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 : 6 [SSC CAPFs SI 2014]
The population of a town increases each year by 4% of its total at the beginning of the year. If the population on 1st January 2001 was 500000, what was it on 1st January, 2004 ?
a) 652432
b) 562432
c) 465223
d) 564232
Answer »Answer: (b)
Using Rule 17,
Required population = $P(1 + R/100)^T$
= $500000(1 + 4/100)^3$
= $500000 × (1 + 1/25)^3$
= $500000 × 26/25 × 26/25 × 26/25$ = 562432
Question : 7
The present price of a scooter is Rs. 7,290. If its value decreases every year by 10%, then its value 3 years back was
a) Rs. 8,000
b) Rs. 10, 500
c) Rs. 10,000
d) Rs. 11,500
Answer »Answer: (c)
Using Rule 18,
A = $P(1 - R/100)^3$
7290 = $P(1 - 10/100)^3 = P(9/10)^3$
7290 = P $× 9/10 × 9/10 × 9/10$
P = ${7290 × 10 × 10 × 10}/{9 × 9 × 9}$ = Rs. 10000
Question : 8 [SSC CGL Prelim 2007]
The population of a village was 9800. In a year, with the increase in population of males by 8% and that of females by 5%, the population of the village became 10458. What was the number of males in the village before increase ?
a) 4410
b) 4200
c) 5600
d) 6048
Answer »Answer: (c)
Let the number of males = x
Number of females= 9800 – x
According to the question,
$x × 108/100 + (9800 - x) × 105/100 = 10458$
108 x + 9800 ×105 – 105x = 1045800
3x + 1029000 = 1045800
3x = 1045800 – 1029000 = 16800
$x = 16800/3 = 5600$
Question : 9 [SSC CGL Prelim 2004]
The population of a town 2 years ago was 62,500. Due to migration to big cities, it decreases every year at the rate of 4%. The present population of the town is:
a) 56,700
b) 57,600
c) 76,000
d) 75,000
Answer »Answer: (b)
Using Rule 18,
Let the present population be P.
P = $62500(1 - 4/100)^2$
= $62500 × 24/25 × 24/25$ = 57600
Question : 10 [SSC MTS 2013]
The value of a machine is Rs.6,250. It decreases by 10% during the first year, 20% during the second year and 30% during the third year. What will be the value of the machine after 3 years?
a) Rs.3,050
b) Rs.2,650
c) Rs.3,150
d) Rs.3,510
Answer »Answer: (c)
Using Rule 28,
Population after ‘n' years = $P(1 ± R_1/100)(1 ± R_2/100)…(1 ± R_n/100)$
Required price of the machine
= $6250(1 - 10/100)(1 - 20/100)(1 - 30/100)$
= $6250 × 90/100 × 80/100 × 70/100$ = Rs.3150
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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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