Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.
If a, b, c, d and e are non-negative integers and p = (2^a) (3^b) and q = (2^c) (3^d) (5^e), is
p/q a terminating decimal?
(1) a > c
(2) b > d
pls clarify on undersating the concept terminating decimals
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We have to find if p/q = (2^a)(3^b)/(2^c)(3^d)(5^e) = 2^(a-c)*3^(b-d)/5^e is a terminating decimal.alltimeacheiver wrote:Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.
If a, b, c, d and e are non-negative integers and p = (2^a) (3^b) and q = (2^c) (3^d) (5^e), is
p/q a terminating decimal?
(1) a > c
(2) b > d
(1) a > c is NOT SUFFICIENT to answer the above question, as we don't know b and d.
(2) b > d is SUFFICIENT.
It can be noticed that 2^(a-c) and 5^e will always be a terminating decimal. Any power of 2 and 5 gives a terminating decimal.
The only info we needed was about the power of 3, and since b > d, so b - d will be positive.
So, p/q will be a terminating decimal
So, (2) is SUFFICIENT.
The correct answer is B.
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If you have 3 or a multiple of three in denominator of a rational number, its decimal form will be non-terminating.
If you take II, all 3's in denominator will be cancelled and therefore its decimal form will be terminating. But expression I does not govern it.
If you take II, all 3's in denominator will be cancelled and therefore its decimal form will be terminating. But expression I does not govern it.
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