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

MOST IMPORTANT quantitative aptitude - 11 EXERCISES

Top 10,000+ Aptitude Memory Based Exercises

The following question based on percentage topic of quantitative aptitude

Questions : The value of a machine depreciates by 5% every year. If its present value is Rs.2,00,000, its value after 2 years will be

(a) Rs.1,99,000

(b) Rs.1,80,500

(c) Rs.1,80,000

(d) Rs.2,10,000

The correct answers to the above question in:

Answer: (b)

Using Rule 18,

$A = P(1 - R/100)^T$

= $200000(1 - 5/100)^T$

= $200000 × 19/20 × 19/20$ = Rs.80500

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

Of the 1000 inhabitants in a town 60% are males of whom 20% are literate. If of all the inhabitants, 25% are literate, then what percentage of the females of the town are ilterate ?

a) 32.5

b) 27.5

c) 37.5

d) 22.5

Answer: (a)

Population of town = 1000

Males ⇒ 600; Females ⇒ 400

Literate males = ${600 × 20}/100$ = 120

Total literate inhabitants

= ${1000 × 25}/100$ = 250

Literate females = 250 – 120 = 130

Required percent = $130/400$ × 100 = 32.5%

Question : 2

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

Question : 3

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 : 4

The value of a machine depreciates every year by 10%. If its present value is Rs.50,000 then the value of the machine after 2 years is _________.

a) Rs.45,000

b) Rs.40,050

c) Rs.40,005

d) Rs.40,500

Answer: (d)

Using Rule 18,

Required value = $50000(1 - 10/100)^2$

= $50000 × {9 × 9}/100$ = Rs.40500

Question : 5

The population of a village increases by 5% annually. If its present population is 4410, then its population 2 years ago was

a) 4000

b) 4500

c) 3800

d) 3500

Answer: (a)

Using Rule 17,

If the population of village two years ago be $P_0$,

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

4410 = $P_0(1 + 5/100)^2$

4410 = $P_0(1 + 1/20)^2$

4410 = $P_0(21/20)^2$

4410 = ${441P_0}/400$

$P_0 = {4410 × 400}/441$ = 4000

Question : 6

In a factory, the production of cycles rose to 48, 400 from 40,000 in 2 years. The rate of growth per annum is

a) 8%

b) 9%

c) 10.5%

d) 10%

Answer: (d)

Using Rule 17,

If the rate of increase per annum be R%, then

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

48400 = 40000$(1 + R/100)^2$

$484/400 = (1 + R/100)^2$

$121/100 = (11/10)^2 = (1 + R/100)^2$

1 + $R/100 = 11/10$

$R/100 = 11/10 - 1 = 1/10$

$R = 100/10$ = 10% per annum

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