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
(a) 105%
(b) 100%
(c) 108%
(d) 111%
The correct answers to the above question in:
Answer: (d)
If the number of men be 100,then
Number of women = 90
Required per cent = $100/90 × 100 ≈ 111%$
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Read more population based Based Quantitative Aptitude Questions and Answers
Question : 1
The population of a town increases by 5% every year. If the present population is 9261, the population 3 years ago was
a) 5700
b) 8000
c) 6000
d) 7500
Answer »Answer: (b)
Using Rule 17,
P = $P_0(1 + R/100)^T$
9261 = $P_0(1 + 5/100)^3$
9261 = $P_0(1 + 1/20)^3$
9261 = $P_0(21/20)^3$
$P_0 = {9261 × 20 × 20 × 20}/{21 × 21 × 21}$ = 8000
Question : 2
In a village panchayat society 574 names are enlisted as ‘below poverty level'. If 14% of the villagers are below poverty level, the total number of villagers is
a) 4200
b) 4100
c) 4000
d) 3800
Answer »Answer: (b)
Let the total population of village be x.
According to the question,${x × 14}/100 = 574$
$x = {574 × 100}/14$ = 4100
Question : 3
The population of a village decreases at the rate of 20% per annum. If its population 2 years ago was 10,000, the present population is
a) 6400
b) 4600
c) 7600
d) 6000
Answer »Answer: (a)
Using Rule 17,
Present population = $10000(1 - 20/100)^2$
= $10000 × 4/5 × 4/5 = 6400$
Question : 4
In a town, the population was 8000. In one year, male population increased by 10% and female population increased by 8% but the total population increased by 9%. The number of males in the town was :
a) 4500
b) 4000
c) 5000
d) 6000
Answer »Answer: (b)
By Alligation Rule
Men : Women = 1 : 1
Number of men = $1/2 × 8000 = 4000$
Question : 5
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 »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 : 6
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 »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,
- After ‘t' years population/cost = $P(1 - r/100)^t$
- 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%
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
percentage Shortcuts and Techniques with Examples
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model 1 simple percentage questions
Defination & Shortcuts … -
model 2 net increase or decrease in %
Defination & Shortcuts … -
model 3 reducing & exceeding prices
Defination & Shortcuts … -
model 4 x & y comparison
Defination & Shortcuts … -
model 5 income & expenditure
Defination & Shortcuts … -
model 6 consumption & remaining
Defination & Shortcuts … -
model 7 marks scored in examinations
Defination & Shortcuts … -
Model 8 population based
Defination & Shortcuts … -
model 9 voters & election
Defination & Shortcuts … -
model 10 percentage with allegations & mixture
Defination & Shortcuts … -
model 11 percentage with ratios
Defination & Shortcuts …
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