3 Marks Question - MATHS STD 7 Questions - Vidyadip

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19 questions · timed · auto-graded

Question 13 Marks

There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?

Answer

Let the number of children be x.
Let the number of donkeys be y.
x + y = 28
y = 28 – x $\quad$……(i)
Children have 2 feet, but donkeys have 4 feet.
According to the question
2x + 4y = 80
⇒ 2x + 4(28 – x) = 80 [From (i)]
⇒ 2x + 112 – 4x = 80
⇒ -2x + 112 = 80
⇒ -2x = 80 – 112
⇒ -2x = -32
⇒ x = 16
There are 16 children, and (28 – 16) = 12 donkeys.

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Question 23 Marks

The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?

Answer

Let the height of a giraffe be x m.
According to the question
$x m=2 \frac{1}{2} m+\frac{x}{2} m$
⇒ $x=\frac{5}{2}+\frac{x}{2}$
$\begin{array}{l}\Rightarrow \quad \frac{x}{1}-\frac{x}{2}=\frac{5}{2} \\ \Rightarrow \quad \frac{2 x-x}{2}=\frac{5}{2}\end{array}$
$\begin{array}{ll}\Rightarrow & \frac{x}{2}=\frac{5}{2} \\ \Rightarrow & x=\frac{5}{2} \times 2 \\ \Rightarrow & x=5\end{array}$
∴ Height of the giraffe = 5 m.

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Question 33 Marks

The Bakhshali Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second, and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?

Answer

Let the amount given to the first person be x.
Amount given to second person = 2x
Amount given to third person = 3(2x) = 6x
Amount given to fourth person= 4(6x) = 24x
Total amount = 132
⇒ x + 2x + 6x + 24x = 132
⇒ 33x = 132
$\Rightarrow x=\frac{132}{4}$
⇒ x = 4
The amount given to the first person is 4.

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Question 43 Marks

Find the measure of the angle of these triangle:

Answer

In ∆ABC,

∠A + ∠B + ∠C = 180° [Sum of angles of a triangle is 180°]
⇒ x + x – 10 + x + 10 = 180°
⇒ 3x = 180°
⇒ x = 60°
∠A = 60°, ∠B = 60° – 10° = 50° and ∠C = 60° + 10° = 70°

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Question 53 Marks

Find the measure of the angle of these triangle:

Answer

In ∆ABC,

AB = AC [Given]
∠B = ∠C [Angles opposite to equal sides in a triangle are equal]
So ∠B = y + 15
∠A + ∠B + ∠C = 180° [Angle sum property of ∆]
⇒ y + y + 15 + y + 15 = 180°
⇒ 3y + 30 = 180°
⇒ 3y = 180 – 30
⇒ 3y = 150
⇒ y = 50
∠A = 50°, ∠B = 50° + 15° = 65°, and ∠C = 50° + 15° = 65°

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Question 63 Marks

The steps to solve equation are shown below. Identify and correct any mistakes:
4x – 5 = 9x + 8
4x = 9x + 8 – 5
4x = 9x + 3
4x – 9x = 3
-5x = 3
$x=\frac{-5}{3}$

Answer

4x – 5 = 9x + 8
4x = 9x + 8 – 5 → Error
4x = 9x + 3
4x – 9x = 3
-5x = 3
$x=\frac{-5}{3} \rightarrow$ Error
Correction
4x – 5 = 9x + 8
4x = 9x + 8 + 5
4x = 9x + 13
4x – 9x = 13
-5x = 13
$x=\frac{-13}{5}$
$x=-2 \frac{3}{5}$

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Question 73 Marks

Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.

Answer

28p – 36 = 98
⇒ 28p = 98 + 36
⇒ 28p = 134
$\Rightarrow p=\frac{134}{28}$
$\Rightarrow p=\frac{67}{14}$
Now to find the value of
(i) 14p – 19 = 14 × $\frac{67}{14}$ – 19
= 67 – 19
= 48
(ii) 28p – 38 = 28 × $\frac{67}{14}$ – 38
= 134 – 38
= 96

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Question 83 Marks

In a restaurant, a fruit juice costs ₹ 15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹ 600, find the cost of the fruit juice and milkshake.

Answer

Let the cost of a chocolate milkshake be ₹ x.
Cost of fruit juice = ₹(x – 15)
Cost of 4 fruit juices and 7 chocolate milkshakes = ₹ 600
According to the question
4(x – 15) + 7x = 600
⇒ 4x – 60 + 7x = 600
⇒ 4x + 7x = 600 + 60
⇒ 11x = 660
⇒ x = 60
Cost of 1 chocolate milkshake = ₹ 60
Cost of 1 fruit juice = ₹ 60 – ₹ 15 = ₹ 45

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Question 93 Marks

The figure shows the diagram for a window with a grill. What is the gap between the two rods in the grill?

Answer

Total height of the window, including the frame = 34 cm
Total height of the window excluding the frame = 34 cm – 3 cm – 3 cm = 28 cm
Width of one rod of grill = 2 cm
No. of rods of the grill = 5
Total width of 5 rods = 5 × 2 cm = 10 cm
Let the height of 1 gap be x cm.
No. of gaps = 6
Total height of gaps = 6x
According to the question
10 + 6x = 28
⇒ 6x = 28 – 10
⇒ 6x = 18
⇒ x = 3
Gap between two rods = 3 cm.

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Question 103 Marks

What are the inputs to these machines?

Answer

(i) $\xrightarrow{45} \div 3 \xrightarrow{15} \div 3=5$
Let the unknown number be a.
(a ÷ 3) ÷ 3 = 5
⇒ a ÷ 3 = 5 × 3
⇒ a ÷ 3 = 15
⇒ a = 15 × 3
⇒ a = 45
The unknown number is 45.
(ii) $\xrightarrow{-3}-4 \xrightarrow{-7}-4=-11$
Let the unknown number be b.
(b – 4) – 4 = -11
⇒ b – 4 – 4 = -11
⇒ b – 8 = -11
⇒ b = -11 + 8
⇒ b = -3
The unknown number is -3.

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Question 113 Marks

In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem.

Answer

Let each black blob hide the blue dots.
There are 3 black blobs and 4 blue dots in the picture.
We are given total blue dots = 25
According to the question,
3u + 4 = 25
⇒ 3u = 25 – 4
⇒ 3u = 21
⇒ u = 7
Each black blob hides 7 blue dots.

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Question 123 Marks

Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer pays her in 50 and 100-rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?

Answer

Let no. of ₹ 50 notes with Ranju be x.
Then no. of ₹ 100 notes with Ranju is also x.
Total money with Ranju = ₹ 750
According to the question,
Value of ₹ 50 notes = ₹ 50x
Value of ₹ 100 notes = ₹ 100x
₹ 50x + ₹ 100x = ₹ 750
⇒ 150x = 750
⇒ x = 5
No. of notes of ₹ 50 is 5.
No. of notes of ₹ 100 is 5.

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Question 133 Marks

One quarter of a number increased by 9 gives the same number. What is the number?

Answer

Let the number be x.
According to the question,
$\begin{array}{l}\frac{1}{4} \times x +9= x \\ \frac{x}{4}+9= x \end{array}$
Subtract $\frac{x}{4}$ from both sides
$\frac{x}{4}+9-\frac{x}{4}=\frac{x}{1}-\frac{x}{4}$
$\Rightarrow 9=\frac{4 x-x}{4}$ [LCM of 1 and 4 = 4]
$\Rightarrow 9=\frac{3 x}{4}$
Divide both sides by $\frac{3}{4}$
$\begin{array}{l}\Rightarrow 9 \div \frac{3}{4}=\frac{3 x}{4} \div \frac{3}{4} \\ \Rightarrow 9 \times \frac{4}{3}=\frac{3 x}{4} \times \frac{4}{3}\end{array}$
⇒ 12 = x
The required number is 12.

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Question 143 Marks

The weight of a brick is 1kg more than half its weight. What is the weight of the brick?

Answer

Let the weight of a brick be x kg.
According to the question,
$X =\frac{x}{2}+1$
Subtract $\frac{x}{2}$ from both sides
$\frac{x}{1}-\frac{x}{2}=\frac{x}{2}+1-\frac{x}{2}$
$\begin{array}{l}\Rightarrow \frac{2 x-x}{2}=1 \\ \Rightarrow \frac{x}{2}=1\end{array}$
Multiply both sides by 2
$\frac{x}{2} \times 2=1 \times 2$
$\Rightarrow x=2$
∴ Weight of the brick = 2 kg

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Question 153 Marks

I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. My ten’s digit is 3 less than my unit’s digit. The sum of all three digits is 15. Who am I?

Answer

Let the ten’s place digit be x.
Then hundred’s place digit = x – 3
Unit’s digit = x + 3
Sum of digits = 15
According to the question,
(x – 3) + x + (x + 3) = 15
⇒ 3x = 15
Divide both sides by 3
$\frac{3 x}{3}=\frac{15}{3}$
⇒ x = 5
Ten’s digit = 5
Hundred’s digit = 5 – 3 = 2
Unit’s digit = 5 + 3 = 8
Number formed = 258

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Question 163 Marks

Solve these equation and check the solution.
$\frac{u}{15}=6$

Answer

$\frac{u}{15}=6$
Multiply both sides by 15
$\frac{u}{15} \times 15=6 \times 15$
$\Rightarrow u=90$
Check:
LHS $=\frac{u}{15}$
$\frac{90}{15}=6$
RHS = 6
LHS = RHS
Hence checked.

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Question 173 Marks

Solve these equation and check the solution.
3u – 7 = 2u + 3

Answer

3u – 7 = 2u + 3
Bring the unknown terms to one side
By subtracting 2u from both sides
⇒ 3u – 7 – 2u = 2u + 3 – 2u
⇒ u – 7 = 3
Add 7 to both sides
⇒ u – 7 + 7 = 3 + 7
⇒ u = 10
Check: For u = 10
LHS = 3u – 7
= 3 × 10 – 7
= 30 – 7
= 23
RHS = 2u + 3
= 2 × 10 + 3
= 20 + 3
= 23
LHS = RHS
Hence checked.

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Question 183 Marks

Solve these equation and check the solution.
5s = 3s

Answer

Bring unknown terms on one side by subtracting 3s
⇒ 5s – 3s = 3s – 3s
⇒ 2s = 0
Divide by 2 on both sides 2
$\Rightarrow \frac{2 s}{2}=\frac{0}{2}$
⇒ s = 0
Check:
LHS = 5s = 5 × 0 = 0
RHS = 3s = 3 × 0 = 0
LHS = RHS
Hence checked.

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Question 193 Marks

Solve these equation and check the solution.
3x – 10 = 35

Answer

3x – 10 = 35
Add 10 on both sides
3x – 10+ 10 = 35 + 10
⇒ 3x = 45
Divide by 3 on both sides
$\Rightarrow \frac{3 x}{3}=\frac{45}{3}$
⇒ x = 15
Check:
LHS = 3x – 10 for x = 15
= 3 × 15 – 10
= 45 – 10
= 35
LHS = RHS
Hence checked.

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3 Marks Question - MATHS STD 7 Questions - Vidyadip