NCTB Class 7 Math Chapter Five Exercise 5.4 Solution

NCTB Class 7 Math Chapter Five Exercise 5.4 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.4“Algebraic Formulaic and Applications” Exercise 5.4 Solution.

Board

NCTB
Class

7

Subject

Math
Chapter

Five

Chapter Name

“Algebraic Formulaic and Applications”
Exercise

5.4 Solution

Exercise:- 5.4

(1) Solution:- (a-5)2 = a2– 2xax5= (5)2

(b) = a2-10a+25

(2) Solution:- (x +y)2+ 2(x +y) (x-y) + (x-y)2

= (x +y + x-y)2

= (2x)2

= 4x2

(3) Solution:- We know, 4ab = (a +b)2 – (a-b)2

=> 4ab = 42-22

= 16-4

= 12

=> a b = 12/4 = 3 (a)

(4) Solution:- If a quantity is divisible without remainder another quantity is called – (c) Multiple

(5) Solution:- Least common multiple of a, a2, a (a +b) is a2 (a +b) (d) Answer.

(6) Solution:- H.C.F of 2a and 3b is 1 (a)

(7) Solution:- If a, b are real numbers – (d) (i), (ii), (iii) are correct.

(8) Solution:- (d) x y (x +y) (x-y)

(9) Solution:- H.C.F if two algebraic expression, (d) x (x-y)

(10) Solution:- The L.C.M is — (d) x y (x +y) (x+2y)

(11) Solution:- L.C.M of 9x2– 25y2  and 15ax-25xy is (d) 5a (9x2-25y2)

(12) Solution:- H.C.F of x3 y5 and a2-b2 is (d) 1.

(13) Solution:- x-1/x = 0; (d) (i), (ii), (iii) Correct.

(14) Solution:- a+1/a= 4 ; a2-4a+1= (d) 0.

(15) Solution:- (a+5)2 = a2+2.a.5+25

= a2+10a+25

(16) Solution:- We know, 4ab= (a +b)2 – (a-b)2

=> 4ab = (8)2 – (4)2

= 64-16

= 48

=> a b= 48/4

= 12

(17) Solution:- 1st, 3a2 b2 c= 32 x a2 x a x a xb2xb2xc2

2nd, 6ab2 c2= 2×32x a2xb2xb2xc2xc

Therefore, H.C.F= 3a2 b2c

(18)Solution:- 1st, 5ab2 x2= 52xa2xb2xbxXxX

2nd, 10a2 by2 = 2×52xa2xaxaxb2xyxy

H.C.F = 5ab

(19) Solution:- 1st, 3a2 x2 = 32xa2x ax X x

2nd, 6axy2 = 2×32x a2x X x y2 x y

3rd, 9 ay2= 3 x32xa2 xy2 x y

H.C.F = 3a

(20) Solution:- 1st, 16a3 x4 y = 2x2x2x2xa3xX4xy

2nd, 40a2 y3 x= 22x 2x 2x5xa2xy3x X

3rd, 28a x3= 22x2 x7 xa2x X3

H.C.F= 2ax

(21) Solution:- 1st, a2+ab = a (a +b)

2nd, a2-b2= (a +b) (a-b)

Therefore, a2- b2= (a +b) (a-b)

Therefore, H.C.F = (a +b)

(22) Solution:- 1st, x3y – xy3

= x y (x2-y)

= x y (x +y) (x-y)

2nd, (x-y)2 = (x-y) (x-y)

Therefore, H.C.F = (x-y)

(23) Solution:- 1st, x2 +7x+12

= x2+4x+3x+12

= x (x+4) +3(x+4)

= (x+4) (x+3)

2nd, x2+9x+20

= x2 + 5x+ 4x+20

= x(x+5) +4 (x+5)

= (x+5) (x+4)

(24) Solution:-  1st, a3-ab2= a (a2-b2)

= a (a +b) (a-b)

2nd, a4+2a3b + a2 b2

= a2 (a2+2ab+b2)

= a2 (a +b)2

= a. a (a +b) (a +b)

Therefore, H.C.F = a (a +b)

(25) Solution:- 1st, a2-16= (a)2 – (4)2

= (a+4) (a-4)

2nd, 3a +12 = 3 (a+4)

3rd, a2+ 5a +4

= a2+ 4a + a+ 4

= a (a +4) +1(a +4)

= (a +4) (a +1)

(27) Solution:- 1st, 6a3 b2 c

= 2x 3x a3x b2xc

2nd, 9a4 bd2= 3 x3 xa4x b x d2

Therefore, L.C.M = 2x 3x 3xa4xb2xc xd2

= 18a4 b2 cd2

(28) Solution:- 1st, 5x2 y2= 5x x2xy2

2nd, 10xz3= 2x 5x X xz3

3rd, 15y3 z4= 3x 5 x y3x z4

Therefore, L.C.M= 2x 3 x5x X2x y3xz4

= 30x2 y3 z4

(29) Solution:- 1st, 2p2 xy2= 2x p2x X x y2

2nd, 3p q2 = 3x p x q2

3rd, 6pqx2 = 2 x 3 x p x q x X2

Therefore, L.C.M= 6p2 q2 x2 y2

(30) Solution:- 1st, (b2-c2)= (b +c) (b-c)

2nd, (b +c)2 = (b +c)

Therefore, L.C.M= (b +c)2 (b-c)

(31) Solution:- 1st, x2=2x

= x( x+2)

2nd, x2+3x+2

= x(x+2) +1(x+2)

= (x+2) (x-1)

Therefore, L.C.M= x (x+1) (x+2)

(32) Solution:- 1st, 9x2-25y2

= (3x)2 – (5y)2

= (3x +5y) (3x -5y)

2nd, 15ax-25ay = 5a (3x-5y)

(33) Solution:- 1st, x2-3x-10

= x2-5x=2x-10

= (x-5) (x+2)

2nd, x2-10x+25

= x2– 5x-5x+25

= (x-5) -5 (x-5)

= (x-5) (x-5)

= (x-5)2

Therefore, L.C.M= (x-5)2 (x+2)

(34) Solution:- 1st, a2-7a+12

= a2-3a-4a+12

= a (a-3) -4 (a-3)

= (a-3) (a-4)

2nd, a2+a-20

= a2+5a-4a-20

= a(a+5) -4(a+5)

= (a-4) (a+5)

3rd, a2+ 2a-15

= a2+5a-3a-15

= (a-5) -3(a+5)

= (a-3) (a+5)

Therefore, L.C.M, (a-3) (a-4) (a+5)

(35) Solution:-  1st, x2-8x+15

= x2-5x-3x+15

= x (x-5) -3(x-5)

= (x-5) (x-3)

2nd, x2-25= (x+5) (x-5)

3rd, x2+2x-25

= x2-3x+5x-15

= x (x-3) +5 (x-3)

= (x-3) (x+5)

Therefore, L.C.M = (x-3) (x2-25)

(36) Solution:- 1st, (x+5)

2nd, x2+5x

= x (x+5)

3rd, x2+7x+10

= (x2+5x+2x+10)

= (x+5) (x+2)

Therefore, L.C.M= x (x+5) (x+2)

(37) (a) Solution:- (a +b)

= 2x-3+2x+5

= (4x+2)

(b) Solution:- a2= (2x-3)2

= (2x)2– 2. 2x.3+ (3)2

= 4x2-12x +9

(c) Solution:- a b

= (2x-3) (2x=50

= 4x2+10x-6x-15

= 4x2+4x-15

If, x= 2;

4x (2)2+ 4×2-15

= 16+8-15

= 24-15

= 9

(38) (a) Solution:- x2+3x-10

= x2+5x-2x-10

= x(x+5)-2(x+5)

= (x+5) (x-2)

(b) Solution:- 1st, x2-625

= (x)2– (25)2

= (x+25) (x-25)

= (x+25){(x)2 – (5)2}

= (x+25) (x+5) (x-5)

2nd, x2+3x-10

= (x=5) (x-2)

L.C.M = (x+5) (x-2) (x-5) (x+25)

(c) Solution:- We get,

From 1st expression, (x+5) (x-2)

2nd, expression (x2+25) (x+5) (x-5)

L.C.M= (x+5) (x-2) (x-5) (x2 +25)

(39) Solution:- (a) (3x-2y+z)2

= {(3x-2y) +z)2

= (3x-2y)2+2(3x-2y) z+ z2

= (3x)2 – 2.3x.2y+ (2y)2+ 6xz- 4yz +z2

= 9x2+ 4y2+z2– 12xy +6x2– 4yz

(b) Solution:- 1st, x2-3x-10

= (x+2)(x-3)

2rd, X3+6x2+8x

= x(x+6x+8)

= x(x2+4x+2x+8)

= x{x (x+4) +2 (x+4)}

= x(x+4) (x+2)

Therefore, H.C.F= (x+2)

(c) Solution:- 3rd, expression, x4-5x3-14x2

= x2(x2-5x-14)

= x2(x2-7x+2x-14)

= x2{x(x-7) +2(x-7)}

= x2 (x-7) (x+2)

Updated: December 17, 2020 — 3:11 pm

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