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Bunuel
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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 \(\frac{1}{2}\)
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can you provide the source and solution of this question. Bunuel

I have a solution though it match non of the options.

a/b=34.06
a=34b+0.06b

now 0.06b is integer, which is multiple of 6.

0.06b=6i in which i is integer.

0.01b=i

in first hundred integer 1-100 as b only b=100 satisfy all the conditions.
similarly in every set of integer as b only 1/100 satisfy all conditions. Ex 200 in 101-200, 300 in 201-300 etc.

hence probability =number of events satisfy the condition/total number of events =1/100

correct me if I am wrong.
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A = 34b + 0.06b. Therefore 0.06b = the remainder = an integer.
What are the possibilities of b so that 0.06b becomes an integer: b = 50 -> 0.06b = 3 / b = 100 -> 0.06b = 6 / b =150 -> 0.06b = 9 and so on.
What is the relation? All the remainders are multiples of 3.
Since 1/2 of all multiples of 3 are multiples of 6. This means that the probability of the remainder being a multiple of 6 is 1/2.
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