# Ex.4.2 Q4 Simple-Equations Solution - NCERT Maths Class 7

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## Question

Solve the following equations:

(a)  \begin{align}10p = 100\end{align}

(b)  \begin{align}10p + 10 = 100\end{align}

(c)   \begin{align}\frac{p}{4} = 5\end{align}

(d)  \begin{align}\frac{{ - p}}{3} = 5\end{align}

(e)  \begin{align}\frac{{3p}}{4} = 6\end{align}

(f)   \begin{align}3s = -9\end{align}

(g)   \begin{align}3s + 12 = 0\end{align}

(h)   \begin{align}3s = 0\end{align}

(i)   \begin{align}2q = 6\end{align}

(j)   \begin{align}2q-6 = 0\end{align}

(k)   \begin{align}2q + 6 = 0\end{align}

(l)   \begin{align}2q + 6 = 12\end{align}

Video Solution
Simple Equations
Ex 4.2 | Question 4

## Text Solution

What is Known?

Equations

What is unknown?

Value of the variable.

Reasoning:

First try to reduce the equation by adding, subtracting, multiplying or dividing by the same number and get the value of variable.

Steps:

(a) $$10p = 100$$

Dividing both the sides by $$10$$ we get,

\begin{align}\frac{{10p}}{{10}}&= \frac{{100}}{{10}}\\p&= 10\end{align}

(b) $$10p + 10 = 100$$

Subtracting $$10$$ from both sides we get,

\begin{align}10p + 10 - 10&= 100 - 10\\10p&= 90\end{align}

Dividing both the sides by $$10$$ we get,

\begin{align}\frac{{10p}}{{10}} &= \frac{{90}}{{10}}\\p &= 9\end{align}

(c) \begin{align}\frac{p}{4} = 5\end{align}

Multiplying both the sides by $$4$$ we get,

\begin{align}\frac{p}{4} \times 4 &= 5 \times 4\\p& = 20\end{align}

(d) \begin{align}\frac{{ - p}}{3} = 5\end{align}

Multiplying both the sides by $$3$$,

\begin{align}\frac{{ - p}}{3} \times 3 &= 5 \times 3\\ - p &= 15\\p &= - 15\end{align}

(e) \begin{align}\frac{{3p}}{4} = 6\end{align}

Multiplying both the sides by $$4,$$

\begin{align}\frac{{3p}}{4} \times 4& = 6 \times 4\\3p&= 24\end{align}

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3p}}{3}&= \frac{{24}}{3}\\p&= 8\end{align}

(f) $$3s = -9$$

Dividing both the sides by $$3$$,

\begin{align}\frac{{3s}}{3}&= \frac{{ - 9}}{3}\\p&= - 3\end{align}

(g) $$3s + 12 = 0$$

Subtracting $$12$$ from both the sides of the equation we get,

\begin{align}3s + 12 - 12 &= 0 - 12\\3s &= - 12\end{align}

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3s}}{3}&= \frac{{ - 12}}{3}\\p&= - 4\end{align}

(h) $$3s = 0$$

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3s}}{3} &= \frac{0}{3}\\p& = 0\end{align}

(i) $$2q = 6$$

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2} &= \frac{6}{2}\\q &= 3\end{align}

(j) $$2q-6 = 0$$

Adding $$6$$ to both sides of the equation we get,

\begin{align}2q - 6 + 6 &= 0 + 6\\2q &= 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2} = \frac{6}{2}\\\end{align}

(k) $$2q + 6 = 0$$

Subtracting $$6$$ from both the sides of the equation we get,

\begin{align}2q + 6 - 6&= 0 - 6\\2q&= - 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2}&= \frac{{ - 6}}{2}\\p&= - 3\end{align}

(l) $$2q + 6 = 12$$

Subtracting $$6$$ from both the sides of the equation we get,

\begin{align}2q + 6 - 6&= 12 - 6\\2q&= 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2}&= \frac{6}{2}\\p&= 3\end{align}

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