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MATHS 5 Marks Questions

Linear Equations in one variable · Rajasthan Board (RBSE) STD 6 (Rajasthan - English Medium). 14 questions.

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14 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
Solve the following equation and verify the answer: $\frac{\text{3x}}{10}-4=14$
Answer
$\frac{\text{3x}}{10}-4=14$
$\Rightarrow\frac{\text{3x}}{10}=14+4$(Transposing -$4 $to $R.H.S.)$
$\Rightarrow\frac{\text{3x}}{10}=18$
$\Rightarrow\frac{\text{3x}}{10}\times=18\times10$(Multiplying both sides by $10)$
$\Rightarrow3x=180$
$\Rightarrow\frac{\text{3x}}{3}=\frac{180}{3}$(Dividing both sides by $3)$
$\Rightarrow \text{x}=60$
So, $\text{x}=60$ is a solution of the given equation.
Check: Substituting $\text{x}=60$ in the given equation,
we get $\text{L.H.S.}=\frac{3\times60}{10}-4=(3\times6)-4$
​​​​​​​ $= 18 - 4 = 14$ and $R.H.S. = 14$
$\therefore$ When $x = 60$
we have $L.H.S. = R.H.S.$
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Question 25 Marks
Solve the following equation and verify the answer:
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
Answer
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
Multiplying each term by $10$, the $L.C.M$. of $5$ and $2,$
we get $\frac{\text{2x}}{5}\times10-\frac{3}{2}\times10$
$=\frac{\text{x}}{2}\times10+1\times10$
$\Rightarrow 4x - 15 = 5x + 10$
$\Rightarrow 4x - 5x = 10 + 15$
(transposing $5x$ to $L.H.S.)$
$\Rightarrow -x = 25$
$\Rightarrow x = -25$
(multiplying both sides by$ -1)$
So, $x = -25$ is a solution of the given equation.
Check: Substituting $x = -25$ in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When$ x = -25$, we have
$L.H.S. = R.H.S.$
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Question 35 Marks
Solve the following equation and verify the answer: $6(1 - 4x) + 7(2 + 5x) = 53$
Answer
$6(1 - 4 x) + 7(2 + 5x) - 53 $
$\Rightarrow 6 - 24x + 14 + 35x = 53$ (Removing brackets)
$\Rightarrow -24x + 35x + 14 + 6 = 53 $
$\Rightarrow 11x + 20 = 53 $
$\Rightarrow 11x = 53 - 20 $
$\Rightarrow 11x = 33$ (Transposing $20$ to $R.H.S.)$
$\Rightarrow\frac{\text{11x}}{11}=\frac{33}{11}$(Dividing both sides by $11)$
$\Rightarrow x = 3$
So, $x = 3$ is a solution of the given equation.
Check: Substituting $x = 3$ in the given equation,
we get, $L.H.S. = 6 (1 - 4 \times 3) + 7(2 + 5 \times 3)$
$= 6(1 - 12) + 7(2 + 15)$
$= 6 \times (-11) + 7 \times 17$
$= -66 + 119 = 53$
$= R.H.S.$
$\therefore$ When $x = 3,$
 we have $L.H.S. = R.H.S.$
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Question 45 Marks
Solve the following equation and verify the answer: $5(x - 1) + 2(x + 3) + 6 = 0$
Answer
$5(x - 1) + 2(x + 3) + 6 = 0$
$\Rightarrow 5(x - 1) + 2(x + 3) = -6$ (Transposing $6$ to $R.H.S.)$
$\Rightarrow 5x - 5 + 2x + 6 = -6$ (Removing brackets)
$\Rightarrow 5x + 2x - 5 + 6 = -6$
$\Rightarrow 7x + 1 = -6$
$\Rightarrow 7x = -6 - 1$ (Transposing $1$ to $R.H.S.)$
$\Rightarrow 7x = -7$
$\Rightarrow\frac{\text{7x}}{7}=\frac{-7}{7}$(Dividing both sides by $7)$
$\Rightarrow x = -1$
So, $x = -1$ is a solution of the given equation.
Check: Substituting $x = -1$ in the given equation,
we get $L.H.S. = 5(-1 - 1) + 2(-1 + 3) + 6$
$= 5 \times (-2) + 2 \times 2 + 6 = 10+ 4 +6$
$= -10 + 10 = 0 = R.H.S$
$\therefore$When $x = -1$, we have
$L.H.S. = R.H.S.$
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Question 55 Marks
Solve the following equation and verify the answer: $\frac{\text{2m}}{3}+8=\frac{\text{m}}{2}-1$
Answer
$\frac{\text{2m}}{3}+8=\frac{\text{m}}{2}-1$
Multiplying each term by $6$, the $L.C.M$. of $2$ and $3,$
we get $\frac{\text{2m}}{3}\times6+8\times6=\frac{\text{m}}{2}\times6-1\times6$
$\Rightarrow 4m + 48 = 3m - 6$
$\Rightarrow 4m - 3m = -6 - 48$ (Transposing $3m$ to $L.H.S$. and $48$ to $R.H.S.)$
$\Rightarrow m = -54$
So, $m = -54$ is a solution of the given equation.
Check: Substituting $m = -54$ in the given equation,
​​​​​​​we get$\text{L.H.S.}=\frac{-54}{2}-1=-27-1=-28$
$=-36+8=-28$
$\text{R.H.S}=\frac{-54}{2}-1=-27-1=-28$
$\therefore$ When $m = -54$, we have $L.H.S. = R.H.S.$
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Question 65 Marks
Solve the following equation and verify the answer: $3(x + 2) - 2(x - 1) = 7$
Answer
$3(x + 2) - 2(x - 1) = 7$
$\Rightarrow 3x + 6 - 2x + 2 = 7$ (Removing brackets)
$3x - 2x + 6 + 2$
$= 7 x + 8$
$= 7 x = 7 - 8$ (Transposing $8$ to $R.H.S.) x = -1$ is a solution of the given equation.
Check: Substituting $x = -1$ in the given equation,
we get $L.H.S. = 3 ( -1 + 2) - 2( -1 - 1)$
$= 3 \times 1 + ( -2 \times - 2)$
$= 3 + 4 = 7$ and $R.H.S. = 7$
When $x = -1$,
we have $L.H.S. = R.H.S.$
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Question 75 Marks
Solve the following equation and verify the answer:$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$
Answer
$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$
Multiplying each term by $12$, the $L.C.M.$ of $4, 6, 2$, we get
$\frac{\text{n}}{4}\times12-5\times12$
$=\frac{\text{n}}{6}\times12+\frac{1}{2}\times12$
$\Rightarrow 3n - 60 = 2n + 6$
$\Rightarrow 3n - 2n = 6 + 60$
(Transposing $2n$ to $L.H.S$. and $-60$ to $R.H.S.)$
$\Rightarrow n = 66$
So, $n = 66$ is a solution of the given equation.
Check: Substituting $n = 66$ in the given equation, we get
$\text{L.H.S.}=\frac{66}{4}-5=\frac{33}{2}-5$
$=\frac{33-10}{2}=\frac{23}{2}$
$\text{R.H.S.}=\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}$
$=\frac{22+1}{2}=\frac{23}{2}$
$\therefore$ When $n = 66,$
we have $L.H.S. = R.H.S.$
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Question 85 Marks
Solve the following equation and verify the answer: $\frac{3}{4}(\text{x}-1)=\text{x}-3$
Answer
$\frac{3}{4}(\text{x}-1)=\text{x}-3$
$\Rightarrow\frac{3}{4}(\text{x}-1)\times4=(\text{x}-3)\times4$
$($Multiplying both sides by $4)$
$\Rightarrow 3(x - 1) = 4(x - 3)$
$\Rightarrow 3x - 3 = 4x - 12$ (Removing brackets)
$\Rightarrow 3x - 4x$
$= -12 + 3 ($Transposing $4x$ to $L.H.S$. and $-3$ to $R.H.S.)$
$\Rightarrow -x = -9$
$\Rightarrow x = 9$
So, $x = 9$ is a solution of the given equation.
Check: Substituting $x = 9$ in the given equation,
we get $\text{L.H.S}=\frac{3}{4}(9-1)=\frac{3}{4}\times8=6$
$\text{R.H.S.}= 9 - 3 = 6$
$\therefore$ When $x = 9,$
we have $L.H.S. = R.H.S.$
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Question 95 Marks
Solve the following equation and verify the answer: $16(3x - 5) - 10(4x - 8) = 40$
Answer
$16(3x - 5) - 10(4x - 8) = 40$
$\Rightarrow 48x - 80 - 40x + 80 = 40 ($ Removing brackets$)$
$\Rightarrow 48x - 40x - 80 + 80 = 40$
$\Rightarrow 8x = 40$
$\Rightarrow\frac{\text{8x}}{8}=\frac{40}{8}$
(Dividing both sides by $8)$
$\Rightarrow x = 5$
So, $x = 5$ is a solution of the given equation
Check: Substituting $x = 5$ in the given equation,
we get $L.H.S. = 16(3 \times 5 - 5) - 10(4 \times 5 - 8)$
$= 16(15 - 5) - 10(20 - 8)$
$= 16 \times 10 - 10 \times 12$
$= 160 - 120$
$= 40 = R.H.S.$
$\therefore$ When $x = 5,$
we have $L.H.S. = R.H.S.$
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Question 105 Marks
Solve the following equation and verify the answer: $3(2 - 5x) - 2(1 - 6x) = 1$
Answer
$3(2 - 5x) - 2(1 - 6x) = 1$
$\Rightarrow 6 - 15x - 2 + 12x = 1$ (Removing brackets)
$\Rightarrow 6 - 2 - 15x + 12x = 1$
$\Rightarrow 4 - 3x$
$= 1 -3x = 1 - 4 ($Transposing $4$ to $R.H.S.)$
$\Rightarrow -3x = -3$
$\Rightarrow\frac{\text{-3x}}{-3}=\frac{-3}{-3}$
(Dividing both sides by $-3)$
$\Rightarrow x = 1$ So, $x = 1$ is a solution of the given equation.
Check: Substituting $x = 1$ in the given equation,
 we get $L.H.S. = 3(2 - 5 \times 1) - 2(1- 6 \times 1)$
$=3(2 - 5) - 2(1 - 6)$
$=[3 \times (-3)] + [-2 \times (-5)]$
$= -9 + 10 = 1 = R.H.S.$
$\therefore$ When x = 1, we have $L.H.S. = R.H.S.$
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Question 115 Marks
Solve the following equation and verify the answer: $\text{2x}-\frac{1}{2}=3$
Answer
$\text{2x}-\frac{1}{2}=3$
$\Rightarrow\text{2x}=3+\frac{1}{2}$ (Transporting $-\frac{1}{2}$ to $R.H.S.)$
$\Rightarrow\text{2x}=\frac{7}{2}$
$\Rightarrow\frac{\text{2x}}{2}=\frac{7}{2}\times\frac{1}{2}$ (Dividing both sides by $2)$
$\Rightarrow\text{x}=\frac{7}{4}$
$\therefore\text{x}=\frac{7}{4}$ is a solution of the given equation.
Check: Substituting $\text{x}=\frac{7}{4}$ in the given equation,
we get $\text{L.H.S.}=2\times\frac{7}{4}-\frac{1}{2}$
$=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3$ and $R.H.S. = 3$
$\therefore$ When $\text{x}=\frac{7}{4},$
we have $L.H.S. = R.H.S.$
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Question 125 Marks
Solve the following equation and verify the answer: $\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
Answer
$\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
$\Rightarrow\frac{\text{x}}{2}-\frac{\text{x}}{3}=1$ (Transpoting $\frac{\text{x}}{3}$
$L.H.S.)$
 Multiplying each term by $6,$ the $L.C.M$. of $2$ and $3,$
we get $\frac{\text{x}}{2}\times6-\frac{\text{x}}{3}\times6=1\times6$
$\Rightarrow3\text{x}-2\text{x}=6$
$\Rightarrow\text{x}=6$
$\therefore\text{x}=6$ is a solution of the given equation.
Check: Substituting $x = 6$ in the given equaation,
we get $\text{L.H.S.}=\frac{6}{2}=3$ and $\text{R.H.S.}=\frac{6}{3}+1=2+1=3$
$\therefore$ When $x = 6$, we have $L.H.S. = R.H.S.$
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Question 135 Marks
Solve the following equation and verify the answer: $3(x + 6) + 2(x + 3) = 64$
Answer
$3(x + 6) + 2 (x + 3) = 64 $
$\Rightarrow 3x + 18 + 2x + 6 = 64$ (Removing brackets) 
$\Rightarrow 3x + 2x + 18 + 6 = 64 $
$\Rightarrow 5x + 24 = 64 $
$\Rightarrow 5x = 64 - 24$ (Transposing $24$ to $R.H.S.) $
$\Rightarrow 5x = 40$
$\Rightarrow\frac{\text{5x}}{5}=\frac{40}{5}$(Dividing both sides by $5)$
$\Rightarrow x = 8$
So, $x = 8$ is a solution of the given equation.
Check: Substituting $x = 8$ in the given equation,
we get $L.H.S. = 3(8 + 6) + 2(8 + 3) = 3 \times 14 + 2 \times 11 = 42 + 22 = 64 = R.H.S.$
$\therefore$ When $x = 8$, we have $L.H.S. = R.H.S$
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Question 145 Marks
Solve the following equation and verify the answer: $\frac{\text{x-3}}{5}=-2=\frac{\text{2x}}{5}$
Answer
$\frac{\text{x-3}}{5}=-2=\frac{\text{2x}}{5}$ multiplying each term by $5$,
we get $\frac{(\text{x}-3)\times5}{5}-(2\times5)=\frac{\text{2x}}{5}\times5$
$\Rightarrow x - 3 - 10 = 2x $
$\Rightarrow x - 13 = 2x $
$\Rightarrow x - 2x = 13$ (Transposing $2x$ to $L.H.S$. and $-13$ to $R.H.S.) $
$\Rightarrow -x = 13 $
$\Rightarrow x = -13$ (Multiplying both sides by -1)
So, $x = -13$ is a solution of the given equation.
Check: Substituting $x = -13$ in the given equation,
we get $\text{L.H.S.}=\frac{-13-3}{5}-2=\frac{-16}{5}-2$$=\frac{-16-10}{5}=\frac{-26}{5}$
$\text{R.H.S.}=\frac{2\times(-13)}{5}=\frac{-26}{5}$
$\therefore$ When $x = -13,$
we have $L.H.S. = R.H.S.$
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