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Question

Let A, B and C be the sets such that AB=AC and AB=AC. Show that B = C

Or

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 three men got medals in all three sports? How many received medals in exactly two of the three sports?

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Solution

We have AB=AC

(AB)C=(AC)C

(AC)(BC)=C

[(AC)C=and using distributive law (AB)C=(AC)(BC)]

(AB)(BC)=C [AC=AB] (i)

Again, AB=AC

(AB)B=(AD)B

B=(AB)(CB)

[(AB)B=B and using distributive law (AC)B=(AB)(CB)]

B=(AB)(BC) (ii)

From Eqs. (i) and (ii), we get.

B = C
or

Let F denotes the set of men who received medals in football, B denotes the set of men who received medals in basketball and C denotes the set of men who received medals in cricket.
Then, we have
n(F)=38,n(B)=15.n(C)=20

n(FBc)=58 and n(FBC)=3

Now, n(FBC)=n(F)+n(B)+n(C)n(FB)n(8C)n(FC)+n(FBC)

58=38+15+20n(FB)n(BC)n(FC)+3

n(FB)+n(BC)+n(FC)=7658=18

Now, number of men who received medals in exactly two of the three sports

=n(FB)+n(BC)+n(FC)3n(FBC)=183×3=189=9

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


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