Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 4 Simple Equations are provided here with simple step-by-step explanations. These solutions for Simple Equations are extremely popular among class 7 students for Maths Simple Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 7 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 104:

#### Question 1:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The solution of the equation *ax* + *b* = 0 is

(a) $\frac{\mathit{a}}{\mathit{b}}$

(b) –*b*

(c) $\mathit{-}\frac{b}{a}$

(d) $\frac{b}{a}$

#### Answer:

*ax + b *= 0

⇒ *ax *=* –b*

⇒ *x *=* $\frac{-b}{a}$*

Hence, the correct answer is option (c).

#### Page No 104:

#### Question 2:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

If *a* and *b* are positive integers, then the solution of the equation *ax* = *b* will always be a

(a) positive number

(b) negative number

(c) 1

(d) 0

#### Answer:

*a *and *b *are positive integers.

∴ *ax *=* b*

⇒ $x=\frac{b}{a}$

Since, *a*, *b > *0

⇒ $\frac{b}{a}>0$

$\therefore x=\frac{b}{a}\mathrm{is}\mathrm{always}\mathrm{positive}$

Hence, the correct answer is option (a).

#### Page No 104:

#### Question 3:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

Which of the following is not allowed in a given equation?

(a) Adding the same number to both sides of the equation.

(b) Subtracting the same number from both sides of the equation.

(c) Multiplying both sides of the equation by the same non-zero number.

(d) Dividing both sides of the equation by the same number.

#### Answer:

Dividing both sides of a equation by the same non-zero number is possible only.

Hence, the correct answer is option (d).

#### Page No 105:

#### Question 4:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The solution of which of the following equations is neither a fraction nor an integer?

(a) 2*x* + 6 = 0

(b) 3*x* – 5 = 0

(c) 5*x* – 8 = *x* + 4

(d) 4*x* + 7 = *x* + 2

#### Answer:

(a) 2*x +* 6 = 0

⇒ 2*x *= –6

⇒ *x *= –3

(b) 3*x* – 5 = 0

⇒ 3*x *= 5

⇒ *x* = $\frac{5}{3}$

(c) 5*x *– 8 = *x *+ 4

⇒ 5*x *– 8 = 4 + 8

⇒ 4*x *= 12

⇒ *x *= 3

(d) 4*x *+ 7 = *x *+ 2

⇒ 4*x – x *= 2 – 7

⇒ 3*x *= –5

⇒ *x = $-\frac{5}{3}$*

Thus, only $\frac{-5}{3}$ is neither a fraction nor an integer.

Hence, the correct answer is option (d).* *

#### Page No 105:

#### Question 5:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The equation which cannot be solved in integers is

(a) 5*y* – 3 = – 18

(b) 3*x* – 9 = 0

(c) 3*z* + 8 = 3 + *z*

(d) 9*y* + 8 = 4*y* – 7

#### Answer:

(a) 5*y *– 3 = –18

⇒ 5*y *= –18 + 3

⇒ *y = $\frac{\u201315}{5}=\u20133$*

(b) 3*x *– 9 = 0

⇒ 3*x *= 9

⇒ *x *= 3

(c) 3*z *+ 8 = 3 + *z*

⇒ 3*z *–* z *= 3 – 8

⇒ 2*z *= –5

⇒ *z = $\frac{-5}{2}$*

(d) 9*y *+ 8 = 4*y *– 7

⇒ 9*y *– 4*y = –*7* – *8

⇒ 5*y *= –15

⇒ *y = *–3

Thus, only *z = $\frac{-5}{2}$ *is not an integer.

Hence, the correct answer is option (c).

#### Page No 105:

#### Question 6:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

If 7*x* + 4 = 25, then x is equal to

(a) $\frac{29}{7}$

(b) $\frac{100}{7}$

(c) 2

(d) 3

#### Answer:

7*x + *4 = 25

⇒ 7*x = *25 – 4

⇒ 7*x *= 21

⇒ *x** *= 3

Hence, the correct answer is option (d).

#### Page No 105:

#### Question 7:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The solution of the equation 3*x* + 7 = – 20 is

(a) $\frac{17}{7}$

(b) –9

(c) 9

(d)$\frac{13}{3}$

#### Answer:

3*x + *7 = –20

⇒ 3*x *= –20 – 7

⇒ 3*x *= –27

⇒ * x *= –9

Hence, the correct answer is option (b).

#### Page No 105:

#### Question 8:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The value of *y* for which the expressions (*y* – 15) and (2*y* + 1) become equal is

(a) 0

(b) 16

(c) 8

(d) – 16

#### Answer:

*y *– 15 = 2*y* + 1

⇒ *y *– 2*y *= 1 + 15

⇒ –*y *= 16

⇒ *y* *= *–16

Hence, the correct answer is option (d).

#### Page No 105:

#### Question 9:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

If *k* + 7 = 16, then the value of 8*k* – 72 is

(a) 0

(b) 1

(c) 112

(d) 56

#### Answer:

*k *+ 7 = 16

⇒ *k *= 16 – 7

⇒ *k *= 9

∴ 8*k *– 72 = 8 × 9 – 72

= 72 – 72

= 0

â€‹Hence, the correct answer is option (a).

#### Page No 105:

#### Question 10:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

If 43*m* = 0.086, then the value of *m* is

(a) 0.002

(b) 0.02

(c) 0.2

(d) 2

#### Answer:

43*m* = 0.086

⇒ *m* = $\frac{0.086}{43}$

⇒ *m = *0.002

Hence, the correct answer is option (a).

#### Page No 106:

#### Question 11:

**In the given question, there are four options out of which, one is correct. Choose the correct one.***x* exceeds 3 by 7, can be represented as

(a) *x* + 3 = 2

(b) *x* + 7 = 3

(c) *x* – 3 = 7

(d) *x* – 7 = 3

#### Answer:

*x *exceeds 3 can be represented as *x *= 3 + 7

⇒ *x *– 3 = 7

Hence, the correct answer is option (c).

#### Page No 106:

#### Question 12:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The equation having 5 as a solution is:

(a) 4*x* + 1 = 2

(b) 3 – *x* = 8

(c) *x* – 5 = 3

(d) 3 + *x* = 8

#### Answer:

(a) 4*x *+ 1 = 2

⇒ 4*x* = 2 – 1 = 1

⇒ $x=\frac{1}{4}$

(b) 3 – *x* = 8* *

⇒ –*x *= 8 – 3

⇒ –*x *= 5

⇒ *x = *–5

(c) *x *– 5 = 3

⇒ *x *= 5 + 3

⇒ *x = *8

(d) 3 + *x *= 8

⇒ *x *= 8 – 3

⇒ *x *= 5

Thus, equation 3 + *x *= 8 has 5 as a solution.

Hence, the correct answer is option (d).

#### Page No 106:

#### Question 13:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

The equation having – 3 as a solution is:

(a) *x* + 3 =1

(b) 8 + 2*x* = 3

(c) 10 + 3*x* = 1

(d) 2*x* + 1 = 3

#### Answer:

(a) *x* + 3 = 1

⇒ *x *= 1 – 3

⇒ *x *= –2

(b) 8 + 2*x *= 3

⇒ 2*x = *–5

⇒ *x = $\frac{-5}{2}$*

(c) 10 + 3*x *=1

⇒ 3*x = *–9

⇒ *x = *–3

(d) 2*x *+ 1 = 3

⇒ 2*x *= 2

⇒ *x* = 1

Thus, equation 10 + 3*x *= 1 has –3 as a solution.

Hence, the correct answer is option (c).

#### Page No 106:

#### Question 14:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

Which of the following equations can be formed starting with* **x* = 0 ?

(a) 2*x* + 1 = – 1

(b)$\frac{x}{2}$ + 5 = 7

(c) 3*x* – 1 = – 1

(d) 3*x* – 1 = 1

#### Answer:

(a) 2*x *+ 1 = –1

⇒ 2*x *= –1 –1

⇒ *x = *–1

(b) $\frac{x}{2}+5=7$

$\Rightarrow \frac{x}{2}=2\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\times 2=4$

(c) 3*x *– 1 = –1

⇒ 3*x *= 0

⇒ *x *= 0

(d) 3*x *– 1 = 1

⇒ 3*x *= 2

⇒$x=\frac{2}{3}$

Thus, equation 3x – = –1 has 0 as a solution.

Hence, the correct answer is option (c).

#### Page No 106:

#### Question 15:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

Which of the following equations cannot be formed using the equation *x* = 7 ?

(a) 2*x* + 1 =15

(b) 7*x* – 1 = 50

(c) *x* – 3 = 4

(d)$\frac{x}{7}$ – 1 = 0

#### Answer:

(a) 2*x *+ 1 = 15

⇒ 2*x *= 15 – 1 = 14

⇒ $x=\frac{14}{2}=7$

(b) 7*x* – 1 = 50

⇒ 7*x *= 50 + 1 = 51

⇒ *x = $\frac{51}{7}$*

(c) *x *– 3 = 4

⇒ *x* = 3 + 4 = 7

(d) $\frac{x}{7}-1=0$

$\Rightarrow \frac{x}{7}=1\phantom{\rule{0ex}{0ex}}\Rightarrow x=7\times 1=7$

Thus, equation 7*x* – 1 = 50 can not be formed with *x *= 7.

Hence, the correct answer is option (b).

#### Page No 106:

#### Question 16:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

If $\frac{x}{2}$= 3, then the value of 3*x* + 2 is

(a) 20

(b) 11

(c) $\frac{13}{2}$

(d) 8

#### Answer:

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

⇒ *x *= 3 × 2

⇒ *x *= 6

∴ 3*x *+ 2 = 3 × 6 + 2

= 18 + 2

= 20

Hence, the correct answer is option (a).

#### Page No 106:

#### Question 17:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

Which of the following numbers satisfy the equation –6 + *x* = –12 ?

(a) 2

(b) 6

(c) – 6

(d) – 2

#### Answer:

–6 + *x* = –12

⇒ *x* = –12 – (–6)

⇒ *x* = –12 + 6

⇒ *x* = –6

Hence, the correct answer is option (c).

#### Page No 106:

#### Question 18:

**In the given question, there are four options out of which, one is correct. Choose the correct one.**

Shifting one term from one side of an equation to another side with a change of sign is known as

(a) commutativity

(b) transposition

(c) distributivity

(d) associativity

#### Answer:

Shifting one term from one side of an equation to another side with a change of sign is known as transposition.

Hence, the correct answer is option (b).

#### Page No 107:

#### Question 19:

**Fill in the blanks to make the statements true.**

The sum of two numbers is 60 and their difference is 30.

(a) If smaller number is *x*, the other number is ___________.(use sum)

(b) The difference of numbers in term of *x* is _________.

(c) The equation formed is ___________.

(d) The solution of the equation is ____________.

(e) The numbers are _________ and _________.

#### Answer:

(a) 60 – *x*

(b) 60 – 2*x*

(c) 60 – 2*x* = 30

(d) 60 – 2*x *= 30

⇒ – 2*x* = 30 – 60

⇒ 2*x *= 30

⇒ *x *= 15

(e) 60 – *x = *60 – 15 = 45

Numbers are 15 and 45.

#### Page No 107:

#### Question 20:

**Fill in the blanks to make the statements true.**

Sum of two numbers is 81. One is twice the other___________.

(a) If smaller number is *x*, the other number is___________ .

(b) The equation formed is ___________.

(c) The solution of the equation is ____________.

(d) The numbers are ___________ and __________.

#### Answer:

(a) 2*x*

(b) *x *+ 2*x =* 81

⇒ 3*x* = 81

(c) We have,

3*x *= 81

⇒ *x *= 27

Thus, the solution of the equation is *x *= 27.

(d) The two numbers are *x *and 2*x.*

Here, *x *= 27

⇒ 2*x *= 2 × 27

= 54

â€‹Thus, the two numbers are 27 and 54.

#### Page No 107:

#### Question 21:

**Fill in the blanks to make the statements true.**

In a test Abha gets twice the marks as that of Palak. Two times Abha's marks and three times Palak's marks make 280.

(a) If Palak gets *x* marks, Abha gets ___________ marks .

(b) The equation formed is ____________.

(c) The solution of the equation is___________ .

(d) Marks obtained by Abha are ____________.

#### Answer:

(a) 2*x*

(b) 2(2*x*) + 3(*x*) = 280

⇒ 4*x *+ 3*x *= 280

⇒ 7*x *= 280

(c) We have,

7*x *= 280

⇒ *x = $\frac{280}{7}$ *= 40

Thus, the solution of the equation is __ x = 40__.

(d) Marks obtained by Abha are

__80.__

#### Page No 107:

#### Question 22:

**Fill in the blanks to make the statements true.**

The length of a rectangle is two times its breadth. Its perimeter is 60 cm.

(a) If the breadth of rectangle is *x* cm, the length of the rectangle is_____________.

(b) Perimeter in terms of* **x* is ____________.

(c) The equation formed is ____________.

(d) The solution of the equation is _____________.

#### Answer:

(a) 2 × *x = *2*x *cm

(b) Perimeter of rectangle = 2 × (length + breadth)

= 2 × (*x *+ 2*x*)

= 2 × 3*x
= *6

*x*

Thus, perimeter of rectangle in terms of

*x*is 6

*x*cm.

(c) The equation formed is 6

*x*= 60.

(d) We have,

6

*x*= 60

⇒

*x*= 10

Thus, the solution of the equation is

*x*= 10.

#### Page No 108:

#### Question 23:

**Fill in the blanks to make the statements true.**

In a bag there are 5 and 2 rupee coins. If they are equal in number and their worth is â‚¹ 70, then

(a) The worth of *x* coins of â‚¹ 5 each ___________.

(b) The worth of *x* coins of â‚¹ 2 each ___________.

(c) The equation formed is_____________ .

(d) There are__________ 5 rupee coins and___________ 2 rupee coins.

#### Answer:

(a) 5*x*

(b) 2*x*

(c) There are equal number of â‚¹5 and â‚¹2 coins in the bag. So,

5*x + *2*x *= 70

⇒ 7*x *= 70

Thus, the equation formed is __7 x = 70.__

(d) We have,

7

*x*= 70

⇒

*x*= 10

Thus, there are

__10__â‚¹5 coins and

__10__â‚¹2 coins.

Hence, the correct answer is option (d).

#### Page No 108:

#### Question 24:

**Fill in the blanks to make the statements true.**

In a Mathematics quiz, 30 prizes consisting of 1st and 2nd prizes only are to be given. 1st and 2nd prizes are worth â‚¹ 2000 and â‚¹ 1000, respectively. If the total prize money is â‚¹ 52,000 then show that:

(a) If 1st prizes are *x* in number the number of 2nd prizes are ___________.

(b) The total value of prizes in terms of *x* are ____________.

(c) The equation formed is ____________.

(d) The solution of the equation is ______________.

(e) The number of 1st prizes are______________ and the number of 2nd prizes are _______________.

#### Answer:

(a) 30 – *x*

(b) 2000*x + *1000(30 – *x*)

(c) We have,

2000*x + *1000(30 – *x*) = 52000

⇒ 2000*x *+ 30000 – 1000*x *= 52000

⇒ 1000*x* = 22000

(d) We have,

1000*x *= 2000

⇒ *x *= 22

â€‹(e) Thus, the number of 1st prizes are __22__ and the number of 2nd prizes are __8.__

#### Page No 108:

#### Question 25:

**Fill in the blanks to make the statements true.**

If *z* + 3 = 5, then *z* = _____________ .

#### Answer:

We have,*z *+ 3 = 5

Subtracting 3 from both sides, (*z* + 3) – 3 = 5 – 3

⇒ *z *= 2

#### Page No 108:

#### Question 26:

**Fill in the blanks to make the statements true.**

___________is the solution of the equation 3*x* – 2 =7.

#### Answer:

We have,

3*x *– 2 = 7

⇒ 3*x *= 7 + 2

⇒ 3*x *= 9

⇒ *x *= 3

Thus, __ x = 3__ is the solution of equation 3

*x*– 2 = 7.

#### Page No 108:

#### Question 27:

**Fill in the blanks to make the statements true.**

____________is the solution of 3*x* + 10 = 7.

#### Answer:

We have,

3*x *+ 10 = 7

⇒ 3*x* = 7 – 10

⇒ 3*x = *–3

⇒ *x *= –1

Thus, *x* = –1 is the solution of 3*x *+ 10 = 7.

#### Page No 108:

#### Question 28:

**Fill in the blanks to make the statements true.**

If 2*x* + 3 = 5, then value of 3*x* + 2 is ______________.

#### Answer:

We have,

2*x* + 3 = 5

⇒ 2*x* = 5 – 3

⇒ 2*x* = 2

⇒ *x* = 1

Now,

3*x *+ 2 = 3(1) + 2

= 3 + 2

= 5

Thus, the value of 3*x *+ 2 is __5.__

#### Page No 108:

#### Question 29:

**Fill in the blanks to make the statements true.**

In integers, 4*x* – 1 = 8 has ___________solution.

#### Answer:

We have,

4*x – *1 = 8

⇒ 4*x *= 8 + 1

⇒ 4*x* = 9

⇒ *x = $\frac{9}{4},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer}$*

Thus. the equation 4*x *– 1 = 8 has __no__ solution in integers.

#### Page No 109:

#### Question 30:

**Fill in the blanks to make the statements true.**

In natural numbers, 4*x* + 5 = – 7 has ___________solution.

#### Answer:

We have.

4*x* + 5 = –7

⇒ 4*x *= –7 – 5

⇒ 4*x *= –12

⇒ *x *= –3, which is not a natural number

Thus, the equation 4*x + *5 = –7 has __no__ solution in natural numbers.

#### Page No 109:

#### Question 31:

**Fill in the blanks to make the statements true.**

In natural numbers, *x* – 5 = – 5 has ____________solution.

#### Answer:

We have.*x* – 5 = –5

⇒ *x *= –5 + 5

⇒ *x *= 0, which is not a natural number

Thus, the equation *x – *5 = –5 has __no__ solution in natural numbers.

#### Page No 109:

#### Question 32:

**Fill in the blanks to make the statements true.**

In whole numbers, *x* + 8 = 12 – 4 has___________ solution.

#### Answer:

We have.*x* + 8 = 12 – 4

⇒ *x *+ 8* *= 8

⇒ *x *= 8 – 8

⇒ *x *= 0, which is a whole number

Thus, the equation *x *+ 8 = 12 – 4 has __one__ solution in natural numbers.

#### Page No 109:

#### Question 33:

**Fill in the blanks to make the statements true.**

If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as _______.

#### Answer:

Let the number be *x*.

When 5 is added to three times a number, it becomes (3*x *+ 5)

When 7 is subtracted from four times of the number, it becomes (4*x *– 7).

Now,

3*x *+ 5 = 4*x *– 7

This fact can be represented as 3*x *+ 5 = 4*x *– 7.

#### Page No 109:

#### Question 34:

**Fill in the blanks to make the statements true.***x* + 7 = 10 has the solution _____________.

#### Answer:

We have,*x *+ 7 = 10

⇒ *x *= 10 – 7

⇒ *x *= 3

Thus, *x* + 7 = 10 has the solution *x* = 3.

#### Page No 109:

#### Question 35:

**Fill in the blanks to make the statements true.***x* – 0 =_____________ ; when 3*x* = 12.

#### Answer:

We have,

3*x *= 12

⇒ *x *= 4

⇒ *x *– 0 = 4 – 0

⇒ *x *– 0 = 4

Thus, *x *– 0 = __4 __ when 3*x *= 12.

#### Page No 109:

#### Question 36:

**Fill in the blanks to make the statements true.***x* – 1=__________ ; when 2*x* = 2.

#### Answer:

We have,

2*x *= 2

⇒ *x *= 1

⇒ *x *– 1 = 1 – 1

⇒ *x *– 1 = 0

Thus, *x *– 1 = __0 __ when 2*x *= 2.

#### Page No 109:

#### Question 37:

**Fill in the blanks to make the statements true.***x* – _________= 15; when $\frac{x}{2}$ = 6.

#### Answer:

$\frac{x}{2}=6$

⇒ *x *= 2 × 6

⇒ *x *= 12

Since, 12 + 3 = 15

⇒ 12 – (–3) = 15

⇒ *x – *

__(–3)__= 15

#### Page No 109:

#### Question 38:

**Fill in the blanks to make the statements true.**

The solution of the equation *x* + 15 = 19 is ___________.

#### Answer:

*x *+ 15 = 19

⇒ *x *= 19 – 15

⇒ *x *= 4

#### Page No 109:

#### Question 39:

**Fill in the blanks to make the statements true.**

Finding the value of a variable in a linear equation that__________ the equation is called a_________ of the equation.

#### Answer:

Finding the value of a variable in a linear equation that __satisfies__ the equation is called a __root__ of the equation.

#### Page No 109:

#### Question 40:

**Fill in the blanks to make the statements true.**

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the ___________of the term.

#### Answer:

Any term of an equation may be transposed from one side of the equation to the other side of the equation by charging the __sign__ of term.

#### Page No 109:

#### Question 41:

**Fill in the blanks to make the statements true.**

$\mathrm{If}\frac{9}{5}x\mathit{}=\frac{18}{5},\mathrm{then}\mathit{}x=\mathrm{\_\_\_\_\_\_\_\_\_\_\_}.$

#### Answer:

$\frac{9}{5}x\mathit{}=\frac{18}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9}{5}x\xf7\frac{9}{5}=\frac{18}{5}\xf7\frac{9}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{18}{5}\times \frac{5}{9}=2$

#### Page No 109:

#### Question 42:

**Fill in the blanks to make the statements true.**

If 3 – *x* = – 4, then *x* =__________.

#### Answer:

3 – *x *= –4

⇒ –*x = *–4 – 3

⇒ –*x *= –7

⇒ *x *= 7

#### Page No 109:

#### Question 43:

**Fill in the blanks to make the statements true.**

If* x* $-\frac{1}{2}$ = $-\frac{1}{2}$, then *x* = ____________.

#### Answer:

$x-\frac{1}{2}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-1}{2}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=0$

#### Page No 110:

#### Question 44:

**Fill in the blanks to make the statements true.**

If $\frac{1}{6}-$ *x* = $\frac{1}{6}$, then *x = ___________.*

#### Answer:

$\frac{1}{6}-x=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow -x=\frac{1}{6}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow -x=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=0$

#### Page No 110:

#### Question 45:

**Fill in the blanks to make the statements true.**

If 10 less than a number is 65, then the number is _____________.

#### Answer:

Let the number be *x*.

∴ *x* – 10 = 65

⇒ *x* = 65 + 10

⇒ *x* = 75

Thus, the number is 75.

#### Page No 110:

#### Question 46:

**Fill in the blanks to make the statements true.**

If a number is increased by 20, it becomes 45. Then the number is ________.

#### Answer:

Let the number be *p*.

∴ *p *+ 20 = 45

⇒ *p *= 45 – 20

⇒ *p *= 25

Thus, the number is 25.

#### Page No 110:

#### Question 47:

**Fill in the blanks to make the statements true.**

If 84 exceeds another number by 12, then the other number is _________.

#### Answer:

Let the number is 25.

∴ 84 – *x *= 12

⇒ –*x* = 12 – 84

⇒ –*x *= –72

⇒ *x *= 72

Thus, the number is 72.

#### Page No 110:

#### Question 48:

**Fill in the blanks to make the statements true.**

If *x *− $\frac{7}{8}=\frac{7}{8}$, then *x* = __________.

#### Answer:

$x-\frac{7}{8}=\frac{7}{8}$

⇒ $x=\frac{7}{8}+\frac{7}{8}$

⇒ $x=\frac{7+7}{8}$

⇒ $x=\frac{14}{8}$

⇒ $\frac{7}{4}$

#### Page No 110:

#### Question 49:

**State whether the statements are True or False.**

5 is the solution of the equation 3*x* + 2 = 17.

#### Answer:

3*x* + 2 = 17

⇒ 3*x *= 17 – 2 = 15

⇒ $\frac{3x}{3}=\frac{15}{3}$

⇒ *x *= 5

Thus, 5 is the solution of 3*x* + 2 = 17.

#### Page No 110:

#### Question 50:

**State whether the statements are True or False.**

$\frac{9}{5}$is the solution of the equation 4*x* – 1 = 8.

#### Answer:

4*x* – 1 = 8

⇒ 4*x *= 8 + 1 = 9

⇒ $\frac{4x}{4}=\frac{9}{4}$

⇒ $\frac{9}{4}$

Thus, the statement is false.

#### Page No 110:

#### Question 51:

**State whether the statements are True or False.**

4*x* – 5 = 7 does not have an integer as its solution.

#### Answer:

4*x* – 5 = 7

⇒ 4*x *= 7 + 5 = 12

⇒ $\frac{4x}{4}=\frac{12}{4}$

⇒ *x *= 3, is an integer

Thus, the statement is false.

#### Page No 110:

#### Question 52:

**State whether the statements are True or False.**

One third of a number added to itself gives 10, can be represented as $\frac{x}{3}+10$ x.

#### Answer:

Let the number be *x*.

∴ One-third of the number = $\frac{x}{3}$

As per the question,

$\frac{x}{3}+x=10$

Thus, the statement is false.

#### Page No 110:

#### Question 53:

**State whether the statements are True or False.**

$\frac{3}{2}$is the solution of the equation 8*x* – 5 = 7.

#### Answer:

8*x* – 5 = 7

⇒ 8*x = *7 + 5 = 12

⇒ $\frac{8x}{8}=\frac{12}{8}$

⇒ $x=\frac{3}{2}$

Thus, the statement is true.

#### Page No 110:

#### Question 54:

**State whether the statements are True or False.**

If 4*x* – 7 = 11, then *x* = 4.

#### Answer:

4*x *– 7 = 11

⇒ 4*x *= 11 + 7 = 18

⇒ $\frac{4x}{4}=\frac{18}{4}$

⇒ $x=\frac{9}{2}$

Thus, the statement is false.

#### Page No 110:

#### Question 55:

**State whether the statements are True or False.**

If 9 is the solution of variable *x* in the equation $\frac{5x-7}{2}=y$ , then the value of *y* is 28

#### Answer:

$\frac{5x-7}{2}=y\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5\times 9\u20137}{2}=y\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{45\u20137}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{38}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=19$

Thus, the statement is false.

#### Page No 111:

#### Question 56:

Match each of the entries in Column I with the appropriate entries in Column II.**Column I Column II**

(i) *x* + 5 = 9 $\left(\mathrm{A}\right)-\frac{5}{3}$

(ii) *x* – 7 = 4 $\left(\mathrm{B}\right)\frac{5}{3}$

$\left(\mathrm{iii}\right)\frac{x}{12}=-5$ (C) 4

(iv) 5*x* = 30 (D) 6

(v) The value of *y* which satisfies 3*y* = 5 (E) 11

(vi) If *p* = 2, then the value of $\frac{1}{3}$(1 – 3p) (F) – 60

(G) 3

#### Answer:

(i) *x* + 5 = 9

⇒ *x *= 9 – 5

⇒ *x *= 4 (C)

(ii) *x *– 7 = 4

⇒ *x *= 7 + 4

⇒ *x *= 11 (E)

(iii) $\frac{x}{12}=\u20135$

⇒ *x *= I2 × (–5)

⇒ *x = *–60 (F)

(iv) 5*x *= 30

⇒ $\frac{5x}{5}=\frac{30}{5}$

⇒ *x* = 6 (D)

(v) 3*y *= 5

⇒ $\frac{3y}{3}=\frac{5}{3}$

⇒ $y=\frac{5}{3}$ (B)

(vi) $\frac{1}{3}\left(1-3p\right)=\frac{1}{3}\left(1-3\times 2\right)$

$=\frac{1}{3}\left(1-6\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\times \left(-5\right)\phantom{\rule{0ex}{0ex}}=\frac{-5}{3}\left(\mathrm{A}\right)$

#### Page No 112:

#### Question 57:

**Express the given statement as an equation.**

13 subtracted from twice of a number gives 3 .

#### Answer:

Let the number be *x*.

∴ 2*x *– 13 = 3

Thus, the given statement can be written in equation as 2*x *– 13 = 3.

#### Page No 112:

#### Question 58:

**Express the given statement as an equation.**

One-fifth of a number is 5 less than that number.

#### Answer:

Let the number be x.

So, one-fifth of the number = $\frac{x}{5}$

Therefore, $\frac{x}{5}=x-5$ is the required equation.

#### Page No 112:

#### Question 59:

**Express the given statement as an equation.**

A number is 7 more than one-third of itself.

#### Answer:

Let the number be x.

So, one-third of number = $\frac{x}{3}$

7 more than $\frac{x}{3}=\frac{x}{3}+7$

Therefore, $x=\frac{x}{3}+7$ is the required equation

#### Page No 112:

#### Question 60:

**Express the given statement as an equation.**

Six times a number is 10 more than the number.

#### Answer:

Let the number be x.

So, six times of the number = 6*x*

Therefore, 6*x* – *x* + 10 is the required equation.

#### Page No 112:

#### Question 61:

**Express the given statement as an equation.**

If 10 is subtracted from half of a number, the result is 4.

#### Answer:

Let the number be x.

So, half of the number $=\frac{x}{2}$

On subtracting 10 from it, we get $\frac{x}{2}-10$

Therefore , $\frac{x}{2}-10=4$ is the required equation.

#### Page No 112:

#### Question 62:

**Express the given statement as an equation.**

Subtracting 5 from *p*, the result is 2.

#### Answer:

The number is *p*.

On subtracting 5 from it, we get *p* – 5.

∴ *p* – 5 = 2 is the required equation.

#### Page No 112:

#### Question 63:

**Express the given statement as an equation.**

Five times a number increased by 7 is 27.

#### Answer:

Let the number be *x*.

So, five times of the number = 5*x*

When increased by 7, it gives the expression 5*x* +7

∴ 5*x* + 7 = 27 is the required equation.

#### Page No 112:

#### Question 64:

**Express the given statement as an equation.**

Mohan is 3 years older than Sohan. The sum of their ages is 43 years.

#### Answer:

Let Sohan is *x* years old.

So, Mohan is *x* + 3 years old.

∴ Sum of their ages be *x* + (*x* + 3).

∴ *x* + (*x* + 3) = 43 is the required equation.

#### Page No 112:

#### Question 65:

**Express the given statement as an equation.**

If 1 is subtracted from a number and the difference is multiplied by $\frac{1}{2}$, the result is 7.

#### Answer:

Let the number be *x*.

On subtracting 1 from it, we get *x* – 1.

Multiplying it by $\frac{1}{2}$ we get $\frac{1}{2}\left(x-1\right)$

∴ $\frac{1}{2}\left(x-1\right)=7$ is the required equation.

#### Page No 112:

#### Question 66:

**Express the given statement as an equation.**

A number divided by 2 and then increased by 5 is 9.

#### Answer:

Let the number be *x*.

Dividing the number by 2, we get $\frac{x}{2}$.

When, increased by 5, it gives the expression $\frac{x}{2}+5$

∴ $\frac{x}{2}+5=9$ is the required equation.

#### Page No 112:

#### Question 67:

**Express the given statement as an equation.**

The sum of twice a number and 4 is 18.

#### Answer:

Let the number be *x*.

So, twice of the number = 2*x*

On adding 4 to it, we get 2*x* + 4

∴ 2*x* + 4 = 18 is the required equation.

#### Page No 112:

#### Question 68:

The age of Sohan Lal is four times that of his son Amit. If the difference of their ages is 27 years, find the age of Amit.

#### Answer:

Let the age of Amit be *x* years.

So, age of Sohan Lal = 4*x* years

According to question, 4*x* – *x* = 27

$\Rightarrow 3x=27\phantom{\rule{0ex}{0ex}}\Rightarrow x=9$

Thus, Amit is 9 years old.

#### Page No 112:

#### Question 69:

A number exceeds the other number by 12. If their sum is 72, find the numbers.

#### Answer:

Let the number be *x*.

∴ Other number = *x* +12

According to question,*x* + *x* + 12 = 72

⇒ 2*x* + 12 = 72

⇒ 2*x* = 72 – 12 = 60

⇒ *x* = 30

∴ *x* + 12 = 30 + 12 = 42

Hence, the numbers are 30 and 42

#### Page No 112:

#### Question 70:

Seven times a number is 12 less than thirteen times the same number. Find the number.

#### Answer:

Let the number be *x*.

So, seven times of the number = 7*x*

Thirteen times of the number = 13*x*

According to question,

7*x* = 13*x* – 12

⇒ 12 = 13*x* – 7*x*

⇒ 6*x* = 12

⇒ *x* = 2

Thus, 2 is the required number.

#### Page No 112:

#### Question 71:

The interest received by Karim is â‚¹ 30 more than that of Ramesh. If the total interest received by them is â‚¹ 70, find the interest received by Ramesh.

#### Answer:

Let the interest received by Ramesh be *x*.

So, interest received by Karim = â‚¹ (30+ *x*)

According to question,*x* + *x* + 30 = 70

⇒ 2*x* = 70 – 30 = 40

⇒ *x* = 20

Thus, â‚¹ 20 is the interest received by Ramesh.

#### Page No 112:

#### Question 72:

Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is â‚¹ 125 more than that of Subramaniam. If

the total money paid by them is â‚¹ 975, find the amount of money donated by Subramaniam.

#### Answer:

Let the amount of money donated by Subramaniam be *x*.

So, the amount paid by Naidu is (*x* + 125).

According to question,*x* + *x* + 125 = 975

⇒ 2*x* = 975 – 125 = 850

⇒ *x* = 425

Thus, â‚¹425 is donated by Subramaniam.

#### Page No 113:

#### Question 73:

In a school, the number of girls is 50 more than the number of boys. The total number of students is 1070. Find the number of girls.

#### Answer:

Let the number of girls be *x*.

So, the number of boys = *x* – 50

According to question,*x* + *x* – 50 = 1070

⇒ 2*x* = 1070 + 50 = 1120

⇒ *x* = 560

Thus, 560 are girls.

#### Page No 113:

#### Question 74:

Two times a number increased by 5 equals 9. Find the number.

#### Answer:

Let the number be *x*.

So, two times of the number = 2*x*

When, increased by 5, it gives the expression 2*x* + 5

∴ 2*x* + 5 = 9

⇒ 2*x* = 9 – 5 = 4

⇒ *x* = 2

Thus, *x* = 2 is the required number.

#### Page No 113:

#### Question 75:

9 added to twice a number gives 13. Find the number.

#### Answer:

Let the number be *x*.

So, twice of the number = 2*x*

On adding 9 to it, we get 2*x* + 9

∴ 2*x* + 9 = 13

⇒ 2*x* = 13 – 9 = 4

⇒ *x* = 2

Thus, *x* = 2 is the required number.

#### Page No 113:

#### Question 76:

1 subtracted from one-third of a number gives 1. Find the number.

#### Answer:

Let the number be *x*.

So, one third of the number $\frac{x}{3}$.

On subtracting 1 from it, we get $\frac{x}{3}-1$

$\frac{x}{3}-1=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{3}=1+1=2\phantom{\rule{0ex}{0ex}}\Rightarrow x=6$

Thus, *x* = 6 is the required number.

#### Page No 113:

#### Question 77:

After 25 years, Rama will be 5 times as old as he is now. Find his present age.

#### Answer:

Let present age of Rama be *x* years.

So, five times of his age = 5*x*.

According to question,

5*x* = *x* + 25

⇒ 5*x* – *x* = 25

⇒ 4*x* = 25

⇒ *x* = $\frac{25}{4}$

Thus, at present Rama is $\frac{25}{4}$ years old.

#### Page No 113:

#### Question 78:

After 20 years, Manoj will be 5 times as old as he is now. Find his present age.

#### Answer:

Let present age of Manoj be *x* years.

So, five times of his age = 5*x*

According to question,

5 = *x* + 20

⇒ 5*x* – *x* = 20

⇒ 4*x* = 20

⇒ *x* = 5

Thus, at present Manoj is 5 years old.

#### Page No 113:

#### Question 79:

My younger sister's age today is 3 times, what it will be 3 years from now minus 3 times what her age was 3 years ago. Find her present age.

#### Answer:

Let present age of my younger sister be *x* years.

After 3 years, her age will be = (*x* + 3) years

Before 3 years, her age was = (*x* – 3) years

According to question,*x* = 3 (*x* + 3) – 3 (*x* – 3)

⇒ *x* = 3 (*x* + 3 – *x* + 3)

⇒ *x* = 3(6) = 18

Thus, at present my younger sister is 18 years old.

#### Page No 113:

#### Question 80:

If 45 is added to half a number, the result is triple the number. Find the number.

#### Answer:

Let the number be *x*.

So, half of the number $\frac{x}{2}$.

On adding 45 to it, we get $\frac{x}{2}+45$.

$\frac{x}{2}+45=3x\phantom{\rule{0ex}{0ex}}3x-\frac{x}{2}=45\phantom{\rule{0ex}{0ex}}\frac{6x-x}{2}=45\phantom{\rule{0ex}{0ex}}\frac{5x}{2}=45\phantom{\rule{0ex}{0ex}}x=18$

Thus, the number is 18.

#### Page No 113:

#### Question 81:

In a family, the consumption of wheat is 4 times that of rice. The total consumption of the two cereals is 80 kg. Find the quantities of rice and wheat consumed in the family.

#### Answer:

Let the quantity of rice consumed in the family be *x* kg.

So, quantity of wheat consumed = 4*x* kg

According to question, *x* + 4*x* = 80

⇒ 5*x* = 80

⇒ *x* = 16

Thus, 16 kg rice consumed in the family and 4 × 16 = 64 kg wheat consumed in the family.

#### Page No 113:

#### Question 82:

In a bag, the number of one rupee coins is three times the number of two rupees coins. If the worth of the coins is â‚¹ 120, find the number of 1 rupee coins.

#### Answer:

Let two rupees coins are *x* in numbers.

So, one rupee coins are 3*x* in numbers.

According to question, 1(3*x*) + 2(*x*) = 120

⇒ 3*x* + 2*x* = 120

⇒ 5*x* = 120

⇒ *x* = 24

Thus, 3 × 24 = 72 coins are of one rupee coins.

#### Page No 113:

#### Question 83:

Anamika thought of a number. She multiplied it by 2, added 5 to the product and obtained 17 as the result. What is the number she had thought of ?

#### Answer:

Let the number Anamika thought be *x*.

Multiplying it by 2, we get 2*x*

On adding 5 to it, we get 2*x* + 5

∴ 2*x* + 5 = 17

⇒ 2*x* = 17 – 5 = 12

⇒ *x* = 6

Thus, she had thought of number 6.

#### Page No 113:

#### Question 84:

One of the two numbers is twice the other. The sum of the numbers is 12. Find the numbers.

#### Answer:

Let the one number be *x*.

So, other number = 2*x*

According to question,*x* + 2*x* = 12

⇒ 3*x* = 12

⇒ *x* = 4

∴ Other number is 2*x* = 2 × 4 = 8.

#### Page No 113:

#### Question 85:

The sum of three consecutive integers is 5 more than the smallest of the integers. Find the integers.

#### Answer:

Let the smallest integer be *x*

So, next two consecutive integer would be *x* + 1 and *x* + 2 respectively.

According to question,*x* + *x* + 1 + *x* + 2 = *x* + 5

⇒ 3*x* + 3 = *x* + 5

⇒ 3*x* – *x* = 5 – 3

⇒ 2*x* = 2

⇒ *x* = 1

Thus, next two consecutive integers are 2 and 3 respectively.

#### Page No 113:

#### Question 86:

A number when divided by 6 gives the quotient 6. What is the number?

#### Answer:

Let the number be *x*.

Dividing it by 6, we get $\frac{x}{6}$.

$\frac{x}{6}=6\phantom{\rule{0ex}{0ex}}x=6\times 6\phantom{\rule{0ex}{0ex}}x=36$

Thus, *x* = 36 is the required number.

#### Page No 113:

#### Question 87:

The perimeter of a rectangle is 40m. The length of the rectangle is 4 m less than 5 times its breadth. Find the length of the rectangle.

#### Answer:

Let the breadth of rectangle be *x* m.

So, five times of breadth = 5*x*

Therefore, length of rectangle = (5*x* – 4) m

Perimeter of rectangle = 40 m

⇒ 2(*x* + 5*x* – 4) = 40

⇒ 2(6*x* – 4) = 40

⇒ 6*x* – 4 = 20

⇒ 6*x* = 20 + 4 = 24

⇒ *x* = 4

Thus, length of rectangle = (5 × 4 – 4) = 16 m

#### Page No 114:

#### Question 88:

Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

#### Answer:

Let length of equal sides of an isosceles triangle be 2*x* cm.

So, the third side of triangle *x* cm.

Perimeter of triangle = 30 cm

⇒ *x* + 2*x* + 2*x* = 30

⇒ 5*x* = 30

⇒ *x* = 6

Thus, 6 cm, 12 cm and 12 cm are the required sides of the triangle.

#### Page No 114:

#### Question 89:

The sum of two consecutive multiples of 2 is 18. Find the numbers.

#### Answer:

Let first multiple of 2 be *x*.

So, next multiple of 2 = *x* + 2

According to question,*x* + *x* + 2 = 18

⇒ 2*x* = 18 – 2 = 16

⇒ *x* = 8

Thus, 8 and 10 are required numbers.

#### Page No 114:

#### Question 90:

Two complementary angles differ by 20°. Find the angles.

#### Answer:

Let one angle be *x*.

So, complement of *x* = 90°– *x*

According to question,*x* – (90°– *x*) = 20°

⇒ *x* – 90° + *x* = 20°

⇒ 2*x* = 20° + 90° = 110°

⇒ *x* = 55°

∴ Complement of *x* = 90°– *x* = 90°– 55° = 35°

Thus, 55° and 35° are required complementary angles.

#### Page No 114:

#### Question 91:

150 has been divided into two parts such that twice the first part is equal to the second part. Find the parts.

#### Answer:

Let first part be *x*.

So, other part = 150 – *x*

According to question, 2*x* = 150 – *x*

⇒ 2*x* + *x* = 150

⇒ 3*x* = 150

⇒ *x* = 50

Thus, other part = 150 – *x* = 150 – 50 = 100

Hence, 150 has been divided into 50 and 100.

#### Page No 114:

#### Question 92:

In a class of 60 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.

#### Answer:

Let the number of boys in class = *x*

So, the number of girls in class = 60 – *x*

According to question,

$60-x=\frac{x}{3}$

⇒ 180 – 3*x* = *x*

⇒ 180 = 3*x* + *x*

⇒ 4*x* = 180

⇒ *x* = 45

Thus, number of boys in the class = 45

And number of girls in the class = 60 – 45 = 15

#### Page No 114:

#### Question 93:

Two-third of a number is greater than one-third of the number by 3. Find the number.

#### Answer:

Let the number be *x*.

So, two-third of numbers $\frac{2}{3}x$.

According to question,

$\frac{2}{3}x=\frac{1}{3}x+3\phantom{\rule{0ex}{0ex}}\frac{2x}{3}-\frac{x}{3}=3\phantom{\rule{0ex}{0ex}}\frac{2x-x}{3}=3\phantom{\rule{0ex}{0ex}}\frac{x}{3}=3\phantom{\rule{0ex}{0ex}}x=9$

Thus, *x* = 9 is the required number.

#### Page No 114:

#### Question 94:

A number is as much greater than 27 as it is less than 73. Find the number.

#### Answer:

Let the number be *x*.

According to question,*x* – 27 = 73 – *x*

⇒ *x* + *x* = 73 + 27

⇒ 2*x* = 100

⇒ *x* = 50

Thus, *x* = 50 is the required number.

#### Page No 114:

#### Question 95:

A man travelled two fifth of his journey by train, one-third by bus, one-fourth by car and the remaining 3 km on foot. What is the length of his total journey?

#### Answer:

Let the length of total journey be *x* km.

∴ Journey by train = *x* km

∴ Joumey by bus = $\frac{2}{5}xkm$

∴ Journey by car = $\frac{x}{3}$ km

And journey on foot = $\frac{x}{4}$ km

$x=\frac{2x}{5}+\frac{x}{3}+\frac{x}{4}+3\phantom{\rule{0ex}{0ex}}x=\frac{24x+20x+15x}{60}+3\phantom{\rule{0ex}{0ex}}x=\frac{59x}{60}+3\phantom{\rule{0ex}{0ex}}x-\frac{59x}{60}=3\phantom{\rule{0ex}{0ex}}\frac{x}{60}=3\phantom{\rule{0ex}{0ex}}x=3\times 60\phantom{\rule{0ex}{0ex}}x=180$

Thus, *x* = 180 km is the length of total journey.

#### Page No 114:

#### Question 96:

Twice a number added to half of itself equals 24. Find the number.

#### Answer:

Let the number be *x*.

According to question,

$2x+\frac{x}{2}=24\phantom{\rule{0ex}{0ex}}\frac{4x+x}{2}=24\phantom{\rule{0ex}{0ex}}5x=2\times 24=48\phantom{\rule{0ex}{0ex}}x=\frac{48}{5}\phantom{\rule{0ex}{0ex}}x=9.6$

Thus, *x* = 9.6 is the required number.

#### Page No 114:

#### Question 97:

Thrice a number decreased by 5 exceeds twice the number by 1. Find the number.

#### Answer:

Let the number be *x*.

So, thrice of the number = 3*x*

When it decreased by 5, we get 3*x* – 5

According to question,

3*x* – 5 = 2*x* + 1

⇒ 3*x* – 2*x* = 1 + 5

⇒ *x* = 6

Thus, *x* = 6 is the required number.

#### Page No 114:

#### Question 98:

A girl is 28 years younger than her father. The sum of their ages is 50 years. Find the ages of the girl and her father.

#### Answer:

Let the age of father be *x* years.

So, age of his girl = (*x* – 28) years

According to question,*x* + *x* – 28 = 50

⇒ 2*x* = 50 + 28 = 78

⇒ *x* = 39

Thus, age of father = 39 years

And age of his girl = 11 years

#### Page No 114:

#### Question 99:

The length of a rectangle is two times its width. The perimeter of the rectangle is 180 cm. Find the dimensions of the rectangle.

#### Answer:

Let the width of rectangle be *x* cm.

So, the length of rectangle = 2*x*

Perimeter of rectangle = 180 cm

⇒ 2(2*x* + *x*) = 180

⇒ 6*x* = 180

⇒ *x* = 30

Thus, width of rectangle = 30 cm

And length of rectangle = 2 × 30 = 60 cm

#### Page No 114:

#### Question 100:

Look at this riddle?

If she answers the riddle correctly how ever will she pay for the pencils?

#### Answer:

Let the cost of one pencil be â‚¹ *x*

So, the cost of 5 pencils = â‚¹ 5*x*

Cost of 7 pencils = â‚¹ 7*x*

According to question,

7*x* = 5*x* + 6

⇒ 7*x* – 5*x* = 6

⇒ 2*x* = 6

⇒ *x* = 3

Cost of 10 pencils = â‚¹ (3 × 10) = â‚¹ 30

#### Page No 115:

#### Question 101:

In a certain examination, a total of 3768 students secured first division in the years 2006 and 2007. The number of first division in 2007

exceeded those in 2006 by 34. How many students got first division in 2006?

#### Answer:

Let number of students got first division in 2006 be *x*.

So, the number of students got first division in 2007 = 3768 – *x*.

According to question,

3768 – *x* = *x* + 34

⇒ 3768 – 34 = *x* + *x*

⇒ 2*x* = 3734

⇒ *x* = 1867

Thus, 1867 students got first division in 2006.

#### Page No 115:

#### Question 102:

Radha got â‚¹ 17,480 as her monthly salary and over-time. Her salary exceeds the over-time by â‚¹ 10,000. What is her monthly salary ?

#### Answer:

Let Radha’s monthly salary = â‚¹ *x*

So, money got by her in over time = â‚¹ (17480 – *x*)

According to question,*x* = 17480 – *x* + 10000

⇒ *x* + *x* = 17480 + 10000

⇒ 2*x* = 27480

⇒ *x* = 13740

Thus, â‚¹13740 is her monthly salary.

#### Page No 115:

#### Question 103:

If one side of a square is represented by 18*x* – 20 and the adjacent side is represented by 42 – 13*x*, find the length of the side of the square.

#### Answer:

Since, square has all sides equal

∴ 18*x* – 20 = 42 – 13*x*

⇒ 18*x* + 13*x* = 42 + 20

⇒ 31*x *= 62

⇒ *x *= 2

∴ Side of square = 18 × 2 – 20 = 36 – 20 = 16

Thus, length of the side of the square is 16 units.

#### Page No 115:

#### Question 104:

Follow the directions and correct the given incorrect equation, written in Roman numerals:

(a) Remove two of these matchsticks to make a valid equation:

(b) Move one matchstick to make the equation valid. Find two different solutions.

graphics needs to change

#### Answer:

(a) Incorrect equation is

After removing two matchsticks the correct equation is

GRAPHIC

(b) Incorrect equation is

GRAPHIC

After removing one matchstick the correct equation is

GRAPHIC

#### Page No 115:

#### Question 105:

What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below:

If i + 69 = 70, then i = ? If 8*u* = 6*u* + 8, then *u* =?

If 4*a* = –5*a* + 45, then *a* = ? if 4*q* + 5 = 17, then *q* =?

If –5t – 60 = – 70, then t = $\frac{1}{4}s$ + 98 = 100, then *s* =?

If $\frac{5}{3}$*p* + 9 = 24, then *p* =_____?

If 3*c* = *c* +12, then *c* =_____?

If 3 (*k* + 1) = 24, then *k* =_____?

For riddle answer : substitute the number for the letter it equals

#### Answer:

i + 69 = 70

⇒ i = 70 – 69 = 1

8u = 6u +8

⇒ 8u – 6u = 8

⇒ 2u = 8

⇒ u = 4

4a = –5a + 45

⇒ 4a + 5a = 45

⇒ 9a – 45

⇒ a = 5

4q + 5 = 17

⇒ 4q = 17 – 5

⇒ 4q = 12

⇒ q = 3

–5t – 60 = – 70

⇒ –5t = –70 + 60

⇒ –5t = –10

⇒ t = 2

$\frac{1}{4}s+98=100\phantom{\rule{0ex}{0ex}}\frac{1}{4}s=2\phantom{\rule{0ex}{0ex}}s=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{5}{3}p+9=24\phantom{\rule{0ex}{0ex}}\frac{5}{3}p=15\phantom{\rule{0ex}{0ex}}p=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3c=c+12\phantom{\rule{0ex}{0ex}}3c-c=12\phantom{\rule{0ex}{0ex}}2c=12\phantom{\rule{0ex}{0ex}}c=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3(k+1)=24\phantom{\rule{0ex}{0ex}}k+1=8\phantom{\rule{0ex}{0ex}}k=7$

Thus, 1 = i, 2 = t, 3 = q, 4 = u, 5 = a, 6 = c, 7 = k, 8 = s, 9 = p.

#### Page No 116:

#### Question 106:

The three scales below are perfectly balanced if • = 3. What are the values of âˆ† and * ?

#### Answer:

Given that • = 3.

$\left(a\right)5*=2\u2206=2\u2022\phantom{\rule{0ex}{0ex}}\left(b\right)2\u2206=2*+2\u2022\phantom{\rule{0ex}{0ex}}\left(c\right)3*+3\u2022=3\u2206$

Solving equations, we get

$5*=2*+2\u2022+2\u2022\phantom{\rule{0ex}{0ex}}5*-2*=4\u2022\phantom{\rule{0ex}{0ex}}3*=4\times 3\phantom{\rule{0ex}{0ex}}*=4$

$\begin{array}{rcl}3\u2206& =& 3\times 4+3\times 3\\ & =& 12+9\\ & =& 21\\ \u2206& =& 7\end{array}$

#### Page No 116:

#### Question 107:

The given figure represents a weighing balance. The weights of some objects in the balance are given. Find the weight of each square and

the circle.

#### Answer:

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