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.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
10 Jun 2020, 04:00
Kudos
Bookmarks
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:
36% (03:11) correct
64%
(02:53)
wrong
based on 105
sessions
History
Date
Time
Result
Not Attempted Yet
The expression \(2a^2 +2b^2\) can be written in the form of \(18x + y\), where a, b, x and y are non-negative integers and \(y < 18\). Is \(|y + 3| = 3\)?
(1) The difference between a and b can be expressed as an even multiple of 3.
(2) b when divided by 3 is an integer.
Are You Up For the Challenge: 700 Level Questions
(1) The difference between a and b can be expressed as an even multiple of 3.
(2) b when divided by 3 is an integer.
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,088
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
10 Jun 2020, 05:27
Kudos
Bookmarks
IMO C
2A^2+2B^2 = 18x+y
Statement 1: The difference between a and b can be expressed as an even multiple of 3.
Therefore A-B = 6,12,18......
Since A,B are nonnegative integers.
Let,
A-B = 6 therefore,
7-1 = 6, 8-2 = 6, 9-3 = 6
The pairs of A and B be (7,1)(8,2)(9,3)(10,4)(11,5)..........
The possible pair of A and B = (7,1)(8,2)(9,3)(10,4)(11,5)....
Putting values in 2A^2+2B^2 = 18x+y
(7,1)
2*7^2+2*1^2 = 18x+y
98+2 = 18x+y
100 = 18x+y
Nearest multiple of 18 to 100 is 90
Therefore X = 5 & y = 10 (not possible)
(8,2)
2*8^2+2*2^2 = 18x+y
128+8 = 18x+y
136 = 18x+y
Nearest multiple of 18 to 136 is 126
Therefore X = 7 & y = 10 (not possible)
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0 (not possible)
(Not Sufficient)
Statement 2: B when divided by 3 leaves an integer.
But we need at least A's value to determine Y(not sufficient)
Together:
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0
(12,6)
2*12^2+2*6^2 = 18x+y
288+72 = 18x+y
360 = 18x+y
Nearest multiple of 18 to 360 is 360
Therefore X = 20 & y = 0
(21,15)
2*21^2+2*15^2 = 18x+y
882+450 = 18x+y
1332 = 18x+y
Nearest multiple of 18 to 1332 is 1332
Therefore X = 74 & y = 0
Therefore for every B which is a multiple of 3 gives out y value as 0.
|y+3| = 3
|0+3| = 3
3 = 3
(Sufficient)
Posted from my mobile device
2A^2+2B^2 = 18x+y
Statement 1: The difference between a and b can be expressed as an even multiple of 3.
Therefore A-B = 6,12,18......
Since A,B are nonnegative integers.
Let,
A-B = 6 therefore,
7-1 = 6, 8-2 = 6, 9-3 = 6
The pairs of A and B be (7,1)(8,2)(9,3)(10,4)(11,5)..........
The possible pair of A and B = (7,1)(8,2)(9,3)(10,4)(11,5)....
Putting values in 2A^2+2B^2 = 18x+y
(7,1)
2*7^2+2*1^2 = 18x+y
98+2 = 18x+y
100 = 18x+y
Nearest multiple of 18 to 100 is 90
Therefore X = 5 & y = 10 (not possible)
(8,2)
2*8^2+2*2^2 = 18x+y
128+8 = 18x+y
136 = 18x+y
Nearest multiple of 18 to 136 is 126
Therefore X = 7 & y = 10 (not possible)
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0 (not possible)
(Not Sufficient)
Statement 2: B when divided by 3 leaves an integer.
But we need at least A's value to determine Y(not sufficient)
Together:
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0
(12,6)
2*12^2+2*6^2 = 18x+y
288+72 = 18x+y
360 = 18x+y
Nearest multiple of 18 to 360 is 360
Therefore X = 20 & y = 0
(21,15)
2*21^2+2*15^2 = 18x+y
882+450 = 18x+y
1332 = 18x+y
Nearest multiple of 18 to 1332 is 1332
Therefore X = 74 & y = 0
Therefore for every B which is a multiple of 3 gives out y value as 0.
|y+3| = 3
|0+3| = 3
3 = 3
(Sufficient)
Posted from my mobile device
MayankSingh
Joined: 08 Jan 2018
Last visit: 20 Dec 2024
Posts: 289
Own Kudos:
Given Kudos: 249
Location: India
Concentration: Operations, General Management
Schools: McDonough '25 (A) Marshall '25 (D) Tepper '25 (WA$) Kenan-Flagler '25 (A) ISB '25 (WL) IIMC '25 (A) Owen '26 (A$$$$)
GMAT 1: 640 Q48 V27
GMAT 2: 730 Q51 V38
GPA: 3.9
WE:Project Management (Manufacturing)
Products:
Schools: McDonough '25 (A) Marshall '25 (D) Tepper '25 (WA$) Kenan-Flagler '25 (A) ISB '25 (WL) IIMC '25 (A) Owen '26 (A$$$$)
GMAT 2: 730 Q51 V38
Posts: 289
Kudos:
277
10 Jun 2020, 06:13
Kudos
Bookmarks
IMO C
(A/Q) 2a^2+2b^2= 18x+y, y<18{a, b, x and y are non-negative integers}
Implies: y = remainder when (2a^2+2b^2) divided by 18 = 2 x [remainder (a^2+b^2) divided by 9]
Question: |y+3|=3 ? => y = 0 ? => [remainder (a^2+b^2) divided by 9]=0 ? (Since y=-6 not possible-cuz negative integer)
(1) The difference between a and b can be expressed as an even multiple of 3.
a-b = 3 x 2K (for some non-negative integer K)
a-b = {0,6,12,18...........}
Since a,b can take any values satisfying above eqn, e.q (3,3) or (7,1)
So, remainder (a^2+b^2) divided by 9 - Cannot be ascertain. (May be zero e.q (3,3) or may not be (7,1))
Not Sufficient
(2) b when divided by 3 is an integer
b= 3P (for some non-negative integer K)
But no idea about "a" -
Not Sufficient
Together:
a-b = 6K
a= 6K + 3P
a= 3(2K+P)
So, a^2 + b^2 = 9 [ (2K+P)^2 + P^2] = 9 x (Some integer)
Clearly, Rem. (a^2+b^2) divided by 9 = 0
Sufficient.
(A/Q) 2a^2+2b^2= 18x+y, y<18{a, b, x and y are non-negative integers}
Implies: y = remainder when (2a^2+2b^2) divided by 18 = 2 x [remainder (a^2+b^2) divided by 9]
Question: |y+3|=3 ? => y = 0 ? => [remainder (a^2+b^2) divided by 9]=0 ? (Since y=-6 not possible-cuz negative integer)
(1) The difference between a and b can be expressed as an even multiple of 3.
a-b = 3 x 2K (for some non-negative integer K)
a-b = {0,6,12,18...........}
Since a,b can take any values satisfying above eqn, e.q (3,3) or (7,1)
So, remainder (a^2+b^2) divided by 9 - Cannot be ascertain. (May be zero e.q (3,3) or may not be (7,1))
Not Sufficient
(2) b when divided by 3 is an integer
b= 3P (for some non-negative integer K)
But no idea about "a" -
Not Sufficient
Together:
a-b = 6K
a= 6K + 3P
a= 3(2K+P)
So, a^2 + b^2 = 9 [ (2K+P)^2 + P^2] = 9 x (Some integer)
Clearly, Rem. (a^2+b^2) divided by 9 = 0
Sufficient.
