Model 8 population based Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 11 EXERCISES

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The following question based on percentage topic of quantitative aptitude

Questions : If the population of a town is 64000 and its annual increase is 10%, then its correct population at the end of 3 years will be :

(a) 85184

(b) 80000

(c) 85000

(d) 85100

The correct answers to the above question in:

Answer: (a)

Using Rule 17,

Population of town = $P(1 + R/100)^T$

= $64000(1 + 10/100)^3$

= $64000 × 11/10 × 11/10 × 11/10$ = 85184

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

In a town, the population was 8000. In one year, male population increased by 10% and female population increased by 8% but the total population increased by 9%. The number of males in the town was :

a) 4500

b) 4000

c) 5000

d) 6000

Answer: (b)

By Alligation Rule

Men : Women = 1 : 1

Number of men = $1/2 × 8000 = 4000$

Question : 2

If population of women in a village is 90% of population of men, what is the population of men as a percentage of population of women ?

a) 105%

b) 100%

c) 108%

d) 111%

Answer: (d)

If the number of men be 100,then

Number of women = 90

Required per cent = $100/90 × 100 ≈ 111%$

Question : 3

The population of a town increases by 5% every year. If the present population is 9261, the population 3 years ago was

a) 5700

b) 8000

c) 6000

d) 7500

Answer: (b)

Using Rule 17,

P = $P_0(1 + R/100)^T$

9261 = $P_0(1 + 5/100)^3$

9261 = $P_0(1 + 1/20)^3$

9261 = $P_0(21/20)^3$

$P_0 = {9261 × 20 × 20 × 20}/{21 × 21 × 21}$ = 8000

Question : 4

The value of an equipment depreciates by 20% each year. How much less will the value of the equipment be after 3 years ?

a) 51.2%

b) 48.8%

c) 54%

d) 60%

Answer: (b)

If the present worth of the equipment be Rs.100, then

its price after 3 years = 100 × $(80/100)^3$ = Rs.51.2

Depriciation = 48.8%

Using Rule 18,

If the population/cost of a town/article is P and it decreases/reduces at the rate of r% annually, then,

  1. After ‘t' years population/cost = $P(1 - r/100)^t$
  2. Before ‘t' years population/cost = $P/{(1 - r/100)^t}$

Let the price of equipment be Rs.100

Its price after 3 years. = $100(1 - 20/100)^3$

= $100 × (80/100)^3$ = Rs.51.2

Depriciation = 48.8%

Question : 5

A TV was bought at a price of Rs.21,000. After one year the value of TV was depreciated by 5%. Find the value of the TV after one year.

a) Rs.20,950

b) Rs.19,950

c) Rs.18,950

d) Rs.17,950

Answer: (b)

Value of TV after one year = 21000 × (100 – 5)%

= ${21000 × 95}/100$ = Rs. 19950

Question : 6

The population of a town is 9000. It the number of females increases by 5% and the males by 7.5%, what will be the total population after increase. The number of females currently is 3000.

a) 9200

b) 9600

c) 10500

d) 9540

Answer: (b)

In the village,

Females = 3000; Males = 9000 – 3000 = 6000

After respective increases,

Population of village

= $3000 × 105/100 + {6000 × 107.5}/100$

= 3150 + 6450 = 9600

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