3 Marks Question

Question 13 Marks

Write atleast one other form for each equation (I), (ii) and (iii).
(i) $5 p=20$
(ii) $3 n+7=1$
(iii) $\frac{m}{5}-2=6$

Answer

(i) Other forms for equation $5 p=20$ are as follows:
(a) Five times a number $p$ is equal to 20.
(b) Multiply a number $p$ by 5 to get 20.
(ii) Other forms for equation $3 n+7=1$ are as follows:
(a) Sum of three times $n$ and 7 is equal to 1.
(b) Three times $n$ plus 7 is equal to 1.
(iii) Other forms for equation $\frac{m}{5}-2=6$ are as follows:
(a) Subtract 2 from one-fifth of a number $m$ to get 6.
(b) One-fifth of $m$ is greater by 2 than 6.

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Question 23 Marks

Solve the following equations by trial and error method.
$3 m-14=4$

Answer

Given equation is $3 m-14=4$.
When $m=0$, then LHS $=3 \times 0-14=0-14=-14$
and$\qquad$RHS $=4$
$\therefore $ LHS $\neq$ RHS
$\begin{array}{l}\text { When } m=1 \text {, then } \\ \text { LHS }=3 \times 1-14=3-14=-11 \\ \text { and }\quad \text { RHS }=4 \\ \therefore \text { LHS } \neq \text { RHS }\end{array}$
When $m=2$, then
LHS $=3 \times 2-14=6-14=-8$
and RHS $=4
\Rightarrow$ LHS $\neq$ RHS
When $m=3$, then
LHS $=3 \times 3-14=9-14=-5$
and RHS $=4
\Rightarrow$ LHS $\neq$ RHS
When $m=4$, then
LHS $=3 \times 4-14=12-14=-2$
and RHS $=4 \Rightarrow$ LHS $\neq$ RHS
When $m=5$, then
LHS $=3 \times 5-14=15-14=1$
and RHS $=4
\Rightarrow \quad$ LHS $\neq$ RHS
When $m=6$, then
LHS $=3 \times 6-14=18-14$
and RHS $=4
\Rightarrow \quad$ LHS $=$ RHS
Hence, $m=6$ is the solution of the given equation.

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Question 33 Marks

Solve the following equations by trial and error method.
$5 p+2=17$

Answer

Given equation is $5 p+2=17$.
When $p=0$, then LHS $=5 \times 0+2=0+2=2$
and$\qquad$RHS $=17$
$\therefore$$\qquad$ LHS $\neq$ RHS
$\begin{array}{l}\text {When } p=1 \text {, then LHS }=5 \times 1+2=5+2=7 \\\text {and } \qquad \text { RHS }=17 \\ \begin{array}{ll}\therefore\text { LHS } \neq \text { RHS }\end{array}\end{array}$
$\begin{array}{l}\text {When } p=2 \text {, then LHS }=5 \times 2+2=10+2=12 \\ \text {and }\qquad \text { RHS }=17 \\ \therefore \text { LHS } \neq \text { RHS } \\ \text {When } p=3 \text {, then LHS }=5 \times 3+2=15+2=17\end{array}$
$
\therefore\text { LHS }=\text { RHS }
$
So, $p=3$ is the solution of the given equation.

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