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28 Mar 2007, 14:24
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I have an inequality :
r/(b+w) > w/(b+r)
cross multiplying :
r(b+r) > w(b+w)
rb + r^2 > wb + w^2
r^2 - w^2 > wb - rb
(r+w)(r-w) > b(w-r) ----- (A)
Now, I have two ways of proceeding from here, and I get two different contradicting answers. I am not sure where I am going wrong
Route 1:
Multiply (A) by -1 on both sides
-1(r+w)(r-w) < -b(w-r)
(r+w)(w-r) < -b(w-r)
Cancelling (w-r) on both sides
r+w < -b ---------- (Result 1)
Route 2:
Multiply (A) by -1 on both sides
-1(r+w)(r-w) < -b(w-r)
-1(r+w)(r-w) < b(r-w)
Cancelling (r-w) on both sides
-1(r+w) < b
-r-w < b
-r < b+w
-r-b < w
-b < w+r ------------(Result 2)
Note that the inequality is reversed between result 1 and result 2. I have gone nuts trying to figure this out.....Any help would be much appreciated.
r/(b+w) > w/(b+r)
cross multiplying :
r(b+r) > w(b+w)
rb + r^2 > wb + w^2
r^2 - w^2 > wb - rb
(r+w)(r-w) > b(w-r) ----- (A)
Now, I have two ways of proceeding from here, and I get two different contradicting answers. I am not sure where I am going wrong
Route 1:
Multiply (A) by -1 on both sides
-1(r+w)(r-w) < -b(w-r)
(r+w)(w-r) < -b(w-r)
Cancelling (w-r) on both sides
r+w < -b ---------- (Result 1)
Route 2:
Multiply (A) by -1 on both sides
-1(r+w)(r-w) < -b(w-r)
-1(r+w)(r-w) < b(r-w)
Cancelling (r-w) on both sides
-1(r+w) < b
-r-w < b
-r < b+w
-r-b < w
-b < w+r ------------(Result 2)
Note that the inequality is reversed between result 1 and result 2. I have gone nuts trying to figure this out.....Any help would be much appreciated.
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28 Mar 2007, 14:31
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prempeter
First of all, welcome to the GMATClub ! :D
U cannot multiply by (b+r) or (b+w) because u do not know their sign... When we mulipty by a negative number, we flip the sign of an inequation
28 Mar 2007, 14:49
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Hi Fig,
That was a fast reply ! Thanks so much for the same.
Can't I cross multiply even when r, b, w are positive numbers(distinct or non distinct)?
I was working out a DS problem from GMATprep and I accidently stumbled on this snippet. The problem involved probabilities of white (w), red (r) and blue(b) balls.
Since this might help others, Here is the problem
There are w white, b blue and r red balls in a bag. Is the probability of getting a white ball greater than getting a red ball?
(1) r/(b+w) > w/(b+r)
(2) Someother condition
Obviously, the number of balls cannot be negative, right ? The answer to this problem is A, and I was able to do it as follows :
r/(b+w) > w/(b+r)
Add 1 on both sides
1 + r/(b+w) > 1 + w/(b+r)
(r+b+w)/(b+w) > (r+b+w)/(b+r)
Taking reciprocal on both sides
(b+w)/(r+b+w) > (b+r)/(r+b+w)
So w > r, (1) is sufficient
Excellent, though be careful with the inequality sign when you take the reciprocal of both sides!
That was a fast reply ! Thanks so much for the same.
Can't I cross multiply even when r, b, w are positive numbers(distinct or non distinct)?
I was working out a DS problem from GMATprep and I accidently stumbled on this snippet. The problem involved probabilities of white (w), red (r) and blue(b) balls.
Since this might help others, Here is the problem
There are w white, b blue and r red balls in a bag. Is the probability of getting a white ball greater than getting a red ball?
(1) r/(b+w) > w/(b+r)
(2) Someother condition
Obviously, the number of balls cannot be negative, right ? The answer to this problem is A, and I was able to do it as follows :
r/(b+w) > w/(b+r)
Add 1 on both sides
1 + r/(b+w) > 1 + w/(b+r)
(r+b+w)/(b+w) > (r+b+w)/(b+r)
Taking reciprocal on both sides
(b+w)/(r+b+w) > (b+r)/(r+b+w)
So w > r, (1) is sufficient
Excellent, though be careful with the inequality sign when you take the reciprocal of both sides!
