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X is directly proportional to Y^2

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X is directly proportional to Y^2

by student4gmat » Mon Oct 06, 2014 6:55 am
X is directly proportional to Y^2 and indirectly proportional to Z. If Z is halved, what percentage change in Y is required to keep X same as before?

My approach -

X/Y^2 = K
X/Z = K

Now Z is halved...so X will be doubled (because we know that X and Z are indirectly proportional)
so now X has become 2X....if we see X and Y relationship now -
X = K * Y^2

Now X is Doubled so we will have to double Y too but X is directly proportional to Y^2 and not Y so -
2X = (root 2 * Y)^2

2X = 2Y^2
Now in order to convert 2X into X we need to divide LHS and RHS by 2
Now how to calculate percentage change in Y?

can somehow please help?

Thanks.
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by GMATGuruNY » Mon Oct 06, 2014 7:09 am
The posted problem is very similar to the following problem from GMAT Prep:
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 percent, 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
x is directly proportional to y and inversely proportional to z implies the following:
x = (y/z)k, where k is a constant.

In the problem above, since the rate is directly proportional to A² and inversely proportional to B, we get:
R = (A²/B)k, where k is a constant.

Original values:
Let A=10, B=1, and k=1.
Then:
R = (A²/B)k = (10²/1)(1) = 100.

New values:
Since the rate is unchanged, R=100.
Since the value of B doubles, new B = 2*1 = 2.
Since k is a constant -- a value that never changes -- k=1.
Plugging these values into R = (A²/B)k, we get:
100 = (A²/2)(1)
A² = 200
A = √200 = 10√2 ≈ (10)(1.4) = 14.

Since the value of A increases from 10 to 14, the percent increase in A = 40%.

The correct answer is D.
Last edited by GMATGuruNY on Tue Oct 07, 2014 1:24 am, edited 1 time in total.
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by student4gmat » Mon Oct 06, 2014 8:49 am
GMATGuruNY wrote:
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 percent, 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
x is directly proportional to y and inversely proportional to z implies the following:
x = (y/z)k, where k is a constant.

In the problem above, since the rate is directly proportional to A² and inversely proportional to B, we get:
R = (A²/B)k, where k is a constant.

Original values:
Let A=10, B=1, and k=1.
Then:
R = (A²/B)k = (10²/1)(1) = 100.

New values:
Since the rate is unchanged, R=100.
Since the value of B doubles, new B = 2*1 = 2.
Since k is a constant -- a value that never changes -- k=1.
Plugging these values into R = (A²/B)k, we get:
100 = (A²/2)(1)
A² = 200
A = √200 = 10√2 ≈ (10)(1.4) = 14.

Since the value of A increases from 10 to 14, the percent increase in A = 40%.

The correct answer is D.
Thanks for replying. Shouldn't an increase in B means decrease in R as both are inversely proportional.

Also, can you please solve the question I posted...the answer to that one is 30%. Thanks.
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by GMATGuruNY » Tue Oct 07, 2014 1:42 am
student4gmat wrote: In the problem above, since the rate is directly proportional to A² and inversely proportional to B, we get:
R = (A²/B)k, where k is a constant.

Thanks for replying. Shouldn't an increase in B means decrease in R as both are inversely proportional.
If value of B increases by a factor of x and the value of A² stays the same or increases by a factor less than x, then the value of R will decrease.
Case 1:
If the value of B doubles and the value of A² stays the same, then the value of R will be multiplied by a factor of 1/2.
As a result, the value of R will DECREASE by 1/2.
Case 2: If the value of B triples and the value of A² doubles, then the value of R will be be multiplied by a factor of 2/3.
As a result, the value of R will DECREASE by 1/3.

If value of B increases by a factor of x and the value of A² increases by a factor greater than or equal to x, then the value of R will NOT decrease.
Case 3:
If the value of B doubles and the value of A² doubles, then the value of R will be multiplied by a factor of 2/2 = 1.
As a result, the value of R will STAY THE SAME.
Case 4:
If the value of B doubles and the value of A² triples, then the value of R will be multiplied by a factor of 3/2.
As a result, the value of R will INCREASE by 1/2.
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by student4gmat » Tue Oct 07, 2014 1:50 am
GMATGuruNY wrote:
student4gmat wrote: In the problem above, since the rate is directly proportional to A² and inversely proportional to B, we get:
R = (A²/B)k, where k is a constant.

Thanks for replying. Shouldn't an increase in B means decrease in R as both are inversely proportional.
If value of B increases by a factor of x and the value of A² stays the same or increases by a factor less than x, then the value of R will decrease.
Case 1:
If the value of B doubles and the value of A² stays the same, then the value of R will be multiplied by a factor of 1/2.
As a result, the value of R will DECREASE by 1/2.
Case 2: If the value of B triples and the value of A² doubles, then the value of R will be be multiplied by a factor of 2/3.
As a result, the value of R will DECREASE by 1/3.

If value of B increases by a factor of x and the value of A² increases by a factor greater than or equal to x, then the value of R will NOT decrease.
Case 3:
If the value of B doubles and the value of A² doubles, then the value of R will be multiplied by a factor of 2/2 = 1.
As a result, the value of R will STAY THE SAME.
Case 4:
If the value of B doubles and the value of A² triples, then the value of R will be multiplied by a factor of 3/2.
As a result, the value of R will INCREASE by 1/2.
Yes, I understood this logic but still cannot solve the question I posted. Can you please solve the original problem I posted which is as follows -

X is directly proportional to Y^2 and indirectly proportional to Z. If Z is halved, what percentage change in Y is required to keep X same as before?

The answer to this is 30%
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by GMATGuruNY » Tue Oct 07, 2014 1:53 am
student4gmat wrote: Also, can you please solve the question I posted...Thanks.
X is directly proportional to Y^2 and indirectly proportional to Z. If Z is halved, what percentage change in Y is required to keep X same as before?
x is directly proportional to y² and inversely proportional to z implies the following:
x = (y²/z)k, where k is a constant.

Original values:
Let y=10, z=10, and k=1.
Then:
x = (y²/z)k = (10²/10)(1) = 10.

New values:
Since the value of x is unchanged, x=10.
Since the value of z decreases by 1/2, new z = (1/2)(10) = 5.
Since k is a constant -- a value that never changes -- k=1.
Plugging these values into x = (y²/z)k, we get:
10 = (y²/5)(1)
y² = 50
y = √50 ≈ 7.

Since the value of y decreases from 10 to 7, the percent decrease in y = [spoiler]30%[/spoiler].
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