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#### Page No 139:

#### 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

#### Page No 140:

#### Question 2:

Write a statement for each of the equations, given below:

(i) *x* − 7 = 14

(ii) 2*y* = 18

(iii) 11 + 3*x* = 17

(iv) 2*x* − 3 = 13

(v) 12*y* − 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.

#### Page No 140:

#### Question 3:

Verify by substitution that

(i) the root of 3*x* − 5 = 7 is *x* = 4

(ii) the root of 3 + 2*x* = 9 is *x* = 3

(iii) the root of 5*x* − 8 = 2*x* − 2 is *x* = 2

(iv) the root of 8 − 7*y* = 1 is *y* = 1

(v) the root of $\frac{z}{7}=8$ is z = 56

#### Answer:

(i)

$3x-5=7\phantom{\rule{0ex}{0ex}}Substitutingx=4inthegivenequation:\phantom{\rule{0ex}{0ex}}L.H.S.\hspace{0.17em}:\phantom{\rule{0ex}{0ex}}3\times 4-5\phantom{\rule{0ex}{0ex}}or,12-5=7=R.H.S.\phantom{\rule{0ex}{0ex}}L.H.S.=R.H.S.\phantom{\rule{0ex}{0ex}}Hence,x=4istherootofthegivenequation.$

(ii)

$3+2x=9\phantom{\rule{0ex}{0ex}}Substitutingx=3inthegivenequation:\phantom{\rule{0ex}{0ex}}L.H.S.\hspace{0.17em}:\phantom{\rule{0ex}{0ex}}3+2\times 3\phantom{\rule{0ex}{0ex}}or,3+6=9=R.H.S.\phantom{\rule{0ex}{0ex}}L.H.S.=R.H.S.\phantom{\rule{0ex}{0ex}}Hence,x=3istherootofthegivenequation.$

(iii)

$5x-8=2x-2\phantom{\rule{0ex}{0ex}}Substitutingx=2inthegivenequation:\phantom{\rule{0ex}{0ex}}L.H.S.\hspace{0.17em}:R.H.S.:\phantom{\rule{0ex}{0ex}}5\times 2-8=2\times 2-2\phantom{\rule{0ex}{0ex}}or,10-8=2=4-2=2\phantom{\rule{0ex}{0ex}}L.H.S.=R.H.S.\phantom{\rule{0ex}{0ex}}Hence,x=2istherootofthegivenequation.$

(iv)

$8-7y=1\phantom{\rule{0ex}{0ex}}Substitutingy=1inthegivenequation:\phantom{\rule{0ex}{0ex}}L.H.S.\hspace{0.17em}:\phantom{\rule{0ex}{0ex}}8-7\times 1\phantom{\rule{0ex}{0ex}}or,8-7=1=R.H.S.\phantom{\rule{0ex}{0ex}}L.H.S.=R.H.S.\phantom{\rule{0ex}{0ex}}Hence,y=1istherootofthegivenequation.$

(v)

$\frac{z}{7}=8\phantom{\rule{0ex}{0ex}}Substitutingz=56inthegivenequation:\phantom{\rule{0ex}{0ex}}L.H.S.\hspace{0.17em}:\phantom{\rule{0ex}{0ex}}\frac{56}{7}=8=R.H.S.\phantom{\rule{0ex}{0ex}}L.H.S.=R.H.S.\phantom{\rule{0ex}{0ex}}Hence,z=56istherootofthegivenequation.$

#### Page No 140:

#### Question 4:

Solve each of the following equations by the trial-and-error method:

(i) *y* + 9 = 13

(ii) *x* − 7 = 10

(iii) 4*x* = 28

(iv) 3*y* = 36

(v) 11 + *x* = 19

(vi) $\frac{x}{3}=4$

(vii) 2*x* − 3 = 9

(viii) $\frac{1}{2}x+7=11$

(ix) 2*y* + 4 = 3*y*

(x) *z* − 3 = 2*z* − 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 |

(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 $\times $ 1 = 4 | 28 | No |

2 | 4 $\times $ 2 = 8 | 28 | No |

3 | 4 $\times $ 3 = 12 | 28 | No |

4 | 4 $\times $ 4 = 16 | 28 | No |

5 | 4 $\times $ 5 = 20 | 28 | No |

6 | 4 $\times $ 6 = 24 | 28 | No |

7 | 4 $\times $ 7 = 28 | 28 | Yes |

(iv) 3

*y*= 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 $\times $ 6 = 18 | 36 | No |

7 | 3 $\times $ 7 = 21 | 36 | No |

8 | 3 $\times $ 8 = 24 | 36 | No |

9 | 3 $\times $ 9 = 27 | 36 | No |

10 | 3 $\times $ 10 = 30 | 36 | No |

11 | 3 $\times $11 = 33 | 36 | No |

12 | 3 $\times $ 12 = 36 | 36 | Yes |

(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 |

(vi) $\frac{x}{3}=4$

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 |

(vii) 2

*x*− 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 $\times $ 1 − 3 = −1 | 9 | No |

2 | 2 $\times $ 2 − 3 = 1 | 9 | No |

3 | 2 $\times $ 3 − 3 = 3 | 9 | No |

4 | 2 $\times $ 4 − 3 = 5 | 9 | No |

5 | 2 $\times $ 5 − 3 = 7 | 9 | No |

6 | 2 $\times $ 6 − 3 = 9 | 9 | Yes |

(viii) $\frac{1}{2}x+7=11$

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 |

(ix) 2

*y*+ 4 = 3

*y*

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 $\times $ 1 + 4 = 6 | 3 $\times $ 1 = 3 | No |

2 | 2 $\times $ 2 + 4 = 8 | 3 $\times $ 2 = 6 | No |

3 | 2 $\times $ 3 + 4 = 10 | 3 $\times $ 3 = 9 | No |

4 | 2 $\times $ 4 + 4 = 12 | 3 $\times $ 4 = 12 | Yes |

(x)

*z*− 3 = 2

*z*− 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 $\times $ 1 − 5 = −3 | No |

2 | 2 − 3 = −1 | 2 $\times $ 2 − 5 = −1 | Yes |

#### Page No 143:

#### 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.

#### Page No 143:

#### 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.

#### Page No 143:

#### 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.

#### Page No 143:

#### 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.

#### Page No 143:

#### Question 5:

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

3*x* − 5 = 13

#### Answer:

3x − 5 = 13

⇒ 3x − 5 + 5 = 13 + 5 [Adding 5 on both the sides]

⇒ 3x = 18

⇒ $\frac{3x}{3}=\frac{18}{3}$ [Dividing both the sides by 3]

⇒ x = 6

Verification:

Substituting x = 6 in the L.H.S.:

⇒ 3 $\times $ 6 − 5 = 18 − 5 = 13 = R.H.S.

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 6:

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

4*x* + 7 = 15

#### Answer:

4*x* + 7 = 15

⇒ 4x + 7 − 7 = 15 − 7 [Subtracting 7 from both the sides]

⇒ 4x = 8

⇒ $\frac{4x}{4}=\frac{8}{4}$ [Dividing both the sides by 4]

⇒ x = 2

Verification:

Substituting x = 2 in the L.H.S.:

⇒ 4$\times $2 + 7 = 8 + 7 = 15 = R.H.S.

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 7:

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

$\frac{x}{5}=12$

#### Answer:

$\frac{x}{5}=12$

⇒ $\frac{x}{5}\times 5=12\times 5$ [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:

#### Question 8:

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

$\frac{3x}{5}=15$

#### Answer:

$\frac{3x}{5}=15$

⇒ $\frac{3x}{5}\times 5=15\times 5$ [Multiplying both the sides by 5]

⇒ 3x = 75

⇒ $\frac{3x}{3}=\frac{75}{3}$

⇒ x = 25

Verification:

Substituting x = 25 in the L.H.S.:

⇒ $\frac{3\times 25}{5}$ = 15 = R.H.S.

⇒ L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 9:

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

5*x* − 3 = *x* + 17

#### Answer:

5*x* − 3 = *x* + 17

⇒ 5*x* − x = 17 + 3 [Transposing x to the L.H.S. and 3 to the R.H.S.]

⇒ 4x = 20

⇒ $\frac{4x}{4}=\frac{20}{4}$ [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:

#### Question 10:

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

$2x-\frac{1}{2}=3$

#### Answer:

$2x-\frac{1}{2}=3$

⇒ 2x $-$$\frac{1}{2}$ + $\frac{1}{2}$ = 3 + $\frac{1}{2}$ [Adding $\frac{1}{2}$ on both the sides]

⇒ 2x = $\frac{6+1}{2}$

⇒ 2x = $\frac{7}{2}$

⇒ $\frac{2x}{2}=\frac{7}{2\times 2}$ [Dividing both the sides by 3]

⇒ x = $\frac{7}{4}$

Verification:

Substituting x = $\frac{7}{4}$ in the L.H.S.:

$2\left(\frac{7}{4}\right)-\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 11:

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

3(*x* + 6) = 24

#### Answer:

3(*x* + 6) = 24

⇒ $3\times x+3\times 6=24$ [On expanding the brackets]

⇒ 3x + 18 = 24

⇒ 3x + 18 $-$ 18 = 24 $-$ 18 [Subtracting 18 from both the sides]

⇒ 3x = 6

⇒ $\frac{3x}{3}=\frac{6}{3}$ [Dividing both the sides by 3]

⇒ x = 2

Verification:

Substituting x = 2 in the L.H.S.:

3(2 + 6) = 3 $\times $8 = 24 = R.H.S.

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 12:

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

6*x* + 5 = 2*x* + 17

#### Answer:

6*x* + 5 = 2*x* + 17

$\Rightarrow $6x $-$ 2x = 17 $-$ 5 [Transposing 2x to the L.H.S. and 5 to the R.H.S.]

$\Rightarrow $4x = 12

$\Rightarrow $$\frac{4x}{4}=\frac{12}{4}$ [Dividing both the sides by 4]

$\Rightarrow $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:

#### Question 13:

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

$\frac{x}{4}-8=1$

#### Answer:

$\frac{x}{4}-8=1$

$\Rightarrow \frac{x}{4}-8+8=1+8$ [Adding 8 on both the sides]

$\Rightarrow \frac{x}{4}=9$

$\Rightarrow \frac{x}{4}\times 4=9\times 4$ [Multiplying both the sides by 4]

or, x = 36

Verification:

Substituting x = 36 in the L.H.S.:

or, $\frac{36}{4}-8$ = 9 − 8 = 1 = R.H.S.

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 14:

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

$\frac{x}{2}=\frac{x}{3}+1$

#### Answer:

$\frac{x}{2}=\frac{x}{3}+1$

$\Rightarrow \frac{x}{2}-\frac{x}{3}=1$ [Transposing $\frac{x}{3}$ to the L.H.S.]

$\Rightarrow \frac{3x-2x}{6}=1$

$\Rightarrow \frac{x}{6}=1$

$\Rightarrow \frac{x}{6}\times 6=1\times 6$ [Multiplying both the sides by 6]

or, x = 6

Verification:

Substituting x = 6 on both the sides:

L.H.S.: $\frac{6}{2}$= 3

R.H.S.: $\frac{6}{3}+1$ = 2 + 1 = 3

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### 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

$\Rightarrow 3\times x+3\times 2-2\times x-2\times (-1)=7$ [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.:

$3(-1+2)-2(-1-1)\phantom{\rule{0ex}{0ex}}or,3\left(1\right)-2(-2)\phantom{\rule{0ex}{0ex}}or,3+4=7=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 16:

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

#### Answer:

5(x-1) +2(x+3) + 6 = 0

$\Rightarrow $5x -5 +2x +6 +6 = 0 (Expanding within the brackets)

$\Rightarrow $7x +7 = 0

$\Rightarrow $x +1 = 0 (Dividing by 7)

$\Rightarrow $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:

#### Question 17:

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

6(1 − 4*x*) + 7(2 + 5*x*) = 53

#### Answer:

6(1 − 4*x*) + 7(2 + 5*x*) = 53

or, $6\times 1-6\times 4x+7\times 2+7\times 5x=53$ [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, $\frac{11x}{11}=\frac{33}{11}$ [Dividing both the sides by 11]

or, x = 3

Verification:

Substituting x = 3 in the L.H.S.:

$6(1-4\times 3)+7(2+5\times 3)\phantom{\rule{0ex}{0ex}}\Rightarrow 6(1-12)+7(2+15)\phantom{\rule{0ex}{0ex}}\Rightarrow 6(-11)+7\left(17\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -66+119=53=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 18:

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

16(3*x* − 5) − 10(4*x* − 8) = 40

#### Answer:

16(3*x* − 5) − 10(4*x* − 8) = 40

or, $16\times 3x-16\times 5-10\times 4x-10\times (-8)=40$ [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.:

$16(3\times 5-5)-10(4\times 5-8)\phantom{\rule{0ex}{0ex}}\Rightarrow 16(15-5)-10(20-8)\phantom{\rule{0ex}{0ex}}\Rightarrow 16\left(10\right)-10\left(12\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 160-120=40=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### 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

$\Rightarrow $3 × x + 3 × 6 + 2 × x + 2 × 3 = 64 [On expanding the brackets]

$\Rightarrow $3x + 18 + 2x + 6 = 64

⇒5x + 24 = 64

⇒5x + 24 $-$ 24 = 64 $-$ 24 [Subtracting 24 from both the sides]

⇒5x = 40

⇒$\frac{5x}{5}=\frac{40}{5}$ [Dividing both the sides by 5]

⇒x = 8

Verification:

Substituting x = 8 in the L.H.S.:

$3(8+6)+2(8+3)\phantom{\rule{0ex}{0ex}}3\left(14\right)+2\left(11\right)\phantom{\rule{0ex}{0ex}}42+22=64=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 20:

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

3(2 − 5*x*) − 2(1 − 6*x*) = 1

#### Answer:

3(2 − 5*x*) − 2(1 − 6*x*) = 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.:

$3(2-5\times 1)-2(1-6\times 1)\phantom{\rule{0ex}{0ex}}\Rightarrow 3(2-5)-2(1-6)\phantom{\rule{0ex}{0ex}}\Rightarrow 3(-3)-2(-5)\phantom{\rule{0ex}{0ex}}\Rightarrow -9+10=1=\mathrm{R}.\mathrm{H}.\mathrm{S}.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### 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:

$\frac{n}{4}-5=\frac{n}{6}+\frac{1}{2}$

or, $\frac{n}{4}-\frac{n}{6}=\frac{1}{2}+5$ [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]

or, $\frac{3n-2n}{12}=\frac{1+10}{2}$

or, $\frac{n}{12}=\frac{11}{2}$

or, $\frac{n}{12}\times 12=\frac{11}{2}\times 12$ [Dividing both the sides by 12]

or, n = 66

Verification:

Substituting n = 66 on both the sides:

L.H.S.:

$\frac{66}{4}-5=\frac{33}{2}-5=\frac{33-10}{2}=\frac{23}{2}=\frac{23}{2}R.H.S.:\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}=\frac{22+1}{2}=\frac{23}{2}$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 22:

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

$\frac{2m}{3}+8=\frac{m}{2}-1$

#### Answer:

$\frac{2m}{3}+8=\frac{m}{2}-1$

or, $\frac{2m}{3}-\frac{m}{2}=-1-8$ [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]

$or,\frac{4m-3m}{6}=-9\phantom{\rule{0ex}{0ex}}or,\frac{m}{6}=-9$

or, $\frac{m}{6}\times 6=-9\times 6$ [Multiplying both the sides by 6]

or, m = $-$54

Verification:

Substituting x = −54 on both the sides:

$L.H.S.:\phantom{\rule{0ex}{0ex}}\frac{2(-54)}{3}+8=\frac{-54}{2}-1\phantom{\rule{0ex}{0ex}}=\frac{-108}{3}+8\phantom{\rule{0ex}{0ex}}=-36+8\phantom{\rule{0ex}{0ex}}=-28\phantom{\rule{0ex}{0ex}}R.H.S.:\phantom{\rule{0ex}{0ex}}\frac{-54}{2}-1\phantom{\rule{0ex}{0ex}}=-27-1\phantom{\rule{0ex}{0ex}}=-28$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### 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:

$\frac{2x}{5}-\frac{3}{2}=\frac{x}{2}+1$

or, $\frac{2x}{5}-\frac{x}{2}=1+\frac{3}{2}$ [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]

$or,\frac{4x-5x}{10}=\frac{2+3}{2}\phantom{\rule{0ex}{0ex}}or,\frac{-x}{10}=\frac{5}{2}$

$or,\frac{-x}{10}(-10)=\frac{5}{2}\times (-10)$ [Multiplying both the sides by −10]

or, x = −25

Verification:

Substituting x = −25 on both the sides:

$L.H.S.:\frac{2(-25)}{5}-\frac{3}{2}\phantom{\rule{0ex}{0ex}}=\frac{-50}{5}-\frac{3}{2}\phantom{\rule{0ex}{0ex}}=-10-\frac{3}{2}=\frac{-23}{2}\phantom{\rule{0ex}{0ex}}R.H.S.:\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 24:

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

$\frac{x-3}{5}-2=\frac{2x}{5}$

#### Answer:

$\frac{x-3}{5}-2=\frac{2x}{5}$

or, $\frac{x}{5}-\frac{3}{5}-2=\frac{2x}{5}$

or, $-\frac{3}{5}-2=\frac{2x}{5}-\frac{x}{5}$ [Transposing x/5 to the R.H.S.]

or, $\frac{-3-10}{5}=\frac{x}{5}$

or, $\frac{-13}{5}=\frac{x}{5}$

or, $\frac{-13}{5}\left(5\right)=\frac{x}{5}\times \left(5\right)$ [Multiplying both the sides by 5]

or, x = −13

Verification:

Substituting x = −13 on both the sides:

$L.H.S.:\frac{-13-3}{5}-2\phantom{\rule{0ex}{0ex}}=\frac{-16}{5}-2=\frac{-16-10}{5}=\frac{-26}{5}R.H.S.:\frac{2\times (-13)}{5}\mathbf{}\mathbf{=}\mathbf{}\frac{-26}{5}$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 25:

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

$\frac{3x}{10}-4=14$

#### Answer:

$\frac{3x}{10}-4=14$

or, $\frac{3x}{10}-4+4=14+4$ [Adding 4 on both the sides]

or, $\frac{3x}{10}=18$

or, $\frac{3x}{10}\times 10=18\times 10$ [Multiplying both the sides by 10]

or, $3x=180$

or, $\frac{3x}{3}=\frac{180}{3}$ [Dividing both the sides by 3]

or, x = 60

Verification:

Substituting x = 60 on both the sides:

$\frac{3\times 60}{10}-4\phantom{\rule{0ex}{0ex}}=\frac{180}{10}-4=18-4=14=R.H.S.$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 143:

#### Question 26:

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

$\frac{3}{4}(x-1)=x-3$

#### Answer:

$\frac{3}{4}\left(x-1\right)=x-3$

$\Rightarrow \frac{3}{4}\times x-\frac{3}{4}\times 1=x-3$ [On expanding the brackets]

$\Rightarrow \frac{3x}{4}-\frac{3}{4}=x-3$

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

$\Rightarrow \frac{3x-4x}{4}=\frac{-12+3}{4}$

$\Rightarrow \frac{-x}{4}=\frac{-9}{4}$

$\Rightarrow \frac{-x}{4}\times \left(-4\right)=\frac{-9}{4}\times \left(-4\right)$ [Multiplying both the sides by -4]

or, x = 9

Verification:

Substituting x = 9 on both the sides:

$L.H.S.:\frac{3}{4}\left(9-1\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{4}\left(8\right)\phantom{\rule{0ex}{0ex}}=6\phantom{\rule{0ex}{0ex}}R.H.S.:9-3=6$

L.H.S. = R.H.S.

Hence, verified.

#### Page No 144:

#### 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.

#### Page No 144:

#### 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, $\frac{4x}{4}=\frac{100}{4}$ [Dividing both the sides by 4]

or, x = 25

Thus, the required number is 25.

#### Page No 144:

#### 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, $\frac{4x}{4}=\frac{80}{4}$ [Dividing both the sides by 4]

or, x = 20

Thus, the required number is 20.

#### Page No 144:

#### 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, $\frac{3x}{3}=\frac{111}{3}$ [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:

#### 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, $\frac{17x}{17}=\frac{221}{17}$ [Dividing both the sides by 17]

or, x = 13

Thus, the required number is 13.

#### Page No 144:

#### 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, $\frac{3x}{3}=\frac{45}{3}$ [Dividing both the sides by 3]

or, x = 15

Thus, the required number is 15.

#### Page No 144:

#### 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, $\frac{2x}{2}=\frac{74}{2}$ [Dividing both the sides by 2]

or, x = 37

Required numbers are:

x = **37**

or, x + 18 = 37 + 18 = **55**

#### Page No 144:

#### 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, $\frac{4x}{4}=\frac{124}{4}$ [Dividing both the sides by 4]

or, x = 31

Thus, the required number is x = **31** and 3x = 3$\times $31 = **93**.

#### Page No 144:

#### 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, $\frac{4x}{4}=\frac{132}{4}$ [Dividing both the sides by 4]

or, x = 33

Thus, the required numbers are x = **33** and 5x = 5$\times $33 =** 165**.

#### Page No 144:

#### 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, $\frac{2x}{2}=\frac{72}{2}$ [Dividing both the sides by 2]

or, x = 36

Thus, the required numbers are x = **36** and x+ 2 =** 38.**

#### Page No 144:

#### 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, $\frac{3x}{3}=\frac{15}{3}$ [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:

#### 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, $\frac{2x}{2}=\frac{22}{2}$ [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:

#### 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$\times $11

=** 22 years**

#### Page No 144:

#### 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$\times $x + 2$\times $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:

#### 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$\times $x + 2$\times $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, $\frac{2x}{2}=\frac{16}{2}$ [Dividing both the sides by 2]

or, x = 8

∴ Present age of the son = **8 years**

Present age of the man = 4x = 4$\times $8

=** 32 years**

#### Page No 145:

#### 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$\times $x $-$ 4$\times $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$\times $15

=** 45 years**

#### Page No 145:

#### 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, $\frac{2x}{2}=\frac{16}{2}$ [Dividing both the sides by 2]

or, x = 8

∴ Present age of Fatima = 8 years

#### Page No 145:

#### 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$\times $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, $\frac{4x}{4}=\frac{72}{4}$ [Dividing both the sides by 4]

or, x = 18

Thus, the present age of Rahim is 18 years.

#### Page No 145:

#### 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, $\frac{1.5x}{1.5}=\frac{30}{1.5}$ [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$\times $20 = 80

#### Page No 145:

#### 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, $\frac{2x}{2}=\frac{17}{2}$ [Dividing both the sides by 2]

or, x = 8.50

∴ Price of one pen = Rs 8.50

#### Page No 145:

#### 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, $\frac{2x}{2}=\frac{238}{2}$ [Dividing both the sides by 2]

or, x = 119

∴ Number of girls in the school = 119

#### Page No 145:

#### 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 3*x* metres.

Perimeter of the park = 2 (Length + Breadth) = 2 ( 3*x** + x* ) m

Given perimeter = 168 m

∴ 2(3x + x) = 168

or, 2 ( 4x ) = 168

or, 8x = 168 [On expanding the brackets]

or, $\frac{8x}{8}=\frac{168}{8}$ [Dividing both the sides by 8]

or, x = 21 m

∴ Breadth of the park =* x *= **21 m**

Length of the park = 3*x** *= 3$\times $21 = **63 m**

#### Page No 145:

#### 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 ( 2*x** *+ 5 ) = 74

or, 2 ×2*x** *+ 2 ×5 = 74 [On expanding the brackets]

or, 4*x* + 10 = 74

or, 4*x** *+ 10 $-$ 10 = 74 $-$ 10 [Subtracting 10 from both the sides]

or, 4*x** *= 64

or, $\frac{4x}{4}=\frac{64}{4}$ [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:

#### 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, $\frac{4x}{4}=\frac{72}{4}$ [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:

#### 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$\times $*x*)

= Rs 25*x*

#### Page No 146:

#### 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 *$\times $ x*

= Rs 16*x*

Similarly, cost of 1 pencil = Rs 5

∴ Cost of '*y'* pencils = Rs 5$\times $*y*

= Rs 5*y*

∴ Total cost of *x* pens and *y* pencils = Rs (16*x** +* 5*y**)*

#### Page No 146:

#### 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 $\times $x

= Rs 30x

Similarly, Lalit's expenditure per day = Rs y

∴ Lalit's expenditure in 30 days = Rs 30 $\times $ y

= Rs 30y

∴ In 30 days, Lalit saves = (Total earnings $-$ Total expenditure)

= Rs (30x $-$ 30y)

= Rs 30(x - y)

#### Page No 146:

#### 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, $\frac{3x}{3}=\frac{12}{3}\phantom{\rule{0ex}{0ex}}$ [Dividing both the sides by 3]

or, x = 4

∴ Required number = 4

#### Page No 146:

#### Question 5:

If *x* = 1, *y* = 2 and *z* = 3, find the value of *x*^{2} + *y*^{2} + 2*xyz*.

#### Answer:

Given:

x =1

y = 2

z = 3

Substituting *x* = 1, *y* = 2 and *z* = 3 in the given equation (*x*^{2} + *y*^{2} + 2*xyz**)*:

${\left(1\right)}^{2}+{\left(2\right)}^{2}+2\left(1\right)\left(2\right)\left(3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1+4+12=17\frac{}{}$

#### Page No 146:

#### Question 6:

Solve: 4*x* + 9 = 17.

#### Answer:

4x + 9 = 17

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

or, 4x = 8

or, $\frac{4x}{4}=\frac{8}{4}$ [Dividing both the sides with 4]

or, x = 2

#### Page No 146:

#### Question 7:

Solve: 3(*x* + 2) − 2(*x* − 1) = 7.

#### Answer:

3(*x* + 2) − 2(*x* − 1) = 7.

or, $3\times x+3\times 2-2\times x-2\times (-1)=7$ [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

#### Page No 146:

#### Question 8:

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

#### Answer:

$\frac{2x}{5}-\frac{x}{2}=\frac{5}{2}$

or, $\frac{4x-5x}{10}=\frac{5}{2}$ [Taking the L.C.M. as 10]

or, $\frac{-x}{10}=\frac{5}{2}$

or, $\frac{-x}{10}\times \left(-10\right)=\frac{5}{2}\times \left(-10\right)$ [Multiplying both the sides by (−10)]

or, x = −25

#### Page No 146:

#### 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

$\frac{3x}{3}=\frac{48}{3}$ [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:

#### 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, $\frac{2x}{2}=\frac{16}{2}$ [Dividing both the sides by 2]

or, x = 8 years

#### Page No 146:

#### Question 11:

By how much does I exceed 2*x* − 3*y* − 4?

(a) 2*x* − 3*y* − 5

(b) 2*x* − 3*y* − 3

(c) 5 − 2*x* + 3y

(d) none of these

#### Answer:

(c) 5 − 2*x* + 3y

1 exceeds 2*x* − 3*y* − 4.

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

= 5 $-$ 2x + 3y

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

#### Page No 146:

#### Question 12:

What must be added to 5*x*^{3} − 2*x*^{2} + 6*x* + 7 to make the sum *x*^{3} + 3*x*^{2} − *x* + 1?

(a) 4x^{3} − 5*x*^{2} + 7*x* + 6

(b) −4*x*^{3}* + *5*x*^{2} − 7*x* − 6

(c) 4*x*^{3} + 5*x*^{2} − 7*x* + 6

(d) none of these

#### Answer:

(b) −4*x*^{3}* + *5*x*^{2} − 7*x* − 6

In order to find what must be added, we subtract (5*x*^{3} − 2*x*^{2} + 6*x* + 7) from (*x*^{3} + 3*x*^{2} − *x* + 1).

* (**x*^{3} + 3*x*^{2} − *x* + 1) − ( 5*x*^{3} − 2*x*^{2} + 6*x* + 7)

or, *x*^{3} + 3*x*^{2} − *x* + 1 − 5*x*^{3} + 2*x*^{2} − 6x − 7

or, *x*^{3} − 5*x*^{3}+ 3*x*^{2}+ 2*x*^{2}− *x *− 6x+ 1 − 7

or, −4*x*^{3} + 5*x*^{2} − 7x − 6

#### Page No 146:

#### Question 13:

2*x* − [3*y* − {2*x* − (*y* − *x*)}] = ?

(a) 5*x* − 4*y*

(b) 4*y* − 5*x*

(c) 5*y* − 4*x*

(d) 4*x* − 5*y*

#### Answer:

(a) 5*x* − 4*y*

2*x* − [3*y* − {2*x* − (*y* − *x*)}]

= 2x − [3y − {2x − y + x}]

= 2x − [3y − {3x − y}]

= 2x − [3y − 3x + y]

= 2x − [4y − 3x]

= 2x − 4y + 3x

= 5x − 4y

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#### Question 14:

The coefficient of *x* in −5*xyz* is

(a) −5

(b) 5*yz*

(c) −5*yz*

(d) *yz*

#### Answer:

(c) −5*yz*

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

#### Page No 146:

#### Question 15:

$\frac{1}{3}(x+7+z)$ 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.

#### Page No 146:

#### Question 16:

If $\frac{x}{5}=1$, then

(a) $x=\frac{1}{5}$

(b) *x* = 5

(c) *x* = (5 + 1)

(d) none of these

#### Answer:

(b) *x* = 5

$\frac{x}{5}=1\phantom{\rule{0ex}{0ex}}or,\frac{x}{5}\times 5=1\times 5[Multiplyingboththesidesby5]$

or, x = 5

#### Page No 146:

#### Question 17:

If *x* = 1, *y* = 2 and *z* = 3 then (*x*^{2} + *y*^{2} + *z*^{2}) = ?

(a) 6

(b) 12

(c) 14

(d) 15

#### Answer:

(c) 14

*Substituting x* = 1, *y* = 2 and *z* = 3 in (*x*^{2} + *y*^{2} + *z*^{2}):

${\left(1\right)}^{2}+{\left(2\right)}^{2}+{\left(3\right)}^{2}$

or, 1 + 4 + 9 = 14

#### Page No 146:

#### Question 18:

If $\frac{1}{3}x+5=8$, then *x* = ?

(a) 3

(b) 6

(c) 9

(d) 12

#### Answer:

(c) 9

$\frac{1}{3}x+5=8$

or, $\frac{1}{3}x+5-5=8-5$ [Subtracting 5 from both the sides]

or, $\frac{1}{3}x=3$

or, $\frac{1}{3}x\times 3=3\times 3$ [Multiplying both the sides by 3]

or, x = 9

#### Page No 146:

#### 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) $3x-5=7-x\Rightarrow x=.......$

(v) $({b}^{2}-{a}^{2})-({a}^{2}-{b}^{2})=.......$

#### 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, $\frac{4x}{4}=\frac{12}{4}$

or, x = 3

(v) 2b^{2} $-$ 2a^{2}

or, b^{2} $-$ a^{2}^{2}^{2}

or, 2b^{2} $-$ 2a^{2}

( b2 − a2) − (a2 − b2)

#### Page No 147:

#### Question 20:

*Write 'T' for true and 'F' for false for each of the statements given below:*

(i) −3*xy*^{2}*z* is a monomial.

(ii) $x=\frac{2}{3}$ is solution of 2*x* + 5 = 8.

(iii) 2*x* + 3 = 5 is a linear equation.

(iv) The coefficient of *x* in 5*xy* is 5.

(v) $8-x=5\Rightarrow x=3$.

#### 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 5*xy* would be 5*y*.

(v) True

8 − x = 5

or, 8 − 5 = x

or, 3 = x

View NCERT Solutions for all chapters of Class 6