Question

Two groups consisting of 108 and 156 students
participate in sports meet. In how many rooms can
all these students be accommodated at the mini-
mum, if all the rooms should have equal number
of students and no two students in a room should
be from different groups?
(a) 22
(b) 20
(c) 19
(d) 13​

Answer 1

Answer: option(a) 22

Step-by-step explanation:

The no. of students situated in each room would be H.C.F of the two quantities above.  

156=2×2×3×13  

108=2×2×3×3×3  

so HCF = 12  

Thus, in each room 12 members can be situated.  

In this way, number of rooms required = Total number of members/12 = (156+108)/12 = 22

hope it helps

Answer 2

Answer:

TO IDENTIFY THE NUMBER OF ROOMS WE NEED TO FIND THE HIGHEST NUBMER OF STUDENTS OF EACH GROUP THAT CAN BE ACCOMODATED IN ONE ROOM.

SO WE WILL FIND ITS HCF:-

156=108×1+48

108=48×2+12

48=12×4+0

So your HCF is 12

Now the total no. of rooms required =

$\frac{total \: number \: of \: students}{hcf}$

SO, TOTAL NO. OF ROOMS =

$\frac{108 + 156}{12} \\ \\ = \frac{264}{12} \\ \\ = 22$

So, your answer is (a)22

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