Any decimal that has only a finite number of nonzero digits

[BTGmoderatorDC]

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4

Source: Official Guide

OA B

[Jay@ManhattanReview]

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4

Source: Official Guide

OA B

A ratio, here, r/s is a terminating decimal if s has only two prime factors: 2 and 5.

Note that there is no role of r, so Statement 1 is insufficient.

From Statement 2, we know that s = 4 has a prime factor 2, it's sufficient to collude that r/s is a terminating decimal.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review GRE Prep

Locations: GMAT Classes San Diego | GRE Prep Course Boston | GRE Prep Chicago | TOEFL Prep Classes NYC | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

[Brent@GMATPrepNow]

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4

Source: Official Guide

OA B

Target question: Is r/s a terminating decimal?

Statement 1: 90 < r < 100
There are several pairs of values that meet this condition. Here are two:
Case a: r = 91 and s = 2, in which case r/s = 91/2 = 45.5 = a terminating decimal
Case b: r = 91 and s = 3, in which case r/s = 91/3 = 30.33333.... = a non-terminating decimal
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: s = 4
Notice that 1/4 = 0.25, 2/4 = 0.5 and 3/4 = 0.75
So, if the denominator is 4, the resulting decimal will definitely be a terminating decimal.
In other words, if s = 4 then r/s must be a terminating decimal.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Aside: There's a nice rule that says something like,
If the prime factorization of the denominator contains only 2's and/or 5's, then the decimal version of the fraction will be a terminating decimal.
Since the denominator, 4 = (2)(2), the rule tells us that r/s must be a terminating decimal.

Answer: B

Cheers,
Brent

[ceilidh.erickson]

Here is further explanation on the logic of terminating decimals:
https://www.beatthegmat.com/ds-decimals ... tml#583298

[fskilnik@GMATH]

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4

Source: Official Guide

\[r,s\,\, \geqslant 1\,\,\,{\text{ints}}\]
\[\frac{r}{s}\,\,\,\mathop = \limits^? \,\,\,\,{\text{terminating}}\]
\[\left( 1 \right)\,\,90 < r < 100\,\,\,\,\left\{ \begin{gathered}
\,\left( {r,s} \right) = \left( {95,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\left( {\frac{{95}}{5} = \operatorname{int} } \right) \hfill \\
\,\left( {r,s} \right) = \left( {91,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\left( {\frac{{91}}{3} = 30\frac{1}{3} = 30.333 \ldots } \right) \hfill \\
\end{gathered} \right.\]

\[\left( 2 \right)\,\,s = 4\]
\[\left( * \right)\,\,\,\,{\text{r/s}}\,\,\,{\text{division}}\,\,{\text{algorithm}}:\,\,\,\left\{ \begin{gathered}
\,r = qs + R\,\,\,\mathop = \limits^{s\,\, = \,\,4} \,\,\,4q + R \hfill \\
\,q\,\,\operatorname{int} \,\,\,,\,\,\,\,0\,\,\, \leqslant \,\,\,R\,\,\operatorname{int} \,\,\, \leqslant \,\,3\,\,\,\,\left( { = s - 1} \right) \hfill \\
\end{gathered} \right.\]
\[\frac{r}{s}\,\,\,\,\mathop = \limits^{\,\left( * \right)} \,\,\,\,\frac{{4q + R}}{4} = q + \frac{R}{4}\,\, = \,\,\operatorname{int} \,\, + \,\,\frac{R}{4}\,\,\,\,\,\,\,\]
\[\frac{R}{4} = \,\,\,\left\{ {\begin{array}{*{20}{c}}
{\,\,\frac{0}{4}} \\
{\,\,\frac{1}{4}} \\
{\,\,\frac{2}{4}} \\
{\,\,\frac{3}{4}}
\end{array}} \right.\begin{array}{*{20}{c}}
{\,\,{\text{if}}\,\,\,R = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} } \right)} \\
{\,\,{\text{if}}\,\,\,R = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.25} \right)} \\
{\,\,{\text{if}}\,\,\,R = 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.5} \right)} \\
{\,\,{\text{if}}\,\,\,R = 3\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.75} \right)}
\end{array}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

[swerve]

For a fraction to be terminating two conditions must satisfy:
1) numerator is an INTEGER.
2) denominator should be of form 2^x 5^y (x,y => integers which also includes 0).

now in this question
the denominator is 2^2 5^0
hence it satisfies.

Regards!

[fskilnik@GMATH]

For a fraction to be terminating two conditions must satisfy:
1) numerator is an INTEGER.
2) denominator should be of form 2^x 5^y (x,y => integers which also includes 0).

Hi swerve!

What about 3/30 ? This number/fraction is terminating (=0.1) but it does not satisfy the conditions you have presented...

Do not forget that for the conditions above, first you must "simplify your fraction"... in other words:

Be sure numerator and denominator are relative prime, then you apply your "laws"!

Regards,
Fabio.

P.S.: curiously, integers x and y MAY be negative... but if so, we are not talking about "denominator", really... therefore I would prefer to add a nonnegative condition on x and y, in your second rule.