Question: Broken Plates

Comment on Broken Plates

Can we follow allegation method to find resultant percentage when 2 different boxes(or volumes) of same plates(or materials) with different percentages are combined?
gmat-admin's picture

Yes, that would work. We can also you the weighted averages formula. However, since this question is part of the first module (Arithmetic) in the course, I didn't get into those other approaches, since most students will not have studied those concepts yet.
For those interested in using the weighted averages approach, watch this video first: https://www.gmatprepnow.com/module/gmat-statistics/video/805

would you plz solve it using the weighted average way
thanks a lot
gmat-admin's picture

Sure thing.

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

So, here we get: average breakage of both boxes = (box A proportion)(box A average) + (box B proportion)(box B average)

There are 150 plates altogether.

So, box A proportion = 60/150 = 2/5
Box A average = 15% breakage

Box B proportion = 90/150 = 3/5
Box B average = 30% breakage

We get: average breakage of both boxes = (2/5)(15) + (3/5)(30)

= 6 + 18

= 24

Hi Brent, I have 2 questions about averages

Question 1: A regular deck of cards contains 52 cards. In a game of Topren, each player must have an equal number of cards
at the start of the game. If not more than 10 players can play Topren at a time, what is the maximum number
players playing the game such that all cards in 2 regular decks are equally distributed among all players?
A) 4
B) 6
C) 8
D) 9
E) 10

and

Question 2: In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book,
10 students each borrowed 2 books, and the rest borrowed at least 3 books. If the average number of books per
student was 2, what is the maximum number of books any single student could have borrowed?
A) 3
B) 5
C) 8
D) 13
E) 15

gmat-admin's picture

Question 1: There are two decks of cards.
2 x 52 = 104
So, there are 104 cards in total.
Let n = the number of players

Key information: All 104 cards must be EQUALLY DISTRIBUTED among the n players.
This tells us that 104 must be DIVISIBLE BY n
104 = (2)(2)(2)(13)
Since n must be less than or equal to 10, we can see that 8 is the greatest possible divisor of 104.

Answer: C

Question 2: Here's my solution: https://gmatclub.com/forum/in-a-class-of-30-students-2-students-did-not-...

Cheers,
Brent

gmat-admin's picture

Question 1: There are two decks of cards.
2 x 52 = 104
So, there are 104 cards in total.
Let n = the number of players

Key information: All 104 cards must be EQUALLY DISTRIBUTED among the n players.
This tells us that 104 must be DIVISIBLE BY n
104 = (2)(2)(2)(13)
Since n must be less than or equal to 10, we can see that 8 is the greatest possible divisor of 104.

Answer: C

Question 2: Here's my solution: https://gmatclub.com/forum/in-a-class-of-30-students-2-students-did-not-...

Cheers,
Brent

Hi,

I found 24 but through what might be a convoluted way. I arrived at 36/150 but didn't find how to simplify it by more than 2, so I arrived at 18/75. From there I know that 75 = 3/4, so I did 18/3/4 = 18*4/3. I simplified 18 by 3 to have 6*4 = 24. Is this a valid approach? Thanks
gmat-admin's picture

I don't quite follow your approach. What you mean when you say "75 = 3/4"?

I probably made a wrong shortcut by thinking 18/75% so 18/3/4...

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