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Q.

In a meet, persons from five different places have assembled in Bangalore High School. From the five places the persons come to represent are $42,60,210,90$ and $84$.

What is the minimum number of rooms that would be required to accommodate so that each room has the same number of occupants and occupants are all from the same places?

 A.

44

 B.

62

 C.

81

 D.

96

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Solution:
Option(C) is correct

All the students from each have to be accommodated in a certain number of rooms.

There should be no person left over (remainder) from any places who can be clubbed together with the persons left over from other places.

To have the minimum number of rooms, the capacity of each room is HCF of all the numbers.
$\text{HCF}(42,60,210,90,84) = 6$

Thus Min number of rooms $= \frac{(42+60+210+90+84)}{6} = 81$


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