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 : 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

The correct answers to the above question in:

Answer: (b)

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

= ${21000 × 95}/100$ = Rs. 19950

Practice percentage (Model 8 population based) Online Quiz

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

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 : 2

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 : 3

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 : 4

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

Question : 5

The population of a village has increased annually at the rate of 25%. If at the end of 3 years it is 10,000, the population in the beginning of the first year was

a) 5000

b) 5120

c) 4900

d) 4500

Answer: (b)

Using Rule 17,

Population in the beginning of the year

=$\text"Population after 3 years"/(1 + \text"Rate"/100)^{ \text"Time"}$

= $10000/(1 + 25/100)^3= 10000/(5/4)^3$

= ${10000 × 64}/125 = 5120$

Question : 6

The value of a property depreciates every year by 10% of its value at the beginning of the year. The present value of the property is Rs.8100. What was its value 2 years ago ?

a) Rs.$(90/100)^2 × 8100$

b) Rs.10,000

c) Rs. $(100/110)^2 × 8100$

d) Rs.9801

Answer: (b)

Using Rule 18,

Suppose the value of property two years ago was Rs.x

According to question $x(1 - 10/100)^2 = 8100$

$x(90/100)^2 = 8100$

$x ={8100 × 10 × 10}/{9 × 9}$ = Rs.10000

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