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#### Question 1:

Write each of the following statements as an equation:
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice  a number is 16.
(vi) If 12 is subtracted from a number, the result is 24.
(vii) Twice a number subtracted from 19 is 11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the number.

#### Answer:

(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

#### Question 2:

Write a statement for each of the equations, given below:
(i) x − 7 = 14
(ii) 2y = 18
(iii) 11 + 3x = 17
(iv) 2x − 3 = 13
(v) 12y − 30 = 6
(vi) $\frac{2z}{3}=8$

#### Answer:

(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.

#### Question 3:

Verify by substitution that
(i) the root of 3x − 5 = 7 is x = 4
(ii) the root of 3 + 2x = 9 is x = 3
(iii) the root of 5x − 8 = 2x − 2 is x = 2
(iv) the root of 8 − 7y = 1 is y = 1
(v) the root of $\frac{z}{7}=8$ is z = 56

(i)

(ii)

(iii)

(iv)

(v)

#### Question 4:

Solve each of the following equations by the trial-and-error method:
(i) y + 9 = 13
(ii) x − 7 = 10
(iii) 4x = 28
(iv) 3y = 36
(v) 11 + x = 19
(vi) $\frac{x}{3}=4$
(vii) 2x − 3 = 9
(viii)
(ix) 2y + 4 = 3y
(x) z − 3 = 2z − 5

#### Answer:

(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

#### Question 1:

Solve each of the following equations and verify the answer in each case:
x
+ 5 = 12

#### Answer:

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.

#### Question 2:

Solve each of the following equations and verify the answer in each case:
x
+ 3 = −2

#### Answer:

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.

#### Question 3:

Solve each of the following equations and verify the answer in each case:
x
− 7 = 6

#### Answer:

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.

#### Question 4:

Solve each of the following equations and verify the answer in each case:
x
− 2 = −5

#### Answer:

x − 2 = −5

Adding 2 on both sides:
⇒ 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.

#### Question 5:

Solve each of the following equations and verify the answer in each case:
3x − 5 = 13

#### Answer:

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.

#### Question 6:

Solve each of the following equations and verify the answer in each case:
4x + 7 = 15

#### Answer:

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.

#### Question 7:

Solve each of the following equations and verify the answer in each case:
$\frac{x}{5}=12$

#### Answer:

[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.

#### Question 8:

Solve each of the following equations and verify the answer in each case:
$\frac{3x}{5}=15$

#### Answer:

[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.

#### Question 9:

Solve each of the following equations and verify the answer in each case:
5x − 3 = x + 17

#### Answer:

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.

#### Question 10:

Solve each of the following equations and verify the answer in each case:
$2x-\frac{1}{2}=3$

#### Answer:

⇒ 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.

#### Question 11:

Solve each of the following equations and verify the answer in each case:
3(x + 6) = 24

#### Answer:

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.

#### Question 12:

Solve each of the following equations and verify the answer in each case:
6x + 5 = 2x + 17

#### Answer:

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.

#### Question 13:

Solve each of the following equations and verify the answer in each case:
$\frac{x}{4}-8=1$

#### Answer:

[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.

#### Question 14:

Solve each of the following equations and verify the answer in each case:
$\frac{x}{2}=\frac{x}{3}+1$

#### Answer:

[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.

#### Question 15:

Solve each of the following equations and verify the answer in each case:
3(x + 2) − 2(x − 1) = 7

#### Answer:

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.

#### Question 16:

Solve each of the following equations and verify the answer in each case:

#### Answer:

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.

#### Question 17:

Solve each of the following equations and verify the answer in each case:
6(1 − 4x) + 7(2 + 5x) = 53

#### Answer:

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.

#### Question 18:

Solve each of the following equations and verify the answer in each case:
16(3x − 5) − 10(4x − 8) = 40

#### Answer:

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.

#### Question 19:

Solve each of the following equations and verify the answer in each case:
3(x + 6) + 2(x + 3) = 64

#### Answer:

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.

#### Question 20:

Solve each of the following equations and verify the answer in each case:
3(2 − 5x) − 2(1 − 6x) = 1

#### Answer:

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.

#### Question 21:

Solve each of the following equations and verify the answer in each case:
$\frac{n}{4}-5=\frac{n}{6}+\frac{1}{2}$

#### Answer:

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.

#### Question 22:

Solve each of the following equations and verify the answer in each case:

#### Answer:

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.

#### Question 23:

Solve each of the following equations and verify the answer in each case:
$\frac{2x}{5}-\frac{3}{2}=\frac{x}{2}+1$

#### Answer:

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.

#### Question 24:

Solve each of the following equations and verify the answer in each case:
$\frac{x-3}{5}-2=\frac{2x}{5}$

#### Answer:

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.

#### Question 25:

Solve each of the following equations and verify the answer in each case:
$\frac{3x}{10}-4=14$

#### Answer:

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.

#### Question 26:

Solve each of the following equations and verify the answer in each case:

#### Answer:

[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.

#### Question 1:

If 9 is added to a certain number, the result is 36. Find the number.

#### Answer:

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.

#### Question 2:

If 11 is subtracted from 4 times a number, the result is 89. Find the number.

#### Answer:

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.

#### Question 3:

Find a number which when multiplied by 5 is increased by 80.

#### Answer:

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.

#### Question 4:

The sum of three consecutive natural numbers is 114. Find the numbers.

#### Answer:

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.

#### Question 5:

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

#### Answer:

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.

#### Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

#### Answer:

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.

#### Question 7:

Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

#### Answer:

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

#### Question 8:

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

#### Answer:

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.

#### Question 9:

Find two numbers such that one of them is five times the other and their difference is 132.

#### Answer:

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.

#### Question 10:

The sum of two consecutive even numbers is 74. Find the numbers.

#### Answer:

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.

#### Question 11:

The sum of three consecutive odd numbers is 21. Find the numbers.

#### Answer:

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

#### Question 12:

Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

#### Answer:

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

#### Question 13:

Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.

#### Answer:

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

#### Question 14:

Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

#### Answer:

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

#### Question 15:

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.

#### Answer:

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

#### Question 16:

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

#### Answer:

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

#### Question 17:

After 16 years, Fatima will be three times as old as she is now. Find her present age.

#### Answer:

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

#### Question 18:

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

#### Answer:

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.

#### Question 19:

A bag contains 25-paisa and 50-paisa coins whose total value is Rs 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

#### Answer:

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

#### Question 20:

Five times the price of a pen is Rs 17 more than three times its price. Find the price of the pen.

#### Answer:

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

#### Question 21:

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

#### Answer:

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

#### Question 22:

The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, fund its dimensions.

#### Answer:

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

#### Question 23:

The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

#### Answer:

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

#### Question 24:

A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

#### Answer:

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

#### Question 1:

A man earns Rs 25 per hour. How much does he earn in x hours?

#### Answer:

Earning of the man per hour = Rs 25

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

#### Question 2:

The cost of 1 pen is Rs 16 and the cost of 1 pencil is Rs 5. What is the total cost of x pens and y pencils.

#### Answer:

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)

#### Question 3:

Lalit earns Rs x per day and spends Rs y per day. How much does he save in 30 days?

#### Answer:

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)

#### Question 4:

Three times a number added to 8 gives 20. Find the number.

#### Answer:

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

#### Question 5:

If x = 1, y = 2 and z = 3, find the value of x2 + y2 + 2xyz.

#### Answer:

Given:
x =1
y = 2
z = 3

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

#### Question 6:

Solve: 4x + 9 = 17.

#### Answer:

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

#### Question 7:

Solve: 3(x + 2) − 2(x − 1) = 7.

#### Answer:

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

#### Question 8:

Solve: $\frac{2x}{5}-\frac{x}{2}=\frac{5}{2}$.

#### Answer:

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

#### Question 9:

The sum of three consecutive natural numbers is 51. Find the numbers.

#### Answer:

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.

#### Question 10:

After 16 years, Seema will be three times as old as she is now. Find her present age.

#### Answer:

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

#### Question 11:

By how much does I exceed 2x − 3y − 4?
(a) 2x − 3y − 5
(b) 2x − 3y − 3
(c) 5 − 2x + 3y
(d) none of these

#### Answer:

(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.

#### Question 12:

What must be added to 5x3 − 2x2 + 6x + 7 to make the sum x3 + 3x2x + 1?
(a) 4x3 − 5x2 + 7x + 6
(b) −4x3 + 5x2 − 7x − 6
(c) 4x3 + 5x2 − 7x + 6
(d) none of these

#### Answer:

(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

#### Question 13:

2x − [3y − {2x − (yx)}] = ?
(a) 5x − 4y
(b) 4y − 5x
(c) 5y − 4x
(d) 4x − 5y

#### Answer:

(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

#### Question 14:

The coefficient of x in −5xyz is
(a) −5
(b) 5yz
(c) −5yz
(d) yz

#### Answer:

(c) −5yz

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

is a
(a) monomial
(b) binomial
(c) trinomial
(d) quadrinomial

#### Answer:

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

#### Question 16:

If $\frac{x}{5}=1$, then
(a) $x=\frac{1}{5}$
(b) x = 5
(c) x = (5 + 1)
(d) none of these

(b) x = 5

or, x = 5

#### Question 17:

If x = 1, y = 2 and z = 3 then (x2 + y2 + z2) = ?
(a) 6
(b) 12
(c) 14
(d) 15

#### Answer:

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

or, 1 + 4 + 9 = 14

If , then x = ?
(a) 3
(b) 6
(c) 9
(d) 12

#### Answer:

(c) 9

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

#### Question 19:

Fill in the blanks.
(i) An expression having one term is called a ...... .
(ii) An expression having two term is called a ...... .
(i) An expression having three term is called a ...... .
(iv)
(v)

#### Answer:

(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)

#### Question 20:

Write 'T' for true and 'F' for false for each of the statements given below:
(i) −3xy2z is a monomial.
(ii) $x=\frac{2}{3}$ is solution of 2x + 5 = 8.
(iii) 2x + 3 = 5 is a linear equation.
(iv) The coefficient of x in 5xy is 5.
(v) .

#### Answer:

(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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