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) 40000
(b) 32000
(c) 31200
(d) 30000
The correct answers to the above question in:
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
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Read more population based Based Quantitative Aptitude Questions and Answers
Question : 1
In a city, 40% of the people are illiterate and 60% are poor. Among the rich, 10% are illiterate. The percentage of the illiterate poor population is
a) 60
b) 36
c) 40
d) 50
Answer »Answer: (a)
Let the population of the city be 100.
Total illiterate people = 40
Poor people = 60; Rich people = 40
Illiterate rich people = ${40 × 10}/100$ = 4
Illiterate poor people = 40 – 4 = 36
Required percent = $36/60 × 100$ = 60%
Question : 2
If a man receives on one-fourth of his capital 3% interest, on two third 5% and on the remainder 11%, the percentage he receives on the whole is
a) 5
b) 4.5
c) 5.5
d) 5.2
Answer »Answer: (a)
Required percent
= $1/4 × 3 + 2/3 × 5 + (1 - 1/4 - 2/3) × 11$
= $3/4 + 10/3 + 11/12 = {9 + 40 + 11}/12$ = 5%
Question : 3
In a factory, the production of cycles rose to 48, 400 from 40,000 in 2 years. The rate of growth per annum is
a) 8%
b) 9%
c) 10.5%
d) 10%
Answer »Answer: (d)
Using Rule 17,
If the rate of increase per annum be R%, then
A = P$(1 + R/100)^T$
48400 = 40000$(1 + R/100)^2$
$484/400 = (1 + R/100)^2$
$121/100 = (11/10)^2 = (1 + R/100)^2$
1 + $R/100 = 11/10$
$R/100 = 11/10 - 1 = 1/10$
$R = 100/10$ = 10% per annum
Question : 4
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 »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 : 5
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 »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 : 6
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
Answer »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
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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