In a certain first grade classroom, the average height is x inches

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In a certain first grade classroom, the average height is x inches for the 5 girls and y inches for the 8 boys. What is the average height for the 13 students in the classroom combined?

(1) 15x + 24y = 1755
(2) x < 40

Answer: A
Source: Veritas Prep
Source: — Data Sufficiency |

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BTGModeratorVI wrote:
Fri Jun 05, 2020 11:51 am
In a certain first grade classroom, the average height is x inches for the 5 girls and y inches for the 8 boys. What is the average height for the 13 students in the classroom combined?

(1) 15x + 24y = 1755
(2) x < 40

Answer: A
Source: Veritas Prep
Target question: What is the average height for the 13 students in the classroom combined?

Given: The average height is x inches for the 5 girls and y inches for the 8 boys.
This is a good candidate for rephrasing the target question.

We can think of this is a weighted averages question. So, we'll apply the following formula:
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...
So, Average weight of classroom = (5/13)(x) + (8/13)(y)
= 5x/13 + 8y/13
= (5x + 8y)/13
So, we can REPHRASE the target question...
REPHRASED target question: What is the value of (5x + 8y)/13?

Aside: Here’s a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Statement 1: 15x + 24y = 1755
Notice that we have (5x + 8y) "hiding" in the above equation.
Take 15x + 24y = 1755 and factor the left side to get: 3(5x + 8y) = 1755
Divide both sides by 3 to get: 5x + 8y = 585
So, (5x + 8y)/13 = (585)/13 = 45
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x < 40
No information about y, which means there's no way to determine the value of (5x + 8y)/13
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A
Brent Hanneson - Creator of GMATPrepNow.com
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