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

7th Grade > Mathematics

COMPARING QUANTITIES MCQs

Total Questions : 120 | Page 7 of 12 pages

Ashin invested an amount of Rs. 13,900 divided into two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B (in rupees)?

___

Discuss Question

Answer: Option A. ->
:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then

$\frac{14 \times x \times 2}{100}$ + $\frac{(13900 - x) \times 11 \times 2}{100}$ =3508

$\frac{28 x + 13900 \times 22 - 22 x}{100}$ = 3508
$28 x + 13900 \times 22 - 22 x = 3508 \times 100$
28x - 22x = 350800 - (13900 × 22)
6x = 45000
x = 7500
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400

Question 62.

The cost price of 20 articles is the same as the selling price of x articles.The value of x is (given profit =20%).

Discuss Question

Answer: Option A. ->
:

Let C.P. of each article be Rs. 1

=> C.P. of x articles=Rs.x.

S.P. of x articles=Rs. 20.

Profit=Rs.(20 - x).

$\frac{20 - x}{x} \times 100$ =25.

2000 - 100x=25x.

125x=2000.

$\Rightarrow$ x=16.

Question 63.

$T h e t a b l e b e l o w s h o w s t h e n u m b e r o f c h i l d r e n w i t h d i f f e r e n t h e i g h t s .$ $\begin{matrix} H e i g h t & 110 c m & 120 c m & 130 c m & 140 c m N o o f c h i l d r e n & 22 & 25 & 21 & 32 \end{matrix}$

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

___

Discuss Question

Answer: Option A. ->
:

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 64.

Ram can type 50 words per minute. If the ratio of words typed by Rahul and Ram is 1 : 2, then Rahul can type _______ words per minute.

Discuss Question

Answer: Option A. ->
:

Given the ratio of words typed by Rahul to Ram = 1:2

Let the number of words that Rahul can type be $x$

∴ $\frac{1}{2}$ = $\frac{N u m b e r o f w o r d s t y p e d b y R a h u l}{N u m b e r o f w o r d s t y p e d b y R a m}$

$\frac{1}{2}$ = $\frac{N u m b e r o f w o r d s t y p e d b y R a h u l}{50}$

The number of words typed by Rahul = $\frac{1}{2}$ × Number of words typed by Ram

= $\frac{1}{2}$ × 50 = 25

Rahul can type 25 words per minute.

Question 65.

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 number of apples he had initially = 700

Question 66.

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 67.

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)?

$\frac{65}{12} %$
$\frac{77}{11} %$
$\frac{60}{11} %$
$\frac{77}{12} %$

Discuss Question

Answer: Option C. ->

\frac{60}{11} %

:
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 68.

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?

50m
500m
600m
1000m

Discuss Question

Answer: Option B. -> 500m
:
B

Since 2cm is equivalent to 100m,

1 cm is equivalent to $\frac{100}{2}$ = 50m.

Therefore, actual length of the line = 10 × 50 = 500m.

Question 69.

$\frac{15}{25}$ is equivalent to ____.

5
$\frac{1}{5}$
$\frac{3}{5}$
$\frac{5}{3}$

Discuss Question

Answer: Option C. ->

\frac{3}{5}

:
C

$\frac{15}{25}$ = $\frac{3}{5}$ . [By dividing both numerator and denominator by 5]

Question 70.

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