Model 8 population based Practice Questions Answers Test with Solutions & More Shortcuts
percentage PRACTICE TEST [11 - 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
Question : 16 [SSC CHSL 2012]
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 : 17 [SSC CAPFs SI 2015]
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 : 18 [SSC CHSL 2014]
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
Question : 19 [SSC CGL Tier-I 2013]
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 : 20 [SSC Constable 2012]
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
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Model 8 population based Shortcuts »
Click to Read...Model 8 population based Online Quiz
Click to Start..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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