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testing the answer choices

by NeilWatson » Tue May 27, 2014 5:33 pm
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

I know this can be solved pretty easily with algebra but can anyone show me how they might use the answer choices to answer this problem? or is this not a good candidate for that
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by Brent@GMATPrepNow » Tue May 27, 2014 7:54 pm
NeilWatson 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 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
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

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by Brent@GMATPrepNow » Tue May 27, 2014 8:13 pm
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
I should mention that we can't really solve this question using regular algebra.
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

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by GMATGuruNY » Wed May 28, 2014 2:39 am
NeilWatson 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 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
We can PLUG IN THE ANSWERS, which represent the total number of apples and bananas.
Since apples are sold for 50 cents each, and bananas are sold for 70 cents each, the average price for all the fruit must be between 50 and 70.
Since 630/70 = 9 and 630/50 = 12.6, the total number of apples and bananas must be between 9 and 12.
Eliminate D and E.

To evaluate the remaining answer choices, we can use ALLIGATION.

Answer choice B: 11 pieces of fruit
Here, the average price per fruit = 630/11.

Step 1: Put the prices over a COMMON DENOMINATOR.
Apple price = 70 = 770/11.
Banana price = 50 = 550/11.
Average price = 630/11.

Step 2: Plot the 3 numerators on a number line, with the numerators for the apples and bananas on the ends and the numerator for the average in the middle.
Apples 770------------------630------------------550 Bananas

Step 3: Calculate the distances between the numerators.
Apples 770-------140--------630--------80-------550 Bananas

Step 4: Determine the ratio of apples to bananas.
To yield an average price of 630/11, the required ratio of apples to bananas is equal to the RECIPROCAL of the distances in red.
Apples : Bananas = 80:140 = 4:7.

Success!
If 4 apples and 7 bananas are purchased -- for a total of 11 pieces of fruit -- the average price per fruit will be 630/11 cents.

The correct answer is B.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
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by Jeff@TargetTestPrep » Thu May 14, 2015 12:36 pm
NeilWatson 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 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

I know this can be solved pretty easily with algebra but can anyone show me how they might use the answer choices to answer this problem? or is this not a good candidate for that
Solution:

I think the algebra approach is a pretty sound method for solving this problem.

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 quantity of apples sold and the quantity of bananas sold.

b = quantity of bananas sold

a = quantity 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

You may now notice that we do not have any other information to set up a second equation as we sometimes do for problems with two variables. So, we must use what we have. Keep in mind that variables a and b MUST be positive whole numbers, because you can't purchase 1.4 apples, for example. You also may notice that "7" and "63" have a factor of 7 in common. Thus, we can move "5b" to the other side of the equation and scrutinize the new equation carefully:

5b = 63 - 7a

5b = 7(9 - a)

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

Remember that a and b MUST be positive whole numbers 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? The only value a can be is 4. And 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 for b.

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

b = [7(5)]/5

b = 35/5

b = 7

Thus a + b = 4 + 7 = 11

The answer B
Last edited by Jeff@TargetTestPrep on Sat May 23, 2015 3:20 am, edited 1 time in total.

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by Aman verma » Fri May 15, 2015 2:44 am
NeilWatson 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 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

I know this can be solved pretty easily with algebra but can anyone show me how they might use the answer choices to answer this problem? or is this not a good candidate for that
Hello NeilWatson,

In this Diophantine Equation, you can take advantage of the fact that the variables are positive integers to set up the algebra and solve:

7A+5B=63
Isolate one variable to one side:
A=9-5B/7,
At B=7 we get an integer value for A. B can be higher multiples of 7,but then we get negative values for A which is inadmissible.
So , total number of fruits=7+4=11. OptionB
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