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Percentage Solved Questions For IBPS, SBI and Other Bank Exams - Page 8

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Percentage Problems Solved Questions (Page 8 of 8)

Results on Population:

Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:

1. Population after n years = P x (1 + R/100)n

2. Population n years ago = P/(1 + R/100)n

Let the population of a town be P now and suppose it decreases at the rate of R% per annum, then:

3. Population after n years = P x (1 - R/100)n

4. Population n years ago = P/(1 - R/100)n


Question 1

The population of a village was 3,20,000 two years ago. If it increased by 6% & 5% respectively in the last two years, then the present population is:

a)400170 b) 296150 c) 356160 d) 382110

Answer : c) 356160

Solution :

Population of the village 2 years ago = P = 3,20,000
After increasing (6%=R1), it becomes
By using the formula (1), we have,=> 320000 x (1+6/100)
Therefore the population after 1 year = 320000x(1+6/100)1

After increasing (5%=R2), population becomes = 320000x(1+6/100) x (1+5/100)1

Required present population = 320000x(1+6/100) x (1+5/100)
= 320000 x 106/100 x 105/100
= 320 x 53 x 21
= 356160

Hence the answer is option c.


Question 2

The population of a city 3 years ago was 1,28,000. If it decreased by 5% on every year, then the present population of the city is:

a) 89750 b) 119035 c) 99744 d) 109744

Answer : d) 109744

Solution :

Population of the city 3 years ago = P = 128000 and n = 3
And, the decreasing percentage = R = 5%

By using the formula (3),
The present population = P x (1 - R/100)n
= 128000 x (1 - 5/100)3
= 128000 x 19/20 x 19/20 x 19/20
= 16x6859 = 109744.
Hence the answer is option d.


Question 3

The present population of a district is 128000. If it increases at the rate of 2.5% per annum then at the end of 2 years, it will be:

a) 134480 b) 194200 c) 154220 d) 182300

Answer : a) 134480

Solution :

Present population of the district = 128000
Increasing rate = R = 2.5% = 5/2 %
By using formula (1),
The present population after 2 years = P x (1 + R/100)n
= 128000 x (1 + 5/2 x 100)2
= 128000 x (205/200)2
= 128000 x 41/40 x 41/40
= 80 x 41 x 41
= 134480.

Hence the required answer is option a.


Question 4

The population of a town in 1991 was 175760. If it increases by 4% annually then what it was 1988?

a) 196520 b) 251100 c) 200950 d) 156250

Answer : d) 156250

Solution :

Population of the district in 1991 = 175760
We have to find the population in 1988. (3 years back).
Increasing rate = R = 4%

By using the formula (2),
Population = 175760/(1 + 4/100)3 = 175760 / (104/100)3
= 175760 x 25/26 x 25/26 x 25/26
= 10 x 25 x 25 x 25 = 156250

Hence the required answer is option d.


Percentage Problems Solved Questions (Page 8 of 8)

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