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buan15
RaghavKhanna
15 is a factor of p as well as q
=> p = 15a
q = 15b

St. 1:

p = 3q

case 1: p = 15, => q = 45
=> hcf(p,q) = 15
=> true

case 2:
p = 30, => q = 90
=> hcf(p,q) = 30
=> false

=> St. 1 is insufficient.

St. 2:
q - 2p = 15
=> p = 2q + 15

case 1: q = 15
=> p = 45

=> hcf(p,q) = 15
=> true

case 2:
q = 120 = 2^3 * 3 * 5
=> p = 255 = 5*3*17

=> hcf(p,q) = 15
=> true

=> St. 2 is sufficient.

=> Answer: B

Is there any alternate way to conclude that option B is sufficient?
Putting values take more time.

chetan2u kindly guide. Thanks.

Absolutely!
I showed cases with values for a better understanding. You can simply use the following method:

Let p = 15m
q = 15n

St. 1:
p = 3q
=> 15m = 45n

The HCF will be 15, only if n = 1.

As, n increases, HCF will increase.

=> St. 1 is insufficient.

St. 2:
q - 2p = 15

=> 15n = 15 + 30m

=> n = 1 + 2m

The HCF will always be 15.

=> St. 2 is sufficient

Posted from my mobile device
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buan15
If 15 is a factor of p as well as q, is 15 the highest common factor of p & q?

1. q=3p.
2. q-2p=15.


Let p=15x and q=15y.
So, surely 15 is the minimum possible answer if x and y have no common factors.

1. q=3p.
So, HCF of p and q, that is HCF (p,3p) will be p.
But we do not know the value of p.

2. q-2p=15.
Substitute values of p and q..
\(15y-2(15x)=15......y-2x=1.....y=2x+1\)
Thus, y is 1 more than a multiple of x. So surely x and y have no factors in common, as y and a multiple of x are consecutive integers.
\(HCF(x,y)=1...HCF(15x,15y)=HCF(p,q)=15\)

B
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chetan2u
buan15
If 15 is a factor of p as well as q, is 15 the highest common factor of p & q?

1. q=3p.
2. q-2p=15.


Let p=15x and q=15y.
So, surely 15 is the minimum possible answer if x and y have no common factors.

1. q=3p.
So, HCF of p and q, that is HCF (p,3p) will be p.
But we do not know the value of p.

2. q-2p=15.
Substitute values of p and q..
\(15y-2(15x)=15......y-2x=1.....y=2x+1\)
Thus, y is 1 more than a multiple of x. So surely x and y have no factors in common, as y and a multiple of x are consecutive integers.
\(HCF(x,y)=1...HCF(15x,15y)=HCF(p,q)=15\)

B

Thanks for the reply.

I spent around 4 minutes in this question & got it wrong.

In the last few mocks I made 1-2 mistake on very difficult question & had to rush in the end , scoring 49.
Will it be better idea to leave these question & move forward even though I see these kind of question in first 10-15 question.

Please advise me to improve. Last time in the real GMAT also I got Q 49. I badly need 50 & have my exam in couple of weeks.
Thanks.
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let p = 15a

1. gcd(p, q) = gcd(p, 3p) = gcd(15a, 3 * 15a) = 15a * gcd(1, 3) = 15a


2. gcd(p, q) = gcd(p, 15 + 2p) = gcd(15a, 15 + 2 * 15a) = gcd(15a, 15 (1 + 2a)) = 15 * gcd(a, 1 + 2a)

with Euclidean algorithm, gcd(x, y) = gcd(y, x % y)

15 * gcd(1 + 2a, a) = 15 * gcd(a, 1) = 15 * 1 = 15
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