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 : 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

The correct answers to the above question in:

Answer: (b)

Using Rule 17,

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

= $100/105 × 1806$ = Rs.1720

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

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 : 2

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 : 3

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 : 4

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 = $180000(1 + 10/100)^2$

= $180000 × 11/10 × 11/10 = 217800$

Question : 5

From 1980-1990, the population of a country increased by 20%. From 1990-2000, the population of the country increased by 20%. From 2000-2010, the population of the country increased by 20%. Then the overall increased population (in percentage) of the country from 1980-2010 was

a) 60 %

b) 72.2 %

c) 72.8 %

d) 62.8 %

Answer: (c)

Using Rule 7,

If a number is first increased by a% and then again increased by b%, then total increase percent is $(a + b + {ab}/100)%$

Single equivalent increase for 20% and 20%

= $(20 + 20 + {20 × 20}/100)%$ = 44%

Single equivalent increase for 44% and 20%

= $(44 + 20 + {44 × 20}/100)%$

= (64 + 8.8)% = 72.8%

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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

model 9 voters & election

model 10 percentage with allegations & mixture

model 11 percentage with ratios
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