A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and ban

[Vincen]

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

Answer: B

Source: Official Guide

[Brent@GMATPrepNow]

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

Answer: B

Source: Official Guide

Here's an approach where we test the POSSIBLE CASES.

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

[swerve]

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

Answer: B

Source: Official Guide

We can try as follows:

\(7a + 5b = 63\)
\(2a+5a+5b=63\)
\(2a+5(a+b)=63\)

Now, substituting the answer choices,

\(2a+5(10)=63 \Rightarrow 2a=13 \Rightarrow a\) has to be an integer. \(\Large{\color{red}\chi}\)
\(2a+5(11)=63 \Rightarrow 2a=8 \Rightarrow a = 4\) \(\Large{\color{green}\checkmark}\)
\(2a+5(12)=63 \Rightarrow 2a=3 \Rightarrow a\) has to be an integer. \(\Large{\color{red}\chi}\)
\(2a+5(13)=63 \Rightarrow 2a=-2 \Rightarrow a\) has to be \(+\)ve. \(\Large{\color{red}\chi}\)
\(2a+5(14)=63 \Rightarrow 2a=-7 \Rightarrow a\) has to be \(+\)ve. \(\Large{\color{red}\chi}\)

Therefore, \(11\) is the correct answer.