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MGMAT - percent change

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MGMAT - percent change

by paridhi » Wed Jul 14, 2010 9:19 am
On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, a close relative, comes up for air on average once every five minutes. The number of times a bottle-nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a beluga whale would come up for air in that same period?
(A) 50%
(B) 100%
(C) 150%
(D) 200%
(E) 250%

[spoiler]What I did was:
no. of times dolphine cmes up in 24 hours = 12 (multiples of 2 between 1 & 24)
No. of times whale cmes up in 24 hours = 4 (multiples of 5 between 1 & 24)

Thus, % change = (12-4)/4 * 100 = 200% => (D)
But this is not the right answer.
OA is C
Can someone tell me how?[/spoiler]
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by kvcpk » Wed Jul 14, 2010 10:07 am
paridhi wrote:What I did was:
no. of times dolphine cmes up in 24 hours = 12 (multiples of 2 between 1 & 24)
No. of times whale cmes up in 24 hours = 4 (multiples of 5 between 1 & 24)

Thus, % change = (12-4)/4 * 100 = 200% => (D)
But this is not the right answer.
OA is C
Can someone tell me how?
I do not quite get what you have done.
Let me tell you my approach:

BND comes up every 2 mins. So in 1 hour, it comes 30 times. in 24 hours -> 720 times.
BW comes up every 5 mins. So in 1 hour, it comes 12 times. in 24 hours -> 288

perc change = (720-288)/288 * 100 = 432/288 * 100 = 150%

Hope this helps!!
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by ayushiiitm » Wed Jul 14, 2010 10:13 am
paridhi wrote:On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, a close relative, comes up for air on average once every five minutes. The number of times a bottle-nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a beluga whale would come up for air in that same period?
(A) 50%
(B) 100%
(C) 150%
(D) 200%
(E) 250%

[spoiler]What I did was:
no. of times dolphine cmes up in 24 hours = 12 (multiples of 2 between 1 & 24)
No. of times whale cmes up in 24 hours = 4 (multiples of 5 between 1 & 24)

Thus, % change = (12-4)/4 * 100 = 200% => (D)
But this is not the right answer.
OA is C
Can someone tell me how?[/spoiler]
number of times it comes in an 24 hrs=number of times in an hour*24==60/2*24 for whale
and 60/5*24

so % will be>>720-288/288*100

=150
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by paridhi » Wed Jul 14, 2010 7:58 pm
:evil: damn!!! careless mistake which I did not even notice while I was trying to figure out what I did wrong. I took the 2 mins interval as 2 hours interval.

Thanks everyone.
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by Testluv » Wed Jul 14, 2010 8:38 pm
Note that we can solve this one without having to funnel the solution through 24 hours. (Computing per day is as unnecessary as computing per week, month, year, etc.)

We know that in a 5 minute span, the beluga will come up for air once while the dolphin will come up 2.5 times. So, the dolphin comes up 2.5 times as much as the dolphin in any time span.

"Percent greater" means we have to use the percent change formula. Because the dolphin comes up 2.5 times as frequently, the number of times it comes up is (2.5-1)/1 = 1.5 times (or 150%) greater than the number of times the beluga comes up.
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by haidgmat » Thu Jul 15, 2010 7:41 am
I found a very easy and quick way to do it::

Btl-Nose in 1 minute comes up 1/2 times
Bgla in 1 minute comes up 1/5 times

So the difference would be 1/2-1/5= 5-2/10= 3/10
% difference= (3/10)/(1/5)*100 = 3/2*100= 150%

So answer is C

Please feel free to comment or correct..

paridhi wrote:On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, a close relative, comes up for air on average once every five minutes. The number of times a bottle-nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a beluga whale would come up for air in that same period?
(A) 50%
(B) 100%
(C) 150%
(D) 200%
(E) 250%

[spoiler]What I did was:
no. of times dolphine cmes up in 24 hours = 12 (multiples of 2 between 1 & 24)
No. of times whale cmes up in 24 hours = 4 (multiples of 5 between 1 & 24)

Thus, % change = (12-4)/4 * 100 = 200% => (D)
But this is not the right answer.
OA is C
Can someone tell me how?[/spoiler]
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by haidgmat » Thu Jul 15, 2010 8:47 am
I found a very easy and quick way to do it::

Btl-Nose in 1 minute comes up 1/2 times
Bgla in 1 minute comes up 1/5 times

So the difference would be 1/2-1/5= 5-2/10= 3/10
% difference= (3/10)/(1/5)*100 = 3/2*100= 150%

So answer is C

Please feel free to comment or correct..

paridhi wrote:On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, a close relative, comes up for air on average once every five minutes. The number of times a bottle-nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a beluga whale would come up for air in that same period?
(A) 50%
(B) 100%
(C) 150%
(D) 200%
(E) 250%

[spoiler]What I did was:
no. of times dolphine cmes up in 24 hours = 12 (multiples of 2 between 1 & 24)
No. of times whale cmes up in 24 hours = 4 (multiples of 5 between 1 & 24)

Thus, % change = (12-4)/4 * 100 = 200% => (D)
But this is not the right answer.
OA is C
Can someone tell me how?[/spoiler]
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by Gurpinder » Thu Jul 15, 2010 9:09 am
i have a very stupid question...... you know in this problem we have to apply the percent change formula....

why is it 288-720/288 and not 720-288/720????? it doesent really say in the question which one is the before event and which one is after.....i thought thats how percent change works....


please explain....thanks all
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by Stuart@KaplanGMAT » Thu Jul 15, 2010 10:27 am
Gurpinder wrote:i have a very stupid question...... you know in this problem we have to apply the percent change formula....

why is it 288-720/288 and not 720-288/720????? it doesent really say in the question which one is the before event and which one is after.....i thought thats how percent change works....


please explain....thanks all
Hi,

Neither of those is actually correct - it should be:

(720-288)/288

% change = (amount of change)/(original amount) * 100%

or

% change = (new- old)/(old) * 100%

The question here is what % greater is the # of dolphin breaths than the whale's, so we take the # of dolphin breaths as the "new" and the whale's as the "old".
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by Gurpinder » Thu Jul 15, 2010 10:39 am
Stuart Kovinsky wrote:
Gurpinder wrote:i have a very stupid question...... you know in this problem we have to apply the percent change formula....

why is it 288-720/288 and not 720-288/720????? it doesent really say in the question which one is the before event and which one is after.....i thought thats how percent change works....


please explain....thanks all
Hi,

Neither of those is actually correct - it should be:

(720-288)/288

% change = (amount of change)/(original amount) * 100%

or

% change = (new- old)/(old) * 100%

The question here is what % greater is the # of dolphin breaths than the whale's, so we take the # of dolphin breaths as the "new" and the whale's as the "old".


AHHH sorry about the wrong numbers but i just wanted to know which one is considered the old and which one is new...


thank you i get it now
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by this_time_i_will » Thu Jul 15, 2010 6:28 pm
we should see that for a "common minutes", how many times do both fish (OK, whale is a mammal! :)) comes up. That common minute would be nothing else, but LCM of 2,5 = 10.
So if you find the percentage change for 10 minutes, it would be applicable for 24*60 minutes as well, since 24*60 is divisible by 10.

so number of time Dolphin would come up in 10 min = 5
& number of time Whale would come up in 10 min =2

change = ((5-2)/5)*100 = 150%
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by Testluv » Thu Jul 15, 2010 6:36 pm
this_time_i_will wrote:we should see that for a "common minutes", how many times do both fish (OK, whale is a mammal! :)) comes up. That common minute would be nothing else, but LCM of 2,5 = 10.
So if you find the percentage change for 10 minutes, it would be applicable for 24*60 minutes as well, since 24*60 is divisible by 10.

so number of time Dolphin would come up in 10 min = 5
& number of time Whale would come up in 10 min =2

change = ((5-2)/5)*100 = 150%
Excellent point....it's also the point I made in my post above!

;)
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by sreak1089 » Fri Jul 16, 2010 2:09 am
Just a small change. It should be ((5-2)/2)*100 = 150% i think.

Testluv wrote:
this_time_i_will wrote:we should see that for a "common minutes", how many times do both fish (OK, whale is a mammal! :)) comes up. That common minute would be nothing else, but LCM of 2,5 = 10.
So if you find the percentage change for 10 minutes, it would be applicable for 24*60 minutes as well, since 24*60 is divisible by 10.

so number of time Dolphin would come up in 10 min = 5
& number of time Whale would come up in 10 min =2

change = ((5-2)/5)*100 = 150%
Excellent point....it's also the point I made in my post above!

;)
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by ahe207 » Sun Jun 17, 2012 11:46 am
I totally understand how to do this problem and completely understagnd why the "answer" is 150%, but I still think this is a VERY sloppy way of asking the question.

Ultimately the question is asking "5 is what percent greater than 2" and while the DIFFERENCE between the 5 and 2 is 150% greater than 2 (the question doesn't actually ask that), 5 is still 250% greater than 2 (2*250%=5).

So whenever I see "__ is what percent greater than __?" should I just assume it means "what is the percent change in the difference of __ and __?" regardless of the fact that it's a sloppy question?
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by sabeer6870 » Sun Jun 17, 2012 12:31 pm
1st whale 1/2 per min second whale 1/5 min..now ques is 1/2 is how much % greater than 1/5

ans is {(1/2)-(1/5)}/(1/5) X 100=150%

24 hrs will not make any diff as it common in both the cases.
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