Comparing Quantities(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

COMPARING QUANTITIES MCQs

Total Questions : 59 | Page 1 of 6 pages

Question 1. Find the compound interest on ₹ 20,000 for 3 years at 10% per annum compounded annually.

₹ 6,620
₹6,000
₹ 26,620
₹26,000

Discuss Question

Answer: Option A. -> ₹ 6,620
:
A

A = P \times [1 + \frac{r}{100}]^{n}

, where

P

is the principal,

r

is the rate of interest and

n

is the time period.
C.I = A - P
Here, P = ₹20,000, r= 10 %,and n = 3 years

A = 20, 000 \times [1 + \frac{10}{100}]^{3}

A = 20, 000 \times [\frac{11 \times 11 \times 11}{10 \times 10 \times 10}]

A =₹26,620
So, C.I= 26,620 - 20,000= ₹6,620

Question 2. According to a survey conducted in a city, 25% were doctors, 30% engineers, 15% businessmen and rest were illiterates. What is the number of illiterates if the total population of the city is 5000?

2000
1500
1000
500

Discuss Question

Answer: Option B. -> 1500
:
B
Given: Out of total population, 25% were doctors, 30% engineers, 15% businessmen and rest were illiterates.
Hence, percentage of illiterates

= 100 - (25 + 30 + 15) = 30 %

So, number of illiterates

= 30 %

of 5000

= \frac{30}{100} \times 5000

= 1500

Question 3. The difference between compound interest and simple interest on an amount of ₹15,000 for 2 years is ₹96. What is the rate of interest per annum?

Discuss Question

Answer: Option C. -> 8%
:
C
Given that, principal,

P = ₹ 15000

and time period,

n = 2

years .
Let the rate of interest per annum be

r

.
Simple interest in 2 years

S . I = \frac{15000 \times r \times 2}{100}

\Rightarrow S . I = 300 r

Compound interest

C . I = A - P = 15000 (1 + \frac{r}{100})^{2} - 15000

, here,

A

is the amount.
Given that

C . I - S . I = 96

(Since C.I > S.I)

\Rightarrow 15000 (1 + \frac{r}{100})^{2} - 15000 - 300 r = 96

\Rightarrow 15000 (1 + \frac{r}{100})^{2} - 300 r = 15096

\Rightarrow 15000 (1 + \frac{r^{2}}{10000} + \frac{2 r}{100}) - 300 r = 15096

\Rightarrow 15000 + \frac{15 r^{2}}{10} + 300 r - 300 r = 15096

\Rightarrow \frac{15 r^{2}}{10} = 96

\Rightarrow 15 r^{2} = 960

\Rightarrow r^{2} = 64

\Rightarrow r = \pm 8

∵

Rate of interest cannot be negative, so

r = 8 %

Question 4. What must be added to each term of the ratio 2 : 3, so that it becomes equal to 4 : 5?

Discuss Question

Answer: Option B. -> 2
:
B
Let the number to be added be

x,

then

(2 + x) : (3 + x) = 4 : 5

\Rightarrow \frac{2 + x}{3 + x} = \frac{4}{5} \Rightarrow 5 (2 + x) = 4 (3 + x) \Rightarrow 10 + 5 x = 12 + 4 x \Rightarrow 5 x - 4 x = 12 - 10 \Rightarrow x = 2

Question 5. A man invests ₹ 5000 at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 5600. The rate of interest per annum is
___ %.

Discuss Question

:
Since the tenure is 1 year,the compound interest will be same as simple interest.
Amount = Principal + Interest
Interest = 5600 - 5000 = 600
Interest for one year

= \frac{P R T}{100} = 600

R = \frac{100 \times 600}{5000 \times 1} = 12 %

Question 6. Ravi has purchased a vehicle at ₹ 60,000. He sold it at a price

30 %

less than its value to Amit. Amit spent

5 %

of the amount he purchased in repairs. Find the percentage decrease of the price of the vehicle, including the repair price also.

30%
26.5%
30.5%
32%

Discuss Question

Answer: Option B. -> 26.5%
:
B
Amount at which Ravi sold the vehicle

= 60, 000 - (60, 000 \times \frac{30}{100}) = 42, 000

Amount spent by Amit on repairs

= 42, 000 \times \frac{5}{100} = 2, 100

Total amount spent by Amit

= 42, 000 + 2, 100 = 44, 100

Decrease in price of vehicle

= 60, 000 - 44, 100 = 15, 900

Percentage decrease

= \frac{15900}{60000} \times 100 = 26.5 %

Question 7. The compound interest on a certain sum of money at

5 %

per annum for two years is ₹246. Calculate the simple interest on the same sum for three years at

6 %

per annum.

₹430
₹432
₹442
₹452

Discuss Question

Answer: Option B. -> ₹432
:
B

We know that, C . I = P [{(1 + \frac{r}{100})}^{n} - 1] where Pis the principal, ris the rate of interest and tis the time period.

Hence, 246 = P [{(1 + \frac{5}{100})}^{2} - 1]

246 = P [(\frac{21}{20}) \times (\frac{21}{20}) - 1] 246 = P \times (\frac{41}{400})

\Rightarrow P = \frac{246 \times 400}{41} = ₹ 2400

Now, P= ₹ 2400, r = 6 % per annum andt = 3 years

S . I = \frac{P \times r \times t}{100} = \frac{2400 \times 6 \times 3}{100} = ₹ 432

Question 8. A table marked at ₹15000 is available for ₹14400. What is the discount %?

Discuss Question

Answer: Option A. -> 4%
:
A

Discount % = \frac{Marked price - Selling price}{Marked price} \times 100

= \frac{15000 - 14400}{15000} \times 100 = 4 %

Question 9. There are 6 apples and 6 oranges. Three more apples should be added so that the ratio of apples to oranges becomes 3 : 2.

True
False
30.5%
32%

Discuss Question

Answer: Option A. -> True
:
A
There are 6 apples and 6 oranges.
The ratio is

\frac{6}{6}

, that is 1:1.
Let the number of apples added be

x .

Then,

\frac{6 + x}{6} = \frac{3}{2}

\Rightarrow 6 + x = \frac{3}{2} \times 6

\Rightarrow x = 9 - 6

\Rightarrow x = 3

Hence, 3 apples should be added.
Thus, the given statement is true.

Question 10. I bought a second-hand car for ₹50,000, had it transported for ₹5,000 and sold it for ₹59,400. What is my profit %?

Discuss Question

Answer: Option D. -> 8%
:
D
Profit is the difference between the selling price and total cost price
Total cost price = Cost price + Transportation costs = ₹50,000 + ₹5,000 = ₹55,000
Selling price = ₹59,400
Profit = ₹59400 - ₹55000 = ₹4400