If 1/2 of the air in a tank is removed with each stroke of a

I get A, too.

Let X = air in the tank.

Plug in 240 = X.

First stroke = 1/2 * 240 = 120
Second stroke = 1/2 * 120 = 60
Third stroke = 1/2 * 60 = 30
Fourth stroke = 1/2 * 30 = 15

So 15 remains, which is the amount of air left.

240 - 15 = 225 (amount of air removed). 225/240 = 15/16

This is a decay-type problem...the tank is halved after each stroke. Therefore:

(0.5)^4 = 1/16 --> 0.5 for halving, 4 for strokes

so 1/16 of air is left in the tank => 15/16 of the air was removed

Oh, I get it..! Thanks..!!!
I think the catch over here is.."the tank is halved after each stroke"....failed to understand the question properly.
Thought that a single stoke halves the air in it...so after 2 strokes guessed there wont be any air in it

A - 15/16..

Left After 1st stroke = 1/2
Left After 2nd stroke = 1/2 * 1/2 = 1/4
Left After 3rd stroke = 1/2 * 1/4 = 1/8
Left After 4th stroke = 1/2 * 1/8 = 1/16

So removed = 1- 1/16 = 15/16

As they say, the devil is in the details!

By the 4th stroke the the air remaining is 1/16 , so what has been removed is 15/16

1st stroke 1/2
2nd stroke 1/4
3rd stroke 1/8
4th stroke 1/16
total: (8+4+2+1)/16=15/16

Better use compound interest formula.

Remaing amount = P ( 1 - R/100)^n

Let P is 100 ( 1 - 50/100)^4
Remaining amount = 100(1/16)

if 1/16 of 100 is remaining means 1-1/16 = 15/16 is lost.

Similar questions to practice:
if-2-3-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-128432.html
if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html
on-january-1-2076-lake-loser-contains-x-liters-of-water-b-82667.html

Hope it helps.

Let say total air = 1

Total air removed

\(= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}\)

\(= \frac{15}{16}\)

Answer = A

let's assign a nice number..16 (divisible by 2 and 4), for total Air..
after 1st stroke, we are left with 8
after 2nd stroke, we are left with 4
after 3rd stroke, we are left with 2
after 4th stroke, we are left with 1

we removed 15 parts out of 16.
A: 15/16

Let initial air be X

after 1st stroke air left x/2
after 2nd stroke air left x/4
after 3rd stroke air left x/8
after 4th stroke air left x/16

air removed = original - air left
=x - x/16 = 15x/16

Hi All,

TESTing Values will work well on this question. A quick look at the answers tells us that the common denominator for all 5 options is 16; THAT is a huge clue that 16 would be a great number to TEST.

We're told that 1/2 of the air in a tank is removed with each stroke. We're asked what fraction has been removed after 4 strokes….

Full = 16
1st = 16-8 = 8 left
2nd = 8-4 = 4 left
3rd = 4-2 = 2 left
4th = 2-1 = 1 left

So 1/16 is LEFT; thus, 15/16 was REMOVED

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Strategy: Since the question is asking us to find a certain fraction (and not the actual volume of air that has been removed), let's assign a nice, easy-to-work-with value for the original volume of air in the tank.
Since we keep halving the volume of air, let's say there are 64 liters of are in the tank at the beginning, and will keep track of the volume after each stroke.

Initial volume: 64 liters
Volume of air remaining after 1 stroke: 32 liters
Volume of air remaining after 2 strokes: 16 liters
Volume of air remaining after 3 strokes: 8 liters
Volume of air remaining after 4 strokes: 4 liters

If 4 liters of air remain in the tank, then the volume of air that was removed = 64 - 4 = 60 liters

So, the fraction of the original amount of air has been removed after 4 strokes = 60/64 = 30/32 = 15/16

Answer: A

What's a fast way to solve such questions? WHat if we had 10 strokes? Pls also explain the strategy avigutman ThatDudeKnows

Elite097 this question is describing a geometric sequence. An analogous question could ask what fraction of the atomic nuclei of a radioactive sample would decay after 100 years, if it has a half-life of 25 years.
The strategy is to figure our how many times we're multiplying by (1/2). In the original question, that's clearly four times, so we have a change factor of (1/2)^4 = 1/16. If that fraction remains, then the remaining 15/16 is gone.
If we had 10 strokes, our change factor would be (1/2)^10 = 1/1,024, so the answer would be 1,023/1,024.
By the way, I have a chapter on geometric sequences in my book. First week access is free with a trial membership at quantreasoning.com

Elite097

I agree with Avi's comments.

One thing to think about is that GMAC will never give us a question that REQUIRES us to do too much math to make sense in the context of a timed exam. If you encounter a question for which you think your solution is going to take far too long, that's an indication that you should take a moment to see if you can find a different path to the finish line. Can you ballpark? Can you use some logic to eliminate incorrect answer choices (e.g. you need a negative odd number, and there is only one)? Can you Plug In The Answers? Can you Plug In for variables? Can you spot a pattern by testing a subset and then apply a rule for the pattern to a larger number (e.g. what if we had 243 strokes?)?

I'm curious what approach you took on this that prompted you to post? I can't come up with a way to solve this question that would take more than just a few seconds. What was your method? And what would be the answer if it were 243 strokes?