# 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?

### Yes, that would work. We can

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

would you plz solve it using the weighted average way
thanks a lot

### Sure thing.

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

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

### Question 1: There are two

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.

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

Cheers,
Brent

### Question 1: There are two

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.

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

Cheers,
Brent

### Hi,

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

### I don't quite follow your

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

### I probably made a wrong

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