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Mathematics

Assertion (A): The present population of a town decreases at the rate of 5% p.a. due to migration. Its present population is 21,600 and 2 years ago it was 24,000.

Reason (R): Let there be an increase in population at R% p.a. Then population after n years is given by P×(1R100)nP \times \Big(1 - \dfrac{R}{100} \Big)^n, where P is the present population.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Compound Interest

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Answer

Given,

P (two years ago) = 24,000

R = 5% p.a.

n = 2 year

By formula,

Population after n years = P(1r100)nP\Big(1 - \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Present population =24000×(15100)2=24000×(1005100)2=24000×(95100)2=24000×(0.95)2=24000×0.9025=21,660\text{Present population }=24000 \times \Big(1 - \dfrac{5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{100 - 5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{95}{100}\Big)^2 \\[1em] =24000 \times (0.95)^2\\[1em] =24000 \times 0.9025\\[1em] =21,660

Assertion is False

Reason is also false as we do not use that formula to calculate the growth in population. We use,

Population growth after n years at R% with initial Population as P is calulated using formula : P×(1+R100)nP \times \Big(1 + \dfrac{R}{100}\Big)^n

Hence, option 4 is correct option.

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