GMAT Prep 2 - Quant 1
This topic has expert replies
Please help me with solution.
- Attachments
-
- Quant 3.doc
- (73 KiB) Downloaded 135 times
-
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
Stmt I
This tells us that m and p are even since they both share a common factor 2
even/even we always get a even remiander if there is a remainder. We know the remiander is not 0 and it cant be 1 so it has to be greater than 1
SUFF
StmtII
x=5 y=6 REMAINDER > 1 NO
x=10 y=15 REMAINDER > 1 YES
INSUFF
A
This tells us that m and p are even since they both share a common factor 2
even/even we always get a even remiander if there is a remainder. We know the remiander is not 0 and it cant be 1 so it has to be greater than 1
SUFF
StmtII
x=5 y=6 REMAINDER > 1 NO
x=10 y=15 REMAINDER > 1 YES
INSUFF
A
Given :
m, p integers &
2 < m < p &
p/m = NQ + r ; N - integer, Q - quotient, r - remainder
Is r > 1 ?
Stmt 1)
GCD (m,p) = 2
Plugging numbers -
m p
GCD( 4, 6) = 2
GCD( 8, 10) = 2
So, m, p have to be consecutive even integers for GCD to be 2 & m not a factor of p. Hence, r = 2.
Stmt 1) sufficient.
Stmt 2)
LCM(m, p ) = 30
Plugging numbers -
LCM( 5, 6) = 30
LCM( 3, 10) = 30
So, r = 1
I guess this is what even Sumit arrived at. I see that the OA is A). So we need to find which combination is not satisfying [ LCM(m,p) as 30 and r = 1] .
Experts, opinions please.
m, p integers &
2 < m < p &
p/m = NQ + r ; N - integer, Q - quotient, r - remainder
Is r > 1 ?
Stmt 1)
GCD (m,p) = 2
Plugging numbers -
m p
GCD( 4, 6) = 2
GCD( 8, 10) = 2
So, m, p have to be consecutive even integers for GCD to be 2 & m not a factor of p. Hence, r = 2.
Stmt 1) sufficient.
Stmt 2)
LCM(m, p ) = 30
Plugging numbers -
LCM( 5, 6) = 30
LCM( 3, 10) = 30
So, r = 1
I guess this is what even Sumit arrived at. I see that the OA is A). So we need to find which combination is not satisfying [ LCM(m,p) as 30 and r = 1] .
Experts, opinions please.
-
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
Look at the possible combinations of factors of 30
30 = 2*3*5
2*5 = 10
3*5 = 15
General Tip:
For number properties questions try factorizing whats given in to their prime factors. This way its easier to see the connections.It will more often help than not.
Good luck, Syr.
Regards,
CR
30 = 2*3*5
2*5 = 10
3*5 = 15
General Tip:
For number properties questions try factorizing whats given in to their prime factors. This way its easier to see the connections.It will more often help than not.
Good luck, Syr.
Regards,
CR
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Yes, that's certainly true. From the definition of quotients and remainders, when you divide n by d, we havemaihuna wrote:Good trick to note here is Even divides Even leaves remainder as even: any mathematical concept to proove this, Ian please?
n = qd + r
where r is the remainder, and q is the quotient. You could rewrite the above:
r = n - qd
and if n is even, and d is even, then r must be even (since on the right side, we have even - even = even).
It's often easier to understand these types of abstract equations by plugging in a few numbers to see how they work.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com