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orange and kiwi

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orange and kiwi

by Sandhya Goud » Sat May 07, 2016 2:54 am
Price of an orange is 90 cents and that of a kiwi is $1. Assuming that Ron buys both the fruits but never buys a partial fruit, how many kiwis does he purchase?

(1) Ron spends exactly $18.
(2) The number of kiwis bought by Ron is less than that of oranges.
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by MartyMurray » Sat May 07, 2016 5:02 am
Key to getting the right answer to this one is understanding that this, "never buys a partial fruit," indicates that the number of each type of fruit will be an integer value.

Statement 1: Ron spends exactly $18.

If Ron buys both of the fruits, he must buy some oranges. Oranges are 90 cents each. So in order to get an integer result, he has to buy oranges in multiples of 10.

10 x $.90 = $9

To get even $18 Ron must therefore buy one of the following combinations of oranges and kiwis.

10 oranges + 9 kiwis: (10 x .9) + (9 x 1) = 18

20 oranges + 0 kiwis: (20 x .9) + (0 x 1) = 18

From the question we know that he buys at least 1 kiwi. So only the first scenario works.

Sufficient.

Statement 2: The number of kiwis bought by Ron is less than that of oranges.

Given what is said in this statement, he could buy any of various numbers of kiwis and oranges.

Insufficient.

The correct answer is A.
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by Matt@VeritasPrep » Fri May 13, 2016 3:19 pm
Another way of solving that might be easier:

Suppose Ron buys x oranges and y kiwis.

S1 tells us that

.9x + 1y = 18

or

9x + 10y = 180

Now, suppose Ron bought ALL oranges. Then x = 20, y = 0.

Now, suppose Ron bought ALL kiwis. Then x = 0, y = 18.

Since we know Ron buys both fruits, neither of these scenarios works. So the only possibility is BETWEEN 18 and 20! That means he must have bought 19, and we have x + y = 19.

Since we now have two equations (x + y = 19 and 9x + 10y = 180), we can solve, and we're set!
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