If 5a=9b=15c, what is the value of a+b+c? 1. 3c-a=5c-3b 2. : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 02:30

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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 5a=9b=15c, what is the value of a+b+c? 1. 3c-a=5c-3b 2.

Author Message
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25

Kudos [?]: 378 [0], given: 0

If 5a=9b=15c, what is the value of a+b+c? 1. 3c-a=5c-3b 2. [#permalink]

Show Tags

08 Oct 2006, 07:09
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:44) correct 0% (00:00) wrong based on 9 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If 5a=9b=15c, what is the value of a+b+c?

1. 3c-a=5c-3b
2. 6cb=10a
Intern
Joined: 21 Jul 2006
Posts: 21
Location: Istanbul, Turkey
Followers: 0

Kudos [?]: 1 [0], given: 0

Show Tags

08 Oct 2006, 07:27
Question gives us the following:
a=3c
3b=5c
5a=9b

Let's see what we can do with 1:

3c-a=5c-3b => 3b-a=2c , if we substitute a with 3c, we find 5c=3b, nothing new...

Now 2:

Substitute a with 3c, the equation becomes;

6cb=30c, divide each side by 6c, we find b=5

=> c=3 and a=9

Hence, a+b+c=9+5+3=17

SVP
Joined: 05 Jul 2006
Posts: 1743
Followers: 6

Kudos [?]: 317 [0], given: 49

Show Tags

08 Oct 2006, 07:46

i agree b this is the only way to solve it.
Senior Manager
Joined: 28 Aug 2006
Posts: 306
Followers: 13

Kudos [?]: 150 [0], given: 0

Show Tags

08 Oct 2006, 09:05
It is B............

Look at this

Given that 5a=9b=15c
Let 5a=9b=15c= k

So a:b:c = 1/5:1/9:1/15
i.e. a:b:c = 9: 5: 3
So a=9x b=5x and c=3x

Statement 1:
if u substitute these values in statment 1 we get nothing..........., I mean the information given in 1 is obvious from the above values...........

Statement 2:

6cb=10a
ie 90x^2 = 90x
ie 90x(x-1) = 0
Clearly x cannot be zero, so x=1
ie a=9 b=5 and c=3
So 2 is sufficient

Hence B
_________________
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25

Kudos [?]: 378 [0], given: 0

Show Tags

08 Oct 2006, 19:29
Well guys, according to the OA/OE, (B) isn't the answer. Any other guesses?
Director
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
Followers: 5

Kudos [?]: 70 [0], given: 1

Show Tags

08 Oct 2006, 19:36
B

Statement 1: Seems to be insufficient

Statement 2:
a = 6bc/10

But, a = 9b/5

c = 3. Other values can be solved for.
Director
Joined: 06 Sep 2006
Posts: 743
Followers: 1

Kudos [?]: 36 [0], given: 0

Show Tags

08 Oct 2006, 20:12
cicerone wrote:
It is B............

Look at this

Given that 5a=9b=15c
Let 5a=9b=15c= k

So a:b:c = 1/5:1/9:1/15
i.e. a:b:c = 9: 5: 3
So a=9x b=5x and c=3x

Statement 1:
if u substitute these values in statment 1 we get nothing..........., I mean the information given in 1 is obvious from the above values...........

Statement 2:

6cb=10a
ie 90x^2 = 90x
ie 90x(x-1) = 0
Clearly x cannot be zero, so x=1
ie a=9 b=5 and c=3
So 2 is sufficient

Hence B

ie 90x(x-1) = 0
Clearly x cannot be zero, so x=1

Why x can not be 0? x has two values doesnt it make B Unsufficient?
Senior Manager
Joined: 11 Jul 2006
Posts: 382
Location: TX
Followers: 1

Kudos [?]: 14 [0], given: 0

Show Tags

08 Oct 2006, 20:46
GMATT73 wrote:
Well guys, according to the OA/OE, (B) isn't the answer. Any other guesses?

Was doublting B , I think its E

b can be 0 or 5
c can be 0 or 3

Example: (Taken from Hong Hu's Math principles in this forum)

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.
Senior Manager
Joined: 05 Oct 2006
Posts: 267
Followers: 1

Kudos [?]: 17 [0], given: 0

Show Tags

09 Oct 2006, 08:04
1. repeats the samething i.e,a:b:c::5:9:15
2.solving we get c=3
hence a=9
b=5

hence,choice b
Senior Manager
Joined: 05 Oct 2006
Posts: 267
Followers: 1

Kudos [?]: 17 [0], given: 0

Show Tags

09 Oct 2006, 08:23
want to make ammendment

while solving for c,i got 10c^2=30c
so 10c(c-3)=0..hence there are two values of namely 0 and 3 for which the above equation holds.
hence there are two value for each of a nad b
and hence there are two values for a+b+c i.e., 17 and zero.
so choice b is not true.
hence choice e is the right answer.
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 325 [0], given: 0

Show Tags

09 Oct 2006, 09:58
GMATT73 wrote:
If 5a=9b=15c, what is the value of a+b+c?

1. 3c-a=5c-3b
2. 6cb=10a

This is a very good question. Tests quite a few skills and traps.

From 1: a-3b+2c=0
You can stop here if you have a feel about the following: Since a,b,c are all related in a fix ratio, this equation tells us that they are all equal to zero.
If you can't see it, do some further analysis:
5a=9b, thus 3b=(5/3)a
5a=15c thus 2c=(2/3)a
In other words a-(5/3)a+(2/3)a=0
a=0
therefore b=0,c=0 and a+b+c=0. Sufficient.

You will waste a little time if you try to make sure the fractions are correct, which is not crucial to get the correct answer. So try to stop as soon as you get the feeling that they are all the same thing and thus are all going to be zero.

(b)6cb=10a, or 3cb=5a.
Substitute back into 5a=9b=15c
3cb=9b
3cb=15c
c=3, b=5, or c=0,b=0
Therefore we cannot uniquely determine what a+b+c is.

Looks like A to me then.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Intern
Joined: 21 Jul 2006
Posts: 21
Location: Istanbul, Turkey
Followers: 0

Kudos [?]: 1 [0], given: 0

Show Tags

09 Oct 2006, 13:13
ivymba wrote:
GMATT73 wrote:
Well guys, according to the OA/OE, (B) isn't the answer. Any other guesses?

Was doublting B , I think its E

b can be 0 or 5
c can be 0 or 3

Example: (Taken from Hong Hu's Math principles in this forum)

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.

Yeah, I think I made a mistake by dividing each side of 6cb=30c by 6c. This could have been:

6cb-30c=0 => 6c(b-5)=0 => c=0 or b=5

Senior Manager
Joined: 11 Jul 2006
Posts: 382
Location: TX
Followers: 1

Kudos [?]: 14 [0], given: 0

Show Tags

09 Oct 2006, 15:20
HongHu wrote:
GMATT73 wrote:
If 5a=9b=15c, what is the value of a+b+c?

1. 3c-a=5c-3b
2. 6cb=10a

This is a very good question. Tests quite a few skills and traps.

From 1: a-3b+2c=0
You can stop here if you have a feel about the following: Since a,b,c are all related in a fix ratio, this equation tells us that they are all equal to zero.
If you can't see it, do some further analysis:
5a=9b, thus 3b=(5/3)a
5a=15c thus 2c=(2/3)a
In other words a-(5/3)a+(2/3)a=0
a=0
therefore b=0,c=0 and a+b+c=0. Sufficient.

You will waste a little time if you try to make sure the fractions are correct, which is not crucial to get the correct answer. So try to stop as soon as you get the feeling that they are all the same thing and thus are all going to be zero.

(b)6cb=10a, or 3cb=5a.
Substitute back into 5a=9b=15c
3cb=9b
3cb=15c
c=3, b=5, or c=0,b=0
Therefore we cannot uniquely determine what a+b+c is.

Looks like A to me then.

HongHu excellent approach but isn't a-(5/3)a+(2/3)a=0 true for any value of a ? a doesn't necessarily have to be 0 to satisfy the equation .
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 325 [0], given: 0

Show Tags

09 Oct 2006, 15:46
Hmmm you are very right. I should have looked at the fractions after all. One does get rusty if one doesn't practise a lot. E it is then.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Senior Manager
Joined: 05 Oct 2006
Posts: 267
Followers: 1

Kudos [?]: 17 [0], given: 0

Show Tags

09 Oct 2006, 21:13
CHOICE B SEEMED a more lucrative offer which it was not.
tx for ur tip on choice a.will keep this in mind.
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25

Kudos [?]: 378 [0], given: 0

Show Tags

12 Oct 2006, 06:19
Good work guys, OA is (E). The last few posts do a good job of proving why either statement alone or together is insuff.
12 Oct 2006, 06:19
Display posts from previous: Sort by