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Best strategy for the following type of problem?

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Best strategy for the following type of problem?

by acchi369 » Tue Oct 09, 2012 6:44 pm
I always trip up on these types of questions and spend way too much time on them. What is a good strategy on these types of questions where you have to pick the right equation? I've tried picking number but I often find that it is very time consuming and you can actually create a pretty messy situation by picking the wrong numbers. Is there a good way to tackle these?

QUESTION:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

Answer:D

Thanks!
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by GMATGuruNY » Tue Oct 09, 2012 7:58 pm
acchi369 wrote:I always trip up on these types of questions and spend way too much time on them. What is a good strategy on these types of questions where you have to pick the right equation? I've tried picking number but I often find that it is very time consuming and you can actually create a pretty messy situation by picking the wrong numbers. Is there a good way to tackle these?

QUESTION:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

Answer:D

Thanks!
Plug in values that make the math EASY.

Let the volume one week ago = 100 liters.

When r% of the CURRENT volume is removed, the RESULTING volume must be 90% of the volume one week ago:
RESULTING volume = .9(100) = 90 liters.

The CURRENT volume should be a value that's easy to decrease to 90.
Since the current volume is x% greater than the volume one week ago -- 100 liters -- let x=80%:
CURRENT volume = 100 + .8(100) = 180 liters.

To decrease from 180 liters to 90 liters, the CURRENT volume must decrease by r=50%. This is our target.

Now we plug x=80 into the answers to see which yields our target of 50.

Only D works:
100(x+10/x+100) = 100 * (80+10)/(80+100) = 100 * 90/180 = 50.

The correct answer is D.
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by acchi369 » Mon Oct 15, 2012 8:03 am
Thanks for the reply. That is helpful. When picking numbers, I find that picking easy numbers that work with all of the equations is quite difficult. Especially when I solve have the problem and realize that the numbers that I thought were easy when I picked them at the beginning are not actually that easy. Do you have any tips for getting started with the right numbers that will be easy throughout the problem?
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by Aishwarya1204 » Mon Oct 15, 2012 10:01 am
acchi369 wrote: QUESTION:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

Answer:D

Thanks!
Hey,
You could try doing it with algebra as well -

Assume 100 is the volume one week ago.

So the volume now will be 100 + x

then 100 + x - r/100(100+x) = 90

100 + x - r - rx/100 = 90

Then take all the r terms on side :

r + rx/100 = 10 + x

100r + rx = 1000 + 100x

r ( 100 + x ) = 1000 + 100x

r = 1000 + 100x/ 100 + x

taking 100 common will give you the answer : 100 ( 10 + x)/100 + x
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by anuprajan5 » Mon Oct 15, 2012 10:10 pm
acchi369 wrote:I always trip up on these types of questions and spend way too much time on them. What is a good strategy on these types of questions where you have to pick the right equation? I've tried picking number but I often find that it is very time consuming and you can actually create a pretty messy situation by picking the wrong numbers. Is there a good way to tackle these?

QUESTION:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

Answer:D

Thanks!

If you can vizualize this, this type of problem becomes a bit easy. Here's how I would do it.

This problem talks of a series of changes on v volume of water. First it is x percent greater than v and then the changed volume is reduced by r%

The formula would become

v*(1+x/100)*(1-r/100) = 90%*v

Once you simplify the above equation to find r, you should have option D
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by mcdesty » Thu Aug 21, 2014 9:59 am
See Image below for how I would tackle this one.
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by GMATinsight » Thu Aug 21, 2014 11:13 am
acchi369 wrote:I always trip up on these types of questions and spend way too much time on them. What is a good strategy on these types of questions where you have to pick the right equation? I've tried picking number but I often find that it is very time consuming and you can actually create a pretty messy situation by picking the wrong numbers. Is there a good way to tackle these?

QUESTION:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

Answer:D

Thanks!
Best strategy for such question is
1) Assuming some value of x,
2) Calculating the result in form of some value as instructed by question
3) Checking Answer Options by substituting value of x assumed and finding option which validates the result obtained.
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by Scott@TargetTestPrep » Wed Dec 13, 2017 12:56 pm
acchi369 wrote: The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

A) x + 10
B) 10x + 1
C) 100(x+10)
D) 100(x+10/x+100)
E) 100(10x + 1/10x + 10)

We can let the volume of water in the tank one week ago = n.

Since the volume is now x percent greater than it was one week ago, the volume today is:

(1 + x/100)n

Next we are given that r percent of the current volume is removed. Thus we now have:

(1 - r/100)(1 + x/100)n

Since the resulting volume will be 90 percent of what it was a week ago, we can create the following equation and isolate r:

(1 - r/100)(1 + x/100)n = 0.9n

(1 - r/100)(1 + x/100) = 0.9

[(100 - r)/100][(100 + x)/100] = 0.9

[10,000 - rx + 100x - 100r]/10,000 = 0.9

10,000 - rx + 100x - 100r = 9,000

1,000 + 100x = rx + 100r

1,000 + 100x = r(x + 100)

100(10 + x) = r(x + 100)

100(10 + x)/(x + 100) = r

Answer: D

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