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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

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?

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.

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.

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.

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:

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 .

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.

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