Question

1244
64. The population of a city first increased by
10% and then decreased by 20%. The net
decrease in the population is
luu
(A) 15%
(B) 20%
10) 12%
(D) 8%​

Answer 1

ANSWER :-

• C) 12%

GIVEN :-

• The population of city first increased by 10% and then decreased by 20%.

TO FIND :-

• Net decrease in population.

SOLUTION :-

Let the population of the city be 'x'.

Firstly , population is increased by 10%.

New population = x + 10% of x

$\\ \implies \sf \: x + \dfrac{10}{100} (x) \\ \\ \implies \sf \: x + \dfrac{1}{10} (x) \\ \\ \implies \boxed{\sf \: \dfrac{11x}{10} }$

Population becomes 11x/10.

Secondly , population is decreased by 20%.

New population = (11x/10) - 20% of (11x/10)

$\\ \implies\sf \: \dfrac{11x}{10} - \dfrac{ \cancel{ {20}} \: ^{2} }{100} \left( \dfrac{11x}{ \cancel{10}} \right) \\ \\ \implies\sf \: \dfrac{11x}{10} - \dfrac{22x}{100} \\ \\ \implies\boxed{ \sf \: \dfrac{88x}{100}}$

Final population is 88x/100.

$\\ \underline{ \boxed{\sf \: \% \: decreased = \dfrac{ initial - final}{initial} \times 100}}$

Putting values , we get...

$\\ \implies\sf \: \dfrac{ x - \dfrac{88x}{100} }{x} \times 100 \\ \\ \implies\sf \: \dfrac{ \dfrac{100x - 88x}{100} }{x} \times 100 \\ \\ \implies\sf \: \dfrac{12x}{100x} \times 100 \\ \\ \implies\sf \: 12\%$

Hence , Population is decreased by 12%. (Option C)