Question: In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?

1. 12
2. 13
3. 14
4. 15
5. 16

Correct Answer: B
Solution and Explanation:
Approach Solution 1:
The total number of medals: 36+12+18= 66.
So, 66 medals are awarded to 45 persons.

12 of them have been awarded to 4 persons.

The other 54 medals have been awarded to 41 persons.

So, there are 54–4= 13 persons with two medals each.

Approach Solution 2:
This problem can be solved through Sets and Venn diagrams.

Although most answers below are accurate, they make use of the Venn diagram with linear equations approach, and I would like to solve it with a more simpler method.

In a Three-Set problem, with say, Set A, B and C –

Let I , II and III be the total number of persons who received exactly one, two and three medals each respectively.

Let S be the total number of medals given in the competition.
Clearly, total number of participants= I + II + III= let's name it X

So needless to say, overall, S number of medals were distributed among X number of individuals where every person can only get one, two or three medals each, and hence S > X

Now for ANY Three-Set Venn diagram we can say that –
S - X = (2*III) + II

The above statement denotes the total number of participants who received more than one medal each, meaning who got two or three medals each, given by S - X
In the given problem statement, III = 4 , S = 36 + 12 + 18 = 66 and X = 45 and we need to find II
Using the relation we can find II = 66 - 45 - 8 = 13

Therefore, 13 participants received exactly two medals.

Approach Solution 3:
Let A= set of persons who got medals in dance.
B= set of persons who got medals in dramatics.
C= set of persons who got medals in music.

Given,
n(A) = 36
n(B) = 12
n(C) = 18
n(A ∪ B ∪ C) = 45
n(A ∩ B ∩ C) = 4

We know that number of elements belonging to exactly two of the three sets A, B, C
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C)
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4 ……..(i)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)

Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C)

From (i) required number
= n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12
= 36 + 12 + 18 + 4 - 45 - 12
= 70 - 57
= 13

“In a competition, a school awarded medals in different categories.”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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