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fractions

by Nycgrl » Thu Aug 21, 2008 6:03 pm
If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?

1.15/16
2. 7/8
3. 1/4
4- 1/8
5. 1/16
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Re: fractions

by Sunny22uk » Thu Aug 21, 2008 6:15 pm
Nycgrl wrote:If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?

1.15/16
2. 7/8
3. 1/4
4- 1/8
5. 1/16
is the OA A?

lets say there were x units of air, x-x/16 were removed i.e. 15/16
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by vishubn » Thu Aug 21, 2008 8:56 pm
Hi ,
-->1/2+(1/2)1/2+1/2(1/4)+1/2(1/8)
-->1/2+1/4+1/8+1/6
-->(64+32+16+8)/120
-->128/120
-->15/16

Hope it helps

Vishu
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by Thouraya » Tue Mar 01, 2011 6:35 am
Are there any easier methods?

I solved it wrong. I assumed that let the total quantity of air present be 100. Then first stroke 100*1/2 is removed, then 50*1/2, then 25*1/2, then 12.5*1/2=6.25==>6.25/100 (to get fraction over original amount). What is it that's wrong in my way of thinking?
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by aleph777 » Tue Mar 01, 2011 6:58 am
Thouraya,

Sure, there's a pretty straight forward way. You just need to multiply the rate of removal by the amount left...

So... If the tank is originally full, and every stroke removes half of the air left, then 1 * 1/2 = 1/2 left after 1 stroke.

1/2 * 1/2 = 1/4 after two strokes.
1/4 * 1/2 = 1/8 after three strokes.
1/8 * 1/2 = 1/16 after four strokes.

But the question asks how much air has been removed. So if there is only 1/16 left, then 15/16 has been removed.
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by Thouraya » Fri Mar 04, 2011 6:16 am
What about if I want to assume that the initial amount present in the tank is 100, and then solve accordingly? can u show me the steps cuz I am still doing the same mistake as my post above.. Thank you!
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by Night reader » Fri Mar 04, 2011 6:38 am
1st stroke 1/2
2nd stroke 1/4
3rd stroke 1/8
4th stroke 1/16
total: (8+4+2+1)/16=15/16

answer A
Nycgrl wrote:If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?

1.15/16
2. 7/8
3. 1/4
4- 1/8
5. 1/16
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by GMATGuruNY » Fri Mar 04, 2011 6:39 am
Nycgrl wrote:If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?

1.15/16
2. 7/8
3. 1/4
4- 1/8
5. 1/16
We can plug in a value for the tank. We should plug in a value that can be divided by 2 four times without yielding a fraction. (The answer choices lead us in the right direction: the biggest denominator is 16.)

Let tank = 16 units.
After the 1st stroke, 1/2*16 = 8 units are removed.
16-8 = 8 units are left.
After the 2nd stroke, 1/2*8 = 4 more units are removed.
8-4 = 4 units are left.
After the 3rd stroke, 1/2*4 = 2 more units are removed.
4-2 = 2 units are left.
After the last stroke, 1/2*2 = 1 more unit is removed.
2-1 = 1 unit is left.

Since only 1 unit is left, 16-1 = 15 total units were removed.
Total removed/Tank = 15/16.

The correct answer is A.
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by Night reader » Fri Mar 04, 2011 6:45 am
:) hi, you just need to find decimal representation of 15/16 and multiply this by 100 with the denominator 100
(15/16)*100/100
you could check below the validity of this method
100 --> 50
50 --> 25
25 --> 12.5
12.5 --> 6.25
total: (50+25+12.5+6.25)/100=93.75/100
Thouraya wrote:What about if I want to assume that the initial amount present in the tank is 100, and then solve accordingly? can u show me the steps cuz I am still doing the same mistake as my post above.. Thank you!
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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