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MATHS 2 Marks Questions

Linear Equations in One Variable · Rajasthan Board (RBSE) STD 8 (Rajasthan - English Medium). 17 questions.

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2 Marks Questions

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17 questions · timed · auto-graded

Question 12 Marks
Simplify and solve the linear equation: $3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17$
Answer
$3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17$
$\therefore 15z – 21 – 18z + 22 = 32z – 52 – 17$
$\therefore – 3z + 1 = 32z – 69$
$\therefore –3z + 32z = –69 – 1 ... [$Transposing $32z$ to $L.H.S.$ and $1$ to $R.H.S.]$
$\therefore –35z = – 70$
$\therefore z = \frac{{ - 70}}{{ - 35}} ... [$Dividing both sides by $–35]$
$\therefore z = 2$ this is the required solution.
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Question 22 Marks
Simplify and solve the linear equation $0.25(4f – 3) = 0.05 (10f – 9).$
Answer
$0.25(4f – 3) = 0.05 (10f – 9)$
$\therefore f – 0.75 = 0.5f – 0.45$
$\therefore f – 0.5f = 0.45 + 0.75 ... [$Transposing $0.5f$ to $L.H.S.$ and $– 0.75$ to $R.H.S.]$
$\therefore 0.5f = 0.30$
$\therefore f = \frac{{0.30}}{{0.5}} ... [$Dividing both sides by $ 0.5]$
$\therefore f = 0.6$ this is the required solution.
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Question 32 Marks
Solve the equations and check your result: $2x - 1 = 14 - x.$
Answer
$2x – 1 = 14 – x$
$2x + x = 14 + 1 ... [$Transposing $–x$ to $L.H.S.$ and $–1$ to $R.H.S.]$
$\therefore 3x = 15$
$\therefore x = \frac{{15}}{3} ... [$Dividing both sides by $3]$
$\therefore x = 5$ this is the required solution.
Verification
$L.H.S. = 2x – 1 = 2(5) – 1 = 10 – 1 = 9$
$R.H.S. = 14 – x = 14 – 5 = 9$
Therefore, $L.H.S. = R.H.S.$
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Question 42 Marks
Solve the equation and check your result: $4z + 3 = 6 + 2z$
Answer
$4z + 3 = 6 + 2z$
$4z – 2z = 6 – 3 ... [$Transposing $2z$ to $L.H.S.$ and $3$ to $R.H.S]$
$\therefore 2z = 3$
$\therefore z = \frac{3}{2} ... [$Dividing both sides by $2]$
This is the required solution.
Verification,
$L.H.S = 4\left( {\frac{3}{2}} \right) + 3 = 6 + 3 = 9$
$R.H.S = 6 + 2z = 6 + 2\left( {\frac{3}{2}} \right) = 6 + 3 = 9$
Therefore, $L.H.S = R.H.S$
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Question 52 Marks
Solve the equation and check your result: $5x + 9 = 5 + 3x$
Answer
$5x + 9 = 5 + 3x$
$5x –3x = 5 – 9 ... [$Transposing $3x$ to $L.H.S.$ and $9$ to $R.H.S]$
$\therefore 2x = –4$
$\therefore x = - \frac{4}{2} ... [$Dividing both sides by $2]$
$\therefore x = –2$ this is the required solution.
Verification
$L.H.S. = 5(–2) + 9 = – 10 + 9 = –1$
$R.H.S. = 5 + 3(–2) = 5 – 6 = –1$
Therefore, $L.H.S = R.H.S$
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Question 62 Marks
Solve the equation and check your result: $5t – 3 = 3t – 5$
Answer
$5t – 3 = 3t – 5$
$5t – 3t = -5 + 3 ... [$Transposing $3t$ to $L.H.S$ and $–3$ to $R.H.S]$
$2t = –2$
$\therefore t = - \frac{2}{2} ... [$Dividing both sides by $2]$
$\therefore t = –1$
this is the required solution.
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Question 72 Marks
Solve the equation and check your result: $3x = 2x + 18$
Answer
$3x = 2x + 18$
$3x – 2x = 18 ... [$Transposing $2x$ to $L.H.S]$
$\therefore x = 18$ this is the required solution.
Verification,
$L.H.S. = 3x = 3 \times 18 = 54$
$R.H.S. = 2x + 18 = 2 \times 18 + 18 = 36 + 18 = 54$
Therefore, $L.H.S = R.H.S$
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Question 82 Marks
What should be added to twice the rational number $ - \frac{7}{3}$ to get $\frac{3}{7}$?
Answer
Let the number be $= x$
According to question,
$2\times (\frac{-7} { 3}) + x = (\frac{3} { 7})$
$(\frac{-14} { 3}) + x = (\frac{3} { 7})$
by transposing
$x = (\frac{3} { 7}) + (\frac{14} { 3})$
by taking $L.C.M$
$x = \frac{9+98} { 21}$
$x = \frac{107} { 21}$
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Question 92 Marks
Solve: $\frac{{15}}{4} - 7x = 9$
Answer
$\frac{{15}}{4} - 7x = 9$
By transposing
$-7x = 9 - \frac {15}{4}$
by $L.C.M$
$-7x = 36 - \frac {15}{4}$
$-7x = \frac {21}{4}$
$-x = \frac {21}{28}$
$x = -\frac {3}{4}$
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Question 102 Marks
Solve: $\frac{x}{3}+\frac{5}{2}=-\frac{3}{2}$
Answer
$\frac{x}{3}+\frac{5}{2}=-\frac{3}{2}$
$\frac{x}{3}=\frac{-3}{2}-\frac{5}{2}=-\frac{8}{2}$
$\frac{x}{3}=-4$
$x = -4\times3$
$x = -12$
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Question 122 Marks
Solve: $5x - 2 (2x - 7) = 2(3x - 1) + \frac {7}{2}$
Answer
$5x - 2 (2x - 7) = 2(3x - 1) + \frac {7}{2}$
$5x – 4x + 14 = 6 x-2+\frac{7}{2}$
$x + 14 = 6x + \frac{3}{2}$
$6 x-x = 14 -\frac{3}{2}$
$5x = \frac{25}{2}$
$x=\frac{25}{2} \times \frac{1}{5}=\frac{5}{2}$
Therefore, the required solution is $x = \frac{5}{2}.$
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Question 132 Marks
Solve: $\frac{6 x+1}{3}+1=\frac{x-3}{6}$
Answer
$\frac{6 x+1}{3}+1=\frac{x-3}{6}$
Multiplying both sides of the equation by $6,$
$\frac{6(6 x+1)}{3}+6 \times 1=\frac{6(x-3)}{6}$
or $2(6x + 1) + 6 = x - 3$
or $12x + 2 + 6 = x - 3$
or $12x + 8 = x - 3$
or $12x – x = -3 -8$
or $11x = -11$
or $x = -1$
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Question 142 Marks
Solve: $5x + \frac{7}{2}=\frac{2}{2}x -14$
Answer
$5x + \frac{7}{2}=\frac{3}{2}x -14$
Multiplying both sides of the equation by $2,$ we get
$2 \times\left(5 x+\frac{7}{2}\right)=2 \times\left(\frac{3}{2} x-14\right)$
$(2 \times 5 x)+\left(2 \times \frac{7}{2}\right)=\left(2 \times \frac{3}{2} x\right)-(2 \times 14)$
$10x + 7 = 3x - 28$
$10x – 3x = -28 - 7$
$7x = -35$
$x = \frac{-35}{7}$
$x = -5$
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Question 152 Marks
Solve: $2x - 3 = x + 2$
Answer
$2x - 3 = x + 2$
$\Rightarrow 2x = x + 2 + 3$
$\Rightarrow 2x = x + 5$
$\Rightarrow 2x - x = x + 5 - x ($subtracting $x$ from both sides$)$
$\Rightarrow x = 5$
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Question 162 Marks
The difference between two whole numbers is $66.$ The ratio of the two numbers is $2 : 5$. What are the two numbers$?$
Answer
Since the ratio of the two numbers is $2 : 5,$ we may take one number to be $2x$ and the other to be $5x.$
The difference between the two numbers is $(5x – 2x).$ It is given that the difference is $66.$ Therefore,
$5x – 2x = 66$
or $3x = 66$ or $x = 22$
Thus the two numbers are $2x = 2 \times 22 = 44$ and $5x = 5 \times 22 = 110.$
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Question 172 Marks
Find the solution of $2x – 3 = 7$
Answer
We have, $2x – 3 = 7$
Step1: Add $3$ to both sides.
$2x – 3 + 3 = 7 + 3 ($The balance is not disturbed$)$
or $2x = 10$​​​​​​​
Step2: Next divide both sides by $2.$
or $x = 5 ($required solution$)$
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