Prasanna

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

by

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

Other Exercises

Factorize each of the following expressions :
Question 1.
qr-pr + qs – ps
Solution:
qr- pr + qs-ps
Arranging in suitable groups = r(q-p) +s (q-p)    {(q – p) is common}
= (q-p) (r + s)

Question 2.
p2q -pr2-pq + r2
Solution:
p2q -pr2-pq + r2
= p2q -pq-pr2 + r2 (Arranging in group)
= pq(p- 1)-r2(p-1) {(p – 1) is common}
= (p – 1) (pq – r2)

Question 3.
1 + x + xy + x2y
Solution:
1 + x + xy + x 2y
= 1 (1 + x) +xy (1 +x)
= (1 + x) (1 + xy) {(1 + x) is common}

Question 4.
ax + ay – bx – by
Solution:
ax + ay – bx – by
= a (x + y) – b (x + y)   {(x + y) is coinmon}
= (x+y) (a- b)

Question 5.
xa2 + xb2 -ya2 – yb2
Solution:
xa2 + xb2 – ya2 – yb2
= x (a2 + b2) -y (a2 + b2)   {(a2 + b2) is common}
= {a2 + b2) (x -y)

Question 6.
x2 + xy + xz + yz
Solution:
x2 + xy + xz + yz
= x (x + y) + z(x + y) {(x + y) is common}
= (x + y) (x + z)

Question 7.
2ax + bx + 2ay + by
Solution:
2ax + bx + 2ay + by
= x {2a + b) + y (2a + b)      {(2a + b) is common}
= (2a + b) (x + y)

Question 8.
ab- by- ay +y2
Solution:
ab – by – ay + y2
= b(a-y)-y(a-y)    {(a -y) is common}
= (a-y) (b – y)

Question 9.
axy + bcxy -az- bcz
Solution:
axy + bcxy – az – bcz
= xy (a + bc) – z (a + bc)       {(a + bc) is common}
= (a + bc) (xy – z)

Question 10.
lm2 – mn2 – lm + n2
Solution:
lm2 – mn2 – lm + n2
= m (lm – n2)- 1 (lm – n2)  {(lm – n2) is common}
= (lm – n2) (m – 1)

Question 11.
x3 – y2 + x – x2y2
Solution:
x3 -y2 + x – x2y2
⇒ x3 + x – x2y2 – y2
= x(x2+ 1)-y2(x2+ 1)        {(x2 + 1) is common}
= (x2 + 1) (x -y2)

Question 12.
6xy + 6 – 9y – 4x
Solution:
6xy + 6 – 9y – 4x
= 6 xy – 4x – 9y + 6
= 2x (3y – 2) – 3 (3y – 2)    {(3y – 2) is common}
= (3y-2) (2x – 3)

Question 13.
x2 – 2ax – 2ab + bx
Solution:
x2 – 2ax – 2ab + bx
⇒ x2 – 2ax + bx – 2ab
= x (x – 2a) + b (x – 2a)   {(x – 2a) is common}
= (x – 2a) (x + b)

Question 14.
x3 – 2x2y + 3xy2 – 6y3
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)     {(x – 2y) is common}
= (x – 2y) (x2 + 3y2)

Question 15.
abx2 + (ay – b) x-y
Solution:
abx2 + (ay – b) x-y
= abx2 + ayx – bx -y 
= ax (bx + y) – 1 (bx + y)               {(bx +y) is common}
= (bx + y) (ax – 1)

Question 16.
(ax + by)2 + (bx – ay)2
Solution:
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + a2y2 + by2
= x2 (a2 + b2) + y2 (a2 + b2)         {(a2 + b2) is common}
= (a2 + b2) (x2 + y2)

Question 17.
16 (a – b)3 -24 (a- b)2
Solution:
16 (a – b)3 -24 (a- b)2
HCF of 16, 24 = 8
and HCF of (a – b)3, (a – b)2 = (a – b)2
∴16 (a – b)3 – 24 (a – b)2
= 8 (a-b)2 {2 (a-b)- 3}
{8 (a – b)2 is common}
= 8 (a – b)2 (2a – 2b – 3)

Question 18.
ab (x2 + 1) + x (a2 + b2)
Solution:
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + b2x + a2x + ab
= bx (ax + b) + a (ax + b)  {(ax + b) is common}
= (ax + b) (bx + a)

Question 19.
a2x2 + (ax2 + 1) x + a
Solution:
a2x2 + (ax2 + 1) x + a
= a2x2 + ax3 + x + a
= ax3 + a2x2 + x + a
= ax2 (x + a) + 1 (x + a) {(x + a) is common}
= (x + a) (ax2 + 1)

Question 20.
a(a- 2b -c) + 2bc
Solution:
a(a- 2b -c) + 2bc
= a2– 2ab -ac +2bc
= a (a – 2b) – c (a – 2b) {(a – 2b) is common}
= (a – 2b) (a – c)

Question 21.
a (a + b – c)- bc
Solution:
a (a + b – c) – bc
= a2 + ab – ac – bc
= a (a + b) – c (a + b)   {(a + b) is common}
= (a + b) (a – c)

Question 22.
x2 – 11xy – x +11y
Solution:
x2 – 11xy-x + 11y
= x2 -x – 11 xy + 11 y
= x (x – 1) – 11y (x – 1)   {(x – 1) is common}
= (x- 1) (x- 11y)

Question 23.
ab – a – b + 1
Solution:
ab – a-b + 1
= a (b – 1) – 1 (b – 1)    {(b – 1) is common}
= (b – 1) (a – 1)

Question 24.
x2 + y – xy – x
Solution:
x2 + y – xy – x
= x2 – x- xy + y
= x (x – 1) – y (x – 1)   {(x – 1) is common}
= (x- 1) (x-y)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B

by

RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B.

Other Exercises

Question 1.
Solution:
(i) 4, 6, 8, 12
If it is in proportion, then
If ad = bc if 4 x 12 = 6 x 8
If 48 = 48
Which is true
∴ 4, 6, 8, 12 are in proportion
(ii) 7, 42, 13, 78
7 : 42 :: 13 : 78
If ad = bc if 7 x 78 = 42 x 13
If 546 = 546
Which is true
∴ 7, 42, 13, 78 are in proportion
(iii) 33, 121, 9, 96 or 33 : 121 :: 9 : 96
are in proportion
If ad = bc
If 33 x 96 = 121 x 9
If 3168 = 1089
Which is not true
∴ 33, 121,9, 96 are not in proportion
(iv) 22, 33, 42, 63 or 22 : 33 :: 42 : 63
are in proportion
If ad = bc
If 22 x 63 = 33 x 42
If 1386 = 1386
Which is true
∴ 22, 33, 42, 63 are in proportion
(v) 32, 48, 70, 210 or 32 : 48 :: 70 : 210
are in proportion
If ad = bc
If 32 x 210 = 48 x 70
If 6720 = 3360
Which is not true
∴ 32, 48, 70, 210 are not in proportion
(vi) 150, 200, 250, 300 or
150 : 200 :: 250 : 300 are in proportion
If ad = bc if 150 x 300 = 200 x 250
If 45000 = 50000
Which is not true
∴ 150, 200,250, 300 are not in proportion

Question 2.
Solution:
(i) We have 60 : 105 :: 84 : 147
Product of means = 105 x 84 = 8820
Product of extremes = 60 x 147 = 8820
∴ Product of means = Product of extremes
Hence 60 : 105 :: 84 : 147 is verified.
(ii) We have 91 : 104 :: 119 : 136
Product of means = 104 x 119 = 12376
Product of extremes = 91 x 136 = 12376
Product of means = Product of extremes
Hence 91 : 104 :: 119 : 136 is verified.
(iii) We have 108 : 72 :: 129 : 86
Product of means = 72 x 129 = 9288
Product of extremes = 108 x 86 = 9288
Product of means = Product of extremes
Hence 108 : 72 :: 129 : 86 is verified.
(iv) We have 39 : 65 :: 141 : 235
Product of means = 65 x 141 = 9165
Product of extremes = 39 x 235 = 9165
∴ Product of means = Product of extremes
Hence 39 : 65 :: 141 : 235 is verified.

Question 3.
Solution:
(i) We have 55 : 11 :: x : 6
Product of means = 11 × x = 11x
Product of extremes = 55 x 6 = 330
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q3.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q3.2

Question 4.
Solution:
(i) We have, 51 : 68 = \(\\ \frac { 51 }{ 68 } \)
= \(\\ \frac { 3 }{ 4 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q4.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q4.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q4.3

Question 5.
Solution:
(i) 25 cm : 1 m and Rs. 40 : Rs. 160
= \(\\ \frac { 25cm }{ 1000cm } \) = \(\\ \frac { 1 }{ 4 } \),
\(\\ \frac { Rs.40 }{ Rs.160 } \) = \(\\ \frac { 1 }{ 4 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q5.1
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q5.2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q5.3

Question 6.
Solution:
Let the third term be x.
Then 51 : 68 :: x : 108
Now, product of means = x × 68
And, product of extremes = 51 × 108
x × 68 = 51 × 108
=> x = \(\\ \frac { 51\times 108 }{ 68 } \)
= 3 × 27 = 81
x = 81
Hence the third term of the given proportion is 81

Question 7.
Solution:
1st term =12, third term = 8 and fourth term = 14
Let 2nd term = x, then
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q7.1

Question 8.
Solution:
(i) The given numbers 48, 60, 75 are in
continued proportion if 48 : 60 :: 60 : 75.
Now, product of means = 60 x 60 = 3600
And, product of extremes = 48 x 75 = 3600
∴ Product of means = Product of extremes
So, 48 : 60 :: 60 : 75
Hence, the numbers 48, 60, 75 are in continued proportion.
(ii) The given numbers 36, 90, 225 are in
continued proportion of 36 : 90 :: 90 : 225
Now, product of means = 90 x 90 = 8100
And, product of extremes = 36 x 225 = 8100
∴ Product of means = Product of extremes
So, 36 : 90 :: 90 : 225
Hence, the numbers 36, 90, 225 are in continued proportion.
(iii)The given numbers 16, 84, 441 are in
continued proportion if 16 : 84 :: 84 : 441.
Now, product of means = 84 x 84 = 7056
And, product of extremes = 16 x 441 = 7056
Product of means = Product of extremes.
So, 16 : 845 :: 84 : 441
Hence 16, 84, 441 are in continued proportion.
(iv) The given numbers 27, 36, 48 are
in continued proportion if 27 : 36 :: 36 : 48
Now, product of means = 36 x 36 = 1296
And, product of extremes = 27 x 48 = 1296
∴ Product of means = Product of extremes.
So, 27 : 36 :: 36 : 48
Hence, the numbers 27, 36, 48 are in continued proportional.

Question 9.
Solution:
It is given that 9, x, x, 49 are in proportion, that is, 9 : x :: x : 49
∴ Product of means = Product of extremes
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q9.1

Question 10.
Solution:
Let the height of the pole be x metres.
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q10.1

Question 11.
Solution:
5 : 3 :: x : 6
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B Q11.1

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

by

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2

Other Exercises

Factorize the following :

Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3)        (HCF of 3, 9 = 3)

Question 2.
5x – 15x2
Solution:
5x- 15x2 = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x2 = x}

Question 3.
20a12b2 – 15a8b4
Solution:
20a12b2 – 15a8b4
{HCF of 20, 15 = 5, a12, a8 = a8, b2, b4 = b2}
= 5a8 b2(4a4 – 3b2)

Question 4.
72xy – 96x7y6
Solution:
72xy – 96x7y6
HCF of 72, 96 = 24 of x6x7 = x6, y7,y6 = y6
∴ 72x7y6 – 96x7y6 = 24x6y6 (3y – 4x)

Question 5.
20X3 – 40x2 + 80x
Solution:
20x3 – 40x2 + 80x
HCF of 20, 40,80 = 20
HCF of x3, x2, x = x
∴ 20x3 – 40x2 + 80x = 20x (x2 – 2x + 4)

Question 6.
2x3y2 – 4x2y3 + 8xy4
Solution:
2x3y2 – 4x2y3 + 8xy4
HCF of 2, 4, 8 = 2
HCF of x3, x2, x = 1
and HCF of y2, y3, y4 = y2
∴ 2x3y2 – 4x2y3 + 8xy4
= 2xy2 (x2 – 2xy + 4y2)

Question 7.
10m3n2 + 15m4n – 20m2n3
Solution:
10m3n2 + 15m4n – 20m2n3
HCF of 10, 15, 20 = 5
HCF of m3, m4, m2 = m2
HCF of n2, n, n3 = n
10m3n2 + 15m4n – 20m2n3
5m2n(2mn + 3m2– 4n2)

Question 8.
2a4b4 – 3a3b5 + 4a2b5
Solution:
2a4b4 – 3a3b5 + 4a2b5
HCF of 2, 3, 4= 1
HCF of a4, a3, a2 = a2
HCF of b4, b5 b5 = b4
∴ 2a4b4 – 3a3b5 + 4a2b5 = a2b4
(2a2 – 3ab + 4b)

Question 9.
28a2 + 14a2b2 – 21a4
Solution:
28a2 + 14a2b2 – 21a4
HCF of 28, 14,21 =7
HCF of a2, a2, a4 = a2
HCF of 1, b2, 1 = 1
∴ 28a2 + 14a2b2-21a4 = 7a2
(4 + 2b2 – 3a2)

Question 10.
a4b – 3a2b2 – 6ab3
Solution:
a4b – 3a2b2 – 6ab3
HCF of 1,3,6 = 1
HCF of a4, a2, a = a
HCF of b, b2, b3 = b
∴ a4b – 3a2b2 – 6ab3 = ab (a3 – 3ab – 6b2)

Question 11.
2l2mn – 3lm2n + 4lmn2
Solution:
2l2mn – 3lm2n + 4lmn2
HCF 2, 3,4 = 1,
HCF of l2,l,l = l
HCF of m, m2, m = m
HCF of n, n, n2 = n
∴ 2l2 mn – 3lm2n + 4lmn2
= lmn (21 -3m + 4n)

Question 12.
x4y2 – x2y4 – x4y4
Solution:
x4y2 – x2y4 – x4y4
HCF of x4, x2, x4 = x2
HCF of y2, y4, y4 =y2
∴ x4y2 – x2y4 – x4y4 = x2y2 (x2 -y2 -x2y2)

Question 13.
9 x2y + 3 axy
Solution:
9 x2y + 3 axy
HCF of 9, 3 = 3
HCF of x2, x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x2y + 3axy = 3xy (3x + a)

Question 14.
16m – 4m2
Solution:
16m – 4m2
HCF of 16, 4 = 4
HCF of m, m2 = m
∴ 16m – 4m2 = 4m (4 – m)

Question 15.
-4a2 + 4ab – 4ca
Solution:
-4a2 + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a2, a, a = a
∴ -4a2 + 4ab – 4ca = -4a (a – b + c)

Question 16.
x2yz + xy2z + xyz2
Solution:
x2yz + xy2z + xyz2
HCF of x2, x, x = x
HCF of y,y2,y=y
HCF of z, z,z2 = z
∴ x2yz + xy2z + xyz2 = xyz (x + y + z)

Question 17.
ax2y + bxy2 + cxyz
Solution:
ax2y + bxy2 + cxyz
HCF of x2, x, x = x,
HCF of y,y2,y = y
ax2y + bxy2 + cxyz = xy (ax + by + cz)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

by

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1

Other Exercises

Find the greatest common factors (GCF / HCF) of the following polynomials : (1 – 14)

Question 1.
2x2 and 12x2
Solution:
2x2 and 12x2
HCF of 2 and 12 =2
HCF of x2,x2=x2
∴ HCF = 2x2

Question 2.
(6xy3 and 18x2y3
Solution:
6x3y and 18xy
HCF of 6, 18 = 6
HCF of x3 and x2 = x2
HCF of y and y3 -y
∴ HCF = 6x2y

Question 3.
7x, 21x2 and 14xy2
Solution:
7x, 21x2 and 14xy2
HCF of 7, 21 and 14 = 7
HCF of x, x2, x = x
∴ HCF = 7x

Question 4.
42x2yz and 63x3y2z3
Solution:
42x2yz and 63x3y2z3
HCF of 42 and 63 = 21
HCF of x2, x3 = x2
HCF of y,y2=y
HCF of z,z3 = z
∴ HCF = 21 x2yz

Question 5.
12ax2,6a2x3 and 2a3 x5
Solution:
12ax2, 6a2x3 and 2a3x5
HCF of 12, 6,2 = 2
HCF of a, a2, a3 = a
HCF of x2, x3, x5 = x2
∴ HCF = 2ax2

Question 6.
9x2, 15x2y3, 6xy2 and 21x2y2
Solution:
9x2, 15xV, 6xy2 and 21x2y2
HCF of 9, 15, 6,21 = 3
HCF of x2, x2, x, x2 = x
HCF of 1, y3, y2, y2 =2
∴ HCF = 3x

Question 7.
4a2b3 -12a3b, 18a4b3
Solution:
4a2b3, -12a3b, 18a4b3
HCF of 4, 12, 18 = 2
HCF of a2, a3, a4 = a2
HCF of b3,b, b3 = b
∴ HCF = 2a2b

Question 8.
6x2y2, 9xy3, 3x3y2
Solution:
6x2y2, 9xy3, 3x3y2
HCF of 6, 9, 3 = 3
HCF of x2, x, x3 = x
HCF of y2,y3,y2=y2
∴ HCF = 3xy2

Question 9.
a2b3, a3b2
Solution:
a2b3, a3b2
HCF of a2, a3 = a2
HCF of b3, b2 = b2
∴ HCF = a2b2

Question 10.
36a2b2c4, 54a5c2,90a4b2c2
Solution:
36a2b2c4, 54a5c2,90a4b2c2
HCF of 36, 54, 90 = 18
HCF of a2, a5, a4 = a2
HCF of b2, 1,b2= 1
HCF of c4,c2,c2 = c2
∴ HCF = 18a2 x 1 x c2 = 18a2c2

Question 11.
x3, – yx2
Solution:
x3, – yx2
HCF of x3, x2 = x2
HCF of 1, y= 1
∴ HCF = x2

Question 12.
15a3, -45a2, -150a
Solution:
15a3,-45a2,-150a
HCF of 15,45, 150 = 15
HCF of a3, a2, a = a
∴ HCF = 15a

Question 13.
2x3y2, 10x2y3, 14xy
Solution:
2x3y2, 10x2y3, 14xy
HCF of 2, 10, 14 = 2
HCF of x3, x2, x = x
HCF of y2,y3,y=y
∴ HCF = 2xy

Question 14.
14x3y5, 10x5y3, 2x2y2
Solution:
14x3y5, 10x5y3, 2x2y2
HCF of 14, 10, 2, = 2
HCF of x3, x5, x2 = x2
HCF of y5,y3,y2=y2
∴ HCF = 2xy

Find the greatest common factor of the terms in each of the following expressions:

Question 15.
5a4 + 10a3 – 15a2
Solution:
5a4 + 10a3– 15a2
HCF of 5, 10, 15 = 5
HCF of a4, a3, a2 = a2
∴ HCF = 5a2

Question 16.
2xyz + 3x2y + 4y2
Solution:
2xyz + 3x2y + 4y2
HCF of 2, 3,4 = 1
HCF of x, x2, 1 = 1
HCF of y,y,y2 =y
HCF of z, 1, 1 = 1
∴ HCF = y

Question 17.
3a2b2 + 4b2c2 + 12a2b2c2
Solution:
3a2b2 + 4b2c2 + 12a2b2c2
HCF of 3, 4, 12 = 1
HCF of a2, 1, a2 = 1
HCF of b2, b2, b2 = b2
HCF of 1, c2, c2 = 1
∴ HCF = b2

 

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C

by

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C.

Other Exercises

Question 1.
Solution:
Let the required number be x.
Then, x + 9 = 36
=> x = 36 – 9
(Transposing 9 to R.H.S.)
=> x = 27
The required number = 27.

Question 2.
Solution:
Let the required number be x. Then,
4x – 11 = 89
=> 4x = 89 + 11
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q2.1

Question 3.
Solution:
Let the required number be x. Then,
x x 5 = x + 80
=> 5x = x + 80
=> 5x – x = 80
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q3.1

Question 4.
Solution:
Let the three consecutive numbers be x,
x + 1 and x + 2. Then,
x + (x + 1) + (x + 2) = 114
=> x – x + 1 + x + 2 = 114
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q4.1

Question 5.
Solution:
Let the required number be x. Then
x x 17 + 4 = 225
=> 17x + 4 = 225
=> 17x = 225 – 4
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q5.1

Question 6.
Solution:
Let the required number be x. Then
3 x + 5 = 50
=> 3 x = 50 – 5
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q6.1

Question 7.
Solution:
Let the smaller number be x.
Then the other number = x + 18
According to question,
x + (x + 18) = 92
=> 2 x + 18 = 92
=> 2 x = 92 – 18
(Transposing 18 to R.H.S.)
=> 2 x = 74
=> x = 37
(Dividing both sides by 2)
One number = 37
Another number = 37 + 18
= 55.

Question 8.
Solution:
Let the smaller number be x.
Then, the other number = 3 x
According to question, x + 3 x = 124
=> 4 x = 124
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q8.1

Question 9.
Solution:
Let the smallest number be x.
Then, the other number = 5 x
According to question,
5 x – x = 132
=> 4 x = 132
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q9.1

Question 10.
Solution:
Let the two consecutive even number be x and x + 2.
Then x + x + 2 = 74
=> 2 x + 2 = 74
=> 2 x = 74 – 2
(Transposing 2 to RHS.)
=> 2x = 72
=> x = 36
(Dividing both sides by 2)
The required numbers are 36 and (36 + 2) i.e. 36 and 38.

Question 11.
Solution:
Let the three consecutive odd numbers be x, (x + 2) and (x + 4).
According to the question,
x + (x + 2) + (x + 4) = 21
3x + 6 = 21
=> 3 x = 21 – 6
(Transposing 6 to R.H.S.)
=> 3 x = 15
=> x = 5
(Dividing both sides by 3)
The required numbers are 5, 5 + 2 and 5 + 4 i.e. 5, 7 and 9.

Question 12.
Solution:
Let the present age of Ajay be x years. Then, the present age of Reena = (x + 6) years
According to the question,
x + (x + 6) = 28
2x + 6 = 28
=> 2x = 28 – 6
(Transposing 6 to R.H.S.)
=> 2 x = 22
=> x = 11
(Dividing both sides by 2)
Present age of Ajay = 11 years
and present age of Reena = 11 + 6
= 17 years.

Question 13.
Solution:
Let the present age of Vikas be x years.
Then, the present age of Deepak = 2x years
According to the question,
2x – x = 11
=> x = 11
Present age of Vikas = 11 years
and present age of Deepak = 2 x 11
= 22 years.

Question 14.
Solution:
Let the present age of Rekha be x years
Then, the present age of Mrs. Goel = (x + 27) years
Rekha’s age after 8 years = (x + 8) years
Mrs. Goel’s age after 8 years = (x + 27 + 8) years
= (x + 35) years
According to the question,
x + 35 = 2 (x + 8)
=> x + 35 = 2x + 16
=> x – 2x = 16 – 35
(Transposing 2x to L.H.S. and 35 to R.H.S.)
=> – x = – 19
=> x = 19
Present age of Rekha =19 years and present age of Mr. Goel = 19 + 27 = 46 years.

Question 15.
Solution:
Let the present age of the son be A years.
Then, the present age of the man
= 4 x years
Son’s age after 16 years = (x + 16) years Man’s age after 16 years
= (4 x + 16) years According to the question,
4x + 16 = 2 (x + 16)
=> 4x + 16 = 2x + 32
=> 4x – 2x = 32 – 16
(Transposing 2 x to L.H.S. and 16 to R.H.S.)
=> 2x = 16
=> x = 8 (Dividing both sides by 2)
Present age of the son = 8 years and present age of the man = 8 x 4 = 32 years.

Question 16.
Solution:
Let the present age of the son be x years.
Then, the present age of the man = 3x years
five years ago, the age of the son = (x – 5) years
five years ago, the age of the man = (3x – 5) years
According to the question, 3x – 5 = 4(x – 5)
=>3x – 5 = 4x – 20
=>3x – 4x = – 20 + 5
(Transposing 4 x to L.H.S. and – 5 to R.H.S.)
=> – x = – 15 => x = 15
Present age of the son = 15 years and present age of the man = 3 x 15 = 45 years.

Question 17.
Solution:
Let the present age of Fatima be A years. According to the question,
x + 16 = 3 x
=> x – 3x = – 16
(Transposing 3 x to L.H.S. and 16 to R.H.S.)
=> – 2 x = – 16
=> x = 8
(Dividing both sides by – 2)
Present age of Fatima = 8 years.

Question 18.
Solution:
Let the present age of Rahim be x years
Rahim’s age after 32 years = (x + 32) years
Rahim’s age 8 years ago = (x – 8) years
According to the question, x + 32 = 5 (x – 8)
=> x + 32 = 5 x – 40
=> x – 5x = – 40 – 32
(Transposing 5 x to L.H.S. and 32 to R.H.S.)
– 4 x = – 72
x= 18
(Dividing both sides by – 4)
Present age of Rahim =18 years

Question 19.
Solution:
Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins = 4x
Total value of 50 paisa coins = 50 x paisa
and total value of 25 paisa coins
= 25 x 4 = 100 x paisa
But total value of both the coins
= Rs. 30 (Given)
= 30 x 100 paisa
= 3000 paisa
According to the question,
50 x + 100 x = 3000
=> 150 x = 3000
\(\\ \frac { 150x }{ 150 } \) = \(\\ \frac { 3000 }{ 150 } \)
(Dividing both sides by 150)
x = 20
Number of 50 paisa coins = 20
and number of 25 paisa coins = 4 x 20
= 80

Question 20.
Solution:
Let the price of the pen be x rupees. According to the question,
5 x = 3 x + 17
5 x – 3 x = 17
(Transposing 3 x to L.H.S.)
=> 2 x = 17
=> x = \(\\ \frac { 17 }{ 2 } \)
(Dividing both sides by 2)
Price of the pen = \(\\ \frac { 17 }{ 2 } \) rupees
= Rs. 8.50.

Question 21.
Solution:
Let the number of girls in the school be x.
Then, the number of boys in the school = (x + 334)
According to the question, x + (x + 334) = 572
=> 2 x + 334 = 572
=> 2 x = 572 – 334
(Transposing 334 to L.H.S.)
2 x = 238
=> x = \(\\ \frac { 238 }{ 2 } \)
(Dividing both sides by 2) => x = 119
Number of girls in the school = 119.

Question 22.
Solution:
Let the breadth of the park be x metres.
Then, the length of die park=3x metres.
According to the question,
Perimeter of the park = 168 metres
=> 2 (x + 3 x) = 168
=> 2 x 4x = 168
=> 8 x = 168
=> x = 21
(Dividing both sides by 8)
Breadth of the park = 21 metres and length of the park = 3 x 21 = 63 metres.

Question 23.
Solution:
Let the breadth of the hall be x metres.
Then, the length of the hall = (x + 5) metres .
According to question,
Perimeter of the hall = 74 metres
=> 2 (x + x + 5) = 74
=> 2 (2 x + 5) = 74
=> 4 x + 10 = 74
=> 4 x = 74 – 10
(Transposing 10 to R.H.S.)
4x = 64
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 64 }{ 4 } \)
(Dividing both sides by 4)
=> x = 16
Breadth of the hall = 16 metres
and length of the hall = 16 + 5 = 21 metres.

Question 24.
Solution:
Since a wire of length 86 cm is bent to form die rectangle, so the perimeter of the rectangle = 86 cm.
Let the breadth of the rectangle = x cm
Then, die length of the rectangle = (x + 7) cm
2 (x + x + 7) = 86
=> 2 (2 x + 7) = 86 .
=> 4 x + 14 = 86
=> 4x = 86 – 14
(Transposing 14 to R.H.S.)
=> 4 x = 72
\(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 72 }{ 4 } \) = 18
(Dividing both sides by 4)
Breadth of the rectangle = 18 cm
Length of the rectangle = (18 + 7) = 25 cm.

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.