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AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.4

Question 1.
Find the errors and correct the following mathematical sentences
(i) 3(x – 9) = 3x – 9
(ii) x(3x+2) = 3x² + 2
(iii) 2x+3x = 5x²
(iv) 2x + x + 3x = sx
(v) 4p + 3p + 2p + p – 9p = 0
(vi) 3x + 2y = 6xy
(vii) (3x)² + 4x +7 = 3x² + 4x +7
(viii) (2x)² + 5x = 4x + 5x = 9x
(ix) (2a + 3)² = 2a² + 6a +9
(x) Substitute x -3 in
(a) x² + 7x + 12 (- 3)² + 7(-3) + 12 = 9 + 4 + 12 = 25
(b) x² – 5x + 6(-3)² – 5(-3) + 69 – 15 + 6 = 0
(c) x² +5x = (-3)² + 5(3) + 6 = -9 – 15 = -24
(xi) (x – 4)² = x² – 16
(xii) (x + 7)² = x² +49
(xiii) (3a + 4b)(a – b)= 3a² – 4a²
(xiv) (x + 4) (x + 2) = x² + 8
(xv) (x – 4) (x – 2) = x²– 8
(xvi) 5x³ ÷ 5 x³ = 0
(xvii) 2x³ + 1 ÷ 2x³ = 1
(xviii) 3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
(xix) 3x + 5 ÷ 3 = 5
(xx) \(\frac{4 x+3}{3}\) = x + 1
Solution:
(i) 3(x – 9) = 3x – 9
3(x – 9) = 3x – 9
⇒ 3x – 3 x 9 = 3x – 9
⇒ 3x – 27 = 3x – 9
⇒ – 27 ≠ – 9
∴ The given sentence is wrong. Correct sentence is 3(x – 9) = 3x – 27.

(ii) x(3x+2) = 3x² + 2
x(3x + 2) = 3x² + 2
⇒ x × 3x + x × 2 = 3x² + 2
⇒ 3x² + 2x ≠ 3x² + 2
∴ The given sentence is wrong.
Correct sentence is x(3x + 2) = 3x² + 2x.

(iii) 2x+3x = 5x²
2x + 3x = 5x²
⇒ 5x = 5x²
⇒ x ≠ x²
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(iv) 2x + x + 3x = 5x
2x + x + 3x = 5x
⇒ 6x = 5x
⇒ 6 ≠ 5
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(v) 4p + 3p + 2p + p – 9p = 0
4p + 3p + 2p + p – 9p = 0
⇒ 10p – 9p = 0
⇒ p = 0
It is not possible
∴ The given sentence is wrong. Correct sentence is
4p + 3p + 2p + p – 9p – p = 0

(vi) 3x + 2y = 6xy
3x + 2y = 6xy
a + b ≠ ab
∴ The given sentence is wrong.
Correct sentence is 3x x 2y = 6xy.

(vii) (3x)² + 4x +7 = 3x² + 4x +7
(3x)² + 4x +7 = 3x² + 4x +7
⇒ (3x)² = 3x²
⇒ 9x² = 3x²
⇒ 9 = 3
It is not possible
∴ The given sentence is wrong. Correct sentence is
(3x)²+ 4x + 7 = 9x² + 4x + 7.

(viii) (2x)² + 5x = 4x + 5x = 9x
(2x)² + 5x = 4x + 5x = 9x
⇒ 4x² + 5x = 4x + 5x
⇒ 4x² = 4x
⇒ x² = x
⇒ x ≠ √x
∴ The given sentence is wrong. Correct sentence is (2x)² + 5x = 4x² + 5x.

(ix) (2a + 3)² = 2a² + 6a +9
(2a + 3)² = 2a² + 6a +9
⇒ (2a)² + 2 × 2a × 3 + 3² = 2a² + 6a + 9
⇒ 4a² + 12a + 9 = 2a²+ 6a + 9
⇒ 4a² – 2a² = 6a – 12a
⇒ 2a² = – 6a
⇒ 2a ≠ 6
∴ The given sentence is wrong.
Correct sentence is
(2a + 3)² = 4a² + 12a + 9.

(x) Substitute x -3 in
(a) x² + 7x + 12 (- 3)² + 7(-3) + 12 = 9 + 4 + 12 = 25
x² + 7x + 12 = (- 3)² + 7 (- 3) + 12
= 9 – 21 + 12
= 21 – 21
= 0 25 (False)

(b) x² – 5x + 6(-3)² – 5(-3) + 69 – 15 + 6 = 0
x² – 5x + 6 = (-3)² – 5 (- 3) + 6
= 9 + 15 + 6
= 30 ≠ 0 (False)

(c) x² +5x = (-3)² + 5(3) + 6 = -9 – 15 = -24
x² + 5x = (- 3)² + 5 (- 3)
= 9 – 15 = – 6 ≠ 24 (False)

(xi) (x – 4)² = x² – 16
(x – 4)² = x² – 16 = (x)² – (4)²
(a – b)² ≠ a² – b²
∴ (x-4)² ≠ (x)² – (4)²
∴ The given sentence is wrong.
Correct sentence is (x – 4)² = x² – 8x + 16.

(xii) (x + 7)² = x² +49
(x + 7)² = x² + 49 = (x)² + (7)²
(a + b)² ≠ a² + b²
∴ (x+7)² ≠ (x)² – (7)²
∴ The given sentence is wrong.
Correct sentence is (x + 7)² = x² + 14x + 49.

(xiii) (3a + 4b)(a – b)= 3a² – 4a²
3a(a – b) + 4b(a – b) = 3a² – 4²
3a² – 3ab + 4ab – 4b² = – a²
3a² + ab – 4b² ≠ a²
∴ The given sentence is wrong. Correct sentence is
(3a + 4b) (a – b) = 3a² + ab – 4b²

(xiv) (x + 4) (x + 2) = x² + 8
(x + 4) (x + 2) = x² + 8
⇒ x² + 6x + 8 = x² + 8
⇒ 6x ≠ 0
Here ’6x’ term is missing in R.H.S.
∴ The given sentence is wrong. Correct sentence is
(x + 4)(x + 2) = x² + 6x + 8.

(xv) (x – 4) (x – 2) = x²– 8
(x – 4) (x – 2) = x² – 8
⇒ x² – 6x + 8 ≠ x² – 8
∴ The given sentence is wrong. Correct sentence is
(x – 4) (x – 2) = x² – 6x + 8

(xvi) 5x³ ÷ 5 x³ = 0
5x³ ÷ 5 x³ = 0
⇒ x^3-3 = 0
⇒ x⁰ = 0
∴ 1 ≠ 0 (∵ but x° = 1)
∴ The given sentence is wrong. Correct sentence is 5x³ ÷ 5x³ = 1.
In the denominator the term T is missing. .•. The given sentence is wrong. Correct sentence is

(xvii) 2x³ + 1 ÷ 2x³ = 1
2x³ + 1 ÷ 2x³ = 1
⇒ \(\frac{2 x^{3}+1}{2 x^{3}}\) = 1
In the denominator the term T is missing.
∴ The given sentence is wrong. Correct sentence is
2x³ + 1 ÷ 2x³ = 1 + \(\frac{1}{2 \mathrm{x}^{3}}\)

(xviii) 3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
3x + 2 ÷ 3x = \(\frac{2}{3 x}\)
⇒ \(\frac{3 x+2}{3 x}=\frac{2}{3 x}\)
⇒ 1 + \(\frac{2}{3 x}=\frac{2}{3 x}\) ⇒ 1 ≠ 0
∴ The given sentence is wrong. Correct sentence is 3x + 2 ÷ 3x = 1 + \(\frac{2}{3 x}\)

(xix) 3x + 5 ÷ 3 = 5
⇒ \(\frac{3 x+5}{3}\) = 5
⇒ \(\frac{3 x}{3}+\frac{5}{3}\) = 5 ⇒ x + \(\frac{5}{3}\) ≠ 5
∴ It is a wrong sentence.
Correct sentence is 3x + 5 ÷ 3 = x + \(\frac{5}{3}\)

(xx) \(\frac{4 x+3}{3}\) = x + 1
\(\frac{4 x+3}{3}\) = x + 1
⇒ \(\frac{4 \mathrm{x}}{3}+\frac{3}{3}\) = x + 1
⇒ \(\frac{4 \mathrm{x}}{3}\) + 1 ≠ x + 1
∴ It is a wrong sentence.
Correct sentence is \(\frac{4 x+3}{3}=\frac{4 x}{3}+1\)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.3

Question 1.
Carry out the following divisions
(i) 48a³ by 6a
(ii) 14x³ by 42x³
(iii) 72a³b⁴c⁵ by 8ab²c³
(iv) 11xy²z³ by 55xyz
(v) -54l⁴m³n² by 9l²m²n²
Solution:
(i) 48a³ by 6a
48a³ ÷ 6a
= \(\frac{6 \times 8 \times a \times a^{2}}{6 \times a}\)
= 8a²

(ii) 14x³ by 42x³
= 14x³ ÷ 42x³

(iii) 72a³b⁴c⁵ by 8ab²c³

(iv) 11xy²z³ by 55xyz
11xy²z³ ÷ 55xyz

(v) -54l⁴m³n² by 9l²m²n²
-54l⁴m³n² ÷ 9l²m²n²

= -6l²m

Question 2.
Divide the given polynomial by the given monomial
(i) (3x² – 2x) ÷ x
(ii) (5a³b – 7ab³) ÷ ab
(iii) (25x⁵ – 15x⁴) ÷ 5x³
(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²
(v) 15 (a³b²c² – a²b³c² + a²b²c³ ) ÷ 3abc
(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)
(vii) (\(\frac{2}{3}\) a² b² c²+ \(\frac{4}{3}\) a b² c³) ÷ \(\frac{1}{2}\)abc
Solution:
(i) (3x² – 2x) ÷ x

(ii) (5a³b – 7ab³) ÷ ab

(iii) (25x⁵ – 15x⁴) ÷ 5x³

= 5x² – 3x (or) x(5x – 3)

(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²

= 2l² – 3l² + 4l = l(2l² – 3l + 4)

(v) 15 (a³ b² c² – a² b³ c² + a² b² c³ ) ÷ 3abc

= 5[a x abc – b x abc + c x abc ]
= 5abc [a – b + c]

(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)

= -[p² – 3pq – 2q²]
= 2² + 3pq – p²

(vii) (\(\frac{2}{3}\) a²b²c²+ \(\frac{4}{3}\) ab²c³) ÷ \(\frac{1}{2}\)abc

Question 3.
Workout the following divisions:
(i) (49x -63) ÷ 7
(ii) 12x (8x – 20,) ÷ 4(2x – 5)
(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)
(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)
(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)
(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)
Solution:
(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)

4 ( x² + 7x + 10)
= 4 ( x² + 5x + 2x + 10)
= 4 [x( x + 5) +2(x + 5)]
= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4.
Factorize the expressions and divide them as directed:
(i) (x² + 7x + 12) ÷ (x + 3)
(ii) (x² – 8x + 12) ÷ (x – 6)
(iii) (p² + 5p + 4,) (p + l)
(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
(v) 151m (2p² – 2q²) ÷ 3l(p + q)
(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
Solution:
(i) (x² + 7x + 12) ÷ (x + 3)
(x² + 7x + 12) ÷ (x + 3)
x² + 7x + 12 = x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)

(ii) (x² – 8x + 12) ÷ (x – 6)
(x² – 8x + 12) ÷ (x – 6)
x² – 8x + 12 = x² – 6x – 2x + 12
= x(x – 6) – 2(x – 6)
= (x – 6) (x – 2)
∴ (x² – 8x + 12) 4 (x – 6)
= \(\frac{(x-6)(x-2)}{(x-6)}\) = x – 2

(iii) (p² + 5p + 4,) (p + 1)
p² + 5p + 4 = p² + p + 4p + 4
= p(p + 1) + 4(p + 1)
= (p + 1) (p + 4)
(p² + 5p + 4) ÷ (p + 1)
= \(\frac{(p+1)(p+4)}{(p+1)}\) = p + 4

(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
15ab (a² – 7a + 10) ÷ 3b (a – 2)
15ab (a² – 7a + 10) = 15ab (a² – 5a – 2a + 10)
= 15ab [(a² – 2a) – (5a -10)]
= 15ab [a(a – 2) – 5(a – 2)]
= 15ab(a – 2)(a – 5)
∴ 15ab (a² – 7a + 10) ÷ 3b (a – 2)

(v) 151m (2p² – 2q²) ÷ 3l(p + q)
15lm (2p² – 2q²) ÷ 3l (p + q)
15lm (2p² – 2q²) = 15lm x 2(p² – q²)
= 30lm (p + q) (p – q)
∴ 15lm(2p² – 2q²) ÷ 3l(p + q)

(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
26z³(32z² – 18) ÷ 13z² (4z – 3)
26z³(32z² – 18) = 26z³ (2 x 16z² – 2 x 9)
= 26z³ x 2 [16z³ – 9]
= 52z³ [(4z)³ – (3)³]
= 52z³ (4z + 3) (4z – 3)
∴ 26z³ (32z² – 18) ÷ 13z² (4z – 3)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.2

Question 1.
If 345 A 7 is divisible by 3,supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
∴ 345A7 ⇒ 3 + 4 + 5 + A + 7 = 19 + A
19 + A = 3 x 7
⇒ A = 21 – 19 = 2 ⇒ A = 24 – 19 = 5

A + 19 = 3 x 8
⇒ A = 24 – 19 = 5

A + 19 = 3 x 9
⇒ A = 27 – 19 = 8

∴ A = {2,5,8}

Question 2.
If 2791 A,is divisible by 9, supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
∴ 2791A = 2 + 7 + 9 + 1 + A = 9 x 3
⇒ 19 + A = 9 x 3 = 27
⇒ A = 27 – 19 = 8
∴ A = 8

Question 3.
Write some numbers which are divisible by 2,3,5,9 and 10 also.
Solution:
90, 180, 270. are divisible by 2, 3, 5, 9 and 10.
[∵ The L.C.M. of 2, 3, 5, 9, 10 is 90]

Question 4.
2A8 is a number divisible by 2, what might be the value of A’?
Solution:
If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.
∴ 2A8 is divisible by 2 for any value of A.
∴ A = (0, 1, 2 ………………….9)

Question 5.
50B is a number divisible by 5, what might be the value of B?
Solution:
Given number is 50B.
The units digit of a number ¡s either ‘0’ or 5, then it is divisible by 5.
∴ 500 → \(\frac { 0 }{ 5 }\) (R = 0)
505 → \(\frac { 5 }{ 5 }\) (R = 0)
∴ B = {0, 5}

Question 6.
2P is a number which is divisible by 2 and 3, what is the value of P
Solution:
The given number is 2P.
If 2P is divisible by 2, 3 then 2P should be a multiple of 6. [ ∵ L.C.M. of 2, 3 is 6]
∴ 2P = 24, 30 ………….
24 → 2 + 4 → \(\frac { 6 }{ 3 }\) (R = 0)
∴ P = 4

Question 7.
54Z leaves remainder 2 when divided by 5 , and leaves remainder 1 when divided by 3, what is the value of Z’?
Solution:
If 54Z is divisible by 3 then the sum of the digits of the number is divisible by 3.
According to problem 54Z is divisible by 3 and leaves remainder 1’.
∴ 5 + 4 + Z = (3 x 4) + 1
= 9 + Z = 13
∴ Z = 4(or)
9 + Z = (3 x 5) + 1
9 + Z = 16
Z = 7
If 54Z is divisible by 5 then Z should be equal to either ‘0’ or ‘5’.
∴ 54(0 + 2) = 542 (Z = 2)
54(0 + 7) = 547 (Z = 7)
∴ From the above two cases
Z = 7
∵ 547 → \(\frac{7}{5}\)(R = 2)

Question 8.
27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?
Solution:
27Q is divided by 5 gives the remainder 3
Le.,27Q = 27 (0 + 3) = 273(Z = 3)(T)
= 27 (0 + 8) = 278 (Z = 8)
27Q is divided by 2 gives the remainder 1.
i.e., 27Q = 27(0 + 1) = 271 (Z = 1)
27Q = 27 (0 + 3) = 273 (Z = 3) (T)
∴ From above situations Z = 3
∴ 27Q = 273→ 2 + 7 + 3 → \(\frac{12}{3}\)(R = 0)
∴ 273 is divisible by 3 and gives the remainder 0’.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.1

Question 1.
Find the common factors of the given terms in each.

(i) 8x, 24
(ii) 3a, 2lab
(iii) 7xy, 35x²y³
(iv) 4m², 6m², 8m³
(v) 15p, 20qr, 25rp
(vi) 4x², 6xy, 8y²x
(vii) 12 x²y, 18xy²
Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x²y³
7xy = 7 × x × y
35x²y³ = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x²y³
= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m², 6m², 8m³
4m² = 2 × 2 × m × m
6m² = 2 × 3 × m × m
8m³ = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m² , 6m² , 8m³
= 2, m, m², 2m, 2m².

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x², 6xy, 8y²x
4x² = 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y²x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x², 6xy, 8xy² = 2, x, 2x.

vii) 12x²y, 18xy²
12x²2y = 2 × 2 × 3 × x × x × y
18xy² = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x²y, 18xy²
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2.
Factorise the following expressions
(i) 5x² – 25xy
(ii) 9a² – 6ax
(iii) 7p² + 49pq
(iv) 36 a²b – 60 a²bc
(v) 3a²bc + 6ab²c + 9abc²
(vi) 4p² + 5pq – 6pq²
(vii) ut + at²
Solution:
(i) 5x² – 25xy
= 5 x × x × – 5 × 5 × x × y
= 5 × x [x – 5 × y]
= 5x [x – 5y]

ii) 9a² – 6ax
= 3 × 3 × a × a – 2 × 3 × a × x
= 3a [3a – 2x]

iii) 7p² + 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a²b – 60a²bc
= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 – 5c]
= 12a²b [3 – 5c]

v) 3a²bc + 6ab²c + 9abc²
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p² + 5pq – 6pq²
= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q
= p [4p + 5q – 6q²]

vii) ut + at²
= u × t + a × t × t = t [u + at]

Question 3.
Factorise the following:
(i) 3ax – 6xy + 8by – 4bx
(ii) x³ + 2x² + 5x + 10
(iii) m² – mn + 4m – 4n
(iv) a³ – a²b² – ab + b³
(v) p²q – pr² – pq + r²
Solution:
i) 3ax – 6xy + 8by – 4ab
= (3ax – 6xy) – (4ab – 8by)
= (3 × a × x – 2 × 3 × x × y)
– (4 ×a × b – 4 × 2 × b × y)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y)(3x – 4b)

ii) x³ + 2x² + 5x + 10
= (x³ + 2x²) + (5x +10)
= (x² × x + 2 × x²) + (5 × x + 5 × 2)
= x²(x + 2) + 5(x + 2)
= (x + 2) (x² + 5)

iii) m² – mn + 4m – 4n
= (m² – mn) + (4m – 4n)
= (m × m – m × n) + (4 × m – 4 × n)
= m(m – n) + 4(m – n)
= (m – n) (m + 4)

iv) a³ – a²b² – ab + b³
= (a³ – a²b²) – (ab – b³)
= (a² × a – a² × b²) – (a × b – b × b²)
= a²(a – b²) – b(a – b²)
= (a – b²) (a² – b)

v) p²1 – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= p(pq – r²) – (pq – r²) × 1
= (p – 1) (pq – r²)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.1

Question 1.
Using divisibility rules, fmd which of the following numbers are divisible by 2,5,10 ( say
yes or no ) in the given table. What do you observe?

Solution:

Question 2.
Using divisibility tests, determine which of following numbers are divisible by 2
(a) 2144 (b) 1258 (c) 4336 (d) 633 (e) 1352
Solution:
If a number is divisible by 2 then the units digit of the number be 0, 2, 4, 6, 8.
∴ a) 2144, b) 1258, c) 4336 e) 1352 are divisible by ‘2’.

Question 3.
Using divisibility tests, determine which of the following numbers are divisible by 5
(a) 438750 (b) 179015 (c) 125 (d) 639210 (e) 17852
Solution:
If a number is divisible by 5 its units digit be either ‘0’ or 5.
∴ a) 438750, b) 179015, c) 125 d) 639210 are divisible by 5.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 10:
(a) 54450 (b) 10800 (c) 7138965 (d) 7016930 (e) 10101010
Solution:
If a number is divisible by 10 then its units digit must be 0’.
a) 54450, b) 10800, d) 7016930, e) 1010100 are divisible by 10.

Question 5.
Write the number of factors of the following’?
(a) 18 (b) 24 (e) 45 (d) 90 (e) 105
Solution:

Number	Factors                              ,	No.of factors
a) 18	1,2,3,6,9,18	6
b) 24	1, 2, 3, 4, 6, 8, 12, 24	8
c) 45	1,3,5,9,15,45	6
d) 90	1, 2, 3, 5, 6, 9,10, 15, 18, 30, 45, 90	12
e) 105	1,3, 5, 7,15,21,35,105	8

Question 6.
Write any 5 numbers which are divisible by 2,5 and 10.
Solution:
10, 20, 30, 40. are divisible by 2, 5 and 10
[∵ The L.C.M. of 2, 5, 10 is 10]

Question 7.
A number 34A is exactly divisible by 2 and leaves a remainder 1, when divided by 5, find A.
Solution:
If 34A Is divisible by 2 then the remainder should be equal to 0.
∴ A should be equal to 0, 2, 4, 6, 8.
∴ 340, 342, 344, 346, 348 are divisible by 2 and gives the remainder ‘0’.
Among these 346 is divisible by 5 and gives the remainder 1.
∴ 346 → \(\frac{6}{5}\) (R = 1)
∴ The value of A = 6.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volumes Exercise 14.2

Question 1.
Find the volume of the cuboid whose dimensions are given below.

Solution:

Question 2.
Find the capacity of the tanks with the following internal dimensions. Express the capacity in cubic meters and litres for each tank.

Solution:

Question 3.
What will happen to the volume of a cube if the length of its edge is reduced to half? Is the volume get reduced? If yes, how much?
Solution:
Volume of a cube of side (s) is V₁ = a³
If the length of the side is reduced by half then
s = \(\frac{\mathrm{a}}{2}\)
∴ Volume of cube (V₂ ) = s³

∴ V₂ = \(\frac { 1 }{ 8 }\) × V₁
∴ V₁ = 8V₂

Question 4.
Find the volume of each of the cube whose sides are.
(i) 6.4 cm
(ii) 1.3 m
(iii) 1.6 m.
Solution:
Volume of a cube(V) = a³

i) a = 6.4 cm
ii) a = 1.3 m
iii) a = 1.6 m

V = (6.4)³
Volume of a cube (V) = a³
= 6.4 × 6.4 × 6.4
= 262.144 cm³

V = (1.3)³
= 1.3 × 1.3 × 1.3
= 2.197 m³

V = (1.6)³
= 1.6 × 1.6 × 1.6
= 4.096 m³

Question 5.
How many bricks will be required to build a wall of 8 m long, 6m height and 22.5 cm thici if each brick measures 25 cm by 11.25 cm by 6 cm?
Solution:
The volume of a wall of measures
8 m × 22.5 cm × 6 m
(V₁) = l₁b₁h₁
= 8 m × 22.5 cm × 6 m
= 800 cm × 22.5 cm × 600 cm
The volume of a brick each measures
25 cm × 11.25 cm × 6 cm
(V₂) = l₂b₂h₂
= 25 × 11.25 × 6 cm³
∴ The no.of bricks will be required

= 32 × 2 × 100 = 6400

Question 6.
A cuboid is 25 cm long, 15 cm broad, and 8 cm high . How much of its volume will differ from that of a cube with the edge of 16 cm’?
Solution:
Volume of a cuboid (V₁) of measures
= 25 cm, b = 15 cm, h = 8 cm.
V₁ = 25 × 15 × 8 = 3000 cm³
Volume of a cube of measure side (s) = 16 cm is
V₂ = (s)³ = (16)³ = 16 × 16 × 16
= 4096 cm³
The difference between their volumes
= V₂ – V₁
= 4096 – 3000
= 1096 cm³

Question 7.
A closed box is made up of wood which is 1cm thick .The outer dimensions of the box is 5 cm × 4 cm × 7 cm. Find the volume of the wood used.
Solution:
The volume of a box formed with outer measures 5 cm × 4 cm × 7 cm
V₁ = l × b × h
= 5 × 4 × 7
∴ V₁ = 140 cm³
Inner measures
= l – 2w, b – 2w, h – 2w
= (5 – 2 × 1), (4 – 2 × 1), (7 – 2 × 1)
= (5 – 2), (4 – 2), (7 – 2)
= 3 cm, 2 cm, 5 cm
∴ Volume of a box formed with inner measures
V₂ = (l – 2w)(b – 2w)(h – 2w)
= 3 × 2 × 5 = 30 cm³
∴ The volume of wood used = V₁ – V₂
= 140 – 30 = 110 cm³

Question 8.
How many cubes of edge 4cm, each can be cut out from cuboid whose length, breadth and height are 20 cm, 18 cm and 16 cm respectively
Solution:
The volume of a cuboid formed with the measures 20 cm × 18 cm × 16 cm
(V₁) = l₁b₁h₁ = 20 × 18 × 16
Volume of a cube (V₂) = s³
s = 4 cm (given)
∴ V₂ = (s)³ = (4)³= 4 × 4 × 4 cm³
∴ No.of cubes are required
= \(\frac{V_{1}}{V_{2}}=\frac{20 \times 18 \times 16}{4 \times 4 \times 4}\)
= 90

Question 9.
How many cuboids of size 4 cm × 3 cm × 2 cm can be made from a cuboid of size 12 cm x 9cm x 6cm?
Solution:
Volume of a cuboid of measures 12 cm × 9 cm × 6 cm
V₁ = l × b × h = 12 × 9 × 6
Volume of the smaller cuboid of measures 4 cm × 3 cm × 2 cm
V₂ = l₂b₂h₂ = 4 × 3 × 2
∴ No.of cuboids are made
= \(\frac{V_{1}}{V_{i}}=\frac{12 \times 9 \times 6}{4 \times 3 \times 2}\) = 27

Question 10.
A vessel in the shape of a cuboid is 30 cm long and 25 cm wide. What should be its height to hold 4.5 litres of water ?
Solution:
Length of a cuboidal vessel (l) = 30 cm
breadth (b) = 25 cm
height (h) = ?
The volume of water m a cuboidal vessel = 4.5 Lts.
= 4.5 × 1000 cm³
= 4500 cm³
∴ l × b ×h = 4500
⇒ 30 × 25 × h = 4500
⇒ h = \(\frac{4500}{30 \times 25}\)
∴ h = 6 cm
∴ Height of the vessel (h) = 6 cm

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D Ex 13.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D Exercise 13.2

Question 1.
Count the number of faces , vertices , and edges of given polyhedra and verify Euler’s formula.

Solution:

Question 2.
Is a square prism and cube are same? explain.
Solution:
All cubes are square prisms, but converse is not true. (i.e.,) All square prisms are either cubes or, not.

Question 3.
Can a polyhedra have 3 triangular faces only? explain.
Solution:
Any polyhedra can’t have 3 triangular faces because the triangular pyramids are formed starts with 4 faces. So it does not exist.

Question 4.
Can a polyhedra have 4 triangular faces only? explain.
Solution:
Yes, a triangular pyramid have 4 triangular faces.

Question 5.
Complete the table by using Euler’s formula.

F	8	5	?
V	6	?	12
E	?	9	30

Solution:

F	8	5	20
V	6	6	12
E	12	9	30

i) E = V + F- 2 = 8 + 6- 2 = 12
ii) V = E + 2- F = 9 + 2- 5 = 6
iii) F = E + 2- V = 30 + 2-12 = 20

Question 6.
Can a polyhedra have 10 faces ,20 edges and 15 vertices?
Solution:
No. of faces = 10
No. of edges = 20
No. of vertices = 15
According to Euler’s formula E = V + F – 2
⇒ 20 = 15 + 10 – 2
20 = 25 – 2
20 = 23 (False)
∴ A polyhedra doesn’t exist with 10 faces, 20 edges, 15 vertices.

Question 7.
Complete the following table

Solution:

Question 8.
Name the 3-D objects or shapes that can be formed from the following nets.

(i) Hexagonal pyramid
(ii) Cuboid
(iii) Pentagonal pyramid
(iv) Cylinder
(v) Cube
(vi) Hexagonal pyramid
(vii) Trapezoid

Question 9.
Draw the following diagram on the check ruled book and fmd out which of the following diagrams makes cube?
(i)

Solution:
The diagrams which makes cubes are a, b, c, e.

(ii) Answer the following questions.
(a) Name the polyhedron which has four vertices, four faces’?
(b) Name the solid object which has no vertex?
(c) Name the polyhedron which has 12 edges’?
(d) Name the solid object which has one surface’?
(e) How a cube is different from cuboid?
(f) Which two shapes have same number of edges, vertices and faces?
(g) Name the polyhedron which has 5 vertices and 5 faces’?
Solution:
(a) Tetrahedron
(b) Sphere
(c) Cube/Cuboid
(d) Sphere
(e) Cube is a regular polyhedron where cuboid is not.
(f) Cube, Cuboid
(g) Square pyramid

(iii) Write the names of the objects given below

Solution:
(a) Octagonal prism
(b) Hexagonal prism
(c) Triangular prism
(d) Pentagonal prism