Word problems

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Word problems

by BTGmoderatorRO » Sun Oct 08, 2017 12:10 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 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. 15
QA is b.

Is there other approach to use to get C. i am somehow confused here. anyone there to help?

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Word problems

by Brent@GMATPrepNow » Sun Oct 08, 2017 1:22 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
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 » Sun Oct 08, 2017 1:23 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 Scott@TargetTestPrep » Sun Nov 24, 2019 5:47 pm
BTGmoderatorRO 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. 15
QA is b.

Is there other approach to use to get C. i am somehow confused here. anyone there to help?
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

We realize 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 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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by [email protected] » Mon Nov 25, 2019 10:09 am
Hi All,

We're told that a certain fruit stand sold apples for $0.70 each and bananas for $0.50 each and that a customer purchased both apples and bananas from the stand for a total of $6.30. We're asked for the TOTAL number of apples and bananas that the customer purchased. Here, the answer choices are relatively small and 'close together', so we can use a bit of 'brute force' to get to the correct answer. We know that there will be no fewer than 10 total pieces of fruit and no more than 14 total pieces of fruit that will total $6.30, so I'm going to list out the first several multiples of apple prices and banana prices:

Apples:
$0.70
$1.40
$2.10
$2.80
$3.50
$4.20
$4.90
$5.60
Etc.

Bananas:
$0.50
$1.00
$1.50
$2.00
$2.50
$3.00
$4.00
$4.50
Etc.

Now we just have to find a pair of numbers (one from each group) that will total $6.30. It's not too much work to find that $2.80 and $3.50 total $6.30, so the total number of pieces of fruit is 7+4 = 11

Final Answer: B

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