a man buys a certain number of mangoes and sells 3/5 of total mangoes. he gives half of the remaining mangoes to his wife, if he still left with 15 mangoes, find total number of mangoes purchased by him?
Answers
Step-by-step explanation:
Given :-
A man buys a certain number of mangoes and sells 3/5 of total mangoes. he gives half of the remaining mangoes to his wife, he still left with 15 mangoes.
To find :-
Find total number of mangoes purchased by him?
Solution :-
Let the number of mangoes bought by the man be X
Number of mangoes sold by him
= 3/5 of the total mangoes
=>(3/5)×X
=> (3×X)/5
=> 3X/5
Remaining number of mangoes
=> Total mangoes - Sold mangoes
=> X-(3X/5)
=> (5X-3X)/5
=> 2X/5
Number of mangoes are given to his wife
= half of the remaining mangoes
= (2X/5)/2
=> 2X/(5×2)
=> 2X/10
=> X/5
The number of mangoes left with him = 15
=> X/5 = 15
=> X = 15×5
=> X = 75
Therefore, X = 75
Answer:-
The total number of mangoes purchased by him is 75
Check:-
Total mangoes = 75
Mangoes sold by him = (3/5)×75
=> (3×75)/5
=> 3×15
=> 45
Remaining mangoes = 75-45 = 30
Mangoes given to his wife = 30/2 = 15
Mangoes were left with him = 15
Verified the given relations in the given problem.
Step-by-step explanation:
Let N be the initial number of mangoes the man had bought,
Given total cost of N mangoes
p---(1)
Let 'x' be the number of mangoes man had sold
Given that by selling x mangoes at Rs 8, man had gained 44 p
So, Cost price of x mangoes = Selling
Price - Gain
= Rs 8-44 p
= Rs 7. 56 p----(2)
From (1) and (2),
Cost Price of N mangoes = Rs 14.4
Cost Price of x mangoes =
So, x/N = 7.56/14.4 = 21/40
So, x = 21N/40
After selling x mangoes, there will be (N x ) mangoes left
= N - 21N/40
= 19N/40
Percentage of mangoes left would be 19N/40N*100 = 47.5%.
Hence, 47.5% of the mangoes which he had initially bought are left.
=
Rs 7.56