A man donated 1/10 of his money to a school, 1/6 of the remaining to a church and the remaining money he distributed equally among his three children .if each child gets 50000, how much money did the man originally have
Answers
He donates 1/10 to a school
1/10 × x = x/10
Remaining money = x - x/10
= 9x/10
He donates 1/6 to a church from the remaining money,
1/6 × 9x/10 = 3x/20
Remaining money = 9x/10 - 3x/20
= 12x/20
= 3x/4
the remaining money he distributed equally among his three children
= 3(x/4)
So,each child gets x/4 amount.
According to the question,
If each child gets 50000 ,
x/4 = 50000
x = 200000
So the amount of money which the man originally have is 2,00,000.
Hope it helps
Answer:
₹2,00,000
Step-by-step explanation:
Let the original money be x
Money donated to a school = 1/10 of x
= x/10
Remaining = x-x/10
= 9x/10
Money donated to a church = 1/6 of 9x/10
= 3x/20
Remaining = 9x/10 - 3x/20
= (18x - 3x) / 20
= 15x/20
= 3x/4
No. of children = 3
Each child's share = ₹ 50,000
So,
3x/4 divided by 3/1 = 50,000
3x/4 multiplied by 1/3 ( i reciprocaled 3/1 to make it multiplication) = 50,000
and,
x = 50,000 multiplied by 4 = ₹2,00,000
So, total money he had = ₹2.00.000