Comparing Quantities(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

COMPARING QUANTITIES MCQs

Total Questions : 120 | Page 3 of 12 pages

Question 21. Ram can run 3 times faster than Rahul. If Rahul's speed is 5 km/hr. Then speed of Ram is ___ km/hr.

Discuss Question

:
Ram can run 3 times faster than Rahul
Rahul's speed = 5 km/hr
So Ram's speed = 3 ×5 = 15km/hr

Question 22. Priya was preparing a cake. She mixed 3 cups of flour, 2 cups butter, and 1 cup sugar. The percentage of flour used in the cake is ___.

Discuss Question

:
Total number of cups of ingredients = 6
Number of cups of flour = 3
Therefore, percentage of flour used in cake=

\frac{3}{6}

× 100 = 50%

Question 23. Ram sold a flower vase at 10% loss. The cost price of the vase was ₹ 120. Find the selling price of the flower vase.

₹ 108
₹ 100
₹ 110
₹ 105

Discuss Question

Answer: Option A. -> ₹ 108
:
A
Loss = 10% of 120

= \frac{10}{100} \times 120

= ₹12
Selling Price = C.P - Loss
Selling Price = 120 - 12 = ₹108

Question 24. Anmol drew a circle and divided it in to four equal parts.Express area of the shaded part to the area of the circle in ratio?

Discuss Question

Answer: Option A. -> 14
:
A
Number of shaded parts = 1
Total number of equal parts = 4
Therefore, Ratio of the area of theshaded part to area of number of equal parts

\frac{Area of the Shaded Part}{Area of the Circle} = \frac{1}{4}

Question 25. Shubendhu buys an old scooter for ₹ 4700 and spends ₹ 800 on its repairs. If he sells the scooter for ₹ 5800, what is his gain (in percentage)?

6512%
7711%
6011%
7712%

Discuss Question

Answer: Option C. -> 6011%
:
C
Cost Price (C.P.) = ₹ (4700+800) = ₹ 5500.
Selling Price (S.P.) = ₹ 5800.
Gain = (S.P.)-(C.P.)= ₹ (5800 - 5500) = ₹ 300
Gain%=

\frac{G a i n}{C . P} \times 100 %

\frac{300}{5500} \times 100 %

\frac{60}{11} %

Question 26.

T h e t a b l e b e l o w s h o w s t h e n u m b e r o f c h i l d r e n w i t h d i f f e r e n t h e i g h t s .

\begin{matrix} H e i g h t & 110 c m & 120 c m & 130 c m & 140 c m N o o f c h i l d r e n & 22 & 25 & 21 & 32 \end{matrix}

What is the percentage of students whose height is more than 125 cm?
___

Discuss Question

:
Number of students whose height is more than 125cm = 21 + 32 = 53
Total number of students = 22 + 25 + 21 + 32 = 100
Percentage of students whose height is more than 125cm =

\frac{53}{100}

× 100 = 53%

Question 27. After recording the height and weight, Ted wanted to know who is taller? Him or Ned. For that, he asked Ned’s height. He told that it was 135 cm and Ted’s height was 4.5 ft. Given 1 foot = 0.3 meters, what will be the value of

\frac{N e d^{'} s h e i g h t}{T e d^{'} s h e i g h t}

100
30
1
0.01

Discuss Question

Answer: Option C. -> 1
:
C
To compare two similar quantities, their units must be same. Ned’s height is in cm and Ted’s height is in ft. In order to compare those heights, we need to convert both the heights to same units.
Thus,

4.5 f t = 4.5 \times 0.3 m = 1.35 m

Now, 1m = 100 cm
Thus,

4.5 f t = 1.35 m = 1.35 \times 100 c m = 135 c m

Thus,

\frac{N e d^{'} s h e i g h t}{T e d^{'} s h e i g h t} = \frac{135}{135} = 1

Question 28. Find the ratio of 5 km to 500 m.

1 : 100
10 : 1
100 : 1
1 : 1000

Discuss Question

Answer: Option B. -> 10 : 1
:
B
Since,
1 km = 1000 m and 5 km = 5000 m
Then, required ratio

= \frac{5000}{500}

= 10 : 1

Question 29. A fruit seller sold 40% of his apples and was left with 420 apples. What is the number of apples he had initially?

Discuss Question

Answer: Option C. -> 700
:
C
Let the number of apples he had initially be x.
Given, (100 - 40)% of x = 420
60% of x = 420

\frac{60}{100} \times x = 420

0.6x = 420
Hence,

x = \frac{420}{0.6}

\frac{4200}{6}

= 700
Therefore, the numberof apples he had initially = 700

Question 30. Ram drew a line segment of length 10 cm. He told Arjun that he has scaled down. The scale used by him is 2cm = 100m. What is the actual length of the line segment?