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 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

The correct answers to the above question in:

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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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

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 : 2

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 : 3

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

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

Question : 4

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 : 5

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 : 6

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

$x = 16800/3 = 5600$

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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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