GRE Prep 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.
Sep 28Live Online Class with Top GRE Expert Mike Butville
08:00 PM EDT
-10:00 PM EDT
Top scores are possible when you enroll in a powerful GRE course, taught live online + 6 months access to TTP GRE OnDemand video courses included! Class starts Mon/Wed Sept. 28, 2026 - Dec. 9, 2026, 8:00pm-10:00pm EST
3k-12j/9 vs k-4j/3
[#permalink]
04 Mar 2018, 13:39
04 Mar 2018, 13:39
1
Expert Reply
00:00
Question Stats:
83% (00:23) correct
16% (00:24) wrong based on 97 sessions
Hide Show timer Statistics
Quantity A |
Quantity B |
\(\frac{3k-12j}{9}\) |
\(\frac{k-4j}{3}\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 1
Question: 2
Page: 493
Question: 2
Page: 493
Re: 3k-12j/9 vs k-4j/3
[#permalink]
04 Mar 2018, 14:05
04 Mar 2018, 14:05
Expert Reply
Explanation
You can find the two quantities to be equal by plugging in values for k and j: If k = 2 and j = 3,
then Quantity A is \(\frac{3(2)-12(3)}{9}\), or \(\frac{-30}{9}\) which can be reduced to \(\frac{-10}{3}\).
Quantity B is \(\frac{2-4(3)}{3}\)or \(\frac{-10}{3}\).
Algebraically, try factoring and canceling a 3 out of the numerator of
Quantity A: \(\frac{3k-12j}{9}=\frac{3(k-4j)}{3 \times 3}=\frac{k-4j}{3}\).
You can find the two quantities to be equal by plugging in values for k and j: If k = 2 and j = 3,
then Quantity A is \(\frac{3(2)-12(3)}{9}\), or \(\frac{-30}{9}\) which can be reduced to \(\frac{-10}{3}\).
Quantity B is \(\frac{2-4(3)}{3}\)or \(\frac{-10}{3}\).
Algebraically, try factoring and canceling a 3 out of the numerator of
Quantity A: \(\frac{3k-12j}{9}=\frac{3(k-4j)}{3 \times 3}=\frac{k-4j}{3}\).
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: 3k-12j/9 vs k-4j/3
[#permalink]
05 Mar 2018, 09:23
05 Mar 2018, 09:23
1
sandy wrote:
Quantity A |
Quantity B |
\(\frac{3k-12j}{9}\) |
\(\frac{k-4j}{3}\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 1
Question: 2
Page: 493
Question: 2
Page: 493
We can solve this question using matching operations
Given:
Quantity A: (3k - 12j)/9
Quantity B: (k - 4j)/3
Eliminate fractions by multiplying both quantities by 9 (the least common multiple of 9 and 3) to get:
Quantity A: 3k - 12j
Quantity B: 3(k - 4j)
Expand Quantity B to get:
Quantity A: 3k - 12j
Quantity B: 3k - 12j
We can see the two quantities are EQUAL
Answer: C
Cheers,
Brent
RELATED VIDEO FROM OUR COURSE
