Q: A certain fruit sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
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A certain fruit stand
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Brent@GMATPrepNow
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Here's an approach where we test the POSSIBLE SCENARIOS.A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas for a total of $6.30, what number of apples and bananas did the customer purchase.
A)10
B)11
C)12
D)13
E)14
FACT #1: (total cost of apples) + (total cost of bananas) = 630 CENTS
FACT #2: total cost of bananas is DIVISIBLE by 50, since each banana costs 50 cents.
Now let's start testing POSSIBLE scenarios.
Customer buys 1 apple.
1 apple costs 70 cents, which means the remaining 560 cents was spent on bananas.
Since 560 is NOT divisible by 50, this scenario is IMPOSSIBLE
Customer buys 2 apples.
2 apple costs 140 cents, which means the remaining 490 cents was spent on bananas.
Since 490 is NOT divisible by 50, this scenario is IMPOSSIBLE
Customer buys 3 apples.
3 apple costs 210 cents, which means the remaining 520 cents was spent on bananas.
Since 520 is NOT divisible by 50, this scenario is IMPOSSIBLE
Customer buys 4 apples.
4 apple costs 280 cents, which means the remaining 350 cents was spent on bananas.
Since 350 IS divisible by 50, this scenario is POSSIBLE
350 cents buys 7 bananas.
So, the customer buys 4 apples and 7 bananas for a total of 11 pieces of fruit
Answer: B
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Brent@GMATPrepNow
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I should mention that we can't really solve this question using regular algebra.A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas for a total of $6.30, what number of apples and bananas did the customer purchase.
A)10
B)11
C)12
D)13
E)14
If we let A = total cost of apples (in cents),
and let B = total cost of bananas (in cents),
we get the equation 70A + 50B = 630
In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to POSITIVE INTEGERS, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables.
Here's a similar question from the Official Guide: https://www.beatthegmat.com/og-13-132-t117594.html
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Exactly. That was the problem i was having. I solved it with your approach by testing values and looking that the remaining cost must be divisible by 0.5. However, i wanted to make the algebraic equation but was unable to do so.
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Matt@VeritasPrep
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While I think Brent's approach is fine, we actually can do a bit of algebra here.
We'll use the fact that a and b MUST BE NONNEGATIVE integers.
70a + 50b = 630
If we sold nothing but apples, we'd have to sell 9 pieces of fruit. (9 * 70¢ = $6.30.) But we sold some bananas, so we MUST have sold at least 10 pieces of fruit.
If we sold nothing but bananas, we'd have to sell at least 12 pieces of fruit (12 * 50¢ is just below $6.30.) So we can't have sold more than 12 pieces of fruit.
Now let's reduce our equation to
7a + 5b = 63
One of these numbers MUST be odd and the other MUST be even, since our sum is odd.
If (a + b) is even, however, this won't happen: a and b will either both be odd or both be even.
Hence 10 and 12 pieces of fruit are impossible, and we must have sold 11. (How slick is that!?
)
We'll use the fact that a and b MUST BE NONNEGATIVE integers.
70a + 50b = 630
If we sold nothing but apples, we'd have to sell 9 pieces of fruit. (9 * 70¢ = $6.30.) But we sold some bananas, so we MUST have sold at least 10 pieces of fruit.
If we sold nothing but bananas, we'd have to sell at least 12 pieces of fruit (12 * 50¢ is just below $6.30.) So we can't have sold more than 12 pieces of fruit.
Now let's reduce our equation to
7a + 5b = 63
One of these numbers MUST be odd and the other MUST be even, since our sum is odd.
If (a + b) is even, however, this won't happen: a and b will either both be odd or both be even.
Hence 10 and 12 pieces of fruit are impossible, and we must have sold 11. (How slick is that!?
)-
Matt@VeritasPrep
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- Posts: 2630
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Thanks! It was dumb luck, but nothing makes me happier than stumbling into something neat like that, then realizing what's going on.shahfahad wrote:@Matt: Awesome approach.
