Srinivas

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 12 Factorisation Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.2

Question 1.
Factorise the following expression
i) a² + 10a +25
ii) l² – 16l + 64
iii) 36x² + 96xy + 64y²
iv) 25x² + 9y² – 30xy
v) 25m²– 40mn + 1 6n²
vi) 81x² – 198 xy + 12ly²
vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
viii) l⁴ + 4l²m² + 4m⁴
Solution:
i) a² + 10a +25
= (a)² + 2 × a × 5 + (5)²
It is in the form of a² + 2ab + b²
a² + 2ab + b²= (a + b)²
∴ a² + 10a + 25 = (a + 5)² = (a + 5) (a + 5)

ii) l² – 16l + 64
l² – 16l + 64
= (l)² – 2 × l × 8 + (8)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ l² – 16l + 64 = (l – 8)² = (l – 8) (l – 8)

iii) 36x² + 96xy + 64y²
36x² + 96xy + 64y²
= (6x)² + 2 × 6x × 8y + (8y)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a + b)²
∴ 36x² + 96xy + 64y²
= (6x + 8y)² = (6x + 8y) (6x + 8y)

iv) 25x² + 9y² – 30xy
25x² + 9y² – 30xy
= (5x)² + (3y)² – 2 × 5x × 3y
It is in the form of a² + b² – 2ab
a² + b² – 2ab = (a – b)²
∴ 25x² + 9y² – 30xy
= (5x – 3y)² = (5x – 3y) (5x – 3y)

v) 25m²– 40mn + 1 6n²
25m² – 40mn + 16n²
= (5m)² – 2 × 5m × 4n + (4n)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 25m² – 40mn + 16n²
= (5m – 4n)²
= (5m – 4n) (5m – 4n)

vi) 81x² – 198 xy + 12ly²
81x² – 198xy + 121y²
= (9x)² – 2 × 9x × 11y + (11y)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 81x² – 198xy + 121y²
= (9x – 11y)² – (9x – 11y) (9x – 11y)

vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
= (x + y)² – 4xy
= x² + y² + 2xy – 4xy
= x² + y² – 2xy = (x – y)² = (x – y)(x – y)

viii) l⁴ + 4l²m² + 4m⁴
l⁴ + 4l²m² + 4m⁴
= (l²)² + 2 × l² × 2m² + (2m²)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a – b)²
∴ l⁴ + 4l²m² + 4m⁴
= (l² + 2m²)² = (l² + 2m²) (l² + 2m²)

Question 2.
Factorise the following
i) x² – 36
ii) 49x² – 25y²
iii) m² – 121
iv) 81 – 64x²
v) x²y² – 64
vi) 6x² – 54
vii) x² – 81
viii) 2x -32 x⁵
ix) 81x⁴ – 121x²
x) (p² – 2pq + q²)-r²
xi) (x+y)² – (x-y)²
Solution:
i) x² – 36
x² – 36
⇒ (x)² – (6)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
∴ x² – 36 = (x + 6) (x – 6)

ii) 49x² – 25y²
= (7x)² – (5y)²
= (7x + 5y) (7x – 5y)

iii) m² – 121
m² -121
= (m)² – (11)²
= (m + 11) (m – 11)

iv) 81 – 64x²
81 – 64x²
= (9)² – (8x)²
= (9 + 8x) (9 – 8x)

v) x²y² – 64
= (xy)² – (8)²
= (xy + 8)(xy – 8)

vi) 6x² – 54
6x² – 54
= 6x² – 6 x 9 ‘
= 6(x² – 9)
= 6[(x)² – (3)²]
= 6(x + 3) (x – 3)

vii) x² – 81
x² – 81
= x² – 9²
= (x + 9 )(x – 9)

viii) 2x – 32 x⁵
2x – 32 x⁵
= 2x – 2x x 16x⁴
= 2 x (1 – 16x⁴)
= 2x [1²) – (4x²)²]
= 2x (1 + 4x²) (1 – 4x²)
= 2x (1 + 4x²) [(1⁵ – (2x)²]
= 2x (1 + 4x²) (1 + 2x) (1 – 2x)

ix) 81x⁴ – 121x²
81x⁴ – 121x²
– x² (81² – 121)
= x²[(9x)² – (11)²]
= x² (9x + 11) (9x -11)

x) (p² – 2pq + q²)-r²
(p² – 2pq + q²) – r²
= (p – q)² – (r)² [∵ p² – 2pq + q² = (p – q)²]
= (p – q + r) (p – q – r)

xi) (x + y)² – (x – y)²
(x + y)² – (x – y)²
It is in the form of a² – b²
a = x + y, b = x- y
∴ a² – b² =(a + b)(a-b)
= (x + y + x – y) [x + y- (x – y)]
= 2x [x + y-x + y]
= 2x x 2y = 4xy

Question 3.
Factorise the expressions
(i) lx² + mx
(ii) 7y² + 35Z²
(iii) 3x⁴ + 6x³y + 9x²Z
(iv) x² – ax – bx + ab
(v) 3ax – 6ay – 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab – b² + 12ac – 2bc
(viii) p²q – pr² – pq + r²
(ix) x (y + z) -5 (y + z)

(i) lx² + mx
lx² + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y² + 35z²
7y²+ 35z²
= 7 × y² + 7 × 5 × z²
= 7(y² + 5z²)

(iii) 3x⁴ + 6x³y + 9x²Z
3x⁴ + 6x³y + 9x²Z
= 3 × x² × x² + 3 × 2 × x × x² × y + 3 × 3 × x² × z
= 3x² (x² + 2xy + 3z)

(iv) x² – ax – bx + ab
x² – ax – bx + ab
= (x² – ax) – (bx – ab)
= x(x – a) – b(x – a)
= (x – a) (x – b)

(v) 3ax – 6ay – 8by + 4bx
3ax – 6ay – 8by + 4bx
= (3ax – 6ay) + (4bx – 8by)
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

(vi) mn + m + n + 1
mn + m + n + 1
= (mn + m) + (n + 1)
= m (n + 1) + (n + 1)
= (n + 1) (m + 1)

(vii) 6ab – b² + 12ac – 2bc
6ab – b² + 12ac – 2bc
= (6ab – b²) + (12ac – 2bc)
= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)
= b [6a – b] + 2c [6a – b]
= (6a – b) (b + 2c)

(viii) p²q – pr² – pq + r²
p²q – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= P(pq – r²) – (pq – r²) × 1
= (pq – r²)(p – 1)

(ix) x (y + z) -5 (y + z)
= x(y + z) – 5(y + z)
= (y + z) (x – 5)

Question 4.
Factorise the following
(i) x⁴ – y⁴
(ii) a⁴ – (b + c)⁴
(iii) l² – (m – n)²
(iv) 49x² – \(\frac{16}{25}\)
(v) x⁴ – 2x²y² + y⁴
(vi) 4 (a + b)² – 9 (a – b)²
Solution:
= (x²)² – (y²)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
x⁴ – y⁴ = (x² + y²)(x² – y²)
= (x² + y²)(x + y)(x – y)

(ii) a⁴ – (b + c)⁴
a⁴ – (b + c)⁴
= (a²)² – [(b + c)²]²
= [a² + (b + c)²] [a² – (b + c)²] ,
= [a² + (b + c)²] (a + b + c) [a – (b + c)]
= [a² + (b + c)²] (a + b + c) (a – b – c)

(iii) l² – (m – n)²
l² – (m – n)²
= (l)² – (m – n)²
= [l + m – n] [l – (m – n)]
= [l + m -n] [l – m + n]

(iv) 49x² – \(\frac{16}{25}\)
= (7x)² – (\(\frac{4}{5}\))²
= (7x+ (\(\frac{4}{5}\)) (7x – (\(\frac{4}{5}\))

(v) x⁴ – 2x² y² + y⁴
= (x² )² – 2x² y² + (y² )²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ x⁴ – 2x² y² + y⁴ = (x² – y² )²
= [(x)² – (y)² ]²
= [(x + y) (x – y)]²
= (x + y)² (x – y)²
[∵ (ab)^m = a ^m . bⁿ ]

(vi) 4 (a + b)² – 9 (a – b)²
4 (a + b)² – 9 (a – b)²
= [2(a + b)]² – [3(a – b)]²
= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]
= (2a + 2b + 3a – 3b) (2a + 2b – 3a + 3b)
= (5a – b) (5b – a)

Question 5.
Factorise the following expressions
(i) a²+ 10a + 24
(ii) x² +9x + 18
(iii) p² – 10q + 21
(iv) x² – 4x – 32
Solution:
(i) a²+ 10a + 24
a² + 10a + 24 .
= a² + 6a + 4a + 24
= a x a + 6a + 4a + 6 × 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a² + 10a + 24

∴ a² + 10a + 24 = (a + 6) (a + 4)

(ii) x² + 9x + 18
x² + 9x + 18
= (x + 3) (x + 6)

∴ x² + 9x + 18 = (x + 3) (x + 6)

(iii) p² – 10q + 21
p² – 10p + 21
= (P – 7) (p – 3)

∴ p² – 10p + 21 = (p – 7)(p – 3)

(iv) x² – 4x – 32
x² – 4x – 32
= (x – 8) (x + 4)

∴ x² – 4x – 32 = (x – 8) (x + 4)

Question 6.
The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Solution:
Perimeter of a triangle
= AB + BC + CA = 48
⇒ c + a + b = 48

The solutions of Harmeet, Rosy are wrong.

∴ Srikar had done it correctly.
⇒ 21 + a + b = 48
⇒ a + b = 48 – 21 = 27
∴ The lengths of a, b should be 10, 17
∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]
∴ 10 + 17 > 2
27 > 21 (T).
∴ The length of the shortest side is 10 cm.

Question 7.
Find the values of ‘m’ for which x² + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.
Solution:
Given equation is x² + 3xy + x + my – m ……….(1)
Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).
Then (x + 3y + a) (x + 0y + b)
= x² + 0xy + bx + 3xy + 0y² + 3by + ax + 0y + ab
= x² + bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x² + 3xy + x + my – m
= x² + (a + b)x + 3xy + 3by + ab
Equating the like terms on both sides,
ab = – m ………….. (3)
(a + b)x = x ⇒ a + b = 1 ……………. (4)
3by = my ⇒ 3b = m ⇒ b = \(\frac{\mathrm{m}}{3}\)
Substitute ‘b’ value in equation (4),
a = \(1-\frac{m}{3}=\frac{3-m}{3}\)
ab = -m
[ ∵ from (3)]
put a & b value then ,
\(\left(\frac{3-m}{3}\right)\left(\frac{m}{3}\right)\) = -m
\(\frac{3 \mathrm{~m}-\mathrm{m}^{2}}{9}\)= -m
⇒ 3m – m² = – 9m
⇒ m² – 12m = 0
⇒ m(m – 12) = 0
⇒ m = 0 (or) m = 12
lf m = 12

∴ b = \(\frac{12}{3}\) = 4&a = \(\frac{3-\mathrm{m}}{3}=\frac{3-12}{3}\)
= \(\frac{-9}{3}\) = -3
∴ Linear factors are (x + 3y – 3), (x + 4) If m = 0
b = \(\frac{0}{3}\) = 0 & a = \(\frac{3-0}{3}=\frac{3}{3}\) = 1
∴ Linear factors are (x + 3y + 1), x.

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 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a – b)² ≡ a² – 2ab+ b² geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)² = a² – 2ab + b² [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)² = 4 sq. units
[∵ (3 – 1 )² = 2² = 4]

ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS = a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a² – b² geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a² – b² = 9 – 4= 5sq . units
[ ∵ 3² – 2² = 9 – 4 = 5]

ii)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a² – b² = 3 sq. units
[∵ (2² – 1²) = 4 – 1 = 3]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.
Select a suitable identity and find the following products
(i) (3k + 4l)(3k + 4l)
(ii) (ax² + by²)(ax² + by²)
(iii) (7d – 9e)(7d – 9e)
(iv) (m² – n²)(m² + n²)
(v) (3t + 9s) (3t – 9s)
(vi) (kl – mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b – a)(2b +c)
Solution:
(3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².
=(3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²
= 3k × 3k + 24kl + 4l × 4l
= 9k² + 24kl + 16l²

ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².
= (ax²)² + 2 × ax² × by² + (by²)² [ ∵ (a + b)² = a² + 2ab + b²]
= ax² × ax² + 2abx²y² + by² × by²
= a²x⁴ + 2ab x²y² + b²y⁴

iii) (7d – 9e) (7d – 9e)
= (7d – 9e)² is in the form of (a – b)².
= (7d)² – 2 × 7d × 9e + (9e)² [ ∵ (a – b)² = a² – 2ab + b²]
= 7d × 7d – 126de + 9e × 9e
= 49d² – 126de + 81e²

iv) (m² – n²) (m² + n²) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a² – b²
∴ (m² + n²) (m² – n²) = (m²)² – (n²)² = m⁴ – n⁴

v) (3t + 9s) (3t – 9s) = (3t)² – (9s)² [ ∵ (a + b) (a – b) = a² – b² ]
= 3t × 3t – 9s × 9s
= 9t² – 81s²

vi) (kl – mn) (kl + mn) = (kl)² – (mn)² [ ∵(a + b) (a – b) = a² – b² ]
= kl × kl – mn × mn
= k²l² – m²n²

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a²x² + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6
= 36x² + 6x × 11 + 30
= 36x² + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a²x² + ax(c – b) – cb
(2b – a) (2b + c) = (2)²(b)² + 2b (c – a) – ca
= 4b² + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 304²
(ii) 509²
(iii) 992²
(iv) 799²
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 304² = (300 + 4)² is in the form of (a + b)².
∵ (a+b)² = a² + 2ab + b²
a = 300, b = 4
(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 509² = (500 + 9)²
a  = 500, b = 9
= (500)² + 2 × 500 × 9 + (9)²
[ ∵ (a + b)² = a² + 2ab + b²]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 992² = (1000 – 8)²
a = 1000, b = 8
= (1000)² – 2 × 1000 × 8 + (8)² [∵ (a-b)² = a² – 2ab + b²]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 799² = (800 – 1)²
a = 800, b = 1
= (800)² – 2 × 800 × 1 + (1)²
= 800 × 800 – 1600 + 1
= 6,40,000 – 1600 + 1
= 6,40,001 – 1600
= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
(a + b) (a – b) = a² – b²
∴ (300 + 4) (300 – 4) = (300)² – (4)²
= 300 × 300 – 4 × 4
= 90,000 – 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)
= (80)² – (3)² [ ∵ (a + b) (a – b) = a² – b²]
= 80 × 80 – 3 × 3
= 6400 – 9
= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)² + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)
= (205)² – (1)² [∵ (a + b)(a-b) = a² – b²]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m² – n² and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a² + 8a – 27a – 36
= 6a² – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x² – xy – 4xy + 2y²
= 2x² – 5xy + 2y²

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k²l – l²k + klm – l²m

iv) m² – n² and m + n
(m² – n²) (m + n) = m²(m + n) – n²(m + n)
= m³ + m²n – n²m – n³

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab² – a²b)
(iv) (p³ + q³)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x² – 5xy + 3x²y + 2xy – 5y² + 3xy²
= 2x² – 5y² – 3xy + 3x²y + 3xy²

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k²l² + k²lm – lm²n + kl²m – klm²

iii) (a – 2b + 3c) (ab² – a²b) = a(ab² – a²b) – 2b(ab² – a²b) + 3c(ab²– a²b)
= a²b² – a³b – 2ab³ + 2a²b² + 3ab²c – 3a²bc
= 3a²b² – a³b – 2ab³ + 3ab²c – 3a²bc

iv) (p³ + q³) (p – 5q + 6r) = p³(p – 5q + 6r) + q³(p – 5q + 6r)
= p⁴ – 5p³q + 6p³r + pq³ – 5q⁴ + 6rq³
= p⁴ – 5q⁴ – 5p³q + 6p³r + pq³ + 6rq³

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m² – mn + n²)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y² – 3xy – 9xy + 9x² + x² – 3xy + xy – 3y²
= 10x² – 14xy

ii) (m + n) (m²– mn + n²)
= m(m² – mn + n²) + n(m² – mn + n²)
= m³ – m²n + n²m + nm² – mn² + n³
= m³ + n³

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a² – 2ab + 5ac – ab + 2b² – 5bc – 2a² + 2ab + 2ac – 3ac + 3bc + 3c² + 12ac – 18a² – 30ab + 2bc – 3ab – 5b²
= – 19a² – 3b² – 34ab + 16ac + 3c²

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p²q² – pq²r + p²qr + pq²r – q²r² + pqr² – p²r – pqr + pr² – p²q – pq² + pqr
= p²q² – q²r² + p²qr + pqr² – p²r + pr² – p²q – pq ²

Question 4.
If a, b, care positive real numbers such that \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) ,find the value of \(\frac{(a+b)(b+c)(c+a)}{a b c}\)
Solution:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) = k then
\(\frac{a+b-c}{c}\) = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.2

Question 1.
Complete the table:

Solution:

Question 2.
Simplify: 4y(3y + 4)
Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y² + 6y

Question 3.
Simplify x(2x² – 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0
Solution:
x(2x² – 7x + 3)
= x × 2x² – x × 7x + x × 3
= 2x³ – 7x² + 3x
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
=9 – 3 – 3 + 1 = 4
∴(a – b)² = 4sq.units
[∵ (3 – 1)² = 2² = 4]

(ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS
= a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
= 5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 4.
Add the product: a(a – b), b(b – c), c(c – a)
Solution:
a(a – b) + b(b – c) + c(c – a)
=a × a – a × b + b × b – b × c + c × c – c × a
=a² – ab + b² – bc + c² – ca
=a² + b² + c² – ab – bc – ca

Question 5.
Add the product: x(x + y – r), y(x – y+r), z(x – y – z)
Solution:
x(x + y – r) +y(x – y + r) + z(x – y – z)
= x² + xy – xr + xy – y² + yr + zx – yz – z²
= x² – y² – z² + 2xy – xr + yr + zx – yz

Question 6.
Subtract the product of 2x(5x – y) from product of 3x(x+2y)
Solution:
3x(x + 2y) – 2x(5x – y)
=(3x × x + 3x × 2y)-(2x × 5x – 2x × y)
= 3x² + 6xy – (10x² – 2xy)
= 3x² + 6xy- 10x² + 2xy
= 8xy – 7x²

Question 7.
Subtract 3k(5k – l + 3rn) from 6k(2k + 3l – 2rn)
Solution:
6k(2k + 3l – 2m) – 3k(5k – l + 3m)
= 12k²+ 18kl – 12km – 15k² + 3kl – 9km
= -3k² + 21kl – 21km

Question 8.
Simplify: a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
Solution:
a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
= a³ – a²b + a²c + ab² + b³ – b²c – ac² + bc² + c³
= a³ + b³ + c³ – a²b + a²c + ab² – b²c – ac² – bc²

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.1

Question 1.
Find the product of the following pairs:
(i) 6, 7k
(ii) – 31, – 2m
(iii) -5t² – 3t²
(iv) 6n, 3m
(v) – 5p², – 2p
Solution:
The product of 6, 7k = 6 × 7k = 42k
ii) The product of – 3l, – 2m = (- 3l) × (- 2m) = 6/m
iii) The product of – 5t², – 3t² = (- 5t²) × (- 3t²) = 15t⁴
iv) The product of 6n, 3m = 6n × 3m = 18mn
v) The product of – 5p², – 2p = (- 5p²) × (- 2p) = 10p³

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials
(i) xy, x²y , xy, x
(ii) a, b, ab, a³ b, ab³
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) – 3a, 4ab, – 6c, d
Solution:
i) The product of xy, x²y, xy, x = xy × x²y × xy × x
= x⁵ × y³= x⁵y³

ii) The product of a, b, ab, a³b, ab³ = a × b × ab × a³b × ab³
= a⁶ × b⁶ = a⁶ b⁶

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k³ × l³ × m³ =k³l³m³

iv) The product of pq, pqr, r = pq × pqr × r
= p² × q² × r² – p²q²r²

v) The product of – 3a, 4ab, – 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a² × b × c × d
= 72a²bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=
Solution:
ABC = xy × yz × zx = x²y²z²

Question 6.
If P = 4x², T = 5x and R = 5y, then \(\frac{\mathrm{PTR}}{100}\) =
Solution:
\(\frac{P^{\prime} \Gamma R}{100}=\frac{4 x^{2} \times 5 x \times 5 y}{100}=\frac{100 x^{3} y}{100}\) = x³ y

Question 7.
Write some monomials of your own and find their products.
Solution:
The product of,some monomials is given below :
i) abc × a²bc = a³b²c²
ii) xy × x²z × yz² = x³y²z³
iii) p × q × r = p³q³r³

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.4

Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ \(\frac{1}{\text { No. of days }}\)

⇒ Compound ratio of 8:12 and 20: 15
= \(\frac{8}{12}=\frac{20}{15}\) = \(\frac{8}{9}\) …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ \(\frac{480}{x}=\frac{8}{9}\)
⇒ x = \(\frac{480 \times 9}{8}\) = ₹540
∴ The cost of required rice is ₹ 540

Method – II :
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = No. of men
D₁ = No .of days
W₁ = Cost of rice
∴ M₁ = 8
D₁ = 20
W₁ = ₹ 480
M₂ = 12
D₂ = 15
W₂ = ? (x)

⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-

Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
∴ M₁ = 10
D₁ = 5
W₁ = 75
M₂ = 15
D₂ = ?
W₂ = 45

∴ x = 2
∴ No. of days are required = 2

Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M₁D₁H₁ = M₂D₂H₂
∴ M₁ = 24
D₁ = 15 days
H₁ = 8 hrs
M₂ = 20
D₂ = ?
H₂ = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9

∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]

Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = 175
D₁ = 36
W₁ = 3150
M₂ = ?
D₂ = 24
W₂ = 3900

∴ No. of workers are required = 325

Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M₁D₁H₁ = M₂D₂H₂
M₁ = 14
D₁ = 12 days
H₁ = 6
M₂ = 4
D₂ = ?
H₂ = 7
⇒ 14 × 12 × 6 = 4 × x × 7

⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x₁y₁ = x₂ y₂
⇒ 8 × 5 = 10 × x
⇒ x = \(\frac{8 \times 5}{10}\) = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x₁y₁ = x₂ y₂ (Let y₂ = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = \(\frac{500 \times 70}{700}\) = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x₁y₁ = x₂ y₂ let y₂ = x (say)
= 36 × 12 = 9 × x
x = \(\frac{36 \times 12}{9}\) = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , Let y₂ = x (say)
⇒ \(\frac{28}{2}\) = \(\frac{56}{x}\)
⇒ x = \(\frac{56}{14}\)
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x₁y₁ = x₂ y_{2 ,} Let x₂ = x (say)
⇒ 16 × 10 = x × 8
⇒ x = \(\frac{16 \times 10}{8}\)= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1\(\frac { 1 }{ 2 }\) hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 5 × 1\(\frac { 1 }{ 2 }\) = x x 1\(\frac { 1 }{ 2 }\)
⇒ 5 × \(\frac { 3 }{ 2 }\) = x x \(\frac { 1 }{ 2 }\)
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 15 × 48 = x × 30
⇒ x = \(\frac{15 \times 48}{30}\) = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 45 × 8 = x × 6
⇒ \(\frac{45 \times 8}{6}\)
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ \(\frac{\mathrm{x}}{\mathrm{y}}\)

Let x₁ = 100x, x₂ = 112x
(∵ It increases 12%)
y₁ = 100y, y₂ = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = \(\frac{(x+1)(x+1)}{(x+2)}\)
∴ k = \(\frac{(x+1)^{2}}{(x+2)}\)

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.
What do you observe? Write your observations in your note books

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1