If p and q are integers and p divided by q is 20.15, then which of the

p=20.15q from which p = 20q + 15q/100 or p = 20q + 3q/20.
Since p and q are integers, 3q/20 must also be an integer. 3 is not divisible by 20, then q must be divisible by 20, and therefore, q/20 is an integer and 3q/20 is an integer which is a multiple of 3. From the given answers, only 15 and 3 are divisible by 3.

Answer C.

General Rule :

p=mq+r
m is multiple
r is remainder
now, we can express 20.15 as 20+0.15

So R =0.15
\(=15/100\)
\(=3/20\)
which implies that the remainder should be a multiple of 3 .
Possible values 3, 15
Hence C.

\(\frac{p}{q} = 20.15 = 20 \frac{15}{100}\)
Notice here that if q = 100, the remainder would be 15

\(\frac{15}{100} = \frac{3}{20}\)
If q = 20, remainder would be 3.

But under no conditions can you cancel off 3 from the numerator and denominator and be left with 5 in the numerator so remainder cannot be 5.

Check this post for detailed discussion of this concept: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/

I feel like I have trouble grasping the concept in general... I mean the next time I see a similar problem I am not exactly sure where to start. Is this always the case where whatever decimal we have we just look at the possible multiples of that decimal? For example, if we had say 20.34 we would take 34q/100 or 17q/50 looking for a multiple of 17 in the answer choice?

And are there a few other similar problems that you might have seen particularly with tweaked premises? Or may be you have a way to tweak the question some other way, let me know.

Appreciate your help guys!

if we had say 20.34 we would take 34q/100 or 17q/50 looking for a multiple of 17 in the answer choice? YES!

The relationship between the dividend (a certain number \(n\)), divisor(\(d\)), quotient(\(q\)) and remainder(\(r\)):
\(n = dq + r,\) all numbers are positive integers, except \(r,\) which can be also 0 (in which case, we say that \(n\) is divisible by \(d\)).
The above can be rewritten as \(n/d = q + r/d.\)
So, \(r/d\) is the 0.something you see in the result of a division.
0.something is the fractional part of the quotient. If you multiply that by the divisor, in our case \(q\), you will get the remainder, which has to be an integer. Always take the fraction in the expression of the fractional part in lowest terms, exactly as you did with \(34q/100\). Since 17 is not divisible by 50, it means \(q\) must be divisible by 50, so \(q/50\) is an integer, and \(17q/50\) is an integer which is a multiple of 17.

Remainder should be of the form 0.15x where x is an integer
3 and 15 are of the form 0.15x

IMO C

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