Question

Ablation of the certain town was 50000 in year the male population increased by 5% and the female population increased by 8% now the population become 53220 what was the number of the male and female in the previous year

Answer 1

Answer:

☆ Number of Females= 24,000

☆ number of Males = 26,000

Step-by-step explanation:

Given:

• Population of town was 50,000
• Male population increased by 5%
• Female population increased by 8%
• Current population as after increasement 53,220

To Find:

• Number of Males and Females previous year.

Solution: Let male population be x and female population be y

According to the question:

✪ Male Population+ Female Population = Total Population

$\small\implies{\sf }$ x + y = 50,000

$\small\implies{\sf }$ y = 50,000 – x ................(1)

Now, x ( Male Population) and y ( Female Population) increased by 5% and 8% respectively and new population is 53,220

∴ ( x + 5% of x ) + ( y + 8% of y )= 53,220

$\small\implies{\sf }$ x + 5x / 100 + y + 8y / 100 = 53,220

=> Take LCM

$\small\implies{\sf }$ 105x / 100 + 108y / 100 = 53,220

$\small\implies{\sf }$ 105x + 108y / 100 = 53,220

$\small\implies{\sf }$ 105x + 108y = 5322000..............(2)

✵ Putting the value of y from equation (1) in the equation (2)

$\small\implies{\sf }$ 100x + 5x + 108 ( 50,000 – x ) = 5322000

$\small\implies{\sf }$ 105x – 108x + 5400000 = 5322000

$\small\implies{\sf }$ 5400000 – 5322000 = 3x

$\small\implies{\sf }$ 78,000 = 3x

$\small\implies{\sf }$ x = 78,000/26 = 26,000

Hence, Number of Males = x = 26,000

and number of females = y = 50,000–x

$\small\implies{\sf }$ y = 50,000 – 26,000

$\small\implies{\sf }$ y = 24,000

Hence, Number of Females = y = 24,000