**NCERT Solutions for Class 7 Maths**

Chapter 4 Simple Equations

Exercise 4.4

Chapter 4 Simple Equations

Exercise 4.4

1. Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

**Answer**

**(a) Add 4 to eight times a number; you get 60.
**Let the number be $x$.

$\therefore$ Eight times of the number $=8 x$

Now, add 4 to eight times a number, we get $8 x+4$

According to the question,

$8 x+4=60$

which is the required equation.

Now, we find the value of unknown number i.e. the value of $x$

We have,

$8 x+4=60$

On transposing $(+4)$ from LHS to RHS, we get

$8 x=60-4 \Rightarrow 8 x=56$

On dividing both sides by 8 , we get

$\frac{8 x}{8}=\frac{56}{8} \Rightarrow x=7$

Hence, the required number is 7

**(b) One-fifth of a number minus 4 gives 3.
**Let the number be $x$

$\therefore$ One-fifth of the number $=\frac{x}{5}$

According to the question,

One-fifth of a number minus $4=3$

i.e. $ \frac{x}{5}-4=3$

which is the required equation.

Now, we find the value of unknown number i.e. the value of $x$ On transposing $(-4)$ from LHS to RHS,

we get $\frac{x}{5}=3+4 \Rightarrow \frac{x}{5}=7$

On multiplying both sides by 5 , we get

$\frac{x}{5} \times 5=7 \times 5 \Rightarrow x=35$

Hence, the required number is 35 .

**(c) If I take three-fourths of a number and add 3 to it, I get 21.
**Let the number be $x$ :

$\therefore$ Three-fourths of the number $=\frac{3}{4} x$

According to the question,

On adding 3 to it, we get 21 .

i.e. $ \frac{3}{4} x+3=21$

which is the required equation.

Now, to solve this equation, transposing $(+3)$ from LHS to RHS, we get

$\frac{3}{4} x=21-3 \Rightarrow \frac{3}{4} x=18$

On multiplying both sides by 4 , we get

$\frac{3}{4} x \times 4=18 \times 4 \Rightarrow 3 x=72$

Again, dividing both sides by 3 , we get

$\frac{3 x}{3}=\frac{72}{3} \Rightarrow x=24$

**(d) When I subtracted 11 from twice a number, the result was 15.
**(d) Let the number be $x$.

$\therefore$ Twice of the number $=2 x$

According to the question,

On subtracting 11 from twice the number, we get 15 .

i.e. $2 x-11=15$

Now, to solve this equation, transposing $(-11)$ from LHS to RHS, we get

$2 x=15+11 \Rightarrow 2 x=26$

Again, dividing both sides by 2 , we get

$\frac{2 x}{2}=\frac{26}{2} \Rightarrow x=13$

Hence, the required number is 13 .

**(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
**(e) Let the number of notebooks =x

$\therefore$ Thrice of the number $=3 x$

According to the question,

On subtracting thrice the number of notebooks from 50 , we get 8

i.e.$50-3 x=8$

Now, to solve this equation, transposing 50 from LHS to RHS, we get

$-8 x=8-50 \Rightarrow-3 x=-42$

Again, dividing both sides by $(-3)$, we get

$\frac{-3 x}{-3}=\frac{-42}{-3} \Rightarrow x=14$

Hence, the required number of notebooks is 14 .

**(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
**(f) Let Ibenhal thinks the number $=x$

On adding 19 to x, we get, x+19

On dividing the sum by 5 , we get $\frac{x+19}{5}$

According to the question, $\frac{x+19}{5}=8$

Now, to solve this equation, multiplying both sides by 5 , we get

$\frac{x+19}{5} \times 5=8 \times 5 \Rightarrow x+19=40$

Again, transposing $+19$ from LHS to RHS, we get

$x=40-19 \Rightarrow x=21$

Hence, the required number is 21 .

**(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
**(g) Let Anwar thinks the number be x

$\therefore \frac{5}{2}$ of the number $=\frac{5 x}{2}$

On subtracting 7 from $\frac{5}{2}$ of the number $=\frac{5 x}{2}-7$

According to the question,

$\frac{5}{2} x-7=\frac{11}{2}$

Now, to solve this equation, transposing $(-7)$ from LHS to RHS, we get

$\Rightarrow\frac{5}{2} x=\frac{11}{2}+7 \Rightarrow \frac{5}{2} x=\frac{11}{2}+\frac{7}{1}$

$\Rightarrow\frac{5}{2} x=\frac{11 \times 1+7 \times 2}{2} \quad$ [:LCM of 2 and $1=21$

$\Rightarrow\frac{5}{2} x=\frac{11+14}{2} \Rightarrow \frac{5}{2} x=\frac{25}{2}$

On multiplying both sides by 2 , we get

$\frac{5}{2} x \times 2=\frac{25}{2} \times 2 \Rightarrow 5 x=25$

Again, dividing both sides by 5 , we get

$\frac{5 x}{5}=\frac{25}{5} \Rightarrow x=5$

Hence, the required number is 5 .

2. Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

**Answer**

Let us assume the lowest score be x

From the question it is given that,

The highest score is = 87

Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7

5/2 of the number = (5/2) x

The given above statement can be written in the equation form as,

Then,

= 2x + 7 = Highest score

= 2x + 7 = 87

By transposing 7 from LHS to RHS it becomes -7

= 2x = 87 – 7

= 2x = 80

Now, Divide both the side by 2

= 2x/2 = 80/2

= x = 40

Hence, the lowest score is 40

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

**Answer**

From the question it is given that,

We know that, the sum of angles of a triangle is 180^{o
}Let base angle be b

Then,

= b + b + 40^{o} = 180^{o
}= 2b + 40 = 180^{o
}By transposing 40 from LHS to RHS it becomes -40

= 2b = 180 – 40

= 2b = 140

Now, Divide both the side by 2

= 2b/2 = 140/2

= b = 70^{o
}Hence, 70^{o} is the base angle of an isosceles triangle.

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

**Answer**

Let us assume Rahul’s score be x

Then, Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short of a double century,

= Rahul’s score + Sachin’s score = 200 – 2

= x + 2x = 198

= 3x = 198

Divide both the side by 3,

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66

and Sachin’s score is 2x = 2 × 66 = 132

3. Solve the following:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

**Answer**

Let us assume number of Parmit’s marbles = m

From the question it is given that,

Then, Irfan has 7 marbles more than five times the marbles Parmit has

= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having

= (5 × m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both the side by 5

= 5m/5 = 30/5

= m = 6

So, Permit has 6 marbles

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

**Answer**

Let Laxmi’s age to be = y years old

From the question it is given that,

Lakshmi’s father is 4 years older than three times of her age

= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father

= (3 × y) + 4 = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 – 4

= 3y = 45

Divide both the side by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmi’s age is 15 years.

(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

**Answer**

Let the number of fruit tress be f.

From the question it is given that,

3 × number of fruit trees + 2 = number of non-fruit trees

= 3f + 2 = 77

By transposing 2 from LHS to RHS it becomes -2

=3f = 77 – 2

= 3f = 75

Divide both the side by 3

= 3f/3 = 75/3

= f = 25

So, number of fruit tree was 25.

4. Solve the following riddle:

I am a number,

Tell my identity!

Take me seven times over

And add a fifty!

To reach a triple century

You still need forty!

**Answer**

Let us assume the number be x.

Take me seven times over and add a fifty = 7x + 50

To reach a triple century you still need forty = (7x + 50) + 40 = 300

= 7x + 50 + 40 = 300

= 7x + 90 = 300

By transposing 90 from LHS to RHS it becomes -90

= 7x = 300 – 90

= 7x = 210

Divide both side by 7

= 7x/7 = 210/7

= x = 30

Hence the number is 30.

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