Examples (Revised) - Chapter 4 - Simple Equations - Ncert Solutions class 7 - Maths

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Chapter 4 - Simple Equations - NCERT Solutions for Class 7 Maths | Comprehensive Guide

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
(i) Three times $x$ is $3 x$.
Sum of $3 x$ and 11 is $3 x+11$. The sum is 32 .
The equation is $3 x+11=32$.
(ii) Let us say the number is $z ; z$ multiplied by 6 is $6 z$.
Subtracting 5 from $6 z$, one gets $6 z-5$. The result is 7 .
The equation is $6 z-5=7$

Example 2

Convert the following equations in statement form:
(i) $x-5=9$
(ii) $5 p=20$
(iii) $3 n+7=1$
(iv) $\frac{m}{5}-2=6$

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 $3 y$ years. Raju's father's age is 5 years more than $3 y$; that is, Raju's father is $(3 y+5)$ years old. It is also given that Raju's father is 44 years old.

Therefore,
$
3 y+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 $\mathrm{m}$, that is, $8 \mathrm{~m}+4$ mangoes. But this is given to be 100 . Thus
$
8 m+4=100
$

You can get the number of mangoes in a small box by solving this equation.

Example 5

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

Solution
(a) We go stepwise to separate the variable $n$ on the LHS of the equation. The LHS is $3 n+7$. We shall first subtract 7 from it so that we get $3 n$. 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,
$
3 n+7-7=25-7
$
or
$
3 n=18
$
(Step 1)

Now divide both sides by 3 ,
$
\frac{3 n}{3}=\frac{18}{3}
$
or $n=6$, which is the solution.
(b) What should we do here? First we shall add 1 to both the sides:
$
2 p-1+1=23+1
$
or
$
2 p=24
$
(Step 1)

Now divide both sides by 2 , we get $\frac{2 p}{2}=\frac{24}{2}$
(Step 2)
or
$
p=12 \text {, which is the solution. }
$

One good practice you should develop is to check the solution you have obtained. Although we have not done this for (a) above, let us do it for this example.
Let us put the solution $p=12$ back into the equation.
$
\begin{aligned}
\mathrm{LHS} & =2 p-1=2 \times 12-1=24-1 \\
& =23=\text { RHS }
\end{aligned}
$

The solution is thus checked for its correctness.
Why do you not check the solution of (a) also?
We are now in a position to go back to the mind-reading game presented by Appu, Sarita, and Ameena and understand how they got their answers. For this purpose, let us look at the equations (4.1) and (4.2) which correspond respectively to Ameena's and Appu's examples.
- First consider the equation $4 x+5=65$.
Subtracting 5 from both sides, $4 x+5-5=65-5$.
i.e. $\quad 4 x=60$

Divide both sides by 4 ; this will separate $x$. We get $\frac{4 x}{4}=\frac{60}{4}$ or $x=15$, which is the solution. (Check, if it is correct.)
- Now consider, $10 y-20=50$
Adding 20 to both sides, we get $10 y-20+20=50+20$ or $10 y=70$
Dividing both sides by 10 , we get $\frac{10 y}{10}=\frac{70}{10}$
or $\quad y=7$, which is the solution. (Check if it is correct.)
You will realise that exactly these were the answers given by Appu, Sarita and Ameena.

They had learnt to set up equations and solve them. That is why they could construct their mind reader game and impress the whole class. We shall come back to this in Section 4.7.

Example 6

Solve: $12 p-5=25$
Solution
- Adding 5 on both sides of the equation, $12 p-5+5=25+5 \quad$ or $\quad 12 p=30$
- Dividing both sides by 12 , $\frac{12 p}{12}=\frac{30}{12}$ or $\quad p=\frac{5}{2}$

Check Putting $p=\frac{5}{2}$ in the LHS of equation 4.12,
$
\begin{aligned}
\mathrm{LHS} & =12 \times \frac{5}{2}-5=6 \times 5-5 \\
& =30-5=25=\mathrm{RHS}
\end{aligned}
$

Note, adding 5 to both sides is the same as changing side of $(-5)$.
$
\begin{aligned}
& 12 p-5=25 \\
& 12 p=25+5
\end{aligned}
$

Changing side is called transposing. While transposing a number, we change its sign.

As we have seen, while solving equations one commonly used operation is adding or subtracting the same number on both sides of the equation. Transposing a number (i.e., changing the side of the number) is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed. What applies to numbers also applies to expressions. Let us take two more examples of transposing.

We shall now solve two more equations. As you can see they involve brackets, which have to be solved before proceeding.
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=\frac{18}{4} \text { or } \quad m+3=\frac{9}{2}
$
or $m=\frac{9}{2}-3$ (transposing 3 to RHS)
or $\quad m=\frac{3}{2} \quad$ (required solution) $\left(\right.$ as $\left.\frac{9}{2}-3=\frac{9}{2}-\frac{6}{2}=\frac{3}{2}\right)$
Check LHS $=4\left[\frac{3}{2}+3\right]=4 \times \frac{3}{2}+4 \times 3=2 \times 3+4 \times 3\left[\right.$ put $\left.m=\frac{3}{2}\right]$
$
=6+12=18=\text { RHS }
$
(b) $-2(x+3)=8$
We divide both sides by ( -2 ), so as to remove the brackets in the LHS, we get, $x+3=-\frac{8}{2} \quad$ or $\quad x+3=-4$
i.e., $x=-4-3 \quad$ (transposing 3 to RHS) $\quad$ or $\quad x=-7 \quad$ (required solution)

Check
$
\begin{aligned}
\text { LHS } & =-2(-7+3)=-2(-4) \\
& =8=\text { RHS as required. }
\end{aligned}
$

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 times the number is $3 x$ and the sum of $3 x$ and 11 is 32 . That is, $3 x+11=32$
- To solve this equation, we transpose 11 to RHS, so that
$3 x=32-11$ or $3 x=21$
Now, divide both sides by 3
So
$
x=\frac{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 $\frac{y}{4}$.
This number $\left(\frac{y}{4}\right)$ is more than 7 by 3 .
Hence we get the equation for $y$ as $\frac{y}{4}-7=3$
- To solve this equation, first transpose 7 to RHS We get, $\frac{y}{4}=3+7=10$. We then multiply both sides of the equation by 4 , to get
$
\frac{y}{4} \times 4=10 \times 4 \quad \text { or } \quad y=40 \quad \text { (the required number) }
$

Let us check the equation formed. Putting the value of $y$ in the equation, $\mathrm{LHS}=\frac{40}{4}-7=10-7=3=$ 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
$
\begin{array}{r}
3 y+5=44 \\
3 y=44-5=39 \\
y=13
\end{array}
$
- To solve it, we first transpose 5, to get
Dividing both sides by 3 , we get
That is, Raju's age is 13 years. (You may check the answer.)