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Recurring decimal---lengthy problem

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Recurring decimal---lengthy problem

by prachich1987 » Sat Dec 11, 2010 2:44 am
I found below problem in one of the book.
Will such type of problems appear in actual GMAT.
If yes, is there any short method of solving the same?Plz guide

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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by Rahul@gurome » Sat Dec 11, 2010 6:58 am
prachich1987 wrote: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
Here is a shortcut method: A fraction in lowest terms with a prime denominator other than 2 or 5 always produces a repeating decimal.

Thus if the denominator of a fraction after canceling all the common factors with the numerator contains any prime other than 2 or 5 in their prime factorization, then the fraction does not have a decimal equivalent that is a terminating decimal.
  • A] 10/189 --> 189 contains 3, 10 doesn't => Repeating Decimal
    B] 15/196 --> 196 contains 7, 15 doesn't => Repeating Decimal
    C] 16/225 --> 225 contains 3, 16 doesn't => Repeating Decimal
    D] 25/144 --> 144 contains 3, 25 doesn't => Repeating Decimal
    E] 39/128 --> 128 contains 2 only => Terminating Decimal
The correct answer is E.
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by shovan85 » Sat Dec 11, 2010 7:01 am
prachich1987 wrote:I found below problem in one of the book.
Will such type of problems appear in actual GMAT.
If yes, is there any short method of solving the same?Plz guide

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
I have not seen till date that GMAT play a Trick on Recurring Decimal... But this problem is not that tough.

If it appears then you cannot blame GMAC ;)

Terminating Decimal

A decimal representation written with a repeating final 0 is said to terminate before these zeros.
13/4 = 3.2500000...
Instead of 3.2500000... we write 3.25

The general formula of a Terminating Decimal is represented as K/[(2)^m * (5)^n]

Non-Terminating Decimal

This can be repeating or non repeating but Never stops. Means we will never get consecutive 0's at any certain point of time in the decimal part (Unlike Terminating One)

As Rahul has shown above only 128 is contains 2 (as 128 = 2^7) it will present a Terminating Decimal.

Thus E is the answer.

PS: https://en.wikipedia.org/wiki/Repeating_decimal
If the problem is Easy Respect it, if the problem is tough Attack it
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