Model 8 population based Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on percentage topic of quantitative aptitude

Questions : The population of a town is 9000. It the number of females increases by 5% and the males by 7.5%, what will be the total population after increase. The number of females currently is 3000.

(a) 9200

(b) 9600

(c) 10500

(d) 9540

The correct answers to the above question in:

Answer: (b)

In the village,

Females = 3000; Males = 9000 – 3000 = 6000

After respective increases,

Population of village

= $3000 × 105/100 + {6000 × 107.5}/100$

= 3150 + 6450 = 9600

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Read more population based Based Quantitative Aptitude Questions and Answers

Question : 1

A TV was bought at a price of Rs.21,000. After one year the value of TV was depreciated by 5%. Find the value of the TV after one year.

a) Rs.20,950

b) Rs.19,950

c) Rs.18,950

d) Rs.17,950

Answer: (b)

Value of TV after one year = 21000 × (100 – 5)%

= ${21000 × 95}/100$ = Rs. 19950

Question : 2

The value of an equipment depreciates by 20% each year. How much less will the value of the equipment be after 3 years ?

a) 51.2%

b) 48.8%

c) 54%

d) 60%

Answer: (b)

If the present worth of the equipment be Rs.100, then

its price after 3 years = 100 × $(80/100)^3$ = Rs.51.2

Depriciation = 48.8%

Using Rule 18,

If the population/cost of a town/article is P and it decreases/reduces at the rate of r% annually, then,

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

Let the price of equipment be Rs.100

Its price after 3 years. = $100(1 - 20/100)^3$

= $100 × (80/100)^3$ = Rs.51.2

Depriciation = 48.8%

Question : 3

If the population of a town is 64000 and its annual increase is 10%, then its correct population at the end of 3 years will be :

a) 85184

b) 80000

c) 85000

d) 85100

Answer: (a)

Using Rule 17,

Population of town = $P(1 + R/100)^T$

= $64000(1 + 10/100)^3$

= $64000 × 11/10 × 11/10 × 11/10$ = 85184

Question : 4

The population of a village has increased annually at the rate of 25%. If at the end of 3 years it is 10,000, the population in the beginning of the first year was

a) 5000

b) 5120

c) 4900

d) 4500

Answer: (b)

Using Rule 17,

Population in the beginning of the year

=$\text"Population after 3 years"/(1 + \text"Rate"/100)^{ \text"Time"}$

= $10000/(1 + 25/100)^3= 10000/(5/4)^3$

= ${10000 × 64}/125 = 5120$

Question : 5

The value of a property depreciates every year by 10% of its value at the beginning of the year. The present value of the property is Rs.8100. What was its value 2 years ago ?

a) Rs.$(90/100)^2 × 8100$

b) Rs.10,000

c) Rs. $(100/110)^2 × 8100$

d) Rs.9801

Answer: (b)

Using Rule 18,

Suppose the value of property two years ago was Rs.x

According to question $x(1 - 10/100)^2 = 8100$

$x(90/100)^2 = 8100$

$x ={8100 × 10 × 10}/{9 × 9}$ = Rs.10000

Question : 6

Of the 1000 inhabitants in a town 60% are males of whom 20% are literate. If of all the inhabitants, 25% are literate, then what percentage of the females of the town are ilterate ?

a) 32.5

b) 27.5

c) 37.5

d) 22.5

Answer: (a)

Population of town = 1000

Males ⇒ 600; Females ⇒ 400

Literate males = ${600 × 20}/100$ = 120

Total literate inhabitants

= ${1000 × 25}/100$ = 250

Literate females = 250 – 120 = 130

Required percent = $130/400$ × 100 = 32.5%

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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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