A vendor sold 1/3 of his fish balls in the afternoon and 2/5 of the remainder in the evening. If he had 150 fish balls left, find the number of fish balls he had at first?
let the number of fish at the start be x
sold (1/3)x in the afternoon, leaving his with (2/3)x
He sold 2/5 of that leaving him with
(3/5) (2/3)x or (2/5)x
(2/5)x = 150
x = 375
check:
sell 1/3 of 375 = 125 , leaves 250
sells 2/5 of that or 100 , leaving him with 150
All is good!
n - n/3 - 2/5 (2n/3) = 150
n - n/3 - 4n/15 = 150
(15 - 5 - 4)n/15 = 150
6n/15 = 150
2n/5 = 150
n = 375
check:
sold 1/3 of 375 = 125, leaving 250
sold 2/5 of 250 = 100, leaving 150
To find the number of fish balls the vendor had at first, we can work backwards from the information given.
Let's start with the number of fish balls he had left, which is 150.
We know that in the evening, the vendor sold 2/5 of the remainder. Therefore, after the evening, he had (1 - 2/5) of the remainder left.
Let's calculate the number of fish balls he had after the evening:
(1 - 2/5) * remainder = 150.
Now, let's solve this equation to find the remainder:
3/5 * remainder = 150.
remainder = 150 * 5/3 = 250.
So after the evening, he had 250 fish balls left.
Now, we know that in the afternoon, he sold 1/3 of the fish balls. Therefore, after the afternoon, he had (1 - 1/3) of the total fish balls left.
Let's calculate the number of fish balls he had at the beginning:
(1 - 1/3) * total = 250.
Now, let's solve this equation to find the total number of fish balls:
2/3 * total = 250.
total = 250 * 3/2 = 375.
Therefore, the vendor had 375 fish balls at first.