NCERT Solutions Class 7th Math Chapter – 4 Simple Equations Examples

Example 1. Write the following statements in the form of equations:
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv) One third of a number plus 5 is 8.

Solution –
(1) Three times x is 3x.
Sum of 3x and 11 is 3x + 11. The sum is 32.
The equation is 3x+11=32.

(ii) Let us say the number is z; z multiplied by 6 is 6z.
Subtracting 5 from 6z, one gets 6z-5. The result is 7.
The equation is 62-5=7

(iii) One fourth of m is m/4
It is greater than 7 by 3. This means the difference (m/4 – 7) is 3.
The equation is m/4 – 7 = 3.

(iv) Take the number to be n. One third of n is – n/3
This one-third plus 5 is n/3 + 5. It is 8.
The equation is n/3 + 5 = 8.

Example 2. Convert the following equations in statement form:
(i) x – 5 = 9
(ii) 5p = 20
(iii) 3n + 7 = 1
(iv) m/5 – 2 = 6

Solution –

(i) Taking away 5 from.x gives 9.
(ii) Five times a number p is 20.
(iii) Add 7 to three times n to get 1.
(iv) You get 6, when you subtract 2 from one-fifth of a number m.
What is important to note is that for a given equation, not just one, but many statement
forms can be given. For example, for Equation (i) above, you can say:
Subtract 5 from x, you get 9.
or The number x is 5 more than 9.
or The number x is greater by 5 than 9.
or The difference between x and 5 is 9, and so on.

Example 3. Consider the following situation:

Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years
old. Set up an equation to find Raju’s age.

Solution – We do not know Raju’s age. Let us take it to be y years. Three times
Raju’s age is 3y years. Raju’s father’s age is 5 years more than 3y; that is, Raju’s father is (3y+5) years old. It is also given that Raju’s father is 44 years old.
Therefore, 3y+5=44
This is an equation in y. It will give Raju’s age when solved.

Example 4. A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100.

Solution – Let a small box contain m mangoes. A large box contains 4 more than 8
times m, that is, 8m + 4 mangoes. But this is given to be 100. Thus
8m + 4 = 100
You can get the number of mangoes in a small box by solving this equation.

Example 5. Solve:
(a) 3n+7=25
(b) 2p-1=23

Solution –

(a) We go stepwise to separate the variable n on the LHS of the equation. The LHS is 3n+7. We shall first subtract 7 from it so that we get 3n. From this, in the next step we shall divide by 3 to get n. Remember we must do the same operation on both sides of the equation. Therefore, subtracting 7 from both sides,
3n + 7 – 7 = 25 -7
or 3n = 18

Now divide both sides by 3,
3n/3 = 18/3
or n = 6, which is the solution.

(b) What should we do here? First we shall add 1 to both the sides:

2p – 1 + 1 = 23 + 1
or 2p=24

Now divide both sides by 2, we get 2p/2 = 24/2
ог p = 12, which is the solution.

Example 6. Solve: 12p-5 = 25

Solution –

• Adding 5 on both sides of the equation,
12p – 5 + 5 = 25 + 5 or 12p = 30

• Dividing both sides by 12,
12p/12 = 30/12 or p = 5/2

Check Puttingpin p = 5/2 the LHS of equation 4.12,
LHS = 12×5/2 -5 =6 × 5 – 5
= 30 – 5 = 25 = RHS

Example 7. Solve
(a) 4 (m+3) = 18
(b) -2 (x + 3 = 8

Solution –
(a) 4 (m + 3) = 18
Let us divide both the sides by 4. This will remove the brackets in the LHS We get,

m + 3 = 18/4 or m + 3 =9/2
or m = 9/2 – 3 (transposing 3 to RHS)
or m = 3/2 (as 9/2 -3 = 9/2 – 6/2 = 3/2

Cheack LHS = 4{3/2 + 3} = 4 x 3/2 + 4 x 3 = 2 x3 + 4 x 3 (Put m = 3/2)
= 16 + 12 = 18 = RHS

(b) -2(x+3)=8
We divide both sides by (-2), so as to remove the brackets in the LHS, we get,
x+3= -8/2 or x+3= -4
i.e., x = -4 – 3 (transposing 3 to RHS) or x = -7 (required solution)
Check LHS= -2(-7 + 3)= -2(-4)
=8= RHS as required.

Example 8. The sum of three times a number and 11 is 32. Find the number.

Solution –
If the unknown number is taken to be x, then three ti the number is 3x and the sum
of 3x and 11 is 32. That is, 3x+11=32
To solve this equation, we transpose 11 to RHS so that
3x=32-11 or 3x=21
Now, divide both sides by 3
So x = 21/3 =7
The required number is 7. (We may check it by taking 3 times 7 and adding 11 to it.
It gives 32 as required.)

Example 9. Find a number, such that one-fourth of the number is 3 more than 7.

Solution –

• Let us take the unknown number to be y; one-fourth of y is y/4
This number) is more than 7 by 3.
Hence we get the equation for y as y/4 -7=3

• To solve this equation, first transpose 7 to RHS We get, y/4 =3+7=10.
We then multiply both sides of the equation by 4, to get
y/4 x 4 = 10 x 4 or y/40 (the required number)
Let us check the equation formed. Putting the value of y in the equation,
LHS = 40/4 -7=10-7=3 RHS, as required.

Example 10. Raju’s father’s age is 5 years more than three times Raju’s age. Find Raju’s age, if his father is 44 years old.

Solution –
• As given in Example 3 earlier, the equation that gives Raju’s age is
3y+5=44

• To solve it, we first transpose 5, to get
3y=44-5=39
Dividing both sides by 3, we get
y = 13
That is, Raju’s age is 13 years. (You may check the answer.)