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Bunuel
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Is it possible to employ the method in which terminating decimal can be determined by expressing the numbers in the form of 2^n.5^m.? If it is correct, then only option E fits the bill. 1/5^3 + 1/2^3.
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Can the trail of 2s and 5s be considered as a factor that will lead to zero? Option E, we have 125=5^3 and 8=2^3.
Bunuel,
Please advice me if my conclusion is wrong.
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Bunuel
Which of the following can be expressed as a terminating decimal?

A. 1/25 + 1/35
B. 1/225 + 1/1,250
C. 1/11,000 + 1/13,000
D. 1/75 + 1/35
E. 1/125 + 1/8

When solving this problem, we should remember that there is a special property about fractions that allows their decimal equivalents to terminate. This property states:

In its most-reduced form, any fraction with a denominator whose prime factorization contains only 2s, only 5s, or both 2s and 5s produces decimals that terminate. A denominator with any other prime factors produces decimals that do not terminate.

Furthermore, the sum of two terminating decimals is also a terminating decimal. Scanning out answer choices, we see that the fractions in answer choice E are 1/125 (which is 1/5^3) and 1/8 (which is 1/2^3). Since each fraction is a terminating decimal, their sum will be also a terminating decimal.

Answer: E
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Bunuel
Which of the following can be expressed as a terminating decimal?

A. 1/25 + 1/35
B. 1/225 + 1/1,250
C. 1/11,000 + 1/13,000
D. 1/75 + 1/35
E. 1/125 + 1/8

This can be answered by just checking prime factors. Your answer will always terminate if the denominator will have either 2, 5 as prime factors AND NOTHING ELSE .

Option A is eliminated because of 7 as prime factor.
Option B is eliminated because of 3 as prime factor
Option C is eliminated because of 11 and 13 as prime factors.
Option D eliminated because of factor 7 and 3 as prime factors.

Saves 40 seconds right away :-P
Option E is correct. (Only 5 and 2 as prime factor)
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