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29 Apr 2020, 01:53
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
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Question Stats:
63% (01:29) correct
37%
(01:42)
wrong
based on 229
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The population of a village is calculated by adding the number of males and the number of females.If this year male population increased by 10% as compared to last year, what is the net percentage change in total population of the village?
(1) The number of females as compared with last year increased this year by 5%.
(2) The ratio of the number of females and the males was 2:3 last year.
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29 Apr 2020, 03:09
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The population of a village is calculated by adding the number of males and the number of females.If this year male population increased by 10% as compared to last year, what is the net percentage change in total population of the village?
(1) The number of females as compared with last year increased this year by 5%.
(2) The ratio of the number of females and the males was 2:3 last year.
Last year: males = 100x, females = 100y
this year: males = 110x, females = ?
net percentage change = \(\frac{(T(P) - L(P))*100}{L(P)}\)
where T(P): total population this year & L(P) : total population last year
statement 1:
this year: males = 110x females = 105y
net percentage change = \(\frac{(110x + 105y - (100x + 100y))*100}{100(x+y)}\) = \(\frac{10x + 5y }{(x+y)}\)
we do not know value of x & y.
not sufficient
statement 2:
ratio of \(\frac{100y}{100x}\) = \(\frac{2}{3}\)
not sufficient
combining both statements,
y = \(\frac{2x}{3}\)
net percentage change =\(5(2x + {\frac{2x}{3}})\)/\((x+\frac{2x}{3})\)
x will get cancelled out leaving us with a unique value
sufficient
Ans: C
(1) The number of females as compared with last year increased this year by 5%.
(2) The ratio of the number of females and the males was 2:3 last year.
Last year: males = 100x, females = 100y
this year: males = 110x, females = ?
net percentage change = \(\frac{(T(P) - L(P))*100}{L(P)}\)
where T(P): total population this year & L(P) : total population last year
statement 1:
this year: males = 110x females = 105y
net percentage change = \(\frac{(110x + 105y - (100x + 100y))*100}{100(x+y)}\) = \(\frac{10x + 5y }{(x+y)}\)
we do not know value of x & y.
not sufficient
statement 2:
ratio of \(\frac{100y}{100x}\) = \(\frac{2}{3}\)
not sufficient
combining both statements,
y = \(\frac{2x}{3}\)
net percentage change =\(5(2x + {\frac{2x}{3}})\)/\((x+\frac{2x}{3})\)
x will get cancelled out leaving us with a unique value
sufficient
Ans: C
29 Apr 2020, 03:26
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Solution
Step 1: Analyse Question Stem
- • The population of males increased by 10% this year.
- • We can use the concept of weighted average to find the answer.
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: The number of females as compared with last year increased this year by 5%.
- • We know the percentage change in males and females, but we don’t know the relation between number of males and females
• Thus, we cannot find the answer.
Statement 2: The ratio of the number of females and the males was 2: 3 last year.
- • Number of females: Number of males = 3: 2
- o We don’t know the percentage change in the population of females.
Step 3: Analyse Statements by combining.
From statement 1: Percentage change in the population of females = 5%
From statement 2: Number of females: Number of males = 3: 2
- • Overall percentage change = \(\frac{2*\frac{10}{100} + 3*\frac{5}{100}}{2+3}\)
Hence, the correct answer is Option C.
