SOLUTION: A fruit seller sold some apples and oranges.
He sold 3/7 of the apples in the morning and 3/8 of the remainder in the afternoon.
He sold 1/3 of the oranges in the morning and 1/4
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-> SOLUTION: A fruit seller sold some apples and oranges.
He sold 3/7 of the apples in the morning and 3/8 of the remainder in the afternoon.
He sold 1/3 of the oranges in the morning and 1/4
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Question 1184047: A fruit seller sold some apples and oranges.
He sold 3/7 of the apples in the morning and 3/8 of the remainder in the afternoon.
He sold 1/3 of the oranges in the morning and 1/4 of the remainder in the afternoon.
He had sole sold the same number of apples and oranges at the end of the day.
If there were 8 more oranges than apples left unsold, how many fruits were there at first? Found 2 solutions by Octo-pie7, MathTherapy:Answer by Octo-pie7(11) (Show Source):
You can put this solution on YOUR website! let a = # of apples he started with
let o = # of oranges he started with
3/7a were sold in the morning, so 4/7a remained.
3/8(4/7a) were sold in the afternoon, so 5/8(4/7a) remained.
1/3 a were sold in the morning, so 2/3 o remained.
1/4(2/3o) were sold in the afternoon, so 3/4(2/3o) remained.
3/8(4/7a) = 1/4(2/3o)
5/8(4/7a) + 8 = 3/4(2/3o)
12/36a = 2/12o => 3/14a = 1/6o
20/36a + 8 = 6/12o => 5/14a + 8 = 1/2o
3/14a = 1/6o
5/14a + 8 = 1/2o
(6)(3/14a) = (1/6o)(6) => 18/14a = o => 9/7a = o
(-2)[(5/14a) + 8] = [1/2o] (-2) => -10/14a - 16 = -o => -5/7a - 16 = -o
9/7a = o
-5/7a - 16 = -o
———————-
4/7a - 16 = o
+ 16 +16
4/7a = 16
7/4 * 4/7a = 16 * 7/4
a = 28
3/14 (28) = 1/6o
6 = 1/6o
6 * 6 = 1/6o * 6
36 = o
You can put this solution on YOUR website! A fruit seller sold some apples and oranges.
He sold 3/7 of the apples in the morning and 3/8 of the remainder in the afternoon.
He sold 1/3 of the oranges in the morning and 1/4 of the remainder in the afternoon.
He had sole sold the same number of apples and oranges at the end of the day.
If there were 8 more oranges than apples left unsold, how many fruits were there at first?
