Q. Various people from five different places have assembled in for a competition. The number of people who have arrived from each place is respectively 42,60,210,90 and 84.What is the minimum number of rooms that would be required so that each room has the same number of occupants and occupants are all from the same places?

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Solution

The correct option is D 81
HCF of 42,60,210,90 and 84
is 6
Required number is $\frac{42}{6} + \frac{60}{6} + \frac{210}{6} + \frac{90}{6} + \frac{84}{6} = 81$

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The correct option is D 81HCF of 42,60,210,90 and 84 is 6 Required number is 426+606+2106+906+846=81

The correct option is D 81
HCF of 42,60,210,90 and 84
is 6
Required number is $\frac{42}{6} + \frac{60}{6} + \frac{210}{6} + \frac{90}{6} + \frac{84}{6} = 81$