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

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[4 marks sum]

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Question 14 Marks

A number is 15 more than the other. The sum of of the two numbers is 195. Find the numbers.

Answer

Let the First number $=x$
and the Second number $=y$
According to the condition
$ \begin{aligned} & x=y+15..(i) \\ & x+7=195..(ii) \end{aligned} $
Substitute the eq. (i) in eq. (ii), we get
$ \begin{aligned} & y+15+7=195 \\ & \Rightarrow 2 y=195-15 \\ & \Rightarrow y=\frac{180}{2}=90 \end{aligned} $
Now, substitute the value of $y$ in eq. (i), we get
$ x=90+15=105 $
$\therefore$ The two numbers are 105 and 90

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Question 24 Marks

The sum of two numbers is 18. If one is twice the other, find the numbers.

Answer

Let the first number $=x$
and the second number $=y$
According to the condition
$ x+y=18 $...(i)
and $x=27$...(ii)
Substitute the eq. (ii) in eq. (i), we get
$ \begin{aligned} & 2 y+y=18 \\ & x=2 y=18 \\ & \Rightarrow 3 y=18 \\ & \Rightarrow y=\frac{18}{3}=6 \end{aligned} $
Now, substitute the value of $y$ in eq. (ii), we get
$ x=2 \times 6=12 $
$\therefore$ The two numbers are 12,6

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Question 34 Marks

Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.

Answer

Let the first conscutive number = x,
Second consecutive number = x + 1<
and Third consecutive number = x + 2
According to the condition,
x + x + 1 = 15 + x + 2
⇒ 2x + 1 = 17 +x
⇒ 2x -x = 17 – 1
⇒ x= 16
∴ The first consecutive number = 16
Second consecutive number =16+1 = 17
Third consecutive number =16 + 2=18

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Question 44 Marks

A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.

Answer

Let the present age of the son = x years
Present age of the father = 4x years
After 20 years,
Son’s age will be (x + 20) years
and Father’s age will be (4x + 20) years
According to the condition,
4x + 20 = 2 (x + 20)
4x + 20 = 2x + 40
4x – 2x = 40 – 20
2x = 20
⇒ x = 10
∴ Present age of the son = 10 years and Present age of the father = 4×10 years = 40 years

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Question 54 Marks

The sum of three consecutive odd numbers is 63. Find the numbers.

Answer

Let the first odd number $=x$
second odd number $=x+2$
and third odd number $=x+4$
According to the condition,
$ \begin{aligned} & x+x+2+x+4=63 \\ & 3 x+6=63 \Rightarrow 3 x=63-6 \\ & \Rightarrow 3 x=57 \Rightarrow x=\frac{57}{3}=19 \end{aligned} $
$\therefore$ First odd number $=19$
Second odd number $=19+2=21$
third odd number $=19+4=23$

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Question 64 Marks

The sum of three consecutive even numbers is 54. Find the numbers.

Answer

Let the first even number $=x$
second even number $=x+2$
and third even number $=x+4$
According to the condition,
$ \begin{aligned} & x+x+2+x+4=54 \\ & \Rightarrow 3 x+6=54 \\ & \Rightarrow 3 x=54-6 \\ & \Rightarrow x=\frac{48}{3}=16 \end{aligned} $
$\therefore$ First even number $=16$
Second even number $=16+2=18$
and third even number $=16+4=20$

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Question 74 Marks

Solve: $\frac{1}{2} \mathrm{~m}+\frac{3}{4} \mathrm{~m}-\mathrm{m}=2.5$

Answer

$\begin{aligned} & \frac{1}{2} \mathrm{~m}+\frac{3}{4} \mathrm{~m}-\mathrm{m}=2.5 \\ & \frac{1}{2} \mathrm{~m}+\frac{3}{4} \mathrm{~m}-\frac{\mathrm{m}}{1}=2.5 \\ & \Rightarrow \frac{2 \mathrm{~m}+3 \mathrm{~m}-4 \mathrm{~m}}{4}=2.5 \\ & \Rightarrow \frac{2 \mathrm{~m}+3 \mathrm{~m}-4 \mathrm{~m}}{4}=4 \times 2.5 \\ & \Rightarrow 2 \mathrm{~m}+3 \mathrm{~m}-4 \mathrm{~m}=10 \\ & \Rightarrow 5 \mathrm{~m}-4 \mathrm{~m}=10 \\ & \Rightarrow \mathrm{m}=10\end{aligned}$

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Question 84 Marks

Solve: $\frac{4 \mathrm{x}}{3}-\frac{7 \mathrm{x}}{3}=1$

Answer

$\begin{aligned} & \frac{4 \mathrm{x}}{3}-\frac{7 \mathrm{x}}{3}=1 \\ & \Rightarrow \frac{4 \mathrm{x}-7 \mathrm{x}}{3}=1 \\ & \Rightarrow \frac{-3 x}{3}=1 \\ & \Rightarrow-3 \mathrm{x}=3 \\ & \Rightarrow x=\frac{3}{-3}=-1 \\ & \therefore x=-1\end{aligned}$

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Question 94 Marks

Solve: $\frac{x}{2}+\frac{x}{5}=14$

Answer

$\begin{aligned} & \frac{x}{2}+\frac{x}{5}=14 \\ & \Rightarrow \frac{5 x+2 x}{10}=14 \\ & \Rightarrow 5 x+2 x=14 \times 10 \\ & \Rightarrow 5 x+2 x=140 \\ & \Rightarrow 7 x=140 \\ & \Rightarrow x=\frac{140}{7}=20 \\ & \therefore x=20\end{aligned}$

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Question 104 Marks

Solve: 2x + 20% of x = 12.1

Answer

$\begin{aligned} & 2 x+20 \% \text { of } x=12.1 \\ & \Rightarrow 2 x+\frac{x \times(20)}{100}=12.1 \\ & \Rightarrow 2 x+\frac{20 x}{100}=12.1 \\ & \Rightarrow 2 x+\frac{2 x}{10}=12.1 \\ & \frac{20 x+2 x=121}{10} \\ & \Rightarrow 22 x=121 \\ & \Rightarrow x=\frac{121}{22}=\frac{11}{2} \\ & \therefore x=\frac{11}{2} \text { or } 5 \frac{1}{2}\end{aligned}$

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Question 114 Marks

Solve: $\frac{3 x}{4}+4 x=38$

Answer

$\begin{aligned} & \frac{3 \mathrm{x}}{4}+4 \mathrm{x}=38 \\ & \Rightarrow \frac{3 \mathrm{x}}{4}+\frac{4 \mathrm{x}}{1}=38 \\ & \Rightarrow \frac{3 \mathrm{x}+16 \mathrm{x}}{4}=38 \\ & \Rightarrow 3 \mathrm{x}+16 \mathrm{x}=38 \times 4 \\ & \therefore 3 \mathrm{x}+16 \mathrm{x}=152 \\ & \Rightarrow 19 \mathrm{x}=152 \\ & \Rightarrow \mathrm{x}=\frac{152}{19}=8\end{aligned}$
$\therefore x=8$

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Question 124 Marks

Solve: $\frac{x-1}{5}-\frac{x}{3}=1-\frac{x-2}{2}$

Answer

$\begin{aligned} & \frac{x-1}{5}-\frac{x}{3}=1-\frac{x-2}{2} \\ & \Rightarrow \frac{6 x-6-10 x=30-15 x+30}{30} \\ & \Rightarrow-4 x-6=60-15 x \\ & \Rightarrow 11 x-4 x=60+6 \\ & \Rightarrow x=\frac{66}{11}=6 \\ & \therefore x=6\end{aligned}$

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Question 134 Marks

Solve: 21 – 3 (x – 7) = x + 20

Answer

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

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Question 144 Marks

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

Answer

$ \begin{aligned} & \frac{2 x+1}{3 x-2}=1 \frac{1}{4} \\ & \Rightarrow \frac{2 x+1}{3 x-2}=\frac{5}{4} \end{aligned} $
By cross multiplication
$ \begin{aligned} & (3 x-2) \times 5=4(2 x+1) \\ & \Rightarrow 15 x-10=8 x+4 \\ & \Rightarrow 15 x-8 x=4+10 \\ & \Rightarrow 7 x=14 \\ & \Rightarrow x=\frac{14}{7}=2 \end{aligned} $
$ \therefore x=2 $

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Question 154 Marks

Solve: $\frac{x+1}{3}+\frac{x+4}{5}=\frac{x-4}{7}$

Answer

$\begin{aligned} & \frac{x+1}{3}+\frac{x+4}{5}=\frac{x-4}{7} \\ & \frac{x+1}{3}+\frac{x+4}{5}-\frac{x-4}{7}=0 \\ & \frac{35(x+1)+21(x+4)-15(x-4)}{105}=0 \\ & 35 x+35+21 x+84-15 x+60=0 \\ & 41 x+179=0 \\ & 41 x=-179 \\ & x=\frac{-179}{41} \\ & x=-4 \frac{15}{41}\end{aligned}$

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Question 164 Marks

Solve: $\frac{x-1}{2}-\frac{x-2}{3}-\frac{x-3}{4}=0$

Answer

$\therefore \frac{x-1}{2}-\frac{x-2}{3}-\frac{x-3}{4}=0 $
$ \Rightarrow \frac{6(x-1)-4(x-2)-3(x-3)}{12}=0 $
$ \Rightarrow 6(x-1)-4(x-2)-3(x-3)=0 \times 12 $
$ \Rightarrow 6 x-6-4 x+8-3 x+9=0 $
$ \Rightarrow 6 x-4 x-3 x-6+8+9=0 $
$ \Rightarrow 6 x-7 x=6-8-9 $
$ \Rightarrow-x=6-17 $
$ \Rightarrow-x=-11 $
$\therefore x=11$

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Question 174 Marks

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

Answer

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

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Question 184 Marks

Solve: $\frac{\mathrm{y}+2}{3}+\frac{\mathrm{y}+5}{4}=6$

Answer

$\begin{aligned} & \frac{y+2}{3}+\frac{y+5}{4}=6 \\ & \Rightarrow \frac{4 y+8+3 y+15}{12}=6 \\ & \Rightarrow 4 y+8+3 y+15=6 \times 12 \\ & \Rightarrow 7 y+23=72 \\ & \Rightarrow 7 y=72-23=49 \\ & \Rightarrow y=\frac{49}{7}=7 \\ & \therefore y=7\end{aligned}$

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Question 194 Marks

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

Answer

$\begin{aligned} & \frac{x+4}{2}+\frac{x}{3}=7 \\ & \Rightarrow \frac{3 x+12 x+2 x}{6}=7 \\ & \Rightarrow 3 x+12+2 x=7 \times 6 \\ & \Rightarrow 3 x+12+2 x=42 \\ & \Rightarrow 5 x=42-12=30 \\ & \Rightarrow x=\frac{30}{5}=6 \\ & \therefore x=6\end{aligned}$

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Question 204 Marks

Solve: x – 30% of x = 35

Answer

$\begin{aligned} & x-30 \% \text { of } x=35 \\ & \Rightarrow x-\frac{30}{100} \times x=35 \\ & \Rightarrow \frac{100 x-30 x}{100}=35 \\ & \Rightarrow 100 x-30 y=35 \times 100 \\ & \Rightarrow 100 x-30 x=3500 \\ & \Rightarrow 70 x=3500 \\ & \Rightarrow x=\frac{3500}{70}=50 \\ & \therefore x=50\end{aligned}$

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Question 214 Marks

Solve: 0.6a +0.2a = 0.4 a + 8

Answer

$\begin{aligned} & 0.6 a+0.2 a=0.4 a+8 \\ & \Rightarrow \frac{6}{10} a+\frac{2}{10} a=\frac{4}{10} a+\frac{8}{1} \\ & \Rightarrow \frac{6 a+2 a}{10}=4 a+80 \\ & \Rightarrow 6 a+2 a=4 a+80 \\ & \Rightarrow 6 a+2 a-4 a=80 \\ & \Rightarrow 4 a=80 \\ & \Rightarrow a=\frac{80}{4}=20 \\ & \therefore a=20\end{aligned}$

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Question 224 Marks

Solve: $\frac{2 \mathrm{p}}{3}-\frac{\mathrm{p}}{5}=35$

Answer

$\begin{aligned} & \frac{2 p}{3}-\frac{p}{5}=35 \\ & \Rightarrow \frac{10 p-3 p}{15}=35 \\ & \Rightarrow 10 p-3 p=35 \times 15 \\ & \Rightarrow 10 p-3 p=525 \\ & \Rightarrow 7 p=525 \\ & \Rightarrow p=\frac{525}{7}=75 \\ & \Rightarrow p=75\end{aligned}$

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Question 234 Marks

A sum of $₹ 500$ is in the form of notes of denominations of $₹ 5$ and $₹ 10$. If the total number of notes is $90,$ find the number of notes of each type.

Answer

Let the number of $₹ 5$ notes $=x$
$\therefore$ The number of $\text₹ 10$ notes $=90-x$
Value of $₹ 10$ notes $= x \times ₹ 5 = ₹ 3 x$
and value of $₹ 10$ notes $= (90-x) x\ ₹10 = (900 - 10x)$
$\therefore$ Total value of all the notes $=₹ 500$
$\therefore 5 x+(900-10 x)=500$
$\Rightarrow 5 x+900-10 x=500$
$\Rightarrow-5 x=500-900$
$\Rightarrow x=\frac{400}{5}$
$\Rightarrow x=80$
$\therefore$ The number of $₹ 5$ notes $= x = 80$
and the number of $₹10$ notes $= 90 - x$
$= 90 - 80 =10$

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Question 244 Marks

A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time. Find their present ages.

Answer

Let the present age of the son = x years
and the present age of the father = 3x years
After 12 years,
Son’s age will be (x + 12) years
and father’s age will be (3x + 12) years
According to the condition,
3x + 12 = 2 (x + 12)
3x + 12 = 2x+ 24
3x – 2x = 24 – 12
x = 12
∴Present age of the son = 12 years
and Present age of the father = 3×12 years
= 36 years

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Question 254 Marks

The numerator of a fraction is four less than its denominator. If 1 is added to both, is numerator and denominator, the fraction becomes $\frac{1}{2}$ Find the fraction.

Answer

Let the numerator of a fraction $=x$
and the denominator of a fraction $=y$
According to the condition,
$ x=y-4 $...(i)
and $\frac{x+1}{y+1}=\frac{1}{2}$
$\Rightarrow 2(x+1)=y+1$
$\Rightarrow 2 x+2=y+1$
$\Rightarrow 2 x-y=-1$...(ii)
Substitute the eq.(i) in eq.(ii)
$2(y-4)-y=-1$
$2 y-8-y=-1$
$y=-1+8$
$y=7$
Now, put the value of $y$ in eq.(i), we get
$ x=7-4=3 $
$\therefore$ The numerator of a fraction is 3
and denominator is 7 and the fraction is $\frac{3}{7}$

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Question 264 Marks

The length of a rectangular plot exceeds its breadth by 5 m. If the perimeter of the plot is 142 m, find the length and the breadth of the plot.

Answer

Let the length of a rectangular plot $=x$
and the breadth of a rectangular plot $=y$
According to the condition,
$ x=y+5 $...(i)
and $2(x+y)=142$
$ \Rightarrow x+y=\frac{142}{2}=71 $
$ \Rightarrow x+y=71 $...(ii)
Now, substitute the value of eq. (i) in eq (ii)
$ \begin{aligned} & y+5+y=71 \\ & \Rightarrow 2 y=71-5 \\ & \Rightarrow y=\frac{66}{2}=33 \end{aligned} $
Now, put the value of $y$ in eq. (i)
$ x=33+5=38 $
$\therefore$ The length of rectangular plot is $38 \mathrm{~m}$ and breadth is $33 \mathrm{~m}$.

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Question 274 Marks

The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

Answer

Let the smallest number $=x$
and the largest number $=\mathrm{y}$
According to the condition,
$ y-x=7 $...(i)
$ \text { and } 6 x+y=77 $....(ii)
From eq. (i)
$ y=7+x $...(ii)
Substitute the eq. (iii) in eq. (ii)
$ \begin{aligned} & 6 x+7+x=77 \\ & \Rightarrow 7 x=77-7 \\ & \Rightarrow x=\frac{70}{7}=10 \end{aligned} $
Now, substitute the value of $x$ in eq. (iii)
$ y=7+10=17 $
$\therefore$ The smallest number 10 and the largest number is 17 .

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