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 present price of a scooter is Rs. 7,290. If its value decreases every year by 10%, then its value 3 years back was

(a) Rs. 8,000

(b) Rs. 10, 500

(c) Rs. 10,000

(d) Rs. 11,500

The correct answers to the above question in:

Answer: (c)

Using Rule 18,

A = $P(1 - R/100)^3$

7290 = $P(1 - 10/100)^3 = P(9/10)^3$

7290 = P $× 9/10 × 9/10 × 9/10$

P = ${7290 × 10 × 10 × 10}/{9 × 9 × 9}$ = Rs. 10000

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

The population of a village was 9800. In a year, with the increase in population of males by 8% and that of females by 5%, the population of the village became 10458. What was the number of males in the village before increase ?

a) 4410

b) 4200

c) 5600

d) 6048

Answer: (c)

Let the number of males = x

Number of females= 9800 – x

According to the question,

$x × 108/100 + (9800 - x) × 105/100 = 10458$

108 x + 9800 ×105 – 105x = 1045800

3x + 1029000 = 1045800

3x = 1045800 – 1029000 = 16800

$x = 16800/3 = 5600$

Question : 2

The population of a town 2 years ago was 62,500. Due to migration to big cities, it decreases every year at the rate of 4%. The present population of the town is:

a) 56,700

b) 57,600

c) 76,000

d) 75,000

Answer: (b)

Using Rule 18,

Let the present population be P.

P = $62500(1 - 4/100)^2$

= $62500 × 24/25 × 24/25$ = 57600

Question : 3

The value of a machine is Rs.6,250. It decreases by 10% during the first year, 20% during the second year and 30% during the third year. What will be the value of the machine after 3 years?

a) Rs.3,050

b) Rs.2,650

c) Rs.3,150

d) Rs.3,510

Answer: (c)

Using Rule 28,

Population after ‘n' years = $P(1 ± R_1/100)(1 ± R_2/100)…(1 ± R_n/100)$

Required price of the machine

= $6250(1 - 10/100)(1 - 20/100)(1 - 30/100)$

= $6250 × 90/100 × 80/100 × 70/100$ = Rs.3150

Question : 4

The population of a town increases each year by 4% of its total at the beginning of the year. If the population on 1st January 2001 was 500000, what was it on 1st January, 2004 ?

a) 652432

b) 562432

c) 465223

d) 564232

Answer: (b)

Using Rule 17,

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

= $500000(1 + 4/100)^3$

= $500000 × (1 + 1/25)^3$

= $500000 × 26/25 × 26/25 × 26/25$ = 562432

Question : 5

An epidemic broke out in a village in which 5% of the population died. Of the remaining, 20% fled out of panic. If the present population is 4655, then the population of the village originally was

a) 6125

b) 6000

c) 5955

d) 5995

Answer: (a)

Using Rule 28,

Population after ‘n' years = $P(1 ± R_1/100)(1 ± R_2/100)…(1 ± R_n/100)$

Original population of village = x (let)

According to the question,

$x × 95/100 × 80/100 = 4655$

$x = {4655 × 100 × 100}/{95 × 80}$ = 6125

Question : 6

The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

a) 54,000

b) 53,900

c) 54,080

d) 54,900

Answer: (c)

Using Rule 17,

Required population = $50000(1 + 4/100)^2$

= $50000 × 26/25 × 26/25$= 54080

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GET percentage PRACTICE TEST EXERCISES

model 1 simple percentage questions

model 2 net increase or decrease in %

model 3 reducing & exceeding prices

model 4 x & y comparison

model 5 income & expenditure

model 6 consumption & remaining

model 7 marks scored in examinations

Model 8 population based

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model 10 percentage with allegations & mixture

model 11 percentage with ratios
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