The population of a town is decreased by 20 % | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the initial population of the town be $100$.After the first year,the population decreases by $20 \%$,so the new population is $100 - 20 = 80$.

Vedclass · Accepted Answer

$40 \%$

Vedclass · Answer

(B) Let the initial population of the town be $100$.After the first year,the population decreases by $20 \%$,so the new population is $100 - 20 = 80$.After the second year,the population decreases by $25 \%$ of the remaining $80$. The decrease is $\frac{25}{100} 	imes 80 = 20$.The final population is $80 - 20 = 60$.The total decrease in population is $100 - 60 = 40$.Therefore,the percentage decrease is $\frac{40}{100} 	imes 100 = 40 \%$.Alternatively,using the formula for successive percentage change: $	ext{Net } \% 	ext{ change} = \left(x + y + \frac{xy}{100}ight) \%$.Here,$x = -20$ and $y = -25$.Net change $= -20 - 25 + \frac{(-20) 	imes (-25)}{100} = -45 + 5 = -40 \%$.The negative sign indicates a decrease of $40 \%$.

The population of a town is decreased by $20 \%$ and $25 \%$ in two successive years. What per cent population is decreased after two years?

Similar Questions

Two numbers are less than a third number by $30 \%$ and $37 \%$,respectively. How much per cent is the second number less than the first?

The sum of two numbers is $\frac{28}{25}$ of the first number. The second number is what percent of the first (in $\%$)?

If the numerator of a fraction is increased by $120 \%$ and the denominator is also increased by $350 \%$,then the fraction becomes $\frac{11}{27}$. What was the original fraction?

The price of a certain article is $Rs. 15000$. Due to a slump in the market, its price decreases by $8 \%$. Find the new price of the article.

$88 \%$ of $370 + 24 \%$ of $210 - ? = 118$

Vedclass Test Series

Exam Paper Generator

Online Exam Module