Model 8 population based Practice Questions Answers Test with Solutions & More Shortcuts

Question : 26 [SSC CISF ASI 2010]

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 : 27

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

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

Question : 28

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 : 29 [SSC CISF Constable 2011]

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 : 30 [SSC CGL Tier-I 2015]

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

IMPORTANT quantitative aptitude EXERCISES

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