Question

A college awarded 38 medals for honesty , 15 for punctuality and 20 for obedience if these medals were bagged by a total of 58 students and only 3 students got medals for all three values, how many students recieved medals for exactly two of the three values ???​

Answer 1

Answer:

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Step-by-step explanation:

Given

38 medals for honesty.

15 for punctuality and

24 for obedience.

Total Medals = 38 + 15+ 24 = 77

only three students got medals in all the three values,

So, the remaining medal = 77- 9 = 68 medals

Total students =58

So, students who receive exactly 2 medals= 68- 58 = 10 medals

Answer 2

A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only 3 men got medals in all the three sports, how many received medals in exactly two of the three sports?

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Answer

Let, F=football, B=basketball and C= cricket.

Then n(F)=38, n(B)=15 and n(C)=20, n(F∪B∪C)=58, n(F∩B∩C)=3.

Since, n(F∪B∪C)= [n(F)+n(B)+n(C)]−[n(F∩B)+n(B∩C)+n(C∩F)]+n(F∩B∩C).

or, 58=(38+15+20)−[n(F∩B)+n(B∩C)+n(C∩F)]+3

or, [n(F∩B)+n(B∩C)+n(C∩F)]=76−58=18

Let, a be the no. of men who play only football and basketball,

b be the no. of men who play only football and cricket,

and c be the no. of men who play only cricket and basketball.

From, venn-diagram,

n(F∩B)=a+3

n(B∩C)=b+3

n(C∩F)=c+3

∴[n(F∩B)+n(B∩C)+n(C∩F)]=a+b+c+9

or, 18=a+b+c+9

a+b+c=9

9 men received medals in exactly two of the three sports.

solution