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Fun Sports (FS) provides training in three sports - Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 3939 students each of whom is enrolled in at least one of the three sports. The following details are known:

  1. The number of students enrolled only in L is double the number of students enrolled in all the three sports.
  1. There are a total of 1717 students enrolled in G.
  1. The number of students enrolled only in G is one less than the number of students enrolled only in L.
  1. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
  1. The maximum student enrollment is in L.
  1. Ten students enrolled in G are also enrolled in at least one more sport.

What is the minimum number of students enrolled in both G and L but not in K?

Entered answer:

Solution

βœ… Correct Answer: 4
Slide 1/9

Understanding the set :-

This is a Venn diagram set.

  1. We’re told:

a) There are 39 students total, and each is in at least one sport.

b) Some are in only one sport, some in exactly two sports, and some in all three.

c) The clues give relationships between these groups.

  1. Draw the Venn diagram and,

Think of three overlapping circles labeled G, K, L.

That gives 7 possible regions:

a) Only G

b) Only K

c) Only L

d) G & K only

e) G & L only

f) K & L only

g) All three

  1. The approach.

a) Represent each region by a letter (say π‘Ž, b, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔).

b) Write equations for each clue.

c) Use totals to find the numbers.

Clue 1 says,

The number of students enrolled only in L is double the number of students enrolled in all the three sports.

Let the number of students enrolled in all the three sports be "x".

=> Number of students enrolled in only L will be "2x".

Clue 2 says,

There are a total of 17 students enrolled in G.

Also, clue 6 says,

Ten students enrolled in G are also enrolled in at least one more sport.

=> Therefore, the number of students enrolled in only G = 17βˆ’10=717 - 10 = 7

Clue 3 says,

The number of students enrolled only in G is one less than the number of students enrolled only in L.

We know, the number of students enrolled in only G = 7.

=> the number of students enrolled only in L = 7+1=87+1=8

=> 2x = 8

=> X = 4

Clue 4 says,

The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.

Here the number of students enrolled in K and L = Number of students enrolled in only K + L and number of students enrolled in all the three.

Now, Let the number of students enrolled in only K + L = y

=> the number of students enrolled in K and L = y + 4

=> number of students enrolled only in K = y + 4

Now, Let us assume that 'z' be the the number of students enrolled in G and K but not L.

Then, the number of students enrolled G and L bot not K = 10βˆ’4βˆ’z10 - 4 - z

= 6βˆ’z6 - z

Now, we know that there are total of 39 enrollments.

=> 7+z+4+6βˆ’z+8+y+y+4=397 + z + 4 + 6 - z + 8 + y + y + 4 = 39

β‡’ y=5y = 5

Now we know,

Number of students enrolled in G = 17

Number of students enrolled in K = 9+4+5+z=18+z9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6βˆ’z+4+5+8=23βˆ’z6 - z + 4 + 5 + 8 = 23 - z

clue 5 says,

The maximum student enrollment is in L.

β‡’ 23βˆ’z>18+z23 - z > 18 + z

β‡’ 2z<52z < 5

β‡’ z<2.5z < 2.5

Therefore, we can say that z can take three values = {0, 1, 2}

we know that, z can take three values = {0, 1, 2}

The number of students enrolled in both G and L but not in K = 6 - z.

This number will be minimum when 'z' is maximum.

and, zmax=2z_{max} = 2

Therefore, the minimum number of students enrolled in both G and L but not in K = 6βˆ’2=46 - 2 = 4

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