Latest IBM Aptitude Question SOLUTION: 48 students have to be seated such that each row has the same number of students as the others. If at least 3 students are to be seated per row and at least 2 rows have to be there

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48 students have to be seated such that each row has the same number of students as the others. If at least 3 students are to be seated per row and at least 2 rows have to be there, how many arrangements are possible?

Asked In IBM (12 years ago)

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lost  (11 years ago)

48 = 2^4 * 3^1
now separating the factors:
x*y = 1*48, 2*24, 3*16, 4*12, 6*8, 8*6, 12*4, 16*3, 24*2, 48*1.
let x=rows and y=no. of students, therefore 10 combinations are possible
according to the question x>=2 and y>=3, therefore 7 combinations is the ans.

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uthra murugesan  (8 years ago)

Three conditions have to be satisfied.
The number of students per row has to be at least 3.
Number of row has to be at least 2.
Equal number of students has to be seated in a row.
The following arrangements satisfy all 3 conditions.

Arrangement 1: 3 students to a row; 16 rows.
Arrangement 2: 4 students to a row; 12 rows.
Arrangement 3: 6 students to a row; 8 rows.
Arrangement 4: 8 students to a row; 6 rows.
Arrangement 5: 12 students to a row; 4 rows.
Arrangement 6: 16 students to a row; 3 rows.
Arrangement 7: 24 students to a row; 2 rows.

You will observe that the number of students in a row is a factor of 48.

So, an alternative and faster approach is to list down factors of 48 – viz., 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

And then start from 3 and quickly find out if the number of rows is at least 2.

Both the conditions are satisfied for the following factors : 3, 4, 6, 8, 12, 16 and 24. i.e., 7 arrangements.

Correct answer choice (4)

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