The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical B present. If the concentration of checmical B is increased by 100%, which of the following is closest to the percent change in the concentration of the checmical A required to keep the reaction rate unchanged?
100% decrease
50% decrease
40% decrease
40% increase
50% increase
Proportions
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GMATPrep: chemical reaction

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
a. 100% decrease
b. 50% decrease
c. 40% decrease
d. 40% increase
e. 50% increase
Yeap question is missing info. Here is the complate question.
Answer would
r > Ca^2/Cb = sqrt(2)^2 *Ca^2/2Cb
Ca *1.4 > 40% increase

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
a. 100% decrease
b. 50% decrease
c. 40% decrease
d. 40% increase
e. 50% increase
Yeap question is missing info. Here is the complate question.
Answer would
r > Ca^2/Cb = sqrt(2)^2 *Ca^2/2Cb
Ca *1.4 > 40% increase
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present.
if chemical A is x and Chemical B is y and rate is r
chemical A is incereased to 2x and rate remains same r
chemical B is changed to Y
r*(4x^2)*y= r*x^2*Y
new concentration of Chemical B Y =4y
there shoud be 300% increase in concentration of B to keep the rate same
I hope 300% increase is the answer
if chemical A is x and Chemical B is y and rate is r
chemical A is incereased to 2x and rate remains same r
chemical B is changed to Y
r*(4x^2)*y= r*x^2*Y
new concentration of Chemical B Y =4y
there shoud be 300% increase in concentration of B to keep the rate same
I hope 300% increase is the answer

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x2suresh has provided the easiest solution to the problem. Another way to look at this is to conceptually 'guestimate' your way through.
Step 1: Decide if the rate will increase or decrease.
Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
Concentration of A now has to change in order to raise the rate back the original. Since the rate is proportional to the square of A's concentration, A's concentration must increase in order to increase the rate.
So far, you can kill the options that show a decrease in the concentration (choices A, B and C). You are now left with D and E.
Step 2: Evaluate the increment
Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
The rate must now double (multiply by 2) to be back at its original value.
Try choices D and E.
If A's concentration is increased by 50% (A > 1.5A), the square of 1.5A is 2.25A^2 (a multiple of 2.25)
If A's concentration is increased by 40% (A > 1.4A), the square of 1.4A is approximately 2A^2 (a multiple of ~2)
So, a 40% increase in A's concentration is equivalent (approx.) to a multiple of 2 of the rate. Choose D.
BM
P.S. I agree this solution is a bit wordy, but when I attempted this question, this was how I thought it through and I think it took me just slightly more than a minute to answer.
Step 1: Decide if the rate will increase or decrease.
Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
Concentration of A now has to change in order to raise the rate back the original. Since the rate is proportional to the square of A's concentration, A's concentration must increase in order to increase the rate.
So far, you can kill the options that show a decrease in the concentration (choices A, B and C). You are now left with D and E.
Step 2: Evaluate the increment
Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
The rate must now double (multiply by 2) to be back at its original value.
Try choices D and E.
If A's concentration is increased by 50% (A > 1.5A), the square of 1.5A is 2.25A^2 (a multiple of 2.25)
If A's concentration is increased by 40% (A > 1.4A), the square of 1.4A is approximately 2A^2 (a multiple of ~2)
So, a 40% increase in A's concentration is equivalent (approx.) to a multiple of 2 of the rate. Choose D.
BM
P.S. I agree this solution is a bit wordy, but when I attempted this question, this was how I thought it through and I think it took me just slightly more than a minute to answer.