[3 marks sum] - MATHS STD 7 Questions - Vidyadip

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[3 marks sum]

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31 questions · timed · auto-graded

Question 13 Marks

Find a number whose one-third part exceeds its one-fifth part by 20.

Answer

Let the number $=x$
According to the condition
$ \begin{aligned} & \frac{1}{3} \mathrm{x}-\frac{1}{5} \mathrm{x}=20 \\ & \Rightarrow \frac{5 \mathrm{x}-3 \mathrm{x}}{15}=20 \ldots(\text { LCM of } 3, \text { and } 5=15) \\ & \Rightarrow \frac{2 \mathrm{x}}{15}=20 \\ & \Rightarrow \mathrm{x}=\frac{20 \times 15}{2} \\ & \Rightarrow \mathrm{x}=150 \end{aligned} $
$\therefore$ Number $=150$

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Question 23 Marks

The difference between a number and one- fourth of itself is 24, find the number.

Answer

Let the number $=x$
According to the condition
$ \begin{aligned} & x-\frac{1}{4} x=24 \\ & \Rightarrow \frac{4 x-x}{x}=24 \\ & \Rightarrow \frac{3 x}{4}=24 \\ & \Rightarrow x=24 \times \frac{4}{3} \\ & \Rightarrow x=8 \times 4 \\ & \Rightarrow x=32 \\ & \therefore \text { Number }=32 \end{aligned} $

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Question 33 Marks

If 5 is subtracted from three times a number, the result is 16. Find the number.

Answer

Let the number $=x$
According to the condition,
$ \begin{aligned} & 3 x-5=16 \\ & \Rightarrow 3 x=16+5 \\ & \Rightarrow 3 x=21 \\ & \Rightarrow x=\frac{21}{3} \\ & \Rightarrow x=7 \end{aligned} $
$\therefore$ The number $=7$

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Question 43 Marks

A man has ₹ x from which he spends ₹6. If twice of the money left with him is ₹86, find x.

Answer

Let the total amount be $x$
According to the condition
$ \begin{aligned} & 2 x=86 \\ & \Rightarrow x=\frac{86}{2} \\ & \Rightarrow x=43 \end{aligned} $
Amount spent by him $=6$
$\therefore$ Total money he have = ₹ 43+ ₹ 6 = ₹ 49

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Question 53 Marks

Solve: $\frac{2 x}{3}+\frac{x}{2}-\frac{3 x}{4}=1$

Answer

$\begin{aligned} & \frac{2 x}{3}+\frac{x}{2}-\frac{3 x}{4}=1 \\ & \Rightarrow \frac{8 x+6 x-9 x}{12}=1 \\ & \Rightarrow 8 x+6 x-9 x=12 \times 1 \\ & \Rightarrow 8 x+6 x-9 x=12 \\ & \Rightarrow 14 x-9 x=12 \\ & \Rightarrow 5 x=12 \\ & \Rightarrow x=\frac{12}{5}=2 \frac{2}{5}\end{aligned}$

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Question 63 Marks

Solve: $\mathrm{y}+\frac{\mathrm{y}}{2}=\frac{7}{4}-\frac{\mathrm{y}}{4}$

Answer

$\begin{aligned} & y+\frac{y}{2}=\frac{7}{4}-\frac{y}{4} \\ & \Rightarrow \frac{y}{1}+\frac{y}{2}=\frac{7}{4}-\frac{y}{4} \\ & \Rightarrow \frac{4 y+2 y=7-y}{4} \\ & \Rightarrow 4 y+2 y=7-y \\ & \Rightarrow 4 y+2 y \Rightarrow 7 y=7 \\ & \Rightarrow y=\frac{7}{7}=1 \\ & \therefore y=1\end{aligned}$

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Question 73 Marks

Solve: $\frac{\mathrm{x}-4}{7}=\frac{\mathrm{x}+3}{7}+\frac{\mathrm{x}+4}{5}$

Answer

$\begin{aligned} & \frac{\mathrm{x}-4}{7}=\frac{\mathrm{x}+3}{7}+\frac{\mathrm{x}+4}{5} \\ & \Rightarrow \frac{5 \mathrm{x}-20=5 x+15+7 x+28}{35} \\ & \Rightarrow 5 \mathrm{x}-5 \mathrm{x}-7 \mathrm{x}=15+28+20 \\ & \Rightarrow-7 \mathrm{x}=63 \\ & \Rightarrow \mathrm{x}=\frac{63}{-7}=-9 \\ & \therefore \mathrm{x}=-9\end{aligned}$

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Question 83 Marks

Solve: $\frac{2 x-3}{3}-(x-5)=\frac{x}{3}$

Answer

$\begin{aligned} & \frac{2 x-3}{3}-(x-5)=\frac{x}{3} \\ & \Rightarrow \frac{2 x-3}{3}-\frac{x-5}{1}=\frac{x}{3} \\ & \Rightarrow \frac{2 x-3-3 x+15=x}{3} \\ & \Rightarrow-x+12=x \\ & \Rightarrow 12=x+x \\ & \Rightarrow 2 x=12 \\ & \therefore x=\frac{12}{2}=6\end{aligned}$

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Question 93 Marks

Solve: $\frac{3 x-2}{7}-\frac{x-2}{4}=2$

Answer

$\begin{aligned} & \frac{3 x-2}{7}-\frac{x-2}{4}=2 \\ & \Rightarrow \frac{12 x-8-7 x+14}{28}=2 \\ & \Rightarrow \frac{5 x+6}{28}=2 \\ & \Rightarrow 5 x+6=28 \times 2 \\ & \Rightarrow 5 x+6=56 \\ & \Rightarrow 5 x=56-6=50 \\ & \Rightarrow x=\frac{50}{5}=10 \\ & \therefore x=10\end{aligned}$

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Question 103 Marks

Solve: 15 – 2 (5 - 3x) = 4 (x - 3) + 13

Answer

$ \begin{aligned} & 15-2(5-3 x)=4(x-3)+13 \\ & \Rightarrow 15-10+6 x=4 x-12+13 \\ & \Rightarrow 6 x-4 x=-12+13-15+10 \\ & \Rightarrow 2 x=23-27=-4 \\ & \Rightarrow x=\frac{-4}{2}=-2 \end{aligned} $
Hence $x=-2$

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Question 113 Marks

Solve: $\frac{1}{2}(x+5)-\frac{1}{3}(x-2)=4$

Answer

$\begin{aligned} & \frac{1}{2}(x+5)-\frac{1}{3}(x-2)=4 \\ & \Rightarrow \frac{3(x+5)-2(x-2)}{6}=4 \\ & \Rightarrow 3 x+15-2 x+4=4 \times 6 \\ & \Rightarrow 3 x+15-2 x+4=24 \\ & \Rightarrow 3 x-2 x=24-15-4 \\ & \Rightarrow x=24-19=5 \\ & \therefore x=5\end{aligned}$

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Question 123 Marks

Solve: $\frac{x}{5}+2 x=33$

Answer

$\begin{aligned} & \frac{\mathrm{x}}{5}+2 \mathrm{x}=33 \\ & \frac{\mathrm{x}+10 \mathrm{x}}{5}=33 \\ & \Rightarrow \frac{11 \mathrm{x}}{5}=33 \\ & \Rightarrow 11 \mathrm{x}=33 \times 5=165 \\ & \Rightarrow x=\frac{165}{11}=15 \\ & \therefore x=15\end{aligned}$

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Question 133 Marks

Solve: y + 20% of y = 18

Answer

$\begin{aligned} & y+20 \% \text { of } y=18 \\ & \Rightarrow y+\frac{20}{100} \times y=18 \\ & \Rightarrow \frac{100 y+20 y}{100}=18 \\ & \Rightarrow 100 y+20 y=18 \times 100 \\ & \Rightarrow 100 y+20 y=1800 \\ & \Rightarrow 120 y=1800 \\ & \Rightarrow y=\frac{1800}{120}=15 \\ & \therefore y=15\end{aligned}$

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Question 143 Marks

Solve: p + 1.4p = 48

Answer

$\begin{aligned} & p+1.4 p=48 \\ & \Rightarrow p+\frac{14}{10} p=48 \\ & \Rightarrow \frac{10 p+14 p}{10}=48 \\ & \Rightarrow 10 p+14 p=48 \times 10 \\ & \Rightarrow 10 p+14 p=480 \\ & \Rightarrow 24 p=480 \\ & \Rightarrow p=\frac{480}{24}=20 \\ & \therefore p=20\end{aligned}$

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Question 153 Marks

Solve: $\frac{3 a}{4}+\frac{a}{6}=66$

Answer

$\begin{aligned} & \frac{3 a}{4}+\frac{a}{6}=66 \\ & \Rightarrow \frac{9 a+2 a}{12}=66 \\ & \Rightarrow(9 a+2 a)=66 \times 12 \\ & \Rightarrow 9 a+2 a=792 \\ & \Rightarrow 11 a=792 \\ & \Rightarrow a=\frac{792}{11}=72 \\ & \Rightarrow a=72\end{aligned}$

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Question 163 Marks

Solve: $\frac{\mathrm{x}}{2}+\mathrm{x}=9$

Answer

$\begin{aligned} & \frac{x}{2}+x=9 \\ & \frac{x}{2}+\frac{x}{1}=9 \\ & \frac{x+2 x}{2}=9 \\ & \Rightarrow x+2 x=2 \times 9 \\ & \Rightarrow 3 x=18 \\ & \Rightarrow x=\frac{18}{3}=6 \\ & \therefore x=6\end{aligned}$

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Question 173 Marks

5y + 8 = 8y – 18

Answer

$\begin{aligned} & 5 y+8=8 y-18 \\ & \Rightarrow 5 y-8 y=-18-8 \\ & \Rightarrow-3 y=-26 \\ & \Rightarrow y=\frac{-26}{-3}=\frac{26}{3} \\ & \therefore y=8 \frac{2}{3}\end{aligned}$

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Question 183 Marks

x (x + 5) = x² +x + 32

Answer

$\begin{aligned} & x(x+5)=x^2+x+32 \\ & \Rightarrow x^2+5 x=x^2+x+32 \\ & \Rightarrow x^2+5 x-x^2-x=32 \\ & \Rightarrow 4 x=32 \\ & \Rightarrow x=\frac{32}{4}=8 \\ & \therefore x=8\end{aligned}$

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Question 193 Marks

9b – 4b + 3b = 16

Answer

$\begin{aligned} & 9 b-4 b+3 b=16 \\ & \Rightarrow(9-4+3) b=16 \\ & \Rightarrow 8 b=16 \\ & \Rightarrow b=\frac{16}{8}=2 \\ & \therefore b=2\end{aligned}$

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Question 203 Marks

21 – 3 (b - 7) = b+ 20

Answer

$\begin{aligned} & 21-3(b-7)=b+20 \\ & \Rightarrow 21-3 b+21=b+20 \\ & \Rightarrow-3 b+42=b+20 \\ & \Rightarrow-3 b-b=20-42 \\ & \Rightarrow-4 b=-22 \\ & \Rightarrow b=\frac{-22}{-4}=\frac{11}{2} \\ & \therefore b=\frac{11}{2}=5 \frac{1}{2}\end{aligned}$

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Question 213 Marks

(x - 4) (2x + 3 ) = 2x²

Answer

$\begin{aligned} & (x-4)(2 x+3)=2 x^2 \\ & \Rightarrow x(2 x+3)-4(2 x+3)=2 x^2 \\ & \Rightarrow 2 x^2+3 x-8 x-12=2 x^2 \\ & \Rightarrow 2 x^2+3 x-8 x-2 x^2=12 \\ & \Rightarrow-5 x=12 \\ & \Rightarrow x=\frac{12}{-5}=\frac{-12}{5} \\ & \therefore x=-2 \frac{2}{5}\end{aligned}$

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Question 223 Marks

7 (x - 2) = 2 (2x - 4)

Answer

$\begin{aligned} & 7(x-2)=2(2 x-4) \\ & \Rightarrow 7 x-14=4 x-8 \\ & \Rightarrow 7 x-4 x=-8+14 \\ & \Rightarrow 3 x=6 \\ & \Rightarrow x=\frac{6}{3}=2 \\ & \therefore x=2\end{aligned}$

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Question 233 Marks

-3 (a - 6) = 24

Answer

$\begin{aligned} & -3(a-6)=24 \\ & \Rightarrow-3 a+18=24 \\ & \Rightarrow-3 a=24-18=6 \\ & \Rightarrow a=\frac{6}{-3}=-2 \\ & \therefore a=-2\end{aligned}$

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Question 243 Marks

4 ( x- 2 ) = 12

Answer

$\begin{aligned} & 4(x-2)=12 \\ & \Rightarrow 4 x-8=12 \\ & \Rightarrow 4 x=12+8=20 \\ & \Rightarrow x=\frac{20}{4}=5\end{aligned}$
$\therefore \mathrm{x}=5$

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Question 253 Marks

9 (a+ 5) + 2 = 11

Answer

$\begin{aligned} & 9(a+5)+2=11 \\ & \Rightarrow 9 a+45+2=11 \\ & \Rightarrow 9 a=11-45-2 \\ & \Rightarrow 9 a=11-47=-36 \\ & \Rightarrow a=\frac{-36}{9}=-4 \\ & \therefore a=-4\end{aligned}$

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Question 263 Marks

6(x+4) = 36

Answer

$\begin{aligned} & 6(x+4)=36 \\ & \Rightarrow 6 x+24=36 \\ & \Rightarrow 6 x=36-24=12 \\ & \Rightarrow x=\frac{12}{6}=2\end{aligned}$
$\therefore x=2$

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Question 273 Marks

0.4x – 1.2 = 0.3x + 0.6

Answer

$\begin{aligned} & 0.4 x-1.2=0.3 x+0.6 \\ & \Rightarrow 0.4 x-0.3 x=0.6+1.2 \\ & \Rightarrow 0.1 x=1.8 \\ & \Rightarrow \frac{1}{10} x=\frac{18}{10} \\ & \Rightarrow x=\frac{18}{10} \times \frac{10}{1}=18 \\ & \therefore x=18\end{aligned}$

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Question 283 Marks

6 – 5m – 1 + 3m = 0

Answer

$\begin{aligned} & 6-5 m-1+3 m=0 \\ & \Rightarrow-5 m+3 m=-6+1 \\ & \Rightarrow-2 m=-5 \\ & \Rightarrow m=\frac{-5}{-2}=\frac{5}{2} \\ & \therefore m=\frac{5}{2}=2 \frac{1}{2}\end{aligned}$

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Question 293 Marks

Solve the following equation:
$\frac{5}{12} m-12=48$

Answer

$\begin{aligned} & \frac{5}{12} \mathrm{~m}-12=48 \\ & \Rightarrow \frac{5}{12} \mathrm{~m}=48+12=60 \\ & \Rightarrow 5 \mathrm{~m}=60 \times 12=720 \\ & \Rightarrow \mathrm{m}=\frac{720}{5}=144 \\ & \therefore \mathrm{m}=144\end{aligned}$

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Question 303 Marks

Solve the following equation:
$\frac{7}{10} x+6=41$

Answer

$\begin{aligned} & \frac{7}{10} x+6=41 \\ & \Rightarrow \frac{7}{10} x=41-6=35 \\ & \Rightarrow 7 x=35 \times 10=350 \\ & \Rightarrow x=\frac{350}{7}=50 \\ & \therefore x=50\end{aligned}$

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Question 313 Marks

Solve the following equation:
$y+5=8 \frac{1}{4}$

Answer

$\begin{aligned} & \mathrm{y}+5=8 \frac{1}{4} \\ & \Rightarrow \mathrm{y}+5=\frac{33}{4} \\ & \Rightarrow \mathrm{y}=\frac{33}{4}-\frac{5}{1} \\ & =\frac{33-20}{4}=\frac{13}{4} \\ & \therefore \mathrm{y}=\frac{13}{4}=3 \frac{1}{4}\end{aligned}$

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[3 marks sum] - MATHS STD 7 Questions - Vidyadip