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MATHS 3 Marks Question

Linear Equations in One Variable · Gujarat Board (GSEB) STD 8 (GSEB - English Medium). 20 questions.

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Question 13 Marks
Simplify and solve the linear equation:$15(y – 4 ) – 2(y – 9) + 5(y + 6) = 0.$
Answer
$15( y – 4 ) – 2( y– 9) + 5(y + 6) = 0$
$ \therefore 15y – 60 – 2y + 18 + 5y + 30 = 0$
$ \therefore 18y – 12 = 0$
$ \therefore 18y = 12 ...$ [Transposing $–12 to R.H.S.]$
$\therefore y = \frac{{12}}{{18}} ...$ [Dividing both sides by $18]$
$\therefore y = \frac{2}{3}$ this is the required solution.
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Question 23 Marks
Simplify and solve the linear equation $3(t – 3) = 5(2t + 1).$
Answer
$3 (t – 3) = 5 (2t + 1)$
$ \therefore 3t – 9 = 10t + 5$
$ \therefore 3t – 10t = 5 + 9 ...$ [Transposing $10t$ to $L.H.S$. and $–9$ to $R.H.S.]$
$ \therefore –7t = 14$
$ \therefore t = -\frac{{14}}{7} ...$ [Dividing both sides by $–7]$
$ \therefore t = –2$
this is the required solution.
Verification,
$L.H.S. = 3(t – 3) = 3(–2 – 3) = 3(–5) = –15$
$R.H.S. = 5(2t + 1) = 5(2 \times (–2) + 1) = 5(– 4 + 1)$
$= 5(–3) = –15 = L.H.S$
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Question 33 Marks
Solve the linear equation: $m - \frac{{m - 1}}{2} = 1 - \frac{{m - 2}}{3}$
Answer
m - $\frac{{m - 1}}{2} = 1 - \frac{{m - 2}}{3}$
It is a linear equation since it involves linear expressions only.
$\therefore$ $m - \frac{m}{2} + \frac{1}{2} = 1 - \frac{m}{3} + \frac{2}{3}$
$\therefore$ $m - \frac{m}{2} + \frac{m}{3} = 1 + \frac{2}{3} - \frac{1}{2}$ ... [Transposing $\frac{{ - m}}{3}$ to $L.H.S$. and $\frac{1}{2}$ to $R.H.S.]$
$\therefore$ $\frac{{6m - 3m + 2m}}{6} = \frac{{6 + 4 - 3}}{6}$
$\therefore$ $\frac{{5m}}{6} = \frac{7}{6}$
$\therefore$ m = $\frac{7}{6} \times \frac{6}{5}$ ... [Multiplying both sides by $\frac{6}{5}$]
$\therefore$ m = $\frac{7}{5}$ this is the required solution.
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Question 43 Marks
Solve the linear equation $\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t$.
Answer
$\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t$ It is a linear equation since it involves linear expressions only.
$\therefore \frac{3}{4}t - \frac{2}{4} - \frac{2}{3}t - \frac{3}{3} = \frac{2}{3} - t$
$\therefore \frac{3}{4}t - \frac{1}{2} - \frac{2}{3}t - 1 = \frac{2}{3} - t$
$\therefore \frac{3}{4}t - \frac{2}{3}t + t = \frac{2}{3} + \frac{1}{2} + 1$ ... [Transposing –t to $L.H.S.$ and $ - \frac{1}{2}$ and –1 to $R.H.S.]$
$\therefore \frac{{9t - 8t + 12t}}{{12}} = \frac{{4 + 3 + 6}}{6}$
$\therefore \frac{{13t}}{{12}} = \frac{{13}}{6}$
$\therefore t = \frac{{13}}{6} \times \frac{{12}}{{13}}$ ... [Multiplying both sides by $\frac{{12}}{{13}}$]
$\therefore t = 2$ this is the required solution.
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Question 53 Marks
Solve the linear equation $\frac{{x - 5}}{3} = \frac{{x - 3}}{5}$.
Answer
$\frac{{x - 5}}{3} = \frac{{x - 3}}{5}$
It is a linear equation since it involves linear expressions only.
$\therefore$ $\frac{x}{3} - \frac{5}{3} = \frac{x}{5} - \frac{3}{5}$
$\therefore$ $\frac{x}{3} - \frac{x}{5} = \frac{3}{5} + \frac{5}{3}$ ... [Transposing $\frac{x}{5}$ to $L.H.S$. and $\frac{{ - 5}}{3}$ to $R.H.S.]$
$\therefore \frac{{5x - 3x}}{{15}} = \frac{{25 - 9}}{{15}}$
$\therefore \frac{{2x}}{5} = \frac{{16}}{{15}}$
$\therefore x = \frac{{16}}{{15}} \times \frac{{15}}{2}$ ... [Multiplying both sides by $\frac{{15}}{2}$]
$\therefore x = 8$
this is the required solution.
Verification,
L.H.S. = $\frac{{8 - 5}}{3} = \frac{3}{3} = 1$
R.H.S. = $\frac{{8 - 3}}{5} = \frac{5}{5} = 1$
Therefore, $L.H.S. = R.H.S.$
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Question 63 Marks
Solve the linear equation $x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}$
Answer
$x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}$
It is a linear equation since it involves linear expressions only.
$\therefore$ $x - \frac{{8x}}{3} + \frac{{5x}}{2} = \frac{{17}}{6} - 7$ ... [Transposing $ - \frac{{5x}}{2}$ to $L.H.S.$ and $7$ to $R.H.S]$
$\therefore$ $\frac{{6x - 16x + 15x}}{6} = \frac{{17 - 42}}{6}$
$\therefore$ $\frac{{5x}}{6} = \frac{{ - 25}}{6}$
$\therefore$ $x = \frac{{ - 25}}{6} \times \frac{6}{5}$ ... [Multiplying both sides by $\frac{6}{5}$]
$\therefore x = -5$ this is the required solution.
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Question 73 Marks
Solve the linear equation $\frac{n}{2} - \frac{{3n}}{4} + \frac{{5n}}{6} = 21$.
Answer
$\frac{n}{2} - \frac{{3n}}{4} + \frac{{5n}}{6} = 21$ It is a linear equation since it involves linear expressions only.
$\therefore \frac{{6n - 9n + 10n}}{{12}}$ $= 21 ... [L.C.M. (2,4,6) = 12]$
$\therefore \frac{{7n}}{{12}}$ = 21
$\therefore n = 2$1 $\times$ $\frac{{12}}{7}$ ... [Multiplying both sides by $\frac{{12}}{7}$]
$\therefore n = 36$ this is the required solution.
Verification,
$L.H.S. = \frac{1}{2} \times 36 - \frac{3}{4} \times 36 + \frac{5}{6} \times 36$
$= 18 – 27 + 30$
$= 21$
$= R.H.S.$
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Question 83 Marks
Solve the linear equation $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$.
Answer
$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
It is a linear equation since it involves linear expressions only.
$\therefore$ $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$ ... [Transposing $\frac{x}{3}$ to $L.H.S.$ and $ - \frac{1}{5}$ to $R.H.S]$
$\therefore$ $\frac{{3x - 2x}}{6} = \frac{{5 + 4}}{{20}}$
$\therefore$ $\frac{x}{6} = \frac{9}{{20}}$
$\therefore$ x = $\frac{9}{{20}} \times 6$ ... [Multiplying both sides by $6]$
$\therefore$ x = $\frac{{27}}{{10}}$
this is the required solution.
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Question 93 Marks
Solve the equation and check your result: $2y + \frac{5}{3} = \frac{{26}}{3} - y$
Answer
$2y + \frac{5}{3} = \frac{{26}}{3} - y$
$2y + y = \frac{{26}}{3} - \frac{5}{3}$ ... [Transposing –y to $L.H.S$. and $\frac{5}{3}$ to $R.H.S]$
$\therefore 3y = \frac{{26 - 5}}{3}$
$\therefore 3y = \frac{{21}}{3}$
$\therefore 3y = 7$
$\therefore$ y = $\frac{7}{3}$ ... [Dividing both sides by $3$]
This is the required solution.
Verification
$L.H.S. = 2\left( {\frac{7}{3}} \right) + \frac{5}{3} = \frac{{14}}{3} + \frac{5}{3} = \frac{{14 + 5}}{3} = \frac{{19}}{3}$
$R.H.S. = \frac{{25}}{3} - \frac{7}{3} - = \frac{{26 - 7}}{3} = \frac{{19}}{3}$
Therefore, $L.H.S = R.H.S$
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Question 103 Marks
Solve the equations and check your result: $\frac{{2x}}{3} + 1 = \frac{{7x}}{{15}} + 3$
Answer
$\frac{{2x}}{3} + 1 = \frac{{7x}}{{15}} + 3$
$\frac{{2x}}{3} - \frac{{7x}}{{15}} = 3 - 1 ...$ [Transposing $\frac{{7x}}{{15}}$ to $L.H.S$. and $1$ to $R.H.S.]$
$\therefore \frac{{2x}}{3} - \frac{{7x}}{{15}} = 2$
$\therefore 15\left( {\frac{{2x}}{3} - \frac{{7x}}{{15}}} \right)$ = 2 $\times 15 ... $[Multiplying both sides by $15]$
$\therefore 10x – 7x = 30$
$\therefore 3x = 30$
$\therefore x = \frac{{30}}{3}$ ... [Dividing both sides by $3]$
$\therefore x = 10$ this is the required solution.
Verification,
$L.H.S. = \frac{{2x}}{3} + 1 = \frac{2}{3}(10) + 1 = \frac{{20 + 3}}{3} = \frac{{23}}{3}$
$R.H.S. = \frac{{7x}}{{15}} + 3 = \frac{7}{{15}}(10) + 3 = \frac{{70}}{{15}} + 3 = \frac{{70 \div 5}}{{15 \div 5}} + 3 = \frac{{14 + 9}}{3} = \frac{{23}}{3}$
Therefore, $L.H.S. = R.H.S.$
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Question 113 Marks
Solve the equations and check your result: $x = \frac{4}{5}(x + 10)$
Answer
$x = \frac{4}{5}(x + 10)$
$5x = 4(x + 10) ... $[Multiplying both sides by $5]$
$\therefore 5x = 4x + 10$
$\therefore 5x – 4x = 40 ... $[Transposing $4x$ to $L.H.S.]$
$\therefore x = 40$ this is the required solution.
Verification
$L.H.S. = 40$
$R.H.S. = \frac{4}{5}(x + 10)$
$\frac{4}{5} = (40 + 10) = \frac{4}{5}(50) = 4(10) = 40$
Therefore, $L.H.S. = R.H.S.$
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Question 123 Marks
Solve the equation and check your result: $8x + 4 = 3(x – 1) + 7$
Answer
$8x + 4 = 3(x – 1) + 7$
$ \therefore 8x + 4 = 3x – 3 + 7$
$ \therefore 8x + 4 = 3x + 4$
$ \therefore 8x – 3x = 4 – 4 ...$ [Transposing $3x$ to $L.H.S.$ and $4$ to $R.H.S.]$
$\therefore 5x = 0$
$ \therefore x = \frac{0}{5} ... $[Dividing both sides by $5]$
$ \therefore x = 0$ this is the required solution.
Verification,
$L.H.S. = 8x + 4 = 8(0) + 4 = 4$
$R.H.S. = 3(x – 1) + 7 = 3(0 – 1) + 7 = 3(–1) + 7 = –3 + 7 = 4$
Therefore, $L.H.S = R.H.S$
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Question 133 Marks
Solve the equation and check your result: $3m = 5m - \frac{8}{5}$
Answer
$3m = 5m - \frac{8}{5}$
$\therefore 3m = 5m - \frac{8}{5}$ ... [Transposing $5m$ to $L.H.S]$
$\therefore -2m = -\frac{8}{5}$
$\therefore m = \frac{{ - 8}}{{5 \times ( - 2)}} = \frac{4}{5}$ ... [Dividing both sides by $–2]$
this is the required solution.
Verification,
$L.H.S = 3 \times \frac{4}{5} = \frac{{12}}{5}$
$R.H.S. = 5\left( {\frac{4}{5}} \right) - \frac{{12}}{5} = 4 - \frac{8}{5} = \frac{{20 - 8}}{5} = \frac{{12}}{5}$
Therefore, $L.H.S = R.H.S$
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Question 143 Marks
The sum of three consecutive multiples of $11$ is $363.$ Find these multiples.
Answer
Let the three consecutive multiples of $11$ as $x, x + 11$ and $x + 22.$
It is given that the sum of these consecutive multiples of $11$ is $363$. Thus,
$x + (x + 11) + (x + 22) = 363$
or $x + x + 11 + x + 22 = 363$
or $3x + 33 = 363$
or $3x = 363 - 33$
or $3x = 330$
or $x = \frac{330}{3} = 110$
Hence, the three consecutive ultiples are $110, 121, 132.$
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Question 153 Marks
Bansi has $3$ times as many two$-$rupee coins as he has five-rupee coins. If he has in all a sum of $₹ 77$, how many coins of each denomination does he have?
Answer
Let the number of five$-$rupee coins that Bansi has be $x$.
Then the number of two$-$rupee coins he has is $3$ times $x$ or $3x.$
The amount Bansi has:
$i.$ from ₹ $5$ rupee coins, ₹ $5 \times x = ₹ 5x$
$ii.$ from ₹ $2$ rupee coins, ₹ $2 \times 3x = ₹ 6x$
Hence the total money he has $= ₹ 11x$
But this is given to be ₹ $77$; therefore,
$11x = 77$
or $x=\frac{77}{11}=7$
Thus, number of five$-$rupee coins $= x = 7$
and number of two$-$rupee coins $= 3x = 21$
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Question 163 Marks
The present age of Sahil’s mother is three times the present age of Sahil. After $5$ years their ages will add to $66$ years. Find their present ages.
Answer
Let Sahil’s present age be $x$ years.
  Sahil Mother
Present age $x$ $3x$
Age after $5$ years $x + 5$ $3x + 5$
It is given that after $5$ years their age sum is $66$ years.
Therefore, $x + 5 + 3x + 5 = 4x + 10 = 66$
$4x = 66 - 10$
or $4x = 56$
or $x=\frac{56}{4}=14$
Thus, Sahil’s present age is $14$ years and his mother’s age is $= 3 \times 14 = 42$ years.
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Question 173 Marks
The perimeter of a rectangle is $13\ cm$ and its width is $2{3\over 4}\ cm$. Find its length.
Answer
Assume the length of the rectangle to be $x \ cm.$
The perimeter of the rectangle $= 2 \times$ (length + width)
$= 2 \times\left(x+2 \frac{3}{4}\right)$
$= 2\left(x+\frac{11}{4}\right)$
The perimeter is given to be $13 \ cm$. Therefore,
$2\left(x+\frac{11}{4}\right)=13$
or $x+\frac{11}{4}=\frac{13}{2}$ (dividing both sides by $2$)
or $x=\frac{13}{2}-\frac{11}{4}$ = $\frac{15}{4}=3 \frac{3}{4}$
The length of the rectangle is $3\frac{3}{4}$cm.
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Question 183 Marks
Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.
Answer
Let us take Shriya’s present age to be $x$ years.
Then Arjun’s present age would be $2x$ years.
Shriya’s age five years ago was $(x - 5)$ years.
Arjun’s age five years ago was $(2x - 5)$ years.
It is given that Arjun’s age five years ago was three times Shriya’s age.
Thus, $2x – 5 = 3(x - 5)$
or $2x – 5 = 3x – 15$
or $15 – 5 = 3x – 2x$
or $10 = x$
So, Shriya’s present age $= x = 10$ years.
Therefore, Arjun’s present age $= 2x = 2 \times 10 = 20$ years.
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Question 193 Marks
The digits of a two-digit number differ by $3$. If the digits are interchanged, and the resulting number is added to the original number, we get $143$. What can be the original number?
Answer
Take, for example, a two-digit number, say, $56$. It can be written as $(10 \times 5) + 6.$
If the digits in 56 are interchanged, we get $65$, which can be written as $(10 \times 6 ) + 5.$
Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by $3$. Let us take it as $b + 3$. So the two-digit number is $10(b + 3) + b = 10b + 30 + b = 11b + 30.$
With interchange of digits, the resulting two-digit number will be $10b + (b + 3) = 11b + 3$
So,
$(11b + 30) + (11b + 3) = 143$
$22b + 33 = 143$
$22b = 143 - 33$
$22b = 110$
$b = \frac{110}{22}$
$b = 5$
The units digit is 5 and therefore the tens digit is $5 + 3$ which is $8$. Thus, the number is $85.$
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Question 203 Marks
Deveshi has a total of $₹ 590$ as currency notes in the denominations of $₹50, ₹20$ and $₹10$. The ratio of the number of $₹50$ notes and $₹20$ notes is $3:5$. If she has a total of $25$ notes, how many notes of each denomination she has?
Answer
Let the number of $₹ 50$ notes and $₹ 20$ notes be $3x$ and $5x$, respectively.
But she has $25$ notes in total.
Therefore, the number of ₹$10$ notes $= 25 – (3x + 5x) = 25 – 8x$
The amount she has
from $₹50$ notes : $3x \times 50 = ₹150x$
from $₹20$ notes : $5x \times 20 = ₹ 100x$
from $₹10$ notes : $(25 – 8x) \times 10 = ₹(250 - 80x)$
Hence the total money she has $= 150x + 100x + (250 – 80x) = ₹(170x + 250)$
But she has ₹ $590$. Therefore, $170x + 250 = 590$
or $170x = 590 – 250 = 340$
or $x = \frac {340}{170} = 2$
The number of ₹ $50$ notes she has $= 3x = 3 \times 2 = 6$
The number of ₹ $20$ notes she has $= 5x = 5 \times 2 = 10$
The number of ₹ $10$ notes she has $= 25 - 8x = 25 – (8 \times 2) = 25 - 16 = 9$
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