A certain fruit stand sold apples for $0.70 each and bananas

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Tue May 28, 2019 1:19 am

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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

OA B

Source: Official Guide

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue May 28, 2019 2:36 am

Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

BTGmoderatorDC 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

Since 70-cent apples and 50-cent bananas yield a total cost of 630 cents, we get:
70A + 50B = 630
7A + 5B = 63

One solution for the equation in blue:
A=9, B=0
In accordance with the rule discussed above, the value of A may change only in increments of 5 (the coefficient for B), while the value of B may change only in increments of 7 (the coefficient for A).
Thus, only one other non-negative integral solution is possible:
A=4, B=7
A+B = 4+7 = 11

The correct answer is B.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue May 28, 2019 5:40 am

BTGmoderatorDC 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

OA B

Source: Official Guide

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed May 29, 2019 5:49 pm

BTGmoderatorDC 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

OA B

Source: Official Guide

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

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