# A class got 48 distinctions in maths, 25 in physics and 30 in chemistry. If these went to a total of 68 students and only 5 students got distinctions in all the three subjects, how many students got distinctions in exactly two of the three subjects?

### Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

### Let M, P and C denote the set of students who got distinction in maths, physics and chemistry respectively.
Then n(M) = 48, n(P) = 25, n(C ) = 30
n(M _{} P_{} C) = 68 and n(M _{} P _{} C) = 5
Therefore,
n(M_{}P_{} C) = n(M) + n(P) + n(C) – n(M_{}P) – n(M_{}C) - n(P_{}C) + n(M_{}P_{}C)
_{} 68 = 48 + 25 + 30 – n(M _{} P) – n(M _{} C) - n(P _{} C) + 5
_{} n(M _{}P) + n(M _{}C) + n(P _{}C) = 108 – 68 = 40
Now consider the following Venn diagram:
Here,
a denotes the number of students who got distinction in maths and physics only.
b denotes the number of students who got distinction in maths and chemistry only.
c denotes the number of students who got distinction in physics and chemistry only.
d denotes the number of students who got distinctions in all the three subjects.
Thus, d = n(M _{} P _{} C) = 5 and
a + d + b +d + c + d = 40
Therefore, a + b + c = 25

_{}P

_{}C) = 68 and n(M

_{}P

_{}C) = 5

_{}P

_{}C) = n(M) + n(P) + n(C) – n(M

_{}P) – n(M

_{}C) - n(P

_{}C) + n(M

_{}P

_{}C)

_{}68 = 48 + 25 + 30 – n(M

_{}P) – n(M

_{}C) - n(P

_{}C) + 5

_{}n(M

_{}P) + n(M

_{}C) + n(P

_{}C) = 108 – 68 = 40

_{}P

_{}C) = 5 and

### Answered by | 4th Jun, 2014, 03:23: PM

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