If 1 < x < 2, is x a terminating decimal? (1) 22x is an integer.

It mentions that the denominator of the fraction is reduced to its lowest prime.

If 1<x<2, as per your analysis of statement 1, x can not be 1/2 or 1/11 and likewise for statement 2 x can not be 1/2 or 1/7. On the other hand x has to be between 23/22 and 43/22 in case of statement 1. In case of statement 2 , x has to be between 57/56 and 111/56. Hence both statements 1 and 2 are insufficient alone.

I would request if Bunuel could explain the solution and clarify if I'm wrong.

Hello, I do not understand the explanation at all. Can anyone please explain why x = 1/2 or 1/11 and not 3/2 in either of the options? The reason being the question states 1<x<2.

THEORY:

A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not.

For example, \(\frac{x}{2^n5^m}\), (where \(x\), \(n\), and \(m\) are integers) will always be a terminating decimal.

(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)

BACK TO THE ORIGINAL QUESTION:

If 1 < x < 2, is x a terminating decimal?

(1) 22x is an integer.

Say that integer is m, so x = m/22.

1 < m/22 < 2
1 < m/(2*11) < 2
22 < m < 44

If m is a multiple of 11, then the 11 in m/(2*11) would cancel out, and x would be a terminating decimal. For instance, if m = 33, then x becomes 3/2 = 1.5, which is a terminating decimal. However, if m is not a multiple of 11, then the 11 in m/(2*11) would not cancel out, and x would not be a terminating decimal. For example, if m = 30, then x becomes 30/22 = 15/11, which is not a terminating decimal.

Not sufficient.

(2) 56x is an integer.

Say that integer is n, so x = n/56.
1 < n/56 < 2
1 < n/(2^3*7) < 2
56 < n < 112

If n is a multiple of 7, then the 7 in n/(2^3*7) would cancel out, and x would be a terminating decimal. For example, if n = 63, then x becomes 9/2^3 = 1.125, which is a terminating decimal. However, if n is not a multiple of 7, then the 7 in n/(2^3*7) would not cancel out, and x would not be a terminating decimal. For example, if n = 60, then x becomes 60/(2^3*7), which is not a terminating decimal.

Not sufficient.

(1)+(2) From x = m/22 = n/56, we get m/n = 22/56 = 11/28. Therefore, m is a multiple of 11. Above, we concluded that if m is a multiple of 11, x will be a terminating decimal. This is sufficient. If interested, since m is a multiple of 11 in the range from 22 to 44, not inclusive, it must be 33, making x equal to 3/2.

Answer: C.

Hope it helps.