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) 32.5
(b) 27.5
(c) 37.5
(d) 22.5
The correct answers to the above question in:
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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Read more population based Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 : 3
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
Answer »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
Question : 4
The value of a machine depreciates by 5% every year. If its present value is Rs.2,00,000, its value after 2 years will be
a) Rs.1,99,000
b) Rs.1,80,500
c) Rs.1,80,000
d) Rs.2,10,000
Answer »Answer: (b)
Using Rule 18,
$A = P(1 - R/100)^T$
= $200000(1 - 5/100)^T$
= $200000 × 19/20 × 19/20$ = Rs.80500
Question : 5
The value of a machine depreciates every year by 10%. If its present value is Rs.50,000 then the value of the machine after 2 years is _________.
a) Rs.45,000
b) Rs.40,050
c) Rs.40,005
d) Rs.40,500
Answer »Answer: (d)
Using Rule 18,
Required value = $50000(1 - 10/100)^2$
= $50000 × {9 × 9}/100$ = Rs.40500
Question : 6
The population of a village increases by 5% annually. If its present population is 4410, then its population 2 years ago was
a) 4000
b) 4500
c) 3800
d) 3500
Answer »Answer: (a)
Using Rule 17,
If the population of village two years ago be $P_0$,
then P = $P_0(1 + R/100)^T$
4410 = $P_0(1 + 5/100)^2$
4410 = $P_0(1 + 1/20)^2$
4410 = $P_0(21/20)^2$
4410 = ${441P_0}/400$
$P_0 = {4410 × 400}/441$ = 4000
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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