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01 Mar 2020, 17:16
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If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?
(1) a > c
(2) b > d
For a fraction to have a terminating decimal, the numerator must be an integer and the denominator must be only powers of 2s and/or 5s. We have the fraction 2^a3^b/2^c3^d5^e. The numerator is okay, because we know it will be an integer. The denominator must be only powers of 2s and/or 5s, so we need to see if we can cancel the 3^d for the fraction to be terminating. The only way to get rid of the 3^d in the denominator would be for the power of 3^b in the numerator to be larger than 3^d in the denominator. When an integer raised to a power in the numerator is larger than the same integer raised to a power in the denominator, we can subtract/cancel the denominator. For instance, 2^4/2^3 is simply 2^1. Thus, we need to see if 3^b is greater than 3^d, so we need to ask: b>d?
So, back to the question.
1) Doesn't tell us anything about b>d. There is still a 3 in the denominator, so we won't be able to prove whether this is a terminating decimal or not. INSUFFICIENT
2) This is exactly what we are looking for. We know that b>d, so we know that the 3^d in the denominator will go away, thus leaving only powers of 2 and 5 in the denominator, making this fraction a terminating decimal. SUFFICIENT.
Answer (B).
(1) a > c
(2) b > d
For a fraction to have a terminating decimal, the numerator must be an integer and the denominator must be only powers of 2s and/or 5s. We have the fraction 2^a3^b/2^c3^d5^e. The numerator is okay, because we know it will be an integer. The denominator must be only powers of 2s and/or 5s, so we need to see if we can cancel the 3^d for the fraction to be terminating. The only way to get rid of the 3^d in the denominator would be for the power of 3^b in the numerator to be larger than 3^d in the denominator. When an integer raised to a power in the numerator is larger than the same integer raised to a power in the denominator, we can subtract/cancel the denominator. For instance, 2^4/2^3 is simply 2^1. Thus, we need to see if 3^b is greater than 3^d, so we need to ask: b>d?
So, back to the question.
1) Doesn't tell us anything about b>d. There is still a 3 in the denominator, so we won't be able to prove whether this is a terminating decimal or not. INSUFFICIENT
2) This is exactly what we are looking for. We know that b>d, so we know that the 3^d in the denominator will go away, thus leaving only powers of 2 and 5 in the denominator, making this fraction a terminating decimal. SUFFICIENT.
Answer (B).
13 Apr 2020, 04:55
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13 Apr 2020, 05:46
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ShivamoggaGaganhs
Hi
As far as 2s and 5s are concerned, it doesn’t matter what is the powers and how much power is in denominator as they will always lead to terminating decimal.
Now if c and e are NEGATIVE, the term would move up and become positive. For example. 1/(2^-3) will become 2^3.
