Model 8 population based Practice Questions Answers Test with Solutions & More Shortcuts

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: (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: (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: (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: (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: (b)

Using Rule 18,

$A = P(1 - R/100)^T$

= $200000(1 - 5/100)^T$

= $200000 × 19/20 × 19/20$ = Rs.80500

IMPORTANT quantitative aptitude EXERCISES

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