Question

the population of a town was 200000 if the number of males increased by 10% and the number of females decreased by 6% the total population would have remained unchanged the number of males is

Answer 1

Answer :

Number of males is 75000

Solution :

Let, the number of males be x

Then, the number of females be (200000 - x)

By the given condition, number of males increased by 10% and the number of females decreased by 6%

So, the current number of males be

= x + ( 10% of x )

= x + $\frac{10x}{100}$

= x + $\frac{x}{10}$

= $\frac{10x+x}{10}$

= $\frac{11x}{10}$

= $\frac{55x}{50}$

and the current number of females be

= (200000 - x) - {6% of (200000 - x)}

= (200000 - x) - $\frac{6(200000-x)}{100}$

= (200000 - x) * $(1 - \frac{6}{100})$

= (200000 - x) * $\frac{100-6}{100}$

= (200000 - x) * $\frac{94}{100}$

= (200000 - x) * $\frac{47}{50}$

According to the question,

$\frac{55x}{50} + (200000 - x) * \frac{47}{50} = 200000$

$\implies \frac{55x + (200000 - x) * 47}{50} = 200000$

➩ 55x + (200000 * 47) - 47x = 50 * 200000

➩ (55 - 47) * x + 9400000 = 10000000

➩ 8x = 10000000 - 9400000

➩ 8x = 600000

➩ x = 600000 ÷ 8

➩ x = 75000

Therefore, the number of males is 75000

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Answer 2

Answer:

Number of Male is 75000

Step-by-step explanation:

Let, the number of males be x

Then, the number of females be (200000 - x)

By the given condition, number of males increased by 10% and the number of females decreased by 6%

So, the current number of males be

= x + ( 10% of x )

= x + \frac{10x}{100}

100

10x

= x + \frac{x}{10}

10

x

= \frac{10x+x}{10}

10

10x+x

= \frac{11x}{10}

10

11x

= \frac{55x}{50}

50

55x

and the current number of females be

= (200000 - x) - {6% of (200000 - x)}

= (200000 - x) - \frac{6(200000-x)}{100}

100

6(200000−x)

= (200000 - x) * (1 - \frac{6}{100})(1−

100

6

)

= (200000 - x) * \frac{100-6}{100}

100

100−6

= (200000 - x) * \frac{94}{100}

100

94

= (200000 - x) * \frac{47}{50}

50

47

According to the question,

\frac{55x}{50} + (200000 - x) * \frac{47}{50} = 200000

50

55x

+(200000−x)∗

50

47

=200000

\implies \frac{55x + (200000 - x) * 47}{50} = 200000⟹

50

55x+(200000−x)∗47

=200000

➩ 55x + (200000 * 47) - 47x = 50 * 200000

➩ (55 - 47) * x + 9400000 = 10000000

➩ 8x = 10000000 - 9400000

➩ 8x = 600000

➩ x = 600000 ÷ 8

➩ x = 75000

Therefore, the number of males is 75000

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