Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 0808:00 AM PDT
-08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.
16 Sep 2014, 01:27
Kudos
Bookmarks
If \(x=0.abcd\), where \(a\), \(b\), \(c\), and \(d\) are digits from 0 to 9, inclusive, is \(x \gt \frac{7}{9}\)?
(1) \(a+b \gt 14\)
(2) \(a-c \gt 6\)
(1) \(a+b \gt 14\)
(2) \(a-c \gt 6\)
16 Sep 2014, 01:27
Kudos
Bookmarks
Official Solution:
If \(x=0.abcd\), where \(a\), \(b\), \(c\), and \(d\) are digits from 0 to 9, inclusive, is \(x \gt \frac{7}{9}\)?
First of all, \(\frac{7}{9}\) is a recurring decimal and can be expressed as \(0.\overline{7}\) (a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely, so \(0.\overline{7}=0.777...\)).
(1) \(a+b \gt 14\).
The least value of \(a\) that satisfies this condition is 6 (since \(6+9=15 \gt 14\)), so in this case, \(x=0.69d \lt 0.\overline{7}\). However, \(a=7\) and \(b=9\) is also possible, and in that case \(x=0.79d \gt 0.\overline{7}\). Not sufficient.
(2) \(a-c \gt 6\).
The least value of \(a\) that satisfies this condition is 7 (since \(7-0=7 \gt 6\)), but we don't know the value of \(b\). Not sufficient.
(1)+(2) The least value of \(a\) that satisfies both conditions is 7, and in this case, from (1), the least value of \(b\) is 8 (since \(7+8=15 \gt 14\)). Hence, the least value of \(x\) is \(0.78d \gt 0.\overline{7}\). Sufficient.
Answer: C
If \(x=0.abcd\), where \(a\), \(b\), \(c\), and \(d\) are digits from 0 to 9, inclusive, is \(x \gt \frac{7}{9}\)?
First of all, \(\frac{7}{9}\) is a recurring decimal and can be expressed as \(0.\overline{7}\) (a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely, so \(0.\overline{7}=0.777...\)).
(1) \(a+b \gt 14\).
The least value of \(a\) that satisfies this condition is 6 (since \(6+9=15 \gt 14\)), so in this case, \(x=0.69d \lt 0.\overline{7}\). However, \(a=7\) and \(b=9\) is also possible, and in that case \(x=0.79d \gt 0.\overline{7}\). Not sufficient.
(2) \(a-c \gt 6\).
The least value of \(a\) that satisfies this condition is 7 (since \(7-0=7 \gt 6\)), but we don't know the value of \(b\). Not sufficient.
(1)+(2) The least value of \(a\) that satisfies both conditions is 7, and in this case, from (1), the least value of \(b\) is 8 (since \(7+8=15 \gt 14\)). Hence, the least value of \(x\) is \(0.78d \gt 0.\overline{7}\). Sufficient.
Answer: C
General Discussion
