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09 Oct 2018, 14:36
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (01:46) correct
67%
(01:03)
wrong
based on 272
sessions
History
Date
Time
Result
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Is x/11 an integer?
(1) 5x/11 is an integer
(2) 7x/11 is an integer
(1) 5x/11 is an integer
(2) 7x/11 is an integer
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
09 Oct 2018, 16:01
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fskilnik
5x/11 = 1
x = 11/5
11/5 * 1/11 = 1/5 not integer.
If x = 11 then yes
Statement 1 is insufficient.
2) follow the same logic.
We get two different answers insufficient
Combine (1) and (2)
x has to be a multiple of 11
Sufficient.
C
Posted from my mobile device
09 Oct 2018, 18:57
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Is x/11 an integer?
(1) 5x/11 is an integer
Let \(\frac{5x}{11}=y.......5x=11y.......x=\frac{11y}{5}\)
If y is a multiple of 5...yes
Otherwise No
Insufficient
(2) 7x/11 is an integer
Let \(\frac{7x}{11}=z......7x=11z.......x=\frac{11z}{7}\)
If z is a multiple of 7..yes
Otherwise No
Insufficient
Combined
\(\frac{11y}{5}\)=\(\frac{11z}{7}\).....7y=5z
So y has to be a multiple of 5 and z a multiple of 7..
Thus x is an integer
Sufficient
C
(1) 5x/11 is an integer
Let \(\frac{5x}{11}=y.......5x=11y.......x=\frac{11y}{5}\)
If y is a multiple of 5...yes
Otherwise No
Insufficient
(2) 7x/11 is an integer
Let \(\frac{7x}{11}=z......7x=11z.......x=\frac{11z}{7}\)
If z is a multiple of 7..yes
Otherwise No
Insufficient
Combined
\(\frac{11y}{5}\)=\(\frac{11z}{7}\).....7y=5z
So y has to be a multiple of 5 and z a multiple of 7..
Thus x is an integer
Sufficient
C
10 Oct 2018, 06:14
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fskilnik
\(\frac{x}{{11}}\,\,\mathop = \limits^? \,\,\operatorname{int}\)
\(\left( 1 \right)\,\,\,\frac{{5x}}{{11}}\,\, = \operatorname{int} \,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,x = 0\,\,\,\,\, \Rightarrow \,\,\,\,\frac{x}{{11}} = \,\,0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,x = \frac{{11}}{5}\,\,\,\,\, \Rightarrow \,\,\,\,\frac{x}{{11}} = \,\,\frac{1}{5}\,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,\frac{{7x}}{{11}}\,\, = \operatorname{int} \,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,x = 0\,\,\,\,\, \Rightarrow \,\,\,\,\frac{x}{{11}} = \,\,0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,x = \frac{{11}}{7}\,\,\,\, \Rightarrow \,\,\,\,\frac{x}{{11}} = \,\,\frac{1}{7}\,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,\,\frac{x}{{11}}\,\,\, = \,\,\,\frac{{15x}}{{11}} - \frac{{14x}}{{11}}\,\,\, = \,\,\,3 \cdot \left( {\frac{{5x}}{{11}}} \right) - 2 \cdot \left( {\frac{{7x}}{{11}}} \right)\,\,\, = \,\,\,3 \cdot \operatorname{int} - 2 \cdot \operatorname{int} \,\,\, = \,\,\,\operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
