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Source: Beat The GMAT — Problem Solving |
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givemeanid
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givemeanid
- Master | Next Rank: 500 Posts
- Posts: 277
- Location: New York, NY
- Thanked: 6 times
I can see the file now.
Rate of the reaction is proportional to A^2/B.
R = k*A^2/B
Now, B is increased by 100%
Lets call it B1. So, B1 = 2B
New rate R1 = k*A1^2/2B
To keep R = R1
k*A^2/B = k*A1^2/2B
A1^2 = 2A^2
A1 = 1.414 * A
That is a little over 40% increase.
Answer is (D).
Rate of the reaction is proportional to A^2/B.
R = k*A^2/B
Now, B is increased by 100%
Lets call it B1. So, B1 = 2B
New rate R1 = k*A1^2/2B
To keep R = R1
k*A^2/B = k*A1^2/2B
A1^2 = 2A^2
A1 = 1.414 * A
That is a little over 40% increase.
Answer is (D).
So It Goes
Amazing technique...But can u pls tell me hw did u get A^2/B pls because I dont know what exactly proportional means in this prob..
Maxx
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givemeanid
- Master | Next Rank: 500 Posts
- Posts: 277
- Location: New York, NY
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moneyman wrote:Amazing technique...But can u pls tell me hw did u get A^2/B pls because I dont know what exactly proportional means in this prob..
The question says chemical reaction is directly proportional to 'square of concentration of A' and 'inversely proportional to the concentration of B'.
When x is proportional to y, it means x = k*y where k is a constant.
When x is inversely proportional to y, it means x = k/y where k is a constant.
So, rate = k * A^2/B
So It Goes
