Decimal

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ILULA08: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Tue Mar 31, 2009 12:17 am; GMAT Score:750+

by ILULA08 » Tue Aug 11, 2009 8:19 pm

Q15:
Which of the following fractions has a decimal equivalent that is a terminating decimal?
A. 10/189
B. 15/196
C. 16/225
D. 25/144
E. 39/128

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Source: Beat The GMAT — Problem Solving | Tue Aug 11, 2009 8:19 pm

prindaroy: Master | Next Rank: 500 Posts; Posts: 113; Joined: Thu Jul 16, 2009 11:23 am; Thanked: 15 times; GMAT Score:730

by prindaroy » Tue Aug 11, 2009 8:42 pm

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

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Tue Aug 11, 2009 9:07 pm

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

The powers of two are bloody impolite!!

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ben wade: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Sun Jun 14, 2009 4:54 am; Thanked: 1 times

by ben wade » Tue Aug 11, 2009 9:13 pm

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.

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tom4lax: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Feb 09, 2009 3:57 pm; Thanked: 4 times

by tom4lax » Wed Aug 12, 2009 2:37 pm

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.

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pink_08: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Wed Dec 31, 2008 7:01 pm

by pink_08 » Wed Aug 12, 2009 6:22 pm

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

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Wed Aug 12, 2009 7:20 pm

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.

The powers of two are bloody impolite!!

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tom4lax: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Feb 09, 2009 3:57 pm; Thanked: 4 times

by tom4lax » Thu Aug 13, 2009 4:35 am

got it, thanks

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