MadeEasy Test Series: General Aptitude - Numerical Computation

Question

Q.10

Given and Q = a,b,c are positive integers. What is the number of ordered triplets (a, b, c) such that a + 2b + c ≤ 40?

Answer 1

P=a/c+a/b+1

or, P=(ab+ac+bc)/bc.................(i)

Q=b/a+b/c+1

or, Q=(ab+ac+bc)/ac....................(ii)

from (i) and (ii)

P/Q =ac/bc =a/b =11

or, a=11b

if  b=1 a=11

equation a + 2b + c ≤ 40

    or, 11+2+c≤40

    or,c≤27

So, C can get 27 values and for that 27 values we get 27 triplets

like(11,2,1),(11,2,2),...................(11,2,27)....................i

Now,when b=2 ,a=22

Now, C value can be upto 14, So 14 triplets.................ii

Now, b=3 then a=33 ,c can be 1 only

So,triplet (33,6,1).......................................................iii

Now,from i,ii,iii we get 

27+14+1=42 triplets

Comments

Recheck after a=11b a=1, b=11???

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yes,thanks

Answer 2

Official Solution given by them. Not sure of correctness !