Rs Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 9 Linear Equation In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equation In One Variable are extremely popular among Class 6 students for Maths Linear Equation In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2020 2021 Book of Class 6 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggrawal 2020 2021 Solutions. All Rs Aggrawal 2020 2021 Solutions for class Class 6 Maths are prepared by experts and are 100% accurate.

#### Page No 139:

(i) Let the required number be x.
So, five times the number will be 5x.
∴ 5x = 40

(ii) Let the required number be x.
So, when it is increased by 8, we get x + 8.
∴ x + 8 = 15

(iii) Let the required number be x.
So, when 25 exceeds the number, we get 25 $-$ x.
∴ 25 $-$ x  = 7

(iv) Let the required number be x.
So, when the number exceeds 5, we get x $-$ 5.
∴ x $-$ 5  = 3

(v) Let the required number be x.
So, thrice the number will be 3x.
∴ 3x $-$ 5 = 16

(vi) Let the required number be x.
So, 12 subtracted from the number will be x $-$ 12.
∴ x $-$ 12 = 24

(vii) Let the required number be x.
So, twice the number will be 2x.
∴ 19 $-$ 2x = 11

(viii) Let the required number be x.
So, the number when divided by 8 will be $\frac{x}{8}$.
∴ $\frac{x}{8}$ = 7

(ix) Let the required number be x.
So, four times the number will be 4x.
∴ 4x $-$ 3 = 17

(x) Let the required number be x.
So, 6 times the number will be 6x.
∴ 6x = x + 5

#### Page No 140:

(i) 7 less than the number x equals 14.
(ii) Twice the number y equals 18.
(iii) 11 more than thrice the number x equals 17.
(iv) 3 less than twice the number x equals 13.
(v) 30 less than 12 times the number y equals 6.
(vi) When twice the number z is divided by 3, it equals 8.

(i)

(ii)

(iii)

(iv)

(v)

#### Page No 140:

(i) y + 9 = 13
We try several values of y until we get the  L.H.S. equal to the R.H.S.

 y L.H.S. R.H.S. Is LHS =RHS ? 1 1 + 9 = 10 13 No 2 2 + 9 = 11 13 No 3 3 + 9 = 12 13 No 4 4 + 9 = 13 13 Yes
∴ y = 4

(ii) x − 7= 10
We try several values of x until we get the  L.H.S. equal to the R.H.S.
 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 10 10 − 7 = 3 10 No 11 11 − 7 = 4 10 No 12 12 − 7 = 5 10 No 13 13 − 7 = 6 10 No 14 14 − 7 = 7 10 No 15 15 − 7 = 8 10 No 16 16 − 7 = 9 10 No 17 17 − 7 = 10 10 Yes

∴ x = 17

(iii) 4x = 28
We try several values of x until we get the  L.H.S. equal to the R.H.S.
 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 1 4 $×$ 1 = 4 28 No 2 4 $×$ 2 = 8 28 No 3 4 $×$ 3 = 12 28 No 4 4 $×$ 4 = 16 28 No 5 4 $×$ 5 = 20 28 No 6 4 $×$ 6 = 24 28 No 7 4 $×$ 7 = 28 28 Yes
∴ x = 7

(iv) 3y = 36
We try several values of x until we get the L.H.S. equal to the R.H.S.
 y L.H.S. R.H.S. Is L.H.S. = R.H.S.? 6 3 $×$ 6 = 18 36 No 7 3 $×$ 7 = 21 36 No 8 3 $×$ 8 = 24 36 No 9 3 $×$ 9 = 27 36 No 10 3 $×$ 10 = 30 36 No 11 3 $×$11 = 33 36 No 12 3 $×$ 12 = 36 36 Yes
∴ y = 12

(v) 11 + x = 19
We try several values of x until we get the L.H.S. equal to the R.H.S.

 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 1 11 + 1 = 12 19 No 2 11 + 2 = 13 19 No 3 11 + 3 = 14 19 No 4 11 + 4 = 15 19 No 5 11 + 5 = 16 19 No 6 11 + 6 = 17 19 No 7 11 + 7 = 18 19 No 8 11 + 8 = 19 19 Yes
∴ x = 8

(vi)
Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.

 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 3 $\frac{3}{3}=1$ 4 No 6 $\frac{6}{3}=2$ 4 No 9 $\frac{9}{3}=3$ 4 No 12 $\frac{12}{3}=4$ 4 Yes
∴ x = 12

(vii) 2x − 3 = 9

We try several values of x until we get the L.H.S. equal to the R.H.S.
 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 1 2 $×$ 1 − 3 = −1 9 No 2 2 $×$ 2 − 3 = 1 9 No 3 2 $×$ 3 − 3 = 3 9 No 4 2 $×$ 4 − 3 = 5 9 No 5 2 $×$ 5 − 3 = 7 9 No 6 2 $×$ 6 − 3 = 9 9 Yes
∴  x = 6

(viii)
Since, R.H.S. is a natural number so L.H.S. must be a natural number Thus, we will try values if x which are multiples of 'x'

 x L.H.S. R.H.S. Is L.H.S. = R.H.S.? 2 2/2 + 7 = 8 11 No 4 4/2 + 7 = 9 11 No 6 6/2 + 7 = 10 11 No 8 8/2 + 7 = 11 11 Yes
∴ x = 8

(ix) 2y + 4 = 3y
We try several values of y until we get the L.H.S. equal to the R.H.S.
 y L.H.S. R.H.S. Is L.H.S. = R.H.S.? 1 2 $×$ 1 + 4 = 6 3 $×$ 1 = 3 No 2 2 $×$ 2 + 4 = 8 3 $×$ 2 = 6 No 3 2 $×$ 3 + 4 = 10 3 $×$ 3 = 9 No 4 2 $×$ 4 + 4 = 12 3 $×$ 4 = 12 Yes
∴ y = 4

(x) z − 3 = 2z − 5
We try several values of z till we get the L.H.S. equal to the R.H.S.
 z L.H.S. R.H.S. Is L.H.S. = R.H.S.? 1 1 − 3 = −2 2 $×$ 1 − 5 = −3 No 2 2 − 3 = −1 2 $×$ 2 − 5 = −1 Yes
∴ z = 2

#### Page No 143:

x + 5 = 12

Subtracting 5 from both the sides:
⇒ x + 5 − 5 = 12 − 5
⇒ x = 7
Verification:
Substituting x = 7 in the L.H.S.:
⇒ 7 + 5 = 12 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

x + 3 = −2

Subtracting 3 from both the sides:
⇒ x + 3 − 3 = −2 − 3
⇒ x = −5

Verification:
Substituting x = −5 in the L.H.S.:
⇒ −5 + 3 =  −2 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

x − 7 = 6
Adding 7 on both the sides:
⇒ x − 7 + 7 = 6 + 7
⇒ x = 13

Verification:
Substituting x = 13 in the L.H.S.:
⇒ 13 − 7 =  6 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

x − 2 = −5

⇒ x − 2 + 2 = −5 + 2
⇒ x = −3
Verification:
Substituting x = −3 in the L.H.S.:
⇒  −3 − 2 = −5 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

3x − 5 = 13
⇒ 3x − 5 + 5 = 13 + 5             [Adding 5 on both the sides]
⇒ 3x = 18
[Dividing both the sides by 3]
⇒ x = 6
Verification:
Substituting x = 6 in the L.H.S.:
⇒  3 $×$ 6 − 5 = 18 − 5 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

4x + 7 = 15
⇒ 4x + 7 − 7 = 15 − 7              [Subtracting 7 from both the sides]
⇒ 4x = 8
[Dividing both the sides by 4]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
⇒  4$×$2 + 7 = 8 + 7 = 15 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

[Multiplying both the sides by 5]
⇒ x = 60
Verification:
Substituting x = 60 in the L.H.S.:
$\frac{60}{5}$ = 12 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

[Multiplying both the sides by 5]
⇒ 3x = 75

⇒ x = 25
Verification:
Substituting x = 25 in the L.H.S.:
= 15 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

5x − 3 = x + 17
⇒ 5x − x = 17 + 3                  [Transposing x to the L.H.S. and 3 to the R.H.S.]
⇒ 4x = 20
[Dividing both the sides by 4]
⇒ x = 5
Verification:
Substituting x = 5 on both the sides:
L.H.S.:  5(5) − 3
⇒ 25 − 3
⇒ 22

R.H.S.:  5 + 17 = 22
⇒ L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

⇒ 2x $-$$\frac{1}{2}$ + $\frac{1}{2}$ = 3 + $\frac{1}{2}$                              [Adding $\frac{1}{2}$ on both the sides]
⇒ 2x  =
⇒ 2x = $\frac{7}{2}$
[Dividing both the sides by 3]
⇒ x = $\frac{7}{4}$
Verification:
Substituting  x = $\frac{7}{4}$ in the  L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

3(x + 6) = 24
[On expanding the brackets]
⇒  3x + 18 = 24
⇒ 3x + 18 $-$ 18 = 24 $-$ 18            [Subtracting 18 from both the sides]
⇒ 3x = 6
[Dividing both the sides by 3]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
3(2 + 6) = 3 $×$8 = 24  = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

6x + 5 = 2x + 17
$⇒$6x  $-$ 2x = 17 $-$ 5                          [Transposing 2x to the L.H.S. and 5 to the R.H.S.]
$⇒$4x = 12
$⇒$                                      [Dividing both the sides by 4]
$⇒$x = 3
Verification:
Substituting x = 3 on both the sides:
L.H.S.: 6(3) + 5
=18 + 5
=23
R.H.S.:   2(3) + 17
= 6 + 17
= 23
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

[Adding 8 on both the sides]

[Multiplying both the sides by 4]
or, x = 36
Verification:
Substituting x = 36 in the L.H.S.:
or, = 9 − 8 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

[Transposing $\frac{x}{3}$ to the L.H.S.]

[Multiplying both the sides by 6]
or, x = 6
Verification:
Substituting x = 6 on both the sides:
L.H.S.: = 3
R.H.S.: =  2 + 1 =  3
L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

3(x + 2) − 2(x − 1) = 7
[On expanding the brackets]
or, 3x + 6 $-$2x + 2 = 7
or, x + 8 = 7
or, x + 8 $-$ 8 = 7 $-$ 8                                        [Subtracting 8 from both the sides]
or, x = $-$1
Verification:
Substituting x = $-$1 in the L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

5(x-1) +2(x+3) + 6 = 0
$⇒$5x -5 +2x +6 +6 = 0        (Expanding within the brackets)
$⇒$7x +7 = 0
$⇒$x +1 = 0       (Dividing by 7)
$⇒$x = -1

Verification:
Putting x = -1 in the L.H.S.:
L.H.S.: 5(-1 -1) + 2(-1 + 3) + 6
= 5(-2) + 2(2) + 6
= -10 + 4 + 6  = 0 = R.H.S.

Hence, verified.

#### Page No 143:

6(1 − 4x) + 7(2 + 5x) = 53
or,               [On expanding the brackets]
or, 6 $-$ 24x + 14 + 35x = 53
or, 11x + 20 = 53
or, 11x + 20 $-$ 20 = 53 $-$ 20                                        [Subtracting 20 from both the sides]
or, 11x = 33
or,                                                       [Dividing both the sides by 11]
or, x = 3
Verification:
Substituting x = 3 in the L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

16(3x − 5) − 10(4x − 8) = 40
or,               [On expanding the brackets]
or, 48x $-$ 80 $-$ 40x + 80 = 40
or, 8x  = 40
or, $\frac{8x}{8}=\frac{40}{8}$                                                      [Dividing both the sides by 8]
or, x = 5
Verification:
Substituting x = 5 in the L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

3(x + 6) + 2(x + 3) = 64
$⇒$3 × x   +   3 × 6 + 2 × x  + 2 × 3   = 64            [On expanding the brackets]
$⇒$3x + 18 +  2x + 6 = 64
⇒5x + 24 = 64
⇒5x + 24 $-$ 24 = 64 $-$ 24                                       [Subtracting 24 from both the sides]
⇒5x = 40
[Dividing both the sides by 5]
⇒x = 8
Verification:
Substituting x = 8 in the L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

3(2 − 5x) − 2(1 − 6x) = 1
or, 3 × 2  + 3 × (−5x) − 2 × 1 − 2 × (−6x) = 1           [On expanding the brackets]
or, 6 − 15x −  2 + 12x = 1
or, 4 - 3x = 1
or,  3  =3x
or, x = 1

Verification:
Substituting x = 1 in the L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

or,                                   [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]
or,
or,
or,                                     [Dividing both the sides by 12]
or, n = 66
Verification:
Substituting n = 66 on both the sides:

L.H.S.:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

or,                       [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]

or,                             [Multiplying both the sides by 6]
or, m = $-$54
Verification:
Substituting x = −54 on both the sides:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

or,                         [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]

[Multiplying both the sides by −10]
or, x = −25
Verification:
Substituting x = −25 on both the sides:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

or,
or,                         [Transposing x/5 to the R.H.S.]
or,
or,
or,                [Multiplying both the sides by 5]
or, x = −13
Verification:
Substituting x = −13 on both the sides:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

or,                         [Adding 4 on both the sides]
or,
or,                               [Multiplying both the sides by 10]
or,
or,                   [Dividing both the sides by 3]
or, x = 60
Verification:
Substituting x = 60 on both the sides:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 143:

[On expanding the brackets]

[Transposing x to the L.H.S. and $-\frac{3}{4}$ to the R.H.S.]

[Multiplying both the sides by -4]
or, x = 9

Verification:
Substituting x = 9 on both the sides:

L.H.S. = R.H.S.
Hence, verified.

#### Page No 144:

Let the required number be x.
According to the question:
9 + x = 36
or, x + 9 $-$ 9 = 36 $-$ 9                        [Subtracting 9 from both the sides]
or, x = 27
Thus, the required number is 27.

#### Page No 144:

Let the required number be x.
According to the question:
4x $-$11 = 89
or, 4x $-$ 11 +11 = 89 + 11                        [Adding 11 on both the sides]
or, 4x = 100
or,                                        [Dividing both the sides by 4]
or, x = 25
Thus, the required number is 25.

#### Page No 144:

Let the required number be x.
According to the question:
or, 5x = x + 80
or, 5x $-$ x = 80                       [Transposing x to the L.H.S.]
or, 4x = 80
or,                                         [Dividing both the sides by 4]
or, x = 20
Thus, the required number is 20.

#### Page No 144:

Let the three consecutive natural numbers be x, (x+1), (x+2).
According to the question:
x + (x + 1) + (x + 2) = 114
or, x + x + 1 + x + 2 = 114
or, 3x + 3 = 114
or, 3x + 3 $-$ 3 = 114 $-$ 3                     [Subtracting 3 from both the sides]
or, 3x = 111
or,                                    [Dividing both the sides by 3]
or, x = 37
Required numbers are:
x = 37
or, x + 1 = 37 + 1 = 38
or ,x + 2 = 37 + 2 = 39
Thus, the required numbers are 37, 38 and 39.

#### Page No 144:

Let the required number be x.
When Raju multiplies it with 17, the number becomes 17x.
According to the question :
17x + 4 = 225
or, 17x + 4 $-$ 4 = 225 $-$ 4                          [Subtracting 4 from both the sides]
or, 17x = 221
or,                                         [Dividing both the sides by 17]
or, x = 13
Thus, the required number is 13.

#### Page No 144:

Let the required number be x.
According to the question, the number is tripled and 5 is added to it
∴ 3x + 5
or, 3x  + 5 = 50
or, 3x + 5 $-$ 5 = 50 $-$ 5                         [Subtracting 5 from both the sides]
or, 3x = 45
or,                               [Dividing both the sides by 3]
or, x = 15
Thus, the required number is 15.

#### Page No 144:

Let one of the number be x.
∴ The other number = (x + 18)
According to the question:
x + (x + 18) = 92
or, 2x + 18 $-$ 18 = 92 $-$ 18                         [Subtracting 18 from both the sides]
or, 2x =74
or,                                            [Dividing both the sides by 2]
or, x = 37
Required numbers are:
x = 37
or, x + 18 = 37 + 18 = 55

#### Page No 144:

Let one of the number be 'x'
∴ Second number = 3x
According to the question:
x + 3x = 124
or, 4x  = 124
or,                       [Dividing both the sides by 4]
or, x = 31
Thus, the required number is x = 31 and 3x = 3$×$31 = 93.

#### Page No 144:

Let one of the number be x.
∴ Second number = 5x
According to the question:
5x $-$ x = 132
or, 4x = 132
or,                         [Dividing both the sides by 4]
or, x = 33
Thus, the required numbers are x = 33 and 5x = 5$×$33 = 165.

#### Page No 144:

Let one of the even number be x.
Then, the other consecutive even number is (x + 2).
According to the question:
x + (x + 2)  = 74
or, 2x + 2  = 74
or, 2x + 2 $-$ 2 = 74 $-$ 2           [Subtracting 2 from both the sides]
or, 2x = 72
or,                          [Dividing both the sides by 2]
or, x = 36
Thus, the required numbers are x = 36 and x+ 2 = 38.

#### Page No 144:

Let the first odd number be x.
Then, the next consecutive odd numbers will be (x + 2) and (x + 4).
According to the question:
x + (x + 2) + (x + 4)  = 21
or, 3x + 6  = 21
or, 3x + 6 $-$ 6 = 21 $-$ 6           [Subtracting 6 from both the sides]
or, 3x = 15
or,                          [Dividing both the sides by 3]
or, x = 5
∴ Required numbers are:
x = 5
x + 2 = 5 + 2 = 7
x + 4 = 5 + 4 = 9

#### Page No 144:

Let the present age of Ajay be x years.
Since Reena is 6 years older than Ajay, the present age of Reena will be (x+ 6) years.
According to the question:
x + (x + 6) = 28
or, 2x + 6 = 28
or, 2x + 6 $-$ 6 = 28 $-$ 6                 [Subtracting 6 from both the sides]
or, 2x = 22
or,                               [Dividing both the sides by 2]
or, x = 11
∴ Present age of Ajay = 11 years
Present age of Reena  = x +6 = 11 + 6
= 17 years

#### Page No 144:

Let the present age of Vikas be x years.
Since Deepak is twice as old as Vikas, the present age of Deepak will be 2x years.
According to the question:
2x $-$ x = 11
x = 11
∴ Present age of Vikas = 11 years
Present age of Deepak  = 2x = 2$×$11
= 22 years

#### Page No 144:

Let the present age of Rekha be x years.
As Mrs. Goel is 27 years older than Rekha, the present age of Mrs. Goel will be (x + 27) years.
After 8 years:
Rekha's age = (x + 8) years
Mrs. Goel's age = (x + 27 + 8)
= (x + 35) years

According to the question:
(x + 35) = 2(x + 8)
or, x + 35 = 2$×$x + 2$×$8                [On expanding the brackets]
or, x + 35 = 2x + 16
or, 35 $-$ 16 = 2x $-$ x                [Transposing 16 to the L.H.S. and x to the R.H.S.]
or, x = 19
∴ Present age of Rekha = 19 years
Present age of Mrs. Goel = x + 27
= 19 + 27
= 46 years

#### Page No 145:

Let the present age of the son be x years.
As the man is 4 times as old as his son, the present age of the man will be (4x) years.
After 16 years:
Son's age  = (x + 16) years
Man's age = (4x + 16) years

According to the question:
(4x + 16) = 2(x + 16)
or, 4x + 16 = 2$×$x + 2$×$16                [On expanding the brackets]
or, 4x + 16 = 2x + 32
or, 4x $-$ 2x = 32 $-$ 16                [Transposing 16 to the R.H.S. and 2x to the L.H.S.]
or, 2x = 16
or,                            [Dividing both the sides by 2]
or, x = 8
∴ Present age of the son = 8 years
Present age of the man  = 4x = 4$×$8
= 32 years

#### Page No 145:

Let the present age of the son be x years.
As the man is 3 times as old as his son, the present age of the man will be (3x) years.

5 years ago:
Son's age = (x $-$ 5) years
Man's age = (3x $-$ 5) years

According to the question:
(3x $-$ 5) = 4(x $-$ 5)
or, 3x $-$ 5 = 4$×$x $-$ 4$×$5                [On expanding the brackets]
or, 3x $-$ 5 = 4x $-$ 20
or, 20 $-$ 5 = 4x $-$ 3x                [Transposing 3x to the R.H.S. and 20 to the L.H.S.]
or, x = 15
∴ Present age of the son = 15 years
Present age of the man  = 3x = 3$×$15
= 45 years

#### Page No 145:

Let the present age of Fatima be x years.

After 16 years:
Fatima's age = (x + 16) years

According to the question:
x + 16 = 3(x)
or, 16 = 3x $-$ x               [Transposing x to the R.H.S.]
or, 16 = 2x
or,                 [Dividing both the sides by 2]
or, x = 8
∴ Present age of Fatima = 8 years

#### Page No 145:

Let the present age of Rahim be x years.
After 32 years:
Rahim's age = (x + 32) years
8 years ago:
Rahim's age = (x $-$ 8) years
According to the question:
x + 32 = 5(x $-$ 8)
or, x + 32  = 5x $-$ 5$×$8
or, x + 32 = 5x $-$ 40
or, 40 + 32 = 5x $-$ x                      [Transposing 'x' to the R.H.S. and 40 to the L.H.S.]
or, 72 = 4x
or,                                [Dividing both the sides by 4]
or, x = 18
Thus, the present age of Rahim is 18 years.

#### Page No 145:

Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins will be 4x.
According to the question:
0.50(x) + 0.25(4x) = 30
or, 0.5x + x = 30
or, 1.5x = 30
or,           [Dividing both the sides by 1.5]
or, x = 20
Thus, the number of 50 paisa coins is 20.
Number of 25 paisa coins = 4x = 4$×$20 = 80

#### Page No 145:

Let the price of one pen be Rs x.
According to the question:
5x = 3x + 17
or, 5x $-$ 3x = 17                    [Transposing 3x to the L.H.S.]
or, 2x = 17
or,                         [Dividing both the sides by 2]
or, x = 8.50
∴ Price of one pen = Rs 8.50

#### Page No 145:

Let the number of girls in the school be x.
Then, the number of boys in the school will be (x + 334).
Total strength of the school = 572

∴ x + (x + 334) = 572
or, 2x + 334 = 572
or, 2x + 334 $-$ 334 = 572 $-$ 334                {Subtracting 334 from both the sides]
or, 2x = 238
or,                                           [Dividing both the sides by 2]
or, x = 119
∴ Number of girls in the school = 119

#### Page No 145:

Let the breadth of the park be x metres.
Then, the length of the park will be 3x metres.
Perimeter of the park = 2 (Length + Breadth) = 2 ( 3x + x ) m
Given perimeter = 168 m

∴ 2(3x + x) = 168
or, 2 ( 4x ) = 168
or, 8x = 168                                      [On expanding the brackets]
or,                                 [Dividing both the sides by 8]
or, x = 21 m
∴ Breadth of the park = x = 21 m
Length of the park = 3x = 3$×$21 = 63 m

#### Page No 145:

Let the breadth of the hall be x metres.
Then, the length of the hall will be (x + 5) metres.
Perimeter of the hall = 2(Length + Breadth) = 2( x + 5 + x) metres
Given perimeter of the rectangular hall = 74 metres

∴ 2( x + 5 + x) = 74
or, 2 ( 2x + 5 ) = 74
or, 2 ×2x + 2 ×5 = 74                              [On expanding the brackets]
or, 4x + 10 = 74
or, 4x + 10 $-$ 10 = 74 $-$ 10                     [Subtracting 10 from both the sides]
or, 4x = 64
or,                                         [Dividing both the sides by 4]
or, x = 16 metres
∴ Breadth of the park = x
= 16 metres
Length of the park = x + 5 = 16 + 5
= 21 metres

#### Page No 145:

Let the breadth of the rectangle be x cm.
Then, the length of the rectangle will be (x + 7) cm.
Perimeter of the rectangle = 2(Length + Breadth) = 2( x + 7 + x) cm
Given perimeter of the rectangle = Length of the wire = 86 cm

∴ 2( x + 7 + x) = 86
or, 2 ( 2x + 7 ) = 86
or, 2 ×2x + 2 × 7 = 86                              [On expanding the brackets]
or, 4x + 14 = 86
or, 4x + 14 - 14 = 86 - 14                   [Subtracting 14 from both the sides]
or, 4x = 72
or,                                    [Dividing by 4 on both the sides]
or, x = 18 metres
Breadth of the hall = x
= 18 metres
Length of the hall = x + 7
= 18 + 7
= 25 metres

#### Page No 146:

Earning of the man per hour = Rs 25

Earning of the man in x hours = Rs (25$×$x)
= Rs 25x

#### Page No 146:

Cost of 1 pen = Rs 16
∴ Cost of 'x' pens = Rs 16 $×$ x
= Rs 16x
Similarly, cost of 1 pencil = Rs 5
∴ Cost of 'y' pencils = Rs 5$×$y
= Rs 5y
∴ Total cost of x pens and y pencils = Rs (16x + 5y)

#### Page No 146:

Lalit's earning per day = Rs x
∴ Lalit's earning in 30 days = Rs 30 $×$x
= Rs 30x

Similarly, Lalit's expenditure per day = Rs y
∴ Lalit's expenditure in 30 days = Rs 30 $×$ y
= Rs 30y
∴ In 30 days, Lalit saves = (Total earnings $-$ Total expenditure)
= Rs (30x $-$ 30y)
= Rs 30(x - y)

#### Page No 146:

Let the required number be x.
Three times this number is 3x.
On adding 8, the number becomes 3x + 8.
3x + 8 = 20
or, 3x + 8 $-$ 8 = 20 $-$ 8                [Subtracting 8 from both the sides]
or, 3x = 12
or,                              [Dividing both the sides by 3]
or, x = 4
∴ Required number = 4

#### Page No 146:

Given:
x =1
y = 2
z = 3

Substituting x = 1, y = 2 and z = 3 in the given equation (x2 + y2 + 2xyz):

#### Page No 146:

4x + 9 = 17
or, 4x + 9 $-$ 9 = 17 $-$ 9                          [Subtracting 9 from both the sides]
or, 4x = 8
or,                                           [Dividing both the sides with 4]
or, x = 2

#### Page No 146:

3(x + 2) − 2(x − 1) = 7.
or,                     [On expanding the brackets]
or, 3x + 6 − 2x + 2 = 7
or, x + 8 = 7
or, x + 8 − 8 = 7 − 8                                            [Subtracting 8 from both the sides]
or, x = −1

#### Page No 146:

or,                                [Taking the L.C.M. as 10]
or,
or,                           [Multiplying both the sides by (−10)]
or, x = −25

#### Page No 146:

Let the three consecutive natural numbers be x, (x + 1) and (x + 2).

∴ x + (x + 1) + (x + 2) = 51
3x + 3 = 51
3x + 3 $-$ 3 = 51 $-$ 3                   [Subtracting 3 from both the sides]
3x = 48
[Dividing both the sides by 3]
x = 16
Thus, the three natural numbers are x = 16, x+1 = 17 and x+2 = 18.

#### Page No 146:

Let the present age of Seema be x years.
After 16 years:
Seema's age = x + 16

After 16 years, her age becomes thrice of her age now
∴ x + 16 = 3x
or, 16 = 3x $-$ x                     [Transposing x to the R.H.S.]
or, 2x = 16
or,                       [Dividing both the sides by 2]
or, x = 8 years

#### Page No 146:

(c) 5 − 2x + 3y

1 exceeds 2x − 3y − 4.

∴1 $-$ (2x − 3y − 4) = 1 $-$2x + 3y + 4
= 5 $-$ 2x + 3y

∴ 1 exceeds 2x − 3y − 4 by 5 $-$ 2x + 3y.

#### Page No 146:

(b) −4x3 + 5x2 − 7x − 6
In order to find what must be added, we subtract (5x3 − 2x2 + 6x + 7) from (x3 + 3x2x + 1).

(x3 + 3x2x + 1)  − ( 5x3 − 2x2 + 6x + 7)
or,  x3 + 3x2x + 1 − 5x3  + 2x2 − 6x − 7
or,  x3 − 5x3+ 3x2+ 2x2x − 6x+ 1 − 7
or, −4x3 + 5x2 − 7x − 6

#### Page No 146:

(a) 5x − 4y

2x − [3y − {2x − (yx)}]
= 2x − [3y − {2x −  y + x}]
= 2x − [3y − {3x −  y}]
= 2x −  [3y −  3x + y]
= 2x −  [4y −  3x]
= 2x −  4y + 3x
= 5x −  4y

#### Page No 146:

(c) −5yz

All the terms in the expression −5xyz barring x will be the coefficient of x, i.e. −5yz.

#### Page No 146:

(b) trinomial
Since it contains three variables, i.e. 'x', 'y' and 'z', it is a trinomial.

(b) x = 5

or, x = 5

#### Page No 146:

(c) 14
Substituting x = 1, y = 2 and z = 3 in (x2 + y2 + z2):

or, 1 + 4 + 9 = 14

#### Page No 146:

(c) 9

or,                                   [Subtracting 5 from both the sides]
or,
or,                                         [Multiplying both the sides by 3]
or, x = 9

#### Page No 146:

(i) monomial
(ii) binomial
(iii) trinomial
(iv) x = 3
3x $-$ 5 = 7 $-$ x
or, 3x + x = 7 + 5                   [Transposing x to the L.H.S. and 5 to the R.H.S.]
or, 4x = 12
or,
or, x = 3
(v) 2b2 $-$ 2a2
or, b2 $-$ a2 $-$ a2 + b2
or, 2b2 $-$ 2a2
( b2  a2)  (a2  b2)

(b2  a2)  (a2  b2
(b2  a2)  (a2  b2)
(b2  a2)  (a2  b2)

#### Page No 147:

(i) True
Since it has one term, it is a monomial.

(ii) False
2x + 5 = 8
or, 2x + 5 $-$ 5 = 8 $-$ 5               [Subtracting 5 from both the sides]
or, 2x = 3
or, x = 3/2 and not x = 2/3

(iii) True
This is because the maximum power of the variable x is 1.

(iv) False
The coefficient of x in 5xy would be 5y.

(v) True
8 − x = 5
or, 8 − 5 = x
or, 3 = x

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