Rs Aggrawal 2019 Solutions for Class 6 Math 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 Math 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 2019 Book of Class 6 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggrawal 2019 Solutions. All Rs Aggrawal 2019 Solutions for class Class 6 Math are prepared by experts and are 100% accurate.

#### 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