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 : 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

The correct answers to the above question in:

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

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

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 : 2

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

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

Question : 3

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 : 4

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

Question : 5

Raman's salary is increased by 5% this year. If his present salary is Rs.1,806, the last year's salary was

a) Rs.1620

b) Rs.1720

c) Rs.1520

d) Rs.1801

Answer: (b)

Using Rule 17,

Required Raman's salary = $100/{100 + 5} × 1806$

= $100/105 × 1806$ = Rs.1720

Question : 6

The present population of a city is 180000. If it increases at the rate of 10% per annum, its population after 2 years will be :

a) 227800

b) 207800

c) 217800

d) 237800

Answer: (c)

Using Rule 17,

If the population/cost of a certain town/ article, is P and annual increament rate is r%, then

  1. After ‘t' years population/cost = $P(1 + r/100)^t$
  2. Before ‘t' years population/cost = $P/{(1 + r/100)^t}$

Required population after two years = $180000(1 + 10/100)^2$

= $180000 × 11/10 × 11/10 = 217800$

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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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