – 4(y – 1) (y – 1) ≥ 0
(y2 + 2y + 1) – 4 (y2 – 2y + 1) ≥ 0
– 3y2 +10y – 3 ≤ 0
3y2 – 10y + 3 ≤ 0
3y2 – 9y – y + 3 ≤ 0
3y (y – 3) – 1 (y – 3) ≤ 0
(3y – 1) (y – 3)≤ 0
\(\frac{1}{3}\) ≤ y ≤ 3.

ii) y = \([\frac{x+2}{2 x^2+3 x+6}/latex]
2x2y + 3xy + 6y – x – 2 = 0
x2 (2y) + x (3y – 1) + 6y – 2 = 0
Discriminant ≥ 0
(3y – 1)2 – 4 (2y) (6y – 2) ≥ 0
9y2 – 6y + 1 – 48y2 + 16y ≥ 0
– 39y2 + 10y + 1 ≥ 0
39y2 – 10y – 1 ≥ 0
39y2 – 13y + 3y – 1 ≤ 0
13y (3y – 1) + (3y – 1) ≤ 0
(13y – 1) (3y – 1) ≤ 0
[latex]-\frac{1}{13}\) ≤ y ≤ \(\frac{1}{3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

iii) y = \(\frac{(x-1)(x+2)}{x+3}\)
y = \(\frac{x^2+x-2}{x+3}\)
yx + 3y = x2 + x – 2
x2 + x(1 – y) – 2 – 3y = 0
Discriminant ≥ 0
(1 – y)2 + 4(2 + 3y) ≥ 0
y2 – 2y + 1 + 8 + 12y ≥ 0
y2 + 10y + 9 ≥ 0
(y + 1) (y + 9) ≥ 0
y ≥ – 1 or y ≤ – 9
(- ∞ < y ≤ – 9) ∪ (- 1 ≤ y < ∞)

iv) y = \(\frac{2 x^2-6 x+5}{x^2-3 x+2}\)
y (x2 – 3x . 2) = 2x2 – 6x + 5
x2 (y – 2) + x(- 3y + 6) + 2y – 5 = 0
Discriminant ≥ 0
(6 – 3y)2 – 4(y – 2) (2y – 5)≥0
9y2 – 36y + 36 – 8y2 + 36y – 40 ≥ 0
y2 -4 ≥ 0
(y – 2) (y + 2) ≥ 0
y ≥ 2 or y ≤ – 2
(- ∞ < y ≤ – 2) ∪ (2 ≤ y < ∞)

Question 2.
Prove that \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4 if x ∈ Real number.
Solution:
y = \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\)
y = \(\frac{4 x+1}{3 x^2+4 x+1}\)
3yx2 + x(4y – 4) + y – 1 = 0
Discriminant ≤ 0
[4 (y – 1)]2 – 4 . 3y (y- 1) ≤ 0
16 (y- 1)2 – 12y(y- 1) ≤ 0
4(y – 1) [4(y – 1) – 3y)]≤ 0
4(3 – 1) [y – 4] ≤ 0
(y – 1) (y – 4) ≤ 0
⇒ y ≤ 4 or y ≥ 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 3.
If x is real, prove that \(\frac{x}{x^2-5 x+9}\) lies between \(\frac{-1}{11}\) and 1.
Solution:
y = \(\frac{x}{x^2-5 x+9}\)
y (x2 – 5x + 9) = x
x2y + x(- 5y – 1) + 9y = 0
Discriminant ≤ 0
(- 5y – 1)2 – 4 . 9y2 ≥ 0
25y2 + 10y + 1 – 36y2 ≥ 0
– 11y2 + 10y + 1 ≥ 0
11y2 – 10y – 1 ≤ 0
11y2 – 11y + y – 1 ≤ 0
11y (y- 1) + (y- 1) ≤ 0
(y – 1) (11y + 1) ≤ 0
\(\frac{-1}{11}\) ≤ y ≤1.

Question 4.
If the expression \(\frac{x-p}{x^2-3 x+2}\) takes all values of x ∈ R, then find the bounds for p.
Solution:
y = \(\frac{x-p}{x^2-3 x+2}\)
y(x2 – 3x + 2) = x – p
x2y + x(- 3y – 1) + p = 0
Discriminant ≥ 0
(- 3y- 1)2 – 4y(2y + p) ≥ 0
9y2 + 6y + 1 – 8y2 – 4p ≥ 0
y2 + y(6 – 4p)+ 1 ≥ 0
Discriminant< 0
(6 – 4p)2 – 4 < 0
16p2 – 48p + 36 – 4< 0
16p2 – 48p + 32 < 0
p2 – 48p + 32 < 0
p2 – 3p + 2 < 0
(p – 2) (p – 10)
1 < p < 2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b)

Question 5.
If c2 ≠ ab and the roots of (c2 – ab) x2 – 2 (a2 – bc) x + (b2 – ac) = 0 are equal, then show that a3 + b3 + c3 = 3abc or a = 0.
Solution:
Roots are equal = Discriminant = 0
4 (a2 – bc)2 – 4 (c2 – ab) (b2 – ac) = 0
a4 + b2c2 – 2a2bc – b2c2 + c3a + b3a – a2bc = 0
a4 – 2a2bc + c3a. b3a – a2bc = 0
a (a3 + b3 + c3 – 3abc = 0
a3 + b3 + c3 – 3abc = 0 or a = 0

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

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Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

I.
Question 1.
Find the number of ways of arranging the letters of the words.
i) INDEPENDENCE
ii) MATHEMATICS
iii) SINGING
iv) PERMUTATION
v) COMBINATION vQ INTERMEDIATE
Solution:
i) Given word is INDEPENDENCE
The given 12 letters word INDEPENDENCE contains 3N’s, 2D’s and 4I’s.
Hence they can be arranged in \(\frac{12 !}{2 ! \times 3 ! \times 4 !}\) ways.

ii) Given word is MATHEMATICS
The given 11 letters word MATHEMATICS contains 2 M’s, 2 A’s and 2 T’s.
Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iii) Given word is SINGING.
The given 7 letters word SINGING contains 2 I’s, 2 N’s and 2 G’s.
Hence they can be arranged in \(\frac{7 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iv) Given Word is PERMUTATION
The given 11 letters word PERMUTATION contains 2T’s.
Hence they can be arranged in \(\frac{11 !}{2 !}\) ways.

v) Given word is COMBINATION.
Given 12 letters word COMBINATION contains 2 O’s, 2 I’s, 3 N’s.
Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

vi) Given word is INTERMEDIATE.
Given 12 letters word INTERMEDIATE’ contains 2 I s, 2 T’s, 3 E’s.
Hence they can be arranged in \(\frac{12 !}{3 ! \times 2 ! \times 2 !}\) ways.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

Question 2.
Find the number of 7- digit numbers that can be formed using 2, 2, 2, 3, 3, 4, 4.
Solution:
The 7 digit numbers that can be formed using three 2’s, two 3’s and two 4’s are \(\frac{7 !}{3 ! \cdot 2 ! \cdot 2 !}\).

II.
Question 1.
Find the number of 4 – letter words that can be formed using the letters of the word RAMANA.
Solution:
Given word is RAMANA.
The given word RAMANA has 6 letters which contains 3 A’s and rest are different.
In forming 4 letter words, these cases arises.

Case – (i) :
When all are different i.e., R, M, N, A.
∴ Number of 4 letter words formed = 4! = 24.

Case – (ii) :
When two are alike (i.e., A, A) and two are different, i.e., selected from R, M, N.
Two different letters are selected in \({ }^3 \mathrm{C}_2\) ways.
∴ Number of 4 letter words formed are \({ }^3 C_2 \times \frac{4 !}{2 !}\) = 36.

Case (iii):
When three are alike (i.e., A, A, A) and one different letter, selected out of R, M, N.
Now one different letters is selected in \({ }^3 \mathrm{C}_1\) ways.
∴ Number of 4 letter words formed = \({ }^3 \mathrm{C}_1 \times \frac{4 !}{3 !}\) = 12
∴ Total number of 4 letter words formed are 24 + 36 + 12 = 72.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

Question 2.
How many numbers can be formed using all digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places?
Solution:
Given digits 1, 2, 3, 4, 3, 2, 1 contains 3 even digits i.e., 2, 4, 2 and 3 even places.
Number of ways of arranging 3 even digits 2, 4, 2 in three even places is \(\frac{3 !}{2 !}\) 3.
The remaining 4 digits 1, 3, 3, 1 in remaining 4 places can be arranged in \(\frac{4 !}{2 ! \cdot 2 !}\) ways.
∴ Number of numbers formed using all the digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places = 3 × \(\frac{4 !}{2 ! \cdot 2 !}\) = 18.

Question 3.
In a library, there are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies each of two different books. Find the number of ways of arranging all these books in a shelf in a single row.
Solution:
There are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies of two different books.
∴ Total number of books = 6 + 4(2) + 5(3) + 3(2) = 35
∴ Number of ways of arranging these 35 books = \(\frac{35 !}{6 ! \times 4 ! \times 4 ! \times 5 ! \times 5 ! \times 5 ! \times 3 ! \times 3 !}\)
= \(\frac{35 !}{6 ! \cdot(4 !)^2 \cdot(5 !)^3 \cdot(3 !)^2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

Question 4.
A book store has ‘m’ copies each of ‘n’ different books. Find the number of ways of arranging these books in a shelf in a single row.
Solution:
Book store has m’ copies each of n’ different books.
∴ Total number of books = mn
∴ Number of ways of arranging these mn books = \(\frac{(m n) !}{m ! \times m ! \times \underbrace{\ldots \ldots m !}_{n \text { times }}}=\frac{(m n) !}{(m !)^n}\)

Question 5.
Find the number of 5-digit numbers that can be formed using the digits 0, 1, 1, 2, 3.
Solution:
Given digits are 0, 1, 1, 2, 3.
The ten thousand’s place of a ‘5’ digit number can be filled by any of non-zero digit in 4 ways.
The remaining 4 places can be filled with remaining 4 digits in 4! ways.
∴ Number of ways = 4 × 4!
Since there are two 1’s in every arrangement, the number of 5 digited numbers formed are \(\frac{4 \times 4 !}{2 !}\) = 48.

Question 6.
In how many ways can the letters of the word CHEESE be arranged so that no two E’s come together ?
Solution:
Given word is ‘CHEESE’.
As no two E’s come together, first arrange letters C, H, S.
Number of ways of arranging C, H, S is 3!.
∴ Number of gaps formed are 4.
Number of ways of arranging 3 E’s in 4 gaps = \(\frac{{ }^4 P_3}{3 !}\)
∴ Required number of ways = 3! × \(\frac{{ }^4 P_3}{3 !}\) = 24.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

III.
Question 1.
Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them
i) all the three S’s together
ii) the two A’s do not come together.
Solution:
Given word is ‘ASSOCIATIONS’.
The given 12 letters word ‘ASSOCIATIONS’ contains 2 A’s, 3 S’s, 2 O’s and 2 I s.
∴ Number of ways of arranging them = \(\frac{12 !}{2 ! \cdot 2 ! \cdot 2 ! \cdot 3 !}\).

i) All the three S’s come together :
Treat 3 S’s as one unit. This unit with remain¬ing 9 letters becomes 10 entities which contain 2 A’s, 2 O’s and 2 I’s.
Number of ways of arranging so that all the three S’s come together = \(\frac{10 !}{2 ! \cdot 2 ! \cdot 2 !}\).

ii) The two A’s do not come together :
As 2 A’s do not come together, arrange remaining 10 letters.
Remaining 10 letters contains 3 S’s, 2 O’s and 2 I’s.
∴ Number of ways of arranging = \(\frac{10 !}{3 ! \cdot 2 ! \cdot 2 !}\)
∴ Number of gaps formed are ’11’.
2 A’s in these 11 gaps can be arranged in up \(\frac{{ }^{11} \mathrm{P}_2}{2 !}\) ways.
∴ Required number of ways of arranging = \(\frac{10 !}{3 ! \times 2 ! \times 2 !} \times \frac{{ }^{11} P_2}{2 !}\).

Question 2.
Find the number of ways of arranging the letters of the word MISSING so that the two S’s are together and the two l’s are together.
Solution:
Given word is MISSING.
Treat 2 S’s as 1 unit and 2 l’s as another unit.
These 2 unit with remaining 3 letters be comes 5 entities.
Number of ways of arranging them = 5!
2 S’s and 21’s can arrange themselves in only 1 way.
∴ Required nunther of ways of arranging = 5! = 120.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

Question 3.
If the letters of the word AJANTA are per muted in all possible ways and the words thus formed are arranged In dictionary order, find the ranks of the words
(i) AJANTA
(ii) JANATA.
Solution:
Given word MANTA in Alphabetic order is AAAJNT.

i) Rank of ‘AJANTA’:
In the dictionary order first comes that words which begin with letter A. Also the second place is to be filled by A.
The remaining 4 places can be filled in 4! ways.
On proceeding like this, we get
The number of word begin with AA is 4! = 24
The number of word begin with AJAA is 2! = 2
The number of word begin with AJANA = 1
Next word is AJANTA
∴ Rank of word AJANTA = 24 + 2 + 1 + 1 = 28.

ii) Rank of JANATA :
In the dictionary order first comes that words which begin with the letter A.
The remaining 5 places to be filled in \(\frac{5 !}{2 !}\) ways.
On proceeding like this
The number of words begin with A = \(\frac{5 !}{2 !}\) = 60.
The number of words begin with JAA = 3! = 6
The number of words begin with JANAA = 1! = 1
Next word is JANATA
∴ Rank of word JANATA = 60 + 6 + 1 + 1 = 68.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

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Telangana TSBIE TS Inter 1st Year English Study Material 13th Lesson The Short-sighted Brothers Textbook Questions and Answers.

TS Inter 1st Year English Study Material 13th Lesson The Short-sighted Brothers

Paragraph Answer Questions (Section – A, Q.No. 5, Marks : 4)

Question 1.
Is the title, “The Short-sighted Brothers” apt to the story ? Explain.
Answer:
Yes. The name “The Short-sighted Brothers” and the content of the story look like made-for- each-other things. The title matches perfectly well with the theme. Three brothers are the lead characters. All the three brothers were very short-sighted. It was a physical disability. And they all suffered from the related mental disability too. They failed to see the possible result of their crooked,plans. They were selfish. They were greedy. They tried to cheat one another. Finally their follies were exposed. As the entire story deals with their physical and mental short- sightedness, the title is appropriate to this ancient chinese folk tale.

అవును. “హ్రస్వదృష్టి కల సోదరులు” అనే పేరు, ఆ కథ ప్రధానాంశం ఒకరికోసం ఒకరు సృష్టించబడినవి లాగా కనిపిస్తాయి. కథ నామము, కథ అంశము సరిగ్గా జతకలుస్తాయి. ముగ్గురు సోదరులు ప్రధాన పాత్రలు. ముగ్గురికీ తీవ్ర హ్రస్వదృష్టి. అది భౌతిక, శారీరక లోపము. అయితే వారికి సంబంధిత మనోవైకల్యం కూడా ఉంది. వారి యొక్క హేయ, దురుద్దేశ పూర్వక ఎత్తుగడల పరిణామాలను ఊహించలేకపోయారు. వారు స్వార్థపరులు. వారిది అత్యాశ. వారు ఒకరినొకరు మోసం చేయాలనుకున్నారు. అంతిమంగా వారి మూర్ఖత్వం బహిర్గతం చేయబడింది. కథ మొత్తం కూడా వారి యొక్క భౌతిక, మనోవైకల్యాల (హ్రస్వదృష్టి) గురించిన చర్చే కాబట్టి, ఈ ప్రాచీన చైనీయుల జానపద గాథకు ఆ పేరు సరిగ్గా సరిపోయింది.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

Question 2.
How did the three brothers try to outsmart one another ?
Answer:
“The Short-sighted Brothers” is a famous Chiness folk story. It exposes the folly of the three brothers. All the three brothers were very short-sighted. Even their personality suffered from the same flaw. They tried to deceive themselves and one another. The youngest brother one day proposed to take charge of their family finances.

He cited his eldest brother’s short- sightedness to support his claim. The second brother too joined the race. The eldest brother proposed a test to prove the power of their sight. They should read the inscription on the newly installed tablet on the door way of the nearest monastery. To outsmart one another, each met the monk secretly and learnt about the writing on it.

“హ్రస్వదృష్టి కల సోదరులు” అనేది ప్రఖ్యాత చైనీయుల జానపద గాథ. అది ఆ ముగ్గురు సోదరుల మూర్ఖత్వాన్ని ఎత్తి చూపుతుంది. ఆ ముగ్గురు సోదరులూ చాలా తీవ్రంగా హ్రస్వదృష్టి కలవారు. వారి వ్యక్తిత్వంలో కూడా అదే లోపము. వారిని వారు మోసం చేసుకుంటూ, ఒకరినొకరు మోసం చేసుకునే ప్రయత్నం చేస్తారు. ఒక రోజున అందరిలో చిన్న సోదరుడు కుటుంబ ఆర్థిక వ్యవహారాల బాధ్యతను తనకు అప్పగించమని ప్రతిపాదన చేస్తాడు.

అందుకు మద్దతుగా పెద్దన్నయ్య హ్రస్వదృష్టిని కారణంగా పేర్కొంటాడు. రెండవ సోదరుడూ పోటీలో జతకలుస్తాడు. వారి చూపుడు శక్తిని పరీక్షించుకోవడానికి పెద్ద సోదరుడు ఒక ప్రతిపాదన చేస్తాడు. వారు కొత్తగా, సమీప సన్యాశ్రమ ద్వారంపై ఏర్పాటు చేసిన శిలాఫలకంపై రాతను చదవాలి. మోసపూరితంగా తమదే పై చేయి అనిపించుకొనుటకు, ప్రతి ఒక్కరూ రహస్యంగా ఒక సన్యాసిని కలిసి, ఆ రాతి పలకపైన ఉన్న రాత గురించి తెలుసుకుంటారు.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

Question 3.
Were the three brothers successful in executing their tricks ? Support your answer.
Answer:
No. The three brothers failed in their efforts. The famous Chinese folk tale “The Short-sighted Brothers”, explains their failure. All the three brothers were very short-sighted. Once, they wanted to prove the power of their sight. They were to read an inscription. Each learnt secretly from the monk about the writing.

They thought they could outsmart the others. They visited the monastery the next day. They started READING from the TABLET. Each one READ out. Then the monk came out. He told them that the TABLET was not yet put up! They READ from the TABLET that WAS NOT there! Their folly was exposed. They realised it!

లేదు. ఆ ముగ్గురు సోదరులు తమ ప్రయత్నాలలో విఫలమయ్యారు. ఈ ప్రఖ్యాత చైనీయుల జానపద గాథ. “హ్రస్వదృష్టి కల సోదరులు” వారి వైఫల్యాన్ని వివరిస్తుంది. ఆ ముగ్గురు సోదరులు అందరిదీ తీవ్ర హ్రస్వదృష్టి లోపము. ఒకసారి వారు తమ దృష్టి శక్తిని నిరూపించుకోదలిచారు. వారు ఒక రాతిపలక మీద రాతను చదవాలి. ఒక సన్యాసి ద్వారా వారిలో ప్రతి ఒక్కరూ రహస్యంగా ఆ రాత గురించి తెలుసుకున్నారు.

వారు ప్రతి ఒక్కరు ఇతరులపై తమదే పై చేయి అని నిరూపించగలము అనుకున్నారు. మరుసటి రోజు వారు ఆ సన్యాసాశ్రమాన్ని సందర్శించారు. వారు ఆ శిలాఫలకము చూసి చదవనారంభించారు. ప్రతి ఒక్కరూ చదివారు. అప్పుడు ఆ సన్యాసి బయటకు వచ్చారు. వారితో ఆ రాతి పలకను ఇంకా ఏర్పాటు చేయలేదు అని వివరిస్తారు. అక్కడ లేని రాతిపలకపై రాతను వారు చదివారు ! వారి మూర్ఖత్వం బహిర్గతమైంది. వారూ గుర్తించారు దాన్ని,

Question 4.
Does the story “The Short-sighted Brothers’ support the wise saying, ‘Honesty is the best policy’ ? Discuss. * (Imp) (Model Paper)
Answer:
Yes. Honesty is undoubtedly the best policy. It is certainly very difficult to practise the policy. The story, The Short-sighted Brothers’ proves both these points clearly. All the three brothers were short-sighted. They were selfish. They had no ethical values. They thought they could easily cheat others.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

They got by heart the inscription ‘Be Honest At All Times’. But they never understood its meaning or its importance. They followed dishonest means to prove the power of their sight. Their follies were exposed. Thus, they learnt that their deceptive tricks failed. Only honesty shines forever! And through the brothers, the readers too get a valuable lesson.

అవును. నిజాయితీ నిస్సందేహంగా అత్యుత్తమ విధానం. ఆ విధానాన్ని పాటించడం నిశ్చయంగా చాలా కష్టము. “హ్రస్వదృష్టి కల సోదరులు” అనే కథ ఈ రెండు విషయాలను స్పష్టంగా నిరూపిస్తుంది. వారికి నైతిక విలువలు లేవు. వారు ఇతరులను తేలికగా మోసం చేయగలము అనుకున్నారు. ‘అన్ని వేళలా నిజాయితీగా ఉండండి’ అనే శిలాక్షరాలను కంఠస్థము అయితే చేశారు.

కాని దాని భావాన్ని అర్థం చేసుకోవటం కాని, దాని ప్రాధాన్యతను గుర్తించటం కానీ ఎప్పుడూ చేయలేదు. వారి దృష్టి శక్తిని నిరూపించుకోవడానికి తప్పుడు మార్గాలు అవలంబించారు. వారి మూర్ఖత్వము బహిర్గతం చేయబడింది. ఈ విధంగా వారు గ్రహించారు తమ మోసపూరిత ఎత్తుగడలు విఫలమయ్యాయని. కేవలం నిజాయితీ మాత్రమే శాశ్వతంగా వర్ధిల్లుతుంది. ఆ ముగ్గురు సోదరుల ద్వారా పాఠకులకు కూడా ఒక విలువైన పాఠం అందుతుంది.

The Short-sighted Brothers Summary in English

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers 1
The folk tale, ‘The Short-sighted Brothers’ makes an interesting reading. With its gripping narration, the story excites the reader thoroughly. In the end takes a sudden twrist, stunning and surprising the reader. Equally shocked were the brothers in the story. It exposes the follies of the brothers, prompting many a reader to introspent.

The three aged brothers were the central characters in the story. They were short sighted, both physically and mentally. They were selfish and greedy. Citing their eldest brother’s short- sightedness as a reason, the youngest brother proposed to manage their family finales. He was blind to his own disability. All of them suffered from the same flow, sight problems and lack of values. Yet, each tried to outmart the other.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

Therefore, they planned to test their own vision by reading the inscription above the doorway of nearby monastery. Each knew that he couldn’t read it. So, they secretly and separately enquired with monk there as to what was written on the tablet. And later, they pretended they were reading the inscription with their own eyes. In the process, they fooled themselves they memorised the quotation, “Be honest at all times”. But, they did not adopt it in their own lives! They failed to see the outcome of their evil plans. It was then, that the monk revealed that the tablet was not put up yet. The brothers realized how foolish they were !

Finally, the story clearly shows the physical weakness of the brothers. It also exposes their follies. Hence, we can very strongly say that the title suits the story well. If at once tells us what we are going to find in it.

The Short-sighted Brothers Summary in Telugu

హ్రస్వదృష్టి కల సోదరులు. (The Short-sighted Brothers) అనే కథ ఒక చైనీయుల జానపద గాథ. అది చక్కని సందేశాన్ని, ఆకట్టుకునే రీతిలో తెలుపుతుంది. మితిమీరిన స్వార్థం, ముందు చూపు లోపించి చేసే తప్పుడు పనులు మనల్ని నవ్వుల పాలు చేస్తాయి అని వివరిస్తుంది ఈ కథ. ఒక నగరంలో (చైనాలో) ముగ్గురు అన్నదమ్ములు నివసించేవారు. ముగ్గురికీ తీవ్ర స్థాయిలో హ్రస్వదృష్టి. చాలా దగ్గరలో ఉంటే మాత్రమే కనిపిస్తాయి. అపరిమిత స్వార్థం. అడ్డదారిలో కోరికలు నెరవేర్చుకోవాలని తపన. ఒక రోజు చిన్న తమ్ముడు అంటాడు : పెద్ద అన్నయ్య చూపు లోపాన్ని ఇతరులు దుర్వినియోగం చేసుకుంటారు.

కావున ఇంటి డబ్బు విషయాలు నేను చూస్తాను అని. ఆ విషయానికి వస్తే, అందరికన్నా నా చూపు మెరుగు, కావున ఆ పెత్తనం నాకు కావాలి అంటాడు, రెండవ సోదరుడు. అలానా ? అయితే, ఎవరి చూపు నిజంగా బాగా ఉందో పరీక్షించుకుందాము అని పెద్ద సోదరుడు ప్రతిపాదిస్తాడు. ఊరి చివరి ఉన్న ఒక సన్యాసాశ్రమం ముఖద్వారంపై ఒక మంచి సూక్తి చెక్కబడిన శిలాఫలకాన్ని ఈ రాత్రి ఏర్పరుస్తారట.

రేపు ఉదయం _ ముగ్గురం అక్కడికి వెళ్ళి, ఎవరు ఎంత చక్కగా చూడగలరో తేల్చుకుందాము అంటాడు. అందరూ ఒప్పుకుంటారు. ఆ రాత్రి ఒకరికి తెలియకుండా ఒకరు రహస్యంగా ఆ ఆశ్రమానికి వెళ్ళి, అక్కడి సన్యాసిని వివరాలు అడుగుతారు. “ఎప్పటికీ నిజాయితీగా ఉండు” అనే సూక్తి, చైనీయుల ప్రసిద్ధ తత్వవేత్త కన్ఫ్యూషన్ చెప్పినది, ఆ రాతి పలక మీద చెక్కించాము అని సన్యాసి వివరిస్తారు పెద్ద అన్నయ్యకు.

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

రెండవ సోదరుడు ఏమైనా అలంకారం ఉందా అని ప్రశ్నించి, పువ్వుల తీగ అలంకారం ఉంది అనే సమాధానాన్ని పొంది, తన తెలివికి తానే సంతోషిస్తూ వెను తిరుగుతాడు. అందరికంటే చిన్నవాడు, మరింత అదనపు సమాచారం కింద ఒక మూలగా, దాత పేరు ఉన్నట్లు సేకరిస్తాడు. మరునాడు ఏమీ ఎరగనట్లు ముగ్గురు సోదరులూ సన్యాశ్రమం చేరుకుంటారు. పెద్దవాడు ముఖ ద్వారం వైపు చూస్తూ, రాత్రి తెలుసుకున్న సూక్తిని “చదువుతాడు”.

దానికి అదనంగా రెండవ వాడు పూవుల తీగ అలంకారం ఉంది అంటాడు. ఆ ఇద్దరికన్నా మిన్నగా దాత పేరు “చదువుతాడు” చిన్నవాడు. ఒకరిని మరొకరు ఓడించగలిగినట్లు సంబరపడుతున్న సమయంలో, రాత్రి వీరు ముగ్గురూ మాట్లాడిన సన్యాసి బయటికి వచ్చి, వారితో “క్షమించండి. రాత్రి ఆ శిలాఫలకం పెట్టలేకపోయాము. ఈ రోజు ఏర్పరుస్తాము. మీరు అది చూడాలని వచ్చారు” అంటాడు. వారి మూర్ఖత్వం బయటపడింది. తమ తెలివితక్కువతనం బయటపడగా, తలలు వంచుకున్నారు. “ఎప్పుడూ నిజాయితీగా ఉండండి” అనే సూక్తిని కంఠస్థం చేశారు. ఆచరించలేకపోయారు.

The Short-sighted Brothers Summary in Hindi

‘हस्व दुष्टि होनेवाले भाई” – The Short- Sights Brootters’ नामक कहानी चीनी देश को लोक कथा है | यह कथ शिक्षा देती है कि अति स्वार्थ, दुर दर्षित के अभाव के कारण होने वाली वाले गलत काम हमें अपहास करते है । एक चीनी शहर में तीन भाई रहते हैं। तीनों ह्रस्व दृष्टि वाले हैं। सभी अपरमित स्वार्थी, बेर्ढतनी से अपनी इच्छाएँ पूरे करने के लिए छटपटानेवाले हैं । एक दिन छोटा भाई कहता है कि वड़े बाई की ह्रस्व दृष्टि का दूसरे लोग दुरुपयोग करते है ।

इसलिए प्यार के पैसे का मामला मैं देख लेता हुँ । दुसरा भाई कहता है कि ऐसा है। बड़ाभाई कहता है कि किसकी दृष्टि अच्छी है, इसकी जाँच करेंगे । गाँव के एक कोने में स्थित सन्यासाश्रम के मुख्यद्वार पर इस रात को सूक्ति अचित शिलाफलक आयोगित किया जाएगा । पालुस करेगे कि कल सबेरे तीनों जाकर कौन कितना अच्छा देख सकता है । सभी मान लेते हैं । उस रात को सभी एक-एक करके रहस्य से अश्रम जाकर वहाँ के सन्यासी से जानाकारी लेते हैं। अगले दिन तीनों जाकर बडा भाई पहली पंक्ति याद रखकर पढता है । दूसरा भाई दूसरी पंक्ति याद रखकर पढता है ।

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

छोटा भाई अंतिम पंक्ति याद रखकर पढता है। हर एक भाई खुशी में है कि उसने बाकी दूसरों को हराया है । इनकी बातों को सुनकर सन्यासी बाहर आकर बताता है कि माफ़ कीजिए, कल रात को शिलाफलक का प्रबंध नहीं हो पाया । आज इसका आयोजन किया जाएगा । इस देखने आप लोग आए होंगे। उनकी मूर्खता मालूम होने पर सब अपना-अपना सिर ॠकाते हैं। शिलाफलक पर उत्कीर्ण सूक्ति ‘हमेशा ईमानदारी से रहो का कंठस्थ करते हैं, लेकिन ने उस सूक्ति का पालन नहीं कर पाते हैं ।

Meanings and Explanations

short-sighted (adj) /sɔ:(r)t saıtıd/ (షో(ర్)ట్ సైటిడ్) (trisyllabic) = not able to see things etc clearly when they are not very near to that person; హ్రస్వదృష్టి లోపం కల; దగ్గరి వస్తువులను మాత్రమే చూడగల.

folklore (n) /fǝuklɔ:(r)/ (ఫఉక్ లో(ర్)) (disyllabic) a collection of conventional tales passed on to posterity orally : మౌఖికంగా భావితరాలకు అందించబడే సాంప్రదాయ కథలు; జానపద గాథలు

to take charge of (phrasal verb) to be in the control of; to take responsibility of : అదుపులో, అధీనంలో ఉంచుకొను; బాధ్యత వహించు

to take advantage of (phrasal verb) = to make use of a situation for one’s selfish interests: ఒక పరిస్థితిని తమ స్వార్థ ప్రయోజనాలకు వాడుకొను

sneer (v) /sniǝ(r)/ (స్నిఅ(ర్)) (monosyllabic) = speak harshly; say something without respect: కఠినంగా, అవమానకరంగా మాట్లాడు, अवहेलना, तिरस्कार

monastery (n) /monǝstri/ (మొనస్ ట్ రి) (trisyllabic) = a place where sages, ascetics, etc. live: సన్యాసాశ్రమము, मट

tablet (n) /tæblet/ (ట్యాబ్ లెట్ ) (disyllabic) = a slab of stone/clay for carving: రాతి (మట్టి) పలక; pill = మాత్ర; an electronic device = ఒక ఎలక్ట్రానిక్ పరికరము; చిన్న కంప్యూటర్, गोली

inscription (n) /inskripsən/ (ఇన్ స్ క్రిప్ షన్) (trisyllabic) = writing or carving on a stone surface : రాతిపలకపైని రాత

strain (n) /strein/ (స్ట్రైఇ న్) (monosyllabic) pressure; difficulty: ఒత్తిడి, కష్టము, कसकर तीनाना

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

in unison (phrase – functions as an adverbial) = all together and at a time: అందరూ కలిసి ఒకే సమయంలో (చెప్పు)

get a few winks (idiom) = sleep for a little while: కొద్దిసేపు నిద్రించు

sneak (v) /sni:k/(స్నేక్ ) (monosyllabic) = go without anyone’s knowledge: ఎవరికీ తెలియకుండా (రహస్యంగా) వెళ్ళు, जाव

monk (n) /maŋk/ (మంక్ ) (monosyllabic) a member of an all male-member religious group : సన్యాసి; అందరూ మగవారు మాత్రమే ఉండే మత బృంద సభ్యుడు, मट

Confucius (proper noun) /kənfjʊ:səs/ (కన్ ఫ్యూషన్) = a very famous philosopher from ancient China (550-479 B.C) : చైనాకు చెందిన ప్రాచీన కాలపు ప్రఖ్యాత తత్వవేత్త

chuckle (v) /tsakǝl/ (చకల్) (disyllabic) = laugh silently and inwardly : తమలో తాము నిశ్శబ్దాన్గా నవ్వుకొను

triumphantly (adv) /trai^mfǝntli/ (ట్రయ్ అమ్ షన్ ట్ లి) (polysyllabic-4 syllables) over the success : విజయ గర్వంతో, ఆనందంగా

applaud (v) /ǝplɔ:d/ (అప్లౌడ్) (disyllabic) = praise; appreciate; పొగడు, మెచ్చుకొను

TS Inter 1st Year English Study Material Chapter 13 The Short-sighted Brothers

besides (preposition) /basardz/ (బసైడ్) (disyllabic) = in addition to : అదనముగా

face falling (phrase) = looking sad; విచారంగా కనిపించు

intone (v) /Intaun/ (ఇన్టఉన్) (disyllabic) = say something and emphatically; నెమ్మదిగా, నొక్కి, నొక్కి అను, చెప్పు

follies (n-pl. of ‘folly’) /foliz/ (ఫొలిజ్) = foolishness; మూర్ఖత్వము, క్లాత్

TS 6th Class Science 5th Lesson Questions and Answers Telangana – Materials and Things

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TS Board Telangana SCERT Class 6 Science Solutions 5th Lesson Materials and Things Textbook Questions and Answers.

Materials and Things – TS 6th Class Science 5th Lesson Questions and Answers Telangana

Improve Your Learning

Question 1.
Name any five objects which are made up of only one material. (Conceptual Understanding) 2M
Answer:
Five objects which are made up of plastic.

  1. Chairs,
  2. Boxes,
  3. Table,
  4. Bottles,
  5. Dolls.

Question 2.
Name any five objects which are made up of more than two materials. (Conceptual Understanding) 2 M
Answer:
Five objects which are made up of more than two materials.

  1. Bicycle,
  2. Bullock cart,
  3. Doors,
  4. Wall clock,
  5. Shuttle bat.

TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things

Question 3.
List five things which we can make using each of the following materials: a) Glass b) Metal c) Plastic d) Wood. (Conceptual Understanding) 4 M
Answer:
Five things which we can make using each of the following material.
(A) Glass Mirror, car window, TV screen, photo frame, dining bowls, plates etc.
(B) Metal Wheels, chairs, cup board, vessels, machines etc.
(C) Plastic Jars, covers (carry bags), chairs, bottles, plates etc.
(D) Wood Tables, chairs, doors, windows, cots, frames etc.

Question 4.
Mary saw a ship travelling on a sea. She knows that iron nail sinks in water. She has many doubts, what are her doubts ? Write them. (Asking Questions and Making Hypothesis) 4 M
Answer:
Mary raises the following questions (doubts) for floating of ship in the sea.

  1. Flow does a ship float on the surface of sea ?
  2. What principle helps the ship to sail on the sea easily ?
  3. Do all material have a chance of floating on the sea water ?
  4. Are there any properties which help the ship floating ?
  5. Can I travel on the surface of sea as ship sails ?

Question 5.
Mary, while examining whether a boiled egg sinks or floats, found that it floats but Vakula made it sink, how is it possible? Guess and write it. (Asking questions and Making hypothesis) 4 M
Answer:

  1. At first Mary used salt water for testing the sinking or floating character of the boiled egg Naturally boiled egg floats on the surface of salt water. Therefore the egg floats on salt water.
  2. But Vakula made the egg sink in by using normal water. She observed that the boiled egg simply sinks In the normal water.

Question 6.
Drop an egg in a beaker of water. Now drop the same egg in another beaker of water in which excessive salt is added. Write your observation. (Experimentation and Field Investigation) 4 M
Answer:
Procedure of the experiment : The egg is dropped in a beaker of water. After sometime, the same egg is dropped in another beaker of water in which excessive salt is added.

Observation : When the egg is placed in the beaker full of water, the egg sinks normally. On the other hand if the same egg is placed in another beaker of salt water, it floats.

Inference : The salt water and normal water exhibit their character to sinking or floating of an egg.

TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things

Question 7.
Do the following activities. Write down your observations. What do you conclude ? (Experimentation and Field Investigation) 8 M
a) Mix chalk powder in water.
b) Place a piece of candle in water.
c) Add some oil drops to a beaker of water.
Answer:
Aim : To observe the nature of substances like chalk powder, candle piece and oil drops in water.

Requirements: Three glass beakers full of water, chalk powder, candle piece and oil drops.

Procedure: Three glass beakers are kept on the table. They are filled with water. Certain amount of chalk powder, candle piece and a few drops of oil are taken into the three beakers 1, 2, 3 respectively.

Observation : It is observed that the chalk powder is dissolved in the first beaker, candle piece is not dissolved in the second beaker. It is observed that the oil drops float on the surface of water.

Conclusion : Water has the capacity of dissolving certain substances like chalk powder. Substances like candle in solid state are insoluble in the water. Oil floats on the water surface.

Question 8.
Make a list of items from your kitchen like utensils, food ingredients etc. Classify them as follows.

ItemSink/float in waterSoluble / insoluble in water

Answer:
Utensils : Glass, saucer, small water vessel, spoon.
Ingredients : Sugar, salt, dal, jeera. Based on the sinking or floating of utensils and solubility of ingredients in water the items are classified as follows.
UTENSILS:

ItemSink/float in water
GlassSlowly sinking
SaucerFloats on water
Water vesselFloats on water
SpoonSinking

INGREDIENTS:

ItemSoluble / insoluble in water
SugarSoluble
SaltSoluble
DalInsoluble
JeeraInsoluble

TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things

Question 9.
Collect different plastic items from your surroundings. Classify them as transparent, opaque and translucent.
Answer:

Item nameTransparent / Opaque / Translucent
Polythene coverTransparent item
Carry bagTranslucent item
BoxOpaque item

Question 10.
Draw different objects made up of wood which we use in our daily life. (Communication through Drawing and Model Making) 8M
Answer:
TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things 3

Question 11.
Make a few models you like using clay. (Communication through Drawing and Model Making) 8M
Answer:
Models made of clay.
TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things 2

Question 12.
We know that a ship, even though it is made up of tonnes of iron, floats on water. How do you feel about the scientists, who found the scientific principles and efforts in making a ship ? (Aesthetic Sence, Values and Application to Daily Life and Concern to Bio-diversity) 4M
Answer:

  1. Invention of ship is a great milestone in the human development.
  2. Iron is a heavy metal which sinks in water. But making a ship made with wood and tonnes of iron floats on water is really an appreciable thing.
  3. We have to appreciate the scientists and their efforts in applying scientific principles for the benefit of mankind.

TS 6th Class Science 5th Lesson Questions and Answers Telangana - Materials and Things

Question 13.
We use so many wooden items in our daily life. Is it good to use wood? What happens by excessive use of it? What is the reason? Is there any
alternative for this ? (Aesthetic Sence, Values and Application to Daily Life and Concern to Bio-diversity) 4M
Answer:
Uses of wood :

  1. Indeed the things made of wood are en vironmentally eco – friendly products.
  2. Wooden items do not cause harm to the environment.
  3. To make the items with wood, we should depend on forests and domestic plants.

Demerits of cutting trees :

  1. Cutting trees for wooden items severely affects the decrease in forests and all the plantations is called deforestation.
  2. Deforestation leads to imbalance in nature and there will be a decrease in rainfall.
  3. Oxygen in the atmosphere decreases.
  4. Deforestation causes top most soil erosion. Thus in turn it results in losing of soil fertility. Due to soil erosion we lose food grain harvestation.

Alternative steps to avoid cutting trees for wooden items :

  1. One way is to grow the wood giving plants in the waste land areas with the help of our society.
  2. The wooden furniture, once we purchase them from the shop, should be used them without damage.
  3. We should not cut down non-wood giving trees along with wood giving plants unnecessarily.
  4. We have to find a solution to convert any waste and already used material into wooden like furniture.

TS 6th Class Science 5th Lesson Notes – Materials and Things

  • In our daily life we use several objects for different acHuif les. These objects are made of different materials.
  • Some objects are made of more than one material.
  • Objects around us are made of large variety of materials :
  • Based on their properties, we use different materials for dfferenf purposes.
  • Material has three important states called solids, liquids and gases.
  • Some materials can sink in water and some materials can float on water.
  • Materials are grouped together on the basis of similarities and differences in their properties.
  • Certain materials change their state from solid to liquid, liquid togas on being heated and from gas to liquid, liquid to solid on being cooled.
  • Material : Materials are the things that you need for a particular activity.
  • Object : An object is anything that has a fixed shape that you can touch or see and that is not alive.
  • Metal : It is a hard substance. Eg : Iron, steel, copper etc.
  • Transparent : We can easily see through some materials. Such materials are said to be transparent. Eg: glass, air, water etc.
  • Opaque : We cannot see through some materials. Such materials are said to be opaque. Eg: wood, steel, card board etc.
  • Traslucent : We can see the objects, but not very clearly are said to be translucent. Eg : oily paper etc.
  • Solid : A solid is a substance that stays in the same shape, whether it is in a container or not. Eg: wood, rock etc.
  • Liquid : A liquid is a substance which flows and can be poured and it. takes the shape of the container. Eg: water, kerosene.
  • Gas : A gas is a substance that is neither liquid nor solid. Eg: air, smoke etc.
  • Soluble : The materials which dissolve in a liquid are said to be soluble in water. Eg : Sugar in water.
  • Insoluble : The materials which do not dissolve in a liquid are said to be insoluble in water. Eg: Kerosene in water.
  • Sink : The material which possess more weight can sink in water. Eg: iron nail, stone etc.
  • Float : The material which possess less weight can float on the water; Eg: dry leaf, sponge etc.

TS 6th Class Science 4th Lesson Questions and Answers Telangana – What Do Animals Eat?

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TS Board Telangana SCERT Class 6 Science Solutions 4th Lesson What Do Animals Eat? Textbook Questions and Answers.

What Do Animals Eat? – TS 6th Class Science 4th Lesson Questions and Answers Telangana

Improve Your Learning

Question 1.
Name some animals in your house which have the same kind of food habit. (Conceptual Understanding) 2 M
Answer:
Goat, sheep, buffalo and cow consume same kind of food, that is grass. Cat and dog depend on meat, milk, curd etc.

Question 2.
Observe your surroundings or go to a nearby field and write about the following : (Information Skills and Projects/C. U.) 4M
a) How does the cow eat grass ?
b) What tools are used while doing so ?
c) In what way can you justify it is a herbivore ?
Answer:
Aim of the project: To observe the way of eating or consuming of grass by cow while grazing.
(a) Eating of grass (Grazing): The cow naturally grazes in the grass land. Before grazing, sometimes it smells the food (grass). It grazes only green leafy plants (grass) only. It ruminates

(b) The tools used while grazing : Cow doesn’t possess upper teeth. Instead, the upper jaw has muscular, strong gums. It plucks the grass by holding with upper muscular gums and lower teeth. The tongue also helps in churning the food.

(c) Since the cow consumes and only depends on green grass, it is called herbivore. It ruminates while in resting stage.

TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat?

Question 3.
Compare the legs and nails of a dog and hen and say why they are different. (Conceptual Understanding) 4 M
(Or)
Differentiate the claws between dog and hen.
Answer:

DogHen
1) It has sharp, curved nails on the small digits.1) It possesses sharp, slightly elongated nails than dog.
2) The legs are muscular and strongly jointed.2) The legs are thin and shorter than dog’s legs.
3) It uses its legs to separate the flesh from bones.3) It uses legs to scratch the ground and eat worms.
4) The nails are also used for tearing the flesh.4) Nails are useful for scratching the soil to pick up worms.

Question 4.
Go to a near by pond where cranes are usually seen. Observe how they catch fish ? Write about the process of catching fish. (Take care of yourself when you are near water places.) (Information Skills and Project) 8 M
Answer:
Aim of the project : To find out the way of food collection and consumption by crane in the water places.

Selection of the place :The selected site is a pond with less depth of water. This enables the cranes to pick up the food (fish) easily.

Procedure that is followed : I went to the pond awaiting the cranes. A couple of cranes came to the pond. Naturally, the crane has long beak to catch the fish. First it flew down to the pond. It started walking and searching for fish with the help of its long beak. When crane found the fish, it held the fish very quickly with its beak. Then it engulfed the fish.

Question 5.
Name some animals which use tongue as a tool for taking in food. (Asking Questions and Making Hypothesis) / Conceptual Understanding) 2 M
Answer:
Wall lizard, garden lizard, chameleon, goat, sheep, cow, dog, frog etc.

TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat?

Question 6.
The butterfly uses to suck honey from flowers. (Conceptual Understanding) 2M
Answer:
The butterfly uses its long hollow tongue to suck honey from flowers.

Question 7.
Do the following and record your observations. Collect one or two earth worms and put them in a bottle containing wet soil. Close it with a lid which has holes. Observe how earthworms get their food. (Experimentation and Field Investigation)
Answer:
Aim : To observ the eating activity of earth worms in the wet soil.

Necessary material: Wet soil, two earthworms, glass bottle (or) thick transparent plastic box.

Procedure for observation : A couple of earthworms are collected and placed in the glass bottle containing wet soil. It is observed that the earthworms felt comfortable to stay in. Then they started swallowing soil in little quantities. While they are swallowing food their food pipe started expanding slightly.

Inference: With all the observations it is concluded that earthworms feed on moist soil that contains minerals and nutrients.

Question 8.
Which animals in the forest depend on only plants or on only animals for food? (Conceptual Understanding) 4 M

Animals that depend on plants for foodAnimals that depend on animals for food
Buffalo, cow, goat, sheep deer, ox, etc.Lion, tiger, fox, wolf, hyna, vulture, eagle, hawk etc.

Question 9.
Fill up the following table. (Information Skills and Projects) 4M
Answer:

Body part used to collect foodExamples
Beakhen, parrot
Tonguelizard, frog
Teethcat, dog
Suckerbutterfly, earth worm
Strong legs with clawstiger, dog, lion

TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat?

Question 10.
Why do most carnivores live in forests ? Give reasons. (Conceptual Understanding) 2M
Answer:

  1. Most of the carnivores live in forests. Because the prey of carnivores is extensively available only in forests than in domestic areas.
  2. The forest is more suitable for capturing the prey.
  3. The animals which eat herbivorous organisms are carnivores.
  4. If carnivores live outside the forest, human beings kill them as they are afraid of carnivores which may kill their animals or even kill them. Eg: Tiger, lion etc.

Question 11.
Make your own food chain and display it in your class room. (Communication through Drawing and Model Making) 4M
Answer:
TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat 1

Question 12.
Prepare a scrap book of animals and separate them into carnivores, omnivores and herbivores. (Information Skills and Project) 2M
Answer:
The photos of the following animals are very easy to collect from the available resources. I collected the photos of the following and classified them as follows.

Herbivores : Goat, sheep, buffalo, cow,
Carnivores : Lion, tiger, wolf, cat etc.
Omnivores : Wild bear, man, monkey.
TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat 2

Question 13.
Identify which of the following statements are wrong and give reasons. (Asking Questions and Making Hypothesis) 4 M
(a) That which lives in water cannot eat animals.
(b) Elephants and deer are the herbivores living in the forest.
(c) Birds’ beaks are designed to suit their food habits.
(d) Sharp claws are useful for hunting.
(e) Most of the food chains end with herbivorous animals.
Answer:
(I) Statements b and c are true to their nature.
(b) The natural living area for elephants and deer are forests.
(c) Different birds feed on different food materials. So, birds’ beaks are designed to suit their food habits.

II) Statements a, d and e are wrong. Reasons :
(a) Certain water animals feed on smaller animals.
Eg: Frog feeds on crustaceAnswer: A blue whale in the sea eats tiny animals called krill.

(d) Sharp claws of some animals meant for tearing the flesh of prey after hunting.
Eg: Lion
In some animals claws help in holding the grip for running while hunting the prey.

(e) Food chains naturally end with decomposers or degraders (microbes).

TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat?

Question 14.
If you want to understand more about food chain what questions would you like to ask ? (Asking Questions and Making Hypothesis) 4M
Answer:
I would like to clarify my doubts by asking the following questions about food chain.

  1. How do we compare animal food chain with human food chain ?
  2. What can we understand from food chain ?
  3. How do we analyse pond food chain ?
  4. What is a food chain ?

Question 15.
Write a play with dialogues between a parrot and a lion about their food habits and organs to get food. Act it with your friends. Send it to school/district children’s magazine. (Aesthetic Sence, Values and Application to Daily Life and Concern to Bio-diversity) 8M
Answer:
Role – play (Dialogues between a parrot and a lion)

Parrot : Good morning, respected king of forest!

Lion : Good morning, how are you ?

Parrot : Fine, thank you king.

Lion : Where are you going beautiful green bird ?

Parrot : I am going to collect food somewhere.

Lion : Oh! You are going to search for food. What food do you like most ?

Parrot : I like fruits such as guava, mango etc. But you people feed on animal’s flesh.

Lion : Yes, we are the carnivores, we like to feed on other animals like deer, goat, sheep etc.

Parrot : Oh! It is your habit to eat other herbivorous animals.

Lion : Yes. I am also going to hunt for food.

Parrot : I will go by my way, you can move by your way. Bye, all the best.

Lion : Bye, all the best. Have a nice day.

Question 16.
Identify the given animal. (Communication through Drawing and Model Making) 4 M
1. What does it eat ?
2. Which part of the body helps it in eating ?
TS 6th Class Science 4th Lesson Questions and Answers Telangana - What Do Animals Eat 3
Answer:

  1. The given animal is called pangolin.
  2. It feeds on ants. Thus it is called spiny ant eater. Pangolin’s tongue is long and has stretching capacity. When it finds ants it expands tongue to capture the prey.

TS 6th Class Science 4th Lesson Notes – What Do Animals Eat?

  • There are a wide variety of animals in the living world and they take a wide variety of food items.
  • Different types of animals that live in our surroundings have their own food habits (way of taking food and type of food taken)
  • Sucking, licking, pecking, chewing, peeling, swallowing are all the ways by which animals take their food in.
  • Beaks of birds differ from one another depending upon the type of food they eat.
  • Most of the wild animals that eat other animals have sharp teeth.
  • Animals are divided into 3 types on the basis of their food. They are carnivores, herbivores and omnivores.
  • Food chain is in connected between animals on the basis of their food habits.
  • Food chain explains the interdependence of diverse organisms in nature.
  • Some animals rely more on one sense than the other and it can be highly developed.
  • Food Habit : Way of taking food and type of food taken is called food habit. Different types of animals that live in our surroundings have their own food habits
  • Food Chain : Food chain is the connection between animals on the basis of their food habits.
  • Sucking : Sucking is a way of taking food by animal.
  • Picking : Picking is another type of way of food habit.
  • Chewing : Churning food is called chewing. :
  • Habitat : The surroundings which meet the needs of a particular organism in the best manner is called habitat.
  • Carnivore : The organism which depends on herbivore for food.
  • Herbivore : The organisms which feeds on plants.
  • Omnivore : The animal which feeds both on carnivores and herbivores.
  • Nocturnal : The organism which is active during night.
  • Rumination : Bringing back food from stomach to mouth for chewing is called rumination.

TS 6th Class Science 3rd Lesson Questions and Answers Telangana – Rain: Where Does it Come From?

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TS Board Telangana SCERT Class 6 Science Solutions 3rd Lesson Rain: Where Does it Come From? Textbook Questions and Answers.

Rain: Where Does it Come From? – TS 6th Class 3rd Science Lesson Questions and Answers Telangana

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Question 1.
How are clouds formed ? Explain. (Conceptual Understanding) 8 M
Answer:

  1. On a sunny day, the sun heats up the ground as well as the water in seas, oceans, rivers, ponds etc. This water converts into water vapour by the process of evaporation.
  2. Evaporation is the process of water changing into water vapour.
  3. Evaporation is a natural process which takes place on the earth.
  4. Water evaporates continuously from the surfaces of water bodies like seas, oceans, rivers, ponds etc., and changes into water vapour due to the heat supplied by sunlight.
  5. The water vapour entered into air through the process of evaporation forms clouds in the sky.

Question 2.
How does the rain water reach from clouds to rivers or oceans?
(Or)
Describe the relationship between oceans and rains. (Conceptual Understanding) 8M
Answer:

  1. The water in the water bodies gets heated up and converts into water vapour by the process of evaporation.
  2. When water vapour reaches higher levels it condenses due to contact with cool air and forms water droplets.
  3. These droplets remain floating in air at higher levels of the atmosphere and appear as clouds.
  4. Sometimes the cool breeze coming along with air makes the clouds cooler.
  5. This leads to water in the clouds condense and form large water drops.
  6. When the size of the water drops increases further it becomes difficult for the cloud to hold them and water drops begin to fall on the earth. This is called rain.
  7. In this way the rain water reaches from clouds to rivers and oce.

TS 6th Class Science 3rd Lesson Questions and Answers Telangana - Rain: Where Does it Come From?

Question 3.
When do clouds become cool ?
(Or)
Explain the changes that take place in clouds before it rain.
(Or)
How do you imagine that it is likely to rain?
(Or)
Water vapour converts into clouds. How clouds turn into rain?
(Or)
Krishnaveni said, “The raining takes place from clouds”. What are the observations you make at the time of raining ? (Conceptual Understanding) 4 M
Answer:

  1. Winds bring the clouds from the sea to the land. Clouds are nothing but evaporated water.
  2. The colder air in the upper layers of the atmosphere cools the clouds.
  3. The clouds moving in air are generally at higher levels.
  4. Sometimes the cool breeze coming along with air makes the clouds cooler.
  5. These cool clouds bring rain.

Question 4.
Explain the relationship between the heat of sun and evaporation. (Conceptual Understanding) 4 M
Answer:

  1. On a warm day, the sun heats the ground as well as the water in the water bodies.
  2. This water converts into water vapour by the process of evaporation.
  3. More the sun heats up the water from water bodies, the more evaporation of water occurs.
  4. The formation of clouds depends on the amount of water that evaporates due to sun heat.

Question 5.
Why do we experience cloud like smoke near our mouth while we speak during the winter season ? (Asking Questions and Making Hypothesis) 4 M
Answer:

  1. In winter, the air in our atmosphere is very cool as compared to the air coming out from our mouth.
  2. Water vapour present in the air coming out from our mouth gets cooled suddenly to form very tiny droplets.
  3. These tiny droplets concentrated in a limited area, appear like smoke or a small cloud near our mouth.

TS 6th Class Science 3rd Lesson Questions and Answers Telangana - Rain: Where Does it Come From?

Question 6.
Correct the given sentence if necessary.
“If the size of water drops decreases in the clouds, they can no longer hold the water drops.” (Asking Questions and Making Hypothesis) 4 M
Answer:

  1. The given sentence is not applicable for the characteristic feature of the appearance of the clouds.
  2. When water vapour reaches higher levels it condenses due to contact with cool air and forms small drops or water droplets.
  3. These tiny droplets remain floating in air at higher levels of the atmosphere and appear as clouds.
  4. Therefore if the size of water drops decreases in the clouds, they can hold water drops.

Question 7.
Which of the following days is more suitable for drying of washed clothes? Explain why. a) Windy day b) Cloudy day (Conceptual Understanding) 4 M
Answer:

  1. Windy day is more suitable than the cloudy day for drying of washed clothes.
  2. The rate of evaporation increases with the wind flow.
  3. Cloudy atmosphere has less capacity of evaporating water into vapour than windy atmosphere. The evaporation will be slow.
  4. Clothes dry faster in windy atmosphere, and slower in cloudy regions.

Question 8.
Which of the following statements are right (Or) wrong ? (Conceptual Understanding) 4 M
a) Evaporation takes place quickly when more heat is supplied.
b) For condensation of water vapour, it should be cooled.
c) Water vapour is obtained from water by its evaporation.
Answer:
The three given statements are true to their nature.
(a) Evaporation takes place quickly when more heat is supplied.

(b) When water vapour gets cooled, it turns into water. Cool air converts water vapour into water droplets which in turn finally into water by the process of condensation.

(c) The process of changing of water into water vapour is called evaporation.

Question 9.
Draw a diagram to explain the water cycle. (Communication through Drawing and Model Making) 4 M
Answer:
TS 6th Class Science 3rd Lesson Questions and Answers Telangana Rain Where Does it Come From 1

Question 10.
How do you feel when you see the beauty of Rainbow ? Express your feelings in the form of a song or a poem. (Aesthetic Sence, Values
and Application to Daily Life and Concern to Bio-diversity) 8M
Answer:

  1. The Rainbow formation during rainy season is a natural phenomenon in the sky.
  2. The refractive index of the light through tiny droplets of rain after rain fall designs the rainbow.
  3. The rainbow is the seven coloured structure of visible light rays of electro magnetic radiation.
  4. It possesses Violet, Indigo, Blue, Green, Yellow, Orange and Red colours which are embedded in visible light.
  5. The beauty of rainbow is beyond our imagination.
    My feelings on rainbow are as follows :
    The beauty of seven colours of crayon
    I am smiling at nature that goes on
    Greet me, I greet you forever and ever
    See me, I see you and leave you never
    Oh God ! What a great creation you make
    For the leisure of creatures and their sake

TS 6th Class Science 3rd Lesson Questions and Answers Telangana - Rain: Where Does it Come From?

Question 11.
Clouds once seen at a particular point, may not be there after sometime? Why? (Asking Questions and Making Hypothesis) 4 M
Answer:

  1. The clouds once seen at a particular point may not be there after sometime.
  2. This is because of the movement of clouds from high pressure areas to lower pressure areas.
  3. Pressure influences the movement of clouds.
  4. The difference in pressures in two different areas leads to the movement of air.
  5. Thus the clouds also move along with air from high pressure area to low pressure area.

Question 12.
Revanth blew air from his mouth onto the mirror while he was getting ready to go to school. He observed that the image in the mirror was not clear. Do you have any doubts to raise in this situation ? Prepare questions on your doubts. (Asking Questions and Making Hypothesis) 8 M
Answer:
Revanth blew air from his mouth onto the mirror. He confused at the appearance of moist layer on the mirror. He doubted and may have questioned himself in the following way.

  1. What is the reason for the formation of some moist layer on the mirror ?
  2. Why did the mirror become unclear after blowing air on it by mouth ?
  3. What comes out of my mouth while blowing ?
  4. Does it happen even for other animals also ?

Question 13.
If it is raining in a village you don’t find rain another village. Why do you think it is happening so? (Asking Questions and Making Hypothesis) 8 M
Answer:

  1. Often we see that there may be rain in some area, where it may not be in its adjacent areas.
  2. This is because of condensation of clouds due cool air which affects on them in the specific area.
  3. Occurance of winds won’t be same in all the areas. Pressure influences on the presence of air.
  4. Clouds along with winds move from high pressure area to low pressure area at which cool air occurs.
  5. That’s why rain is seen only in some areas, where we can’t see it in its adjacent areas.

Question 14.
If condensation fails to occur innature what happens? (Asking Questions and Making Hypothesis) 8 M
Answer:
If there is no occurance of condensation in the water, water cycle stops due to lack of rains. If there are no rains water sources will not be filled with water. Even ground water level decreases. Low percentage of water levels in the natural water bodies cause damage to the living kingdom, thus it finally lead to destruction of nature. Hence condensation process is very essential in the water cycle.

TS 6th Class Science 3rd Lesson Questions and Answers Telangana - Rain: Where Does it Come From?

Question 15.
Why does the driver of a vehicle wipe the glass inside, even if the wiper is working on outer surface of the glass when he drives in rain ?
(Aesthetic Sence, Values and Application to Daily Life and Concern to Bio-diversity) 8 M
Answer:

  1. The driver wipes the glass inside, even wiper is working on outer surface of the glass when he drives in rain.
  2. Because of the natural process called condensation rain that falls on the glass of the cabin cools its surface.
  3. Air inside the driver’s cabin contains water vapour which is warmer than the outside surface of the glass.
  4. Due to the cold glass, air close to its inner surface will also become cooler.
  5. This changes the water vapour in the air of the inner surface of the glass into water and forms small drops on the inner surface.

TS 6th Class Science 3rd Lesson Notes – Rain: Where Does it Come From?

  • Water is available in nature in three forms, ice (solid form), water (liquid form) and vapour (gaseous form).
  • Solid form of water is ice. Snow occurs naturally.
  • The three forms of water are interchangeable.
    TS 6th Class Science 3rd Lesson Questions and Answers Telangana - Rain Where Does it Come From 2
  • Evaporation : The process of changing of water into water vapour is called evaporation.
  • Condensation : The process of conversion of water vapour into water is called condensation.
  • Water cycle : The conversion of water into water vapour, water vapour to clouds and clouds to rain is known as water cycle.
  • Cloud : Clouds are formed from tiny droplets of water vapour.
  • Water vapour : The gaseous form of water is water vapour.
  • Atmosphere : Atmosphere is the main factor on the earth.
  • Stream : A narrow flow of water.
  • Droplets : Tiny particLes of water.
  • Dew : Small drops of water that form on outdoor surfaces.
  • Rain : Small drops of water falling from clouds.
  • Hails : Winds on the earth are called hails.
  • Breeze : Cool winds are called breeze.
  • Wind : Movement of air from high pressure area to low pressure area.

TS 6th Class Science 2nd Lesson Questions and Answers Telangana – Playing with Magnets

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TS Board Telangana SCERT Class 6 Science Solutions 2nd Lesson Playing with Magnets Textbook Questions and Answers.

Playing with Magnets – TS 6th Class Science 2nd Lesson Questions and Answers Telangana

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Question 1.
Predict which of the following material are magnetic and non-magnetic material. Test with a bar magnet and check your predictions. What do you say after testing all material ?
Plastic, Iron, Stainless Steel, Wood, Aluminium, Gold, Silver, Copper, (Experimentation and Field Investigation) 8 M
Answer:
Aim : To classify the given material as magnetic and non – magnetic substances by testing with bar magnet.
Apparatus: Bar magnet

Given material for Testing : Plastic, iron, stainless steel, wood, aluminium, gold, silver, copper, paper, cloth.

Procedure : A bar magnet is taken and started keeping each substance of the given material close to the bar magnet. The same method is followed for all the material.

Observation : Some substances are attracted by magnet and some are not attracted.

Result: The material attracted by magnet – Iron.
The following materials are not attracted by bar magnet a) Stainless steel b) Wood, c) Plastic, d) Aluminium, e) Gold, f) Silver, g) Copper, h) Paper,i) Cloth.

Inference : The substances which are attracted by magnet are called magnetic substances.

Eg: Iron.
The substances which are not attracted by magnets are non-magnetic substances.

Eg: Paper, Aluminium etc.

TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets

Question 2.
List out the magnetic and non – magnetic materials in your classroom. (Conceptual Understanding) 2 M
Answer:

  1. Magnetic materials : Iron rods of the window, pins, bolt of the door, binding wire, nails etc.
  2. Non – magnetic materials: Paper, note book, plastic pen, rubber band, eraser etc.

Question 3.
For which purposes do people use magnets in their daily life ? Ask your family members and other elders and collect the information and prepare a list of uses of magnets. (Information Skills and Projects) 2 M
Answer:
Aim : Aim of the project is to collect the information of uses of magnets in our daily life.
Uses :

  1. Pure stainless steel is not attracted by magnet. The quality of stainless steel can be checked with a magnet while buying.
  2. It would be easy to handle the pins easily if a magnet is placed in the lid of a pin box.
  3. Magnets are used as door stoppers. One magnet is placed on the door and the other on wall. The door is attracted to wall and doors will not move for the wind.
  4. Magnets are used in refrigerator doors, toys, magnetic stickers, fans, loud speakers, microphones, automobile dynamos, audio and video tapes and computer hard disks.

Question 4.
Draw a bar magnet and locate the poles. (Communication through Drawing and Model Making) 2 M
Answer:
TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets 1

Question 5.
Observe and locate North and South poles for the second bar magnet shown in the figure given below. (Communication through Drawing and Model Making) 2 M
Answer:
TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets 2

Question 6.
Think and say, in which direction your house is facing ? Use the compass and find out the exact direction of your house and compare it with your prediction. Similarly predict and find out in which direction you keep your head while sleeping at night, the directions you face while you are reading, eating etc. (Asking Questions and Making Hypothesis) 4 M
Answer:

  1. Naturally any architecture professional designs the house blue print in East – West direction. This is because of one reason by allowing morning sun-rays into the house and better ventilation.
  2. Compass is used to find out the specific direction of its construction. I clearly found that it is in East – West direction.
  3. Head is kept either in East – West (or) West – East direction while sleeping. This is to avoid North – South direction of the magnetic influence.
  4. Our elders advise us to sit for study in East-West direction, we avoid sitting North – South direction. We do the same thing while taking meals. For the above reasons we avoid North – South direction of the magnetic influence.

TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets

Question 7.
Prepare a toy using magnets and write the procedure of preparation briefly. (Experimentation and Field Investigation) 8 M
Answer:
Aim : To prepare a toy using magnets.
Apparatus : Two bar magnets, a toy (car or doll)

Procedure:

  1. A toy car is brought from the shop which is in good condition.
  2. A couple of bar – magnets are taken. One is inserted into the front portion of the toy facing north pole towards front and south pole – towards back. The preparation is kept on the floor.
  3. Another bar magnet is taken into hands keeping its south pole towards front part of the car. It tries to change the direction of the magnet towards the toy car.

Observation : While keeping the south pole of hand’s bar magnet towards the car, the car comes to the hand. While keeping north pole of the hand’s bar – magnet towards the car, the car moves away from hand.

Inference : Like poles of bar magnet repel each other, unlike poles of bar – magnet attract each other. This principle helps us making toy cars and dolls.

Question 8.
Think and say where the poles will be located in a ring magnet ? Try to find out its poles using a bar magnet and check your prediction.
(Experimentation and Field Investigation) 4 M
Answer:

Aim : To find out the North – South poles of the ring magnet by using bar magnet.

Apparatus : Ring magnet, bar magnet

Procedure : A ring magnet is kept on the table. A bar magnet is brought very close to the ring magnet. Amazingly the bar magnet’s north pole is attracted on upper portion of the ring magnet.

Observation: The direction of the bar magnet is changed towards upper portion. But both the magnets repelled. That means the south.

Result: To a ring magnet the poles are located at upper and lower sides.

Question 9.
Magnetise a needle using a bar magnet. Make a compass with that needle by following the process explained in activity 10. (Experimentation and Field Investigation) 4 M
Answer:
Aim : To make a compass with my own magnetised needle.

Apparatus : Needle, glass, round light weight cork, water etc.

Procedure:
A) Preparation of magnetised needle :

  1. A needle is taken and kept on the table. A bar magnet is placed, one of its poles near one edge of the needle.
  2. Without lifting the bar magnet, it is moved along the length of the needle.
  3. Then the magnet is brought to the first end of the needle and is moved along the length. This is repeated 20-30 times.
  4. Now the bar magnet is removed and some iron filings are brought to the magnetised needle.

Observation :

  1. Iron filings are attracted by the needle.
  2. Therefore I succeeded in making my own magnet by magnetising needle.

B) i) The magnetised needle is kept on the cork and is placed on the water surface in the glass.
ii) In order to make free floating of the cork a little detergent is added.

Observation : We observed the magnetised needle pointing north and south directions.

Result: We can prepare a compass with a magnetised needle.

Precaution : We should not drag the bar magnet back and forth on the needle, it should be moved in only one direction.

TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets

Question 10.
Sometimes people use magnets to keep the doors open and sometimes to close the doors firmly. Think and say how it is possible and how we should arrange the magnets in each case, (or) Magnets are used in closing and opening of doors and windows. How is it possible ? Think and write. (Asking Questions and Making Hypothesis) 4 M
Answer:

  1. It is possible to close and open the doors firmly.
  2. The facility of this is based on the magnetic property of two magnets.
  3. The magnetic property says unlike poles of magnets attract, like poles of magnets repel.
  4. We can do this by changing the direction of table magnet. The suspended one also changes its direction.

Application of Magnetic property to the doors :
(a) When doors are opened firmly, it is because of placing two magnets facing each other with like poles in two doors. This results in firm opening of doors.

(b) When doors are closed firmly, it is because of placing of two magnets facing each other with unlike poles in two doors. This results in firm closing of the doors.

Question 11.
Does the Earth behave as a magnet? How do you prove it? (Experimentation and Field Investigation) 4M
Answer:
Aim : To prove the magnetic behaviour of the Earth.

Apparatus : Two bar magnets, thread, table etc.

Procedure:

  1. A bar magnet is placed on the table. Another bar magnet is suspended very close to the first one kept on the table.
  2. It is observed that the north pole of the suspended bar magnet points towards the south pole of the magnet placed on the table.
  3. The south pole of the suspended bar magnet points towards the north pole of the bar magnet kept on the table.
  4. Later the first bar magnet is removed from the table. But the suspended magnet is still hanging on the table.

Observation:

  1. The suspended magnet comes to rest in the North-South direction.
  2. It is said that there is some magnet below the suspended one, which makes it to come to rest in that particular direction.

Result: It is evident that the earth possesses magnetic property which acts upon the suspended bar magnet.

Inference : The earth exhibits greater magnetic property with north and south poles.

Question 12.
If you have two similar bars, one a magnet and another a piece of iron, can you find out which one of these is a magnet ? Explain the process.
(Experimentation and Field Investigation) 2 M
Answer:

  1. The substance which exhibits magnetic property attracts the other magnetic material. This occurs in case of magnet.
  2. Because it attracts the other bar, which is iron.
  3. When both the bars are kept closely, the true magnet bar attracts the iron bar.
  4. On the other hand iron bar doesn’t show attractive property on true magnet.
  5. By the following way we can recognise true magnet and magnetic substance.

Question 13.
Teacher said that earth is a magnet. But Sree Vidya has some doubts and she asked her teacher some questions. What may be the questions? (Asking Questions and Making Hypothesis) 2M
Answer:

  1. If Earth possesses magnetic property why don’t all the iron material attract towards any one of the poles of the Earth ?
  2. How can we prove the magnetic property of the Earth ?
  3. What is the advantage of magnetic property of Earth for living kind ?
  4. Where are south and north poles of Earth’s magnet ?

TS 6th Class Science 2nd Lesson Questions and Answers Telangana - Playing with Magnets

Question 14.
Surya was wonderstruck to know that Earth is a big magnet and appreci¬ated efforts of scientists to discover this. Do you notice any such things in magnets to appreciate? Explain. (Aesthetic Sence, Values and Application to Daily Life and Concern to Bio-diversity) 4M
Answer:

  1. Isaac Newton discovered that the Earth possesses magnetic property. He proved it by observing fall of apple from the plant.
  2. Indeed we should appreciate the great power of magnetism.
  3. Because of the magnetic property, the Earth revolves round the sun in the name of gravitational force.
  4. The moon revolves round the Earth with specific magnetic property influenced by Earth.
  5. In favour of living kind the earth sustains all the organisms.
  6. Due to above reasons we can appreciate the magnets and the concept of magnetism which exist in the nature.

Question 15.
Kiran wants to prepare a toy using some magnets to make people understand the slogan “Reject bad food and accept only good food.” Can you help him to prepare the toy? If yes, how? (Communication through Drawing and Model Making) 4M
Answer:

  1. Applying the magnetic property for making a toy magnet to conduct the programme is interesting.
  2. The junk food box is placed on one side of the table. Good food box is kept on the other side.
  3. Each item of the junk food with bar magnet facing its specific pole is arranged. Likewise to the good food items having magnet facing its specific pole.
  4. Moving the toy magnet in front of junk food, keeps away from them. Because of the facing pole of the toy magnet is as same as pole of the junk food items.
  5. Moving the toy magnet in front of the good food items, attract them, because of unlike poles of the food items magnet and toy  magnet. We should emphasise the magnetic property of the two bar magnets in our day to day life. To conduct an awareness programme on food items, toy magnet game will be very interesting.

TS 6th Class Science 2nd Lesson Notes – Playing with Magnets

  • The cap of the pin holder contains a material known as magnet which attracts sub¬stances like iron pins, iron nails etc.
  • The stone which Magnus pulled out is called lode stone.
  • Lode stone is a natural magnet
  • Magnets are of different shapes i.e., bar magnets, horse shoe magnets, ring type magnets etc.
  • Each magnet has two magnetic poles : North and South A freely suspended magnet always aligns in the North – South direction.
  •  Magnet : The material which attracts substances like iron pins, iron nails etc. is known as a magnet.
  • Magnetic materials : The materials that are attracted by magnets are called magnetic materials. Eg: iron pins, nails etc.
  • Non-magnetic materials : The materials that are not attracted by the magnets are called non-magnetic materials. Eg: paper, wood, plastic etc.
  • North Pole, South Pole : A bar magnet always have two ends whose attracting capacity is more than other parts of it, these are called poles. Each magnet has two poies – north pole and south pole.
  • Magnetic Compass : A magnetic compass is used to find directions. It is mostly used in ships and aeroplanes, by mountaineers and army people. It works on the “directional property of magnets”.
  • Like poles : North pole – North pole or South pole – South pole of two magnets are called like poles. These are repelled by each other.
  • Unlike poles : North pole – South pole or South pole – North pole of two magnets are called ‘unlike poles’. These are attracted by each other.
  • Attraction : Unlike poles (N-S, S-N) attract each other.
  • Repulsion : Like poles (N-N, S-S) repel with each other.
  • Magnetic induction : Magnetic property possessed by a magnetic substance due to the presence of a magnet near to it is called magnetic induction.

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

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Regular practice with TS 6th Class Science Bits with Answers 4th Lesson What Do Animals Eat? improves students’ confidence and readiness for assessments and examinations.

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 1.
The animals which eat the flesh of other animals
A) Herbivores
B) Carnivores
C) Omnivores
D) None
Answer:
B) Carnivores

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 2.
Example for Omnivores
A) cow
B) lion
C) man
D) deer
Answer:
C) man

Question 3.
Read the names of the following animals
1. Cow
2. Dog
3. Man
4. Tiger
Which are omnivores from the above?
A) 1, 2
B) 1, 3, 4
C) 3, 4
D) 2, 3
Answer:
D) 2, 3

Question 4.
The animals which eat only plants are …
A) Herbivores
B) Carnivores
C) Omnivores
D) All
Answer:
A) Herbivores

Question 5.
According to their food habits animals are divided into … types.
A) 1
B) 3
C) 4
D) 2
Answer:
B) 3

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 6.
The insect which feeds on other insects is
A) Cockroach
B) Pond skater
C) Butterfly
D) Bug
Answer:
B) Pond skater

Question 7.
Match the following basing on food capturing organs.
List-I — List-Il
I Hen — a) Tongue
2. Camel — b) Beak
3. Frog — c) Mouth
A) 1-a, 2-b, 3-c
B) 1-b, 2-c, 3-a
C) 1-a, 2-c, 3-h
D) 1-c, 2-b, 3-a
Answer:
B) 1-b, 2-c, 3-a

Question 8.
Which part of the body in frog is mainly useful for food capturing/ingestion?
A) Legs
B) Mouth
C) Cloaca
D) Tongue
Answer:
D) Tongue

Question 9.
Which of the following animals uses its tongue to capture food?
A) Frog
B) Lizard
C) Garden Lizard
D) All the above
Answer:
D) All the above

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 10.
a) Mostanimals are motile
b) Sponges are sedentary
A) ‘a’ and ‘b’ are true
B) ‘a’ is true ‘b’ is false
C) ‘a’ is fake ‘b’ is true
D) a and h are false
Answer:
A) ‘a’ and ‘b’ are true

Question 11.
Vultures are good example for ……………
A) predators
B) herbivores
C) natural scavamer
D) omnivores
Answer:
C) natural scavamer

Question 12.
The feeding habit of a cow is called ……….
A) crushing
B) churning
C) sucking
D) rumination
Answer:
D) rumination

Question 13.
Most of the animals that eat other animals have teeth.
A) long
B) short
C) sharp
D) small
Answer:
C) sharp

Question 14.
Which animal is not a ruminant?
A) Cow
B) Tiger
C) Buffalo
D) Camel
Answer:
B) Tiger

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 15.
Which part of the body is helpful for crane to pick up the food?
A) Long beak
B) Claws
C) Legs
D) Wings
Answer:
A) Long beak

Question 16.
Identify the ruminate animal.
A) Camel
B) Pig
C) Monkey
D) Dog
Answer:
A) Camel

Question 17.
Ducks use the teeth for the food.
A) churning
B) masticating
C) filtering
D) All
Answer:
C) filtering

Question 18.
Leeches get their food by the process of
A) sucking
B) absorbing
C) eating
D) masticating
Answer:
A) sucking

Question 19.
The animals which search for their food during nights are called
A) diurnals
B) tetrahedrals
C) miurnals
D) nocturnals
Answer:
D) nocturnals

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 20.
In butterfly is used to suck honey from flowers.
A) tube like mouth
B) needle like tongue
C) teeth of mouth
D) legs of body
Answer:
A) tube like mouth

Question 21.
Example of nocturnals
A) man
B) parrot
C) bats
D) sparrow
Answer:
C) bats

Question 22.
Natural scavengers
A) Crane
B) Honey bird
C) Wood pecker
D) Crow
Answer:
D) Crow

Question 23.
Suckers are present in
A) snail
B) earthworm
C) housefly
D) leech
Answer:
D) leech

Question 24.
The leopard is a member of family.
A) crane
B) cat
C) lizard
D) rat
Answer:
B) cat

Question 25.
The nocturnal animal you see in your locality
A) Crow
B) Cow
C) Owl
D) Sheep
Answer:
C) Owl

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 26.
Match the following.
1. Camel — a) Nocturnal
2. Lion — b) Herbivore
3. Bat — c) Carnivore
A) 1-c, 2-b, 3-a
B) 1-a, 2-b, 3-c
C) 1-b, 2-c, 3-a
D) 1-c, 2-a, 3-b
Answer:
C) 1-b, 2-c, 3-a

Question 27.
Frog uses these parts to collect the food materials
A) Legs
B) Hands
C) Tongue
D) Nose
Answer:
C) Tongue

Question 28.
What are the specific organs present in leech to suck the blood?
A) Suckers
B) Tongue
C) Glands
D) Teeth
Answer:
A) Suckers

Question 29.
Which animal search for food at night?
A) Snake
B) Tiger
C) Crane
D) Owl
Answer:
D) Owl

Question 30.
Identify the wrong sentence.
A) Mode of food taking in leech is sucking
B) Deer is a carnivore
C) Man belongs to omnivore
D) Omnivores eat both plants and animals
Answer:
B) Deer is a carnivore

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 31.
Food chain is the connection between animals on the basis of
A) living place
B) water facility
C) food habits
D) air availability
Answer:
C) food habits

Question 32.
The relation between animals in the food chain is based on
A) Living place
B) Water facility
C) Food habits
D) Availability of air
Answer:
C) Food habits

Question 33.
Complete the following food chain.
TS 6th Class Science Bits 4th Lesson What Do Animals Eat 1
A) man
B) deer
C) hen
D) snake
Answer:
B) deer

Question 34.
Read the following sentences.
1. Eggs and larvae are eaten by fish and frog.
2. Camel is a ruminant animal.
A) Both 1 & 2 sentences are correct
B) 1 is wrong 2 is correct
C) 1 is correct 2 is wrong
D) Both 1 & 2 sentences are wrong
Answer:
A) Both 1 & 2 sentences are correct

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 35.

Body part used for collecting foodExample
Sucker
Strong legs with claws		Leech
?

A) Goat
B) Crow
C) Vulture
D) Cow
Answer:
C) Vulture

Question 36.
Birds are
A) mammals
B) annelids
C) arthropods
D) vertebrates
Answer:
D) vertebrates

Question 37.
An example for carnivorous animal
A) Cow
B) Elephant
C) Buffalo
D) Fox
Answer:
D) Fox

Question 38.
Sense organs used by bats in finding food:
A) Eyes
B) Ears
C) Nose
D) Skin
Answer:
B) Ears

Question 39.
Identify the animal which search their food during night
A) Tiger
B) Dog
C) Cockroach
D) Cow
Answer:
C) Cockroach

Question 40.
The main man made mistake that show Impacts on food chains
A) Using fertilizers sufficiently
B) Using inseticides and pesticides in large quantities
C) Using bio-fertilizers in large quantities.
D) B or C
Answer:
B) Using inseticides and pesticides in large quantities

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 41.
Several food chains connecting to each other in the form of
A) Food web
B) Habitat
C) Food net
D) Environment
Answer:
A) Food web

Question 42.
Why do we say that ants are good farmers?
A) They bring manure to the crop field.
B) Ants manufacture compost
C) They help in growing a type of fungus which they eat
D) All the ants die to make the soil fertile
Answer:
C) They help in growing a type of fungus which they eat

Question 43.
Which one of the following collects food by virtue of vision
A) Eagle
B) Dog
C) Cat
D) Bat
Answer:
A) Eagle

Question 44.
Goat: herbivore : Lion :
A) omnivore
B) carnivore
C) herbivore
D) suctivore
Answer:
B) carnivore

Question 45.
Find out the odd one
A) sheep
B) cow
C) bullock
D) dog
Answer:
D) dog

Question 46.
lf sharp teeth are absent in dog
A) it can’t take grass
B) it can’t eat juicy food
C) it can’t eat meat
D) all the above
Answer:
C) it can’t eat meat

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 47.
Find the correct pair regarding food collection
A) Frog – Teeth
B) Leech-Suckers
C) Earthworm – Tongue
D) Snake – Tail
Answer:
B) Leech-Suckers

Question 48.
Grass → Insects (crickets) → Frog → Snake
What will happen to the food chain if all frogs die? :
A) Insects population increase
B) Snakes suffer from lack of food
C) Grass plants grow more
D) A & B
Answer:
D) A & B

Question 49.
There is a great balance in nature among plants and animals regarding food habits.
What will happen if all animals ate plants?
A) Animals suffer from lack of breathing air
B) Plants will disappear from earth. Then animals will struggle for food
C) Earth will become hot
D) All the above
Answer:
D) All the above

Question 50.
Giraffe: Herbivore:: Crane: … …….
A) Carnivore
B) Plant eating
C) Omnivore
D) Herbivore
Answer:
A) Carnivore

Question 51.
The parts of hen used for picking up food
A) claws
B) beak
C) mouth
D) A and B
Answer:
B) beak

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 52.
Example of omnivore that we see daily in our surroundings
A) Hen
B) Crow
C) Sheep
D) A and B
Answer:
D) A and B

Question 53.
What is the major difference between dog and rat in case of eating the food material?
A) Dog uses canine to tear the food and rat uses front teeth (incisors) to eat the food.
B) Dog depends on incisors to eat the food and rat uses incisors to eat
C) Both dog and rat do not use teeth
D) Dog uses big teeth. Rat uses canine teeth.
Answer:
A) Dog uses canine to tear the food and rat uses front teeth (incisors) to eat the food.

Question 54.
Which is a carnivore among butterfly, garden lizard, grass hopper in your school garden that you observed.
A) Garden lizard
B) Butterfly
C) Grasshopper
D) Lion
Answer:
A) Garden lizard

Question 55.
Arrange the following animals based on their food habits.
A) Cow, Goat – Omnivore
Fox, Eagle, Tiger – Herbivore.
Fish, Man – Carnivore

B) Eagle, Tiger – Herbivore
Cow Goat – Carnivore
Fox, Eagle, Tiger – Herbivore.

C) Fish, Man – Omnivore
Cow, Goat – Herbivore.
Fox, Eagle, Tiger – Carnivore

D) Fish, Man – Omnivore
Cow, Goat – Carnivore
Fox, Eagle, Tiger – Herbivore.
Answer:
C) Fish, Man – Omnivore
Cow, Goat – Herbivore.
Fox, Eagle, Tiger – Carnivore

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 56.
Read the following table and give suitable answer for below question.

AnimalOrgan which helps in food collection
1. Hen / cockBeak
2. ManHands

Hen collects its food through
A) Tongue
B) Wings
C) Beak
D) Eyes
Answer:
C) Beak

Question 57.
Read the following paragraph and answer the question.
Vultures use sharp claws along with strong hooked beak to tear the flesh. While humming bird that sucks nectar need a long thin beak. What are the parts of vulture help in capturing food ?
A) claws
B) hooked beak
C) wings
D) A and B
Answer:
D) A and B

Question 58.
Read the para and answer the given question.
Ducks and fish have teeth. They act as filters to get food from water. Teeth these animals are not useful for grinding. The use of teeth in duck and fish.
A) for filtering
B) for grinding
C) for sucking
D) for swallowing
Answer:
A) for filtering

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 59.

Food groupExample
Only plants Only animals BothCow, Goat Fox, Tiger Man, Duck

What we call the animals if they eat only animals ?
A) Carnivores
B) Omnivores
C) Herbivores
D) Detrivores
Answer:
A) Carnivores

Question 60.

Body part used to collect foodExamples
BeakHen
TongueFrog
TeethDog

Name some other animal that uses tongue to eat food
A) Tiger
B) Lion
C) Duck
D) A & B
Answer:
C) Duck

Question 61.
Find out the following diagram.
TS 6th Class Science Bits 4th Lesson What Do Animals Eat 2
A) Mosquito
B) Dragonfly
C) Pond skater
D) Grass hopper
Answer:
C) Pond skater

Question 62.
In the given diagram, body part used in taking food….
TS 6th Class Science Bits 4th Lesson What Do Animals Eat 3
A) Teeth
B) Beak
C) Tongue
D) Mouth
Answer:
D) Mouth

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 63.
The given flow chart shows
Grass → Rabbit → Wolf
A) Food web
B) Food chain
C) Pyramid
D) Pyramid number
Answer:
B) Food chain

Question 64.
Find out the missing one in the given flow chart
Grass → Deer → ?
A) Rat
B) Rabbit
C) Cockroach
D) Lion
Answer:
D) Lion

Question 65.
Choose possible food chain based on the given figure.
TS 6th Class Science Bits 4th Lesson What Do Animals Eat 4
A) Smaller plants → Frog → Fish → Crane
B) Smaller grass → Fish → Frog →Crane
C) Frog → Fish → Crane → Smaller grass
D) Smaller seeds → Frog → Fish → Crane
Answer:
B) Smaller grass → Fish → Frog →Crane

Question 66.
Carrot plant → Rabbit → Tiger
Producer in the above food chain
A) Carrot plant
B) Rabbit
C) Tiger
D) Carrot plant and Rabbit
Answer:
A) Carrot plant

Question 67.
The following flow chart indicates
Grass → Insects → Frog → Snake
A) Food web
B) Pyramid
C) Food chain
D) Above all
Answer:
C) Food chain

Question 68.
Grains → Rat →  Cat → ……….. Lion
A) Deer
B) Fox
C) Rabbit
D) Man
Answer:
B) Fox

Question 69.
Complete the following food chain
Grass → …………… → Tiger
A) Man
B) Deer
C) Hen
D) Snake
Answer:
B) Deer

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 70.
After knowing about food chain, student can
A) kill snakes
B) cut plants
C) grow plants
D) rear frogs
Answer:
C) grow plants

Question 71.
What we learn from ants?
A) An ideal social life
B) Enmity among ants
C) They are dangerous animals
D) None of these
Answer:
A) An ideal social life

Question 72.
What can we find a valuable theme (idea) from food chains?
A) Animals struggle to live
B) An animal depends on itself for food
C) Interdependence of diverse organisms
D) Animals eat their own races
Answer:
C) Interdependence of diverse organisms

Question 73.
Just we keep cows for milk, ants keep a type of insect called aphids for
A) leaf juice
B) honey dew
C) honey bees
D) stem water
Answer:
B) honey dew

TS 6th Class Science Bits 4th Lesson What Do Animals Eat?

Question 74.
We should allow crows and vultures to live on the earth. Because they help us as
A) Predators
B) Natural omnivores
C) Natural destroyers
D) Natural scavengers
Answer:
D) Natural scavengers

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life 

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These TS 6th Class Science Important Questions 11th Lesson Changes Around Us are crafted to align with the curriculum, ensuring students are well-prepared for assessments.

TS 6th Class Science Important Questions 11th Lesson Changes Around Us

Question 1.
How is water level in the reservoirs measured?
Answer:
In feet.

Question 2.
How is the water released from dams and projects during floods, measured?
Answer:
In cusecs (cubic centimeters/sec)

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life

Question 3.
How is sea water?
Answer:
Salty

Question 4.
What are the drought-prone areas In our state?
Answer:
Anantapuram and Mahaboobnagar.

Question 5.
Mention a natural hazard.
Answer:
Floods
(Some other natural hazards are – droughts, volcanoes, forest fires, earth-quakes, etc.)

Question 6.
Can we drink the water available ¡n the sea?
Answer:
No. It is not fresh water. Sea water is salty. So we cannot drink it.

Question 7.
Do plants and animals also require water like us?
Answer:
For their survival and growth, plants and animals require water.

Question 8.
What are the main water sources of villages and cities ?
Answer:
Wells, canals, tanks, ponds, rivers, bore-wells are the main water sources in villages and cities.

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life

Question 9.
Explain the stages involved in safe drinking water supply. (or)
Prepare a flow chart showing the different stages of protected drinking water schemes.
Answer:
Stages of safe drinking water supply.

TS-6th-Class-Science-Important-Questions-11th-Lesson-Water-in-Our-Life-2

  • Water is collected into tank: Untreated water is collected from rivers or lakes into tanks.
  • Filtration: The water is filtered through sand particles and grills and then sent to another aerated tank. Nearly 99% of solid wastes are filtered through this process.
  • Aeration: Filtered water is exposed to sunlight. As a result some bacteria present in the water bodies will die due to the Sun’s heat.
  • Chlorination: Chlorine compounds are used to destroy all the disinfectants present in the water. Now the water is available to be used for drinking.
  • Overhead tank: Chlorinated water is pumped into overhead tank before supplying.
  • Supply of water: Pure water is supplied to the people through taps.

Question 10.
What is the percentage of fresh water on the earth?
Answer:
One percent

Question 11.
Write the occupancy ratio of the water on the earth.
Answer:
3/4th of the earth is occupied by water.

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life

Question 12.
Are the sources from where you get water for your daily needs and crop same or not? Give your reasons.
Ans.
For our daily needs, we get fresh water, supplied by the town authoritie under Protected Drinking Water Scheme.’ For the crops, we depend on rains, canals, rivers, bore – wells etc. It is n fresh water.

Question 13.
We know that nearly 314th of the surface of the earth is occupied by water. Is this water useful for us?
Answer:
Although 3/4th of our earth is covered with water, about 97% of the water available in the seas and oceans is salt water. About 2% of water is in the fon of ice at the poles. This means, only 1% of water on the earth is useful for us. Therefore we should not waste water.

Question 14.
What happens if there is less rain fall or too much rain fall?
Answer:

  • If there is less rain fall, there will be scarcity of water.
  • If there is no rain for 4 or 5 years continuously, it causes droughts.
  • Too much rain fall may cause floods.

Question 15.
What would happen if there was no rainfall of five years?
Answer:

  • Severe drought and famine will occur in the area.
  • People will face drastic problems such as food and water scarcity.
  • Cattle will die due to unavailability of grass or non grazing fields.
  • Plants will die and those remaining will enter the jaws of death.
  • As a result people will migrate to places with food and water resources.
  • Nutritional problems will arise in the children of poor families.

Question 16.
Drought occurred in Ramanna’s village. Guess the problems faced by
Ramanna due to drought.
Answer:
Ramanna faced many problems due to drought.

  • For the last five years, there were no rains.
  • Their fields dried and cracks developed on them.
  • His father invested money on bore – wells with no results.
  • Some water they could get from a bore – well lying kilometers away from their village.
  • Several people sold their cattle and migrated to distant cities like Hyderabad and Bangalore.

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life

Question 17.
What problems can arise due to water scarcity in a particular place?
Answer:
Loss of harvestation, drinking problems, domestic water problems etc., will arise.

Question 18.
If several bore wells are dug and underground water is tapped constantly what will happen to the source of ground water?
Answer:

  • Ground water levels will decrease.
  • Soil will lose moist conditions.
  • Grass lands will not grow.
  • Plants will die due to dryness in soil.
  • Soil organisms such as earthworms, useful insects will die due to lack of moisture in the soil.
  • People who depend on ground water for drinking and other purposes will face serious problems.

Question 19.
What are the possible and relevant questions you will ask on droughts – water supply?
Answer:

  • What will happen if rainfall this year is less than that of last year?
  • What would happen if there was no rainfall for five years?
  • What could be the possible reasons for water scarcity?
  • What problems can arise due to water scarcity?

Question 20.
A student named Virat had a curiosity to know how the water is being supplied to the villages with a govt. scheme i.e. ‘Safe drinking water’. The river water is filtered by separating all the unwanted materials and aerating the water with the process of sprinkling & chlorination. The purified water is then sent to the area tanks, then the purified water is supplied through the pipe lines to the individual houses of that particular areas.
1. What is the process used to separate unwanted materials and microorganisms?
2. What is the main source of water for village people?
3. Show the pumping/supplying of purified water in the form of flowchart.
4. What is meant by safe drinking water?
Answer:

  1. Sprinkling and chlorination.
  2. The river water
  3. Tank → Filtration → Aeration → Chlorination → Overhead tank → Taps
  4. The purified water purely free from impurities and microbes useful for drinking is called safe drinking water.

Question 21.
How do polythene bags, and disposable plates cause floods?
Answer:

  • The polythene bags and disposable plates do not decompose in the soil.
  • They obstruct water flow in canals and drainages during heavy rains and cause floods.
  • People become shelterless.

TS Board 6th Class Science Important Questions 11th Lesson Water in Our Life

Question 22.
Collect information about ‘cyclones’. In which season do they occur?
Write about effects of cyclones that occur in the coastal area of our state. (or)
Write about the information that you collected on cyclones. In which season they occur? Write about the effects of cyclones in coastal A.P.
Answer:
Cyclones frequently occur in the months of October and November.
Often we see cyclones in the month of May due to high pressure in the atmosphere.
The consequences seen after cyclones:

  • Roof tops and houses collapse due to heavy winds.
  • Hoarding boards and street light poles collapse.
  • Heavy loss of population and death of animals occurs.
  • Falling of trees is one of the effects of heavy flood.
  • People suffer from heavy damage of property.