Bunuel
If a and b are both positive integers \(\frac{a}{b}=34.06\), what is the probability that the remainder of \(\frac{a}{b}\) is a multiple of 6?
A. 1/3
B. 9/20
C. 1/2
D. 2/3
E. 3/4
a=34.06b { a=bq+r ; where q is the quotient and r is the remainder; here, q=34.06, r=0}
a=34b+0.06b
When we divide a by b, any remainder will finally only be divided from the decimal portion of the equation (i.e. 0.06b) and not from the term before the decimal (i.e. 34b).Now consider only 0.06b the portion that determines the value of the remainder.
Hence, 0.06b could be any integer 1,2,3,4,5,6,7,8 or 9.
Now, as per the question, we are given that b is a positive integer
The only values that satisfy this are 3 and 6.
0.06b = 3 or 6.
This gives us 2 outcomes.
The probability of selecting 6 as a multiple of \(\frac{a}{b}\) among 3 and 6 is \(\frac{1}{2}\).
Answer is C \(\frac{1}{2}\)