Decimal

1) 10/189 = 189 = 3*9*7, neither of the three digits cause a terminating decimal

2) 15/196 = 196 = 2 * 2 * 7 * 7 = again 2 causes a terminating decimal but 4 doesnt

3) 16/225 = 225 = 3*5*3*5 = 3 doesn't cause a terminating decimal

4) 25/144 = 144 = 2*2*2*2*3*3 = 3 doesn't cause a terminating decimal

5) 39/128 = 128 = 2*2*2*2*2*2*2 = 2 causes a terminating decimal so E is the answer

if the denominator contains only 2's, or 5's or 1's or multiples of them, the resulting value will be a terminating decimal.

answer E. rest all have 3 or 7

thanks for the answer prindaroy,
but can you elaborate a bit on how you are concluding that certain numbers do not cause terminating decimals?
My method was to split the numerator and denominators and then calculate for one of them.
But this again takes much time.

For a question like this would you ever look to the numerator? From the answers above it appears those the answer is no, but i just want to confirm.

The numbers 3,6,9 in denominator produce repeating #s

You can test if a number is divisible by 3,6,9 by counting the number of digits in the numerator. For e.g 234/3 is some integer. (2+3+4 =9/3=3)


So as far as the above problem goes, I could eliminate all the choices except E and B which you can check quickly

tom4lax wrote:For a question like this would you ever look to the numerator? From the answers above it appears those the answer is no, but i just want to confirm.

of course you need to check the numerator.
If you get more than 1 options with denominator 3,7, or some prime number or multiples of them, you need to check if the same is there in the numerator and they cancel out.

got it, thanks