2 Marks Questions

Question 512 Marks

Solve the following equations.
$\frac{m}{4}=16$

Answer

Given $\frac{m}{4}=16$
On multiplying both sides by 4 , we get
$
\frac{m}{4} \times 4=16 \times 4
$
$\begin{array}{ll}\Rightarrow m=64 \\ \therefore m=64,\end{array}$
which is the required solution.

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Question 522 Marks

Solve the following equations.
$4 I=44$

Answer

Given, $4 l=44$
On dividing both sides by 4 ,
$
\frac{4 l}{4}=\frac{44}{4} \Rightarrow l=11
$
$
\therefore
l=11,
$
which is the required solution.

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Question 532 Marks

Solve the following equations
$34-6(x-5)=0$

Answer

Given, equation $34-6(x-5)=0$
On subtracting 34 from both sides, we get
$
\begin{aligned}
-34+34-6(x-5) & =0-34 \\
-6(x-5) =-34
\end{aligned}
$
On dividing by -6 both sides, we get
$
\frac{-6(x-5)}{-6}=\frac{-34}{-6} \Rightarrow x-5=\frac{+34}{6}
$
On adding $5$ to both sides. we get
$
\begin{array}{l}
x-5+5-\frac{34}{6}=5 \\
x=\frac{64}{6}=\frac{32}{3}
\end{array}
$
So, $x =\frac{64}{6}=\frac{32}{3}$, which is the required solution

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Question 542 Marks

Solve the following equations
$-2(3 x+3)=8$

Answer

Given, equation -2(3x+3)=8
Dividing both sides by -2, we get
$(3 x+3)=-4$
$\Rightarrow
3 x=-4-3 \Rightarrow x=-\frac{7}{3}$
$\therefore x=-\frac{7}{3}$ is the required solution.

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Question 552 Marks

Solve the following equations
$3(2 x+1)=6$

Answer

Given, equation $3(2 x+1)=6$
On dividing both sides by 3 , we get
$
\begin{array}{l}
(2 x+1)=\frac{6}{3} \Rightarrow 2 x+1=2 \\
2 x=2-1 \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}
\end{array}
$
$\therefore x=\frac{1}{2}$ is the required solution.

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Question 562 Marks

Solve the following equations.
$\frac{-2 p}{3}=5$

Answer

Given, $\frac{-2 P}{3}=5$
On multiplying both sides by 3 , we get
$
\frac{-2 P}{3} \times 3=5 \times 3 \Rightarrow-2 P=15
$
Now, on dividing both sides by -2 , we get
$
\frac{-2 P}{-2}=-\frac{15}{2} \Rightarrow P=-\frac{15}{2}
$
$\therefore P=-\frac{15}{2}$ is the required solution.

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Question 572 Marks

Solve the following equations.
$\frac{n}{4}=4$

Answer

Given, $\frac{n}{4}=4$
On multiplying both sides by 4 , we get
$
\frac{n}{4} \times 4=4 \times 4 \Rightarrow n=16
$
$\therefore n=16$ is the required solution.

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Question 582 Marks

Solve the following equations.
$5 p+10=100$

Answer

Given, $5 P+10=100$
Subtracting 10 from the both sides, we get
$
5 P+10-10=100-10 \Rightarrow 5 P=90
$
Now, on dividing both sides by 5 , we get
$
\frac{5 P}{5}=\frac{90}{5} \Rightarrow P=18
$
$\therefore P=18$ is the required solution.

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Question 592 Marks

Solve the following equations.
$20 m=100$

Answer

Given, $20 m=100$
On dividing both sides by 20 , we get
$
\frac{20 m}{20}=\frac{100}{20} \Rightarrow m=5
$
$\therefore m=5$ is the required solution.

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Question 602 Marks

Solve the following equations by trial and error method.
$7 P+6=27$

Answer

Given, 7 P + 6=27
On putting P=0 in7P+6 = 27, we get
(7×0)+6= 27
$\Rightarrow
0+6=27 \Rightarrow 6 \neq 27$
Now, on putting P=1 in 7P+6 = 27, we get
(7×1)+6=27
$\Rightarrow
7+6=27 \Rightarrow 13 \neq 27$
Again, on putting P = 2 in 7P+6 = 27, we get
(7 x 2) +6 = 27
$\Rightarrow 14 + 6=27 \Rightarrow 20 \neq 27$
Again, on putting P = 3 in 7P +6 = 27, we get
(7x3) +6= 27
$\Rightarrow 21+6=27 \Rightarrow 27=27$
Hence, P = 3 is the solution of given equation 7p+6=27.

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Question 612 Marks

Solve the following equations by trial and error method.
$6 m+2=14$

Answer

Given, $6 m+2=14$
On putting $m=0$ in $6 m+2=14$, we get
$(6 \times 0)+2=14$
$\Rightarrow
0+2=14$
$\Rightarrow
2 \neq 14$
Now, on putting m =1 in 6m + 2 =14, we get
$(6 \times 1)+2=14$
$\Rightarrow
6+2=14$
$\Rightarrow \quad 8 \neq 14$
Again, on putting m = 2 in 6m + 2 =14, we get
(6 x 2) +2=14
$\Rightarrow\qquad$ 12+2=14
$\Rightarrow\qquad$14=14
So, m = 2 is the solution of the given equation 6m+2=14.

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Question 622 Marks

Check whether the values given in the brackets is a solution to the given equations.
$6 n-5 n=3,(n=3)$

Answer

Given, $6 n-5 n=3,(n=3)$
On putting $n=3$ in $6 n-5 n=3$, we get
$(6 \times 3)-(5 \times 3)=3$
$\Rightarrow 18-15=3$
$\Rightarrow
3=3$
So, the value given in the bracket is a solution of equation.

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Question 632 Marks

Check whether the values given in the brackets is a solution to the given equations.
$7 n+6=20,(n=2)$

Answer

Given, $7 n+6=20,(n=2)$
On putting $n=2$ in $7 n+6=20$, we get
$
(7 \times 2)+6=20
$
$\Rightarrow 14+6=20$
$\Rightarrow 20=20$
So, the value given in the bracket is a solution of equation.

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MATHS 2 Marks Questions