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

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)