Free Comparing Quantities 03 Practice Test - 7th grade

Question 1

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)?

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SOLUTION

Solution :
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 2

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

SOLUTION

Solution :
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 3

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

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

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SOLUTION

Solution :
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 4

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.

SOLUTION

Solution :
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 5

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

708

600

700

605

SOLUTION

Solution : 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 6

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)?

$\frac{65}{12} %$

$\frac{77}{11} %$

$\frac{60}{11} %$

$\frac{77}{12} %$

SOLUTION

Solution : 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 7

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

SOLUTION

Solution : 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 8

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

$\frac{1}{5}$

$\frac{3}{5}$

$\frac{5}{3}$

SOLUTION

Solution : C

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

Question 9

Find the ratio of 5 km to 500 m.

1 : 100

10 : 1

100 : 1

1 : 1000

SOLUTION

Solution : B

Since,
1 km = 1000 m and 5 km = 5000 m

Then, required ratio
$= \frac{5000}{500}$ = 10 : 1

Question 10

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

C. 1

D. 0.01

SOLUTION

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

Free Comparing Quantities 03 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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