Question : The total number of males and females in a town is 70,000. If the number of males is increased by 6% and that of females is increased by 4%, then the total number of males and females in the town would become 73,520. What is the difference between the number of males and females in the town, in the beginning?
Option 1: 1,500
Option 2: 1,800
Option 3: 2,000
Option 4: 1,400
Correct Answer: 2,000
Solution :
Given: The total number of males and females in a town is 70,000.
The number of males is increased by 6% and that of females is increased by 4%, the total number of males and females in the town would become 73,520.
Let $x$ be the number of males.
The number of females = $(70000 - x)$
The town's overall male and female population rose by = 73520 – 70000 = 3,520
According to the question,
$x\times \frac{6}{100}+[(70000-x)\times \frac{4}{100}]=3520$
⇒ $0.06x+2800-0.04x=3520$
⇒ $0.06x-0.04x=3520-2800$
⇒ $0.02x=720$
⇒ $x=36,000$
There are 36,000 males in total.
The number of females = 70000 – 36000 = 34,000
The difference between the number of males and females in the town, in the beginning,
= 36000 – 34000 = 2,000
Hence, the correct answer is 2,000.
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