Apples and Oranges
A fruit seller buys a certain number of apples and half that many pairs of oranges. (In effect equal number of apples and oranges). He paid $2 for each of the apples and $1 for each of the oranges. On every fruit, he placed a retail price that was 10% more than the cost price.
After some time, all but 7 of the fruits had been sold. The fruit seller found that his revenue is equal to his cost price. His potential profit therefore, was represented by the combined retail value of the 7 remaining fruits.
What is his potential profit?
This puzzle is not an original. It is adapted from a problem posed by the great Martin Gardener.
The answer is 13.20.
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1 solution
L e t n b e n u m b e r o f a p p l e s a n d n u m b e r o f o r a n g e s b o u g h t . S o $ 3 n w e r e p a i d . ∴ p r o f i t = $ 0 . 3 n . L e t o r a n g e s l e f t = x . T h e n p r o f i t e = 1 . 1 ∗ x + 2 . 2 ∗ ( 7 − x ) = 1 5 . 4 − 1 . 1 x . ∴ 0 . 3 n = 1 5 . 4 − 1 . 1 x . ⟹ n = . 3 1 5 . 4 − . 3 1 . 1 x = 5 1 3 1 − . 3 1 . 1 x B u t n a n d x a r e i n t e g e r s a n d n o r a n g e s a r e i n p a i r s t h a t i s e v e n . ⟹ 0 ≤ x ≤ 7 , a n d n i s e v e n . B u t n c a n b e e v e n i n t e g e r o n l y i f . 3 1 . 1 x i s a n o d d i n t e g e r + 3 1 . S o x = 2 i s t h e o n l y s o l u t i o n . . P r o f i t e = 0 . 3 ∗ 4 4 = $ 1 3 . 2 .
I missed since I over looked "half that many pairs".
Let X be the number of apples bought. The number of oranges bought is also X since its "half that many pairs" of oranges. Let Y be number of Apples left after selling. So Number of oranges left is (7-Y) since we know 7 fruits are left.
Number of apples sold = X-Y Number of oranges sold = X- (7-Y) The price of each apple is 2 $ and price of each orange is 1$. So total cost price is 2X + 1X=3X.
The selling price is 10 % over cost price. So selling price of each apple is 2.2 $ and that of each orange is 1.1$ We know that after he has sold all but 7 remaining fruits he retrieves his cost price so here goes---
3X = 2.2 (X-Y) + 1.1 (X-7+Y)
3X= 2.2X - 2.2Y + 1.1X - 7.7 +1.1Y
0.3X= 1.1Y + 7.7
3X= 11Y+77
Now we have 2 variables. When we try to slot in all possible integer values up to 7 as possible values for Y to give a whole number for X , we find there are only 2 possibilities. Y=2 or Y = 5. But if Y =2, X is 33.
And this is the Crux. We know that X cannot be an odd number since the oranges bought were in "pairs" hence Y=5 and X = 44. So apples left are 5 and oranges left are 2. 5(2.2)+ 2(1.1) = $ 13.20. And that's his potential profit.