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MATHS 2 Marks Questions

Linear Equations in One Variable · Gujarat Board (GSEB) STD 8 (GSEB - English Medium). 18 questions.

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2 Marks Questions

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18 questions · timed · auto-graded

Question 12 Marks
Solve the following linear equations :
$\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$.
Answer
We have, $\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$
$\Rightarrow \quad \frac{6 n-9 n+10 n}{12}=21 \quad[\because$ LCM of 2,4 and 6 is 12$]$
$\Rightarrow \quad \frac{16 n-9 n}{12}=21 \Rightarrow \frac{7 n}{12}=21$
$\Rightarrow \quad \frac{7}{12} n \times \frac{12}{7}=\frac{21 \times 12}{7}\quad$ [multiplying both sides by $\frac{12}{7}$]
$\Rightarrow \quad n=\frac{21 \times 12}{7}$
$\Rightarrow n=3 \times 12=36$, which is the required solution.
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Question 22 Marks
Solve the following linear equations :
$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}.$
Answer
We have, $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$
$\Rightarrow \quad \frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\quad$ [transposing $-\frac{1}{5}$ to RHS and $\frac{x}{3}$ to LHS]
$\Rightarrow \quad \frac{3 x-2 x}{6}=\frac{5+4}{20}$
$[\because \text{LCM}$ of 2 and 3 is 6 and LCM of 4 and 5 is 20]
$\Rightarrow \quad \frac{x}{6}=\frac{9}{20}$
$\Rightarrow \quad x=\frac{9}{20} \times 6 \quad$ [multiplying both sides b]
$\Rightarrow \quad x=\frac{9 \times 3}{10}=\frac{27}{10}$,
which is the required solution.
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Question 32 Marks
Solve the following equations :
$x=\frac{4}{5}(x+10)$
Answer
We have, $\quad x=\frac{4}{5}(x+10)$
$\Rightarrow \quad 5 x=4(x+10) \quad$ [multiplying both sides by 5]
$\Rightarrow \quad 5 x=4 x+40$
$\Rightarrow \quad 5 x-4 x=40 \quad$ [transposing $4 x$ to LHS]
$\therefore x=40$, which is the required solution.
Check For $x=40$,
$LHS =40$
$RHS =\frac{4}{5}(40+10)=40$
$\therefore \quad$ LHS $=$ RHS $\quad$ [as required]
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Question 42 Marks
Solve the following equations :
$2 x-1=14-x.$
Answer
We have,
$5 t-3=3 t-5$
$\Rightarrow \quad 5 t-3 t=-5+3\quad$ [transposing 3t to LHS and -3 to RHS]
$\Rightarrow \quad 2 t=-2$
$\Rightarrow \quad t=\frac{-2}{2} \quad$ [dividing both sides by 2]
$\therefore t=-1$, which is the required solution.
Check For $t=-1$
LHS $=5(-1)-3=-5-3=-8$
RHS $=3(-1)-5=-3-5=-8$
$\therefore \quad LHS = RHS \quad$ [as required]
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Question 52 Marks
Solve the following equations :
$4 z+3=6+2 z$
Answer
We have, $5 x+9=5+3 x$
$\Rightarrow \quad 5 x-3 x=5-9\quad$ [transposing $3 x$ to LHS and 9 to RHS]
$\Rightarrow \quad 2 x=-4 \Rightarrow x=\frac{-4}{2}\quad$ [dividing both sides by 2]
$\therefore x=-2$, which is the required solution.
Check For $x=-2$
$LHS =5(-2)+9=-1$
RHS $=5+3(-2)=-1$
$\therefore \quad LHS = RHS \quad$ [as required]
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Question 62 Marks
Solve the following equations :
$5 x+9=5+3 x.$
Answer
We have, $5 x+9=5+3 x$
$\Rightarrow \quad 5 x-3 x=5-9\quad$ [transposing $3 x$ to LHS and 9 to RHS]
$\Rightarrow \quad 2 x=-4 \Rightarrow x=\frac{-4}{2}\quad$ [dividing both sides by 2]
$\therefore x=-2$, which is the required solution.
Check For $x=-2$
$LHS =5(-2)+9=-1$
RHS $=5+3(-2)=-1$
$\therefore \quad LHS = RHS \quad$ [as required]
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Question 72 Marks
Solve the following equations :
$5 t-3=3 t-5$
Answer
We have,
$5 t-3=3 t-5$
$\Rightarrow \quad 5 t-3 t=-5+3\quad$ [transposing 3t to LHS and -3 to RHS]
$\Rightarrow \quad 2 t=-2$
$\Rightarrow \quad t=\frac{-2}{2} \quad$ [dividing both sides by 2]
$\therefore t=-1$, which is the required solution.
Check For $t=-1$
LHS $=5(-1)-3=-5-3=-8$
RHS $=3(-1)-5=-3-5=-8$
$\therefore \quad LHS = RHS \quad$ [as required]
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Question 82 Marks
Solve the following equations :
$3 x=2 x+18$
Answer
We have,
$3 x=2 x+18$
$\Rightarrow 3 x-2 x=18 \quad[$transposing $2 x$ to LHS$]$
$\therefore x=18$, which is the required solution.
Check For $x=18$,
$3 x=2 x+18$
$\Rightarrow \quad 3 x-2 x=18$
LHS = $3(18)-2(18)$
$=54-36=18=$ RHS
$\therefore \quad \text{LHS = RHS} \quad$ [as required]
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Question 102 Marks
Solve $0.44 t-1.05=2(0.71 t-0.01)+1.11$
Answer
We have,
$0.44 t-1.05=2 \times 0.71 t-2 \times 0.01+1.11$
$\Rightarrow \quad 0.44 t-1.05=1.42 t-0.02+1.11$
$\Rightarrow \quad 0.44 t-1.42 t=1.05-0.02+1.11\quad$ [transposing 1.42 t to LHS and -1.05 to RHS]
$\Rightarrow \quad-0.98 t=2.14$
$\Rightarrow \quad t=\frac{2.14 \times 100}{-0.98 \times 100} \Rightarrow t=\frac{214}{-98}=\frac{-107}{49}$
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Question 112 Marks
Find the solution of $\frac{1}{x}-\frac{3}{x}=\frac{5}{2 x}-3$.
Answer
We have, $\frac{1}{x}-\frac{3}{x}=\frac{5}{2 x}-3$
$\Rightarrow \quad \frac{1-3}{x}-\frac{5}{2 x}=-3 \quad$ [transposing $\frac{5}{2 x}$ to LHS]
$\Rightarrow \quad \frac{-2}{x}-\frac{5}{2 x}=-3 \Rightarrow \frac{-4-5}{2 x}=-3$
$\Rightarrow \quad \frac{-9}{2 x}=-3 \Rightarrow \frac{2 x}{-9}=\frac{-1}{3}$
$\Rightarrow \quad x=\frac{-9}{2} \times\left(\frac{-1}{3}\right)\quad$ [multiplying both sides by $\frac{-9}{2}$ ]
$\Rightarrow \quad x=\frac{3}{2}$
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Question 122 Marks
Find the solution of $3 x+4=2 x+5$.
Answer
We have, $(3 x+4)=(2 x+5)$
$\Rightarrow \quad 3 x-2 x=5-4 \Rightarrow x=1\quad$ [transposing $2 x$ to LHS and 4 to RHS]
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Question 132 Marks
Solve $x-\frac{3}{7}=\frac{3}{7}$
Answer
We have, $x-\frac{3}{7}=\frac{3}{7}$
$\Rightarrow \quad x=\frac{3}{7}+\frac{3}{7} \Rightarrow x=\frac{6}{7} \quad\left[\right.$ transposing $\frac{-3}{7}$ to RHS]
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Question 142 Marks
Solve $\frac{3 x}{4}+5=\frac{1}{2}$
Answer
We have, $\frac{3}{4} x+5=\frac{1}{2}$
$\Rightarrow \quad \frac{3}{4} x=\frac{1}{2}-5=\frac{1-10}{2} \Rightarrow \frac{3}{4} x=\frac{-9}{2}$
$\Rightarrow \quad x=\frac{-9}{2} \times \frac{4}{3}$ [multiplying both sides by $\frac{4}{3}$ ]
$\Rightarrow \quad x=-6$
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Question 152 Marks
Solve $0.25(8 a-0.5)=7.5$.
Answer
We have, $0.25(8 a-0.5)=7.5$
$\Rightarrow 0.25 \times 8 a-0.25 \times 0.5=7.5$
$\Rightarrow \quad 0.25 \times 8 a=7.5+0.25 \times 0.5\quad$ [transposing $-0.25 \times 0.5$ to RHS]
$\Rightarrow \quad 2.0 \times a=7.5+0.125$
$\Rightarrow \quad a=7.625 \times \frac{1}{2}=3.8125\quad$ [dividing both sides by 2]
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Question 162 Marks
Solve $\frac{-39}{4}-x=\frac{5}{8}$
Answer
We have, $\frac{-39}{4}-x=\frac{5}{8} \Rightarrow \frac{-39}{4}-\frac{5}{8}=x\quad$ [transposing $\frac{5}{8}$ to LHS and $-x$ to RHS]
$\Rightarrow \quad \frac{-78-5}{8}=x \Rightarrow x=-\frac{83}{8}$
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Question 172 Marks
Solve $\frac{13}{5}-5 x=13$
Answer
We have, $\frac{13}{5}-5 x=13$
$\Rightarrow \quad \frac{13}{5}-13=5 x\quad$ [transposing 13 to LHS and $-5 x$ to RHS]
$\Rightarrow \quad \frac{13-65}{5}=5 x$
$\Rightarrow \quad \frac{-52}{5}=5 x$
$\Rightarrow \quad x=-\frac{52}{25} \quad$ [dividing both sides by 5]
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Question 182 Marks
Find the solution of $2 y+18=30$.
Answer
We have, $2 y+18=30$
$\Rightarrow \quad 2 y=30-18 \quad$ [transposing 18 to RHS]
$\Rightarrow \quad 2 y=12$
$\Rightarrow \quad y=12 \div 2 \quad$ [dividing both sides by 2]
$\Rightarrow \quad y=6$
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