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 : The population of a city is 20000. It increases by 20% during the first year and 30% during the second year. The population after two years will be:

(a) 40000

(b) 32000

(c) 31200

(d) 30000

The correct answers to the above question in:

Answer: (c)

Population of city after two years

= $P(1 + R_1/100)(1 + R_2/100)$

= $20000(1 + 20/100)(1 + 30/100)$

= $20000 × 120/100 × 130/100$ = 31200

Practice percentage (Model 8 population based) Online Quiz

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

In a city, 40% of the people are illiterate and 60% are poor. Among the rich, 10% are illiterate. The percentage of the illiterate poor population is

a) 60

b) 36

c) 40

d) 50

Answer: (a)

Let the population of the city be 100.

Total illiterate people = 40

Poor people = 60; Rich people = 40

Illiterate rich people = ${40 × 10}/100$ = 4

Illiterate poor people = 40 – 4 = 36

Required percent = $36/60 × 100$ = 60%

Question : 2

If a man receives on one-fourth of his capital 3% interest, on two third 5% and on the remainder 11%, the percentage he receives on the whole is

a) 5

b) 4.5

c) 5.5

d) 5.2

Answer: (a)

Required percent

= $1/4 × 3 + 2/3 × 5 + (1 - 1/4 - 2/3) × 11$

= $3/4 + 10/3 + 11/12 = {9 + 40 + 11}/12$ = 5%

Question : 3

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

Question : 4

Present population of a village is 67600. It has been increasing annually at the rate of 4%. What was the population of the village two years ago ?

a) 63000

b) 62500

c) 64756

d) 65200

Answer: (b)

Using Rule 17,

Population of the village two years ago

= $P/(1 + R/100)^2 = 67600/(1 + 4/100)^2$

= ${67600 × 25 × 25}/{26 × 26}$ = 62500

Question : 5

A man received Rs.8,80,000 as his annual salary of the year 2007 which was 10% more than his annual salary in 2006. His annual salary in the year 2006 was

a) Rs.8,00,000

b) Rs.4,80,000

c) Rs.4,00,000

d) Rs.8,40,000

Answer: (a)

Using Rule 17,

Let the man's annual salary in 2006 be Rs.x.

${110x}/100$ = 880000

$x = {880000 × 100}/110$ = Rs.800000

Question : 6

The value of a property decreases every year at the rate of 5%. If its present value is Rs.4,11,540, what was its value 3 years ago ?

a) Rs.4,60,000

b) Rs.4,50,000

c) Rs.4,75,000

d) Rs.4,80,000

Answer: (d)

Using Rule 18,

Value of the property 3 years ago

= $P/{(1 - R/100)^T} = 411540/{(1 - 5/100)^3}$

= ${411540 × 20 × 20 × 20}/{19 × 19 × 19}$ = Rs.480000

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