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A certain fruit stand sold apples for $0.70 each and

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A certain fruit stand sold apples for $0.70 each and

by BTGmoderatorLU » Tue Apr 17, 2018 2:44 pm

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A certain fruit stand 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

The OA is B.

I solve this PS question as follows,

0.7a+0.5b = 6.3

7a+5b= 63

Now we know that the unit's place of b has to be 5 or 0 (since it is a multiple of 5).
So, 7*a has to have a unit's place of 8 or 3 to get a total sum of 63

One possibility is a=9, b=0
But since the questions states that he bought BOTH, and since 9 is not an answer choice, this is ruled out.

The next possibility is trying to find a multiple of 7 that has unit's place = 8 ; 28
a=4
7*4 = 28
28 + 5*7 = 63

So, a = 4, b = 7 ; a+b=11. Option B.

Please, can anyone explain another way to solve this question? Thanks in advance!
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by Brent@GMATPrepNow » Tue Apr 17, 2018 3:25 pm
BTGmoderatorLU wrote:A certain fruit stand 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
Here's an approach where we test the POSSIBLE SCENARIOS.

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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by Jeff@TargetTestPrep » Thu Apr 19, 2018 4:43 pm
BTGmoderatorLU wrote:A certain fruit stand 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
We are given that apples were sold for $0.70 each and that bananas were sold for $0.50 each. We can set up variables for the number of apples sold and the number of bananas sold.

b = number of bananas sold

a = number of apples sold

With these variables, it follows that:

0.7a + 0.5b = 6.3

We can multiply this equation by 10 to get:

7a + 5b = 63

5b = 63 - 7a

5b = 7(9 - a)

b = [7(9 - a)]/5

Remember that a and b MUST be positive integers here. Thus, 5 must evenly divide into 7(9 - a). Since we know that 5 DOES NOT divide evenly into 7, it MUST divide evenly into (9 - a). We can ask the question: What must a equal so that 5 divides into 9 - a? Of course, a could equal 9; but that would produce a zero for b and since the question states that both apples AND bananas were purchased, b cannot equal zero. The only other value a can be is 4. We can check this:

(9 - a)/5 = ?

(9 - 4)/5 = ?

5/5 = 1

Since we know a = 4, we can use that to determine the value of b.

b = [7(9 - 4)]/5

b = [7(5)]/5

b = 35/5

b = 7

Thus a + b = 4 + 7 = 11.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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