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Mike has been biking through the Rockies for 2 days

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Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 29 May 2019, 03:38
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Mike has been biking through the Rockies for 2 days and has used up 4/9 of his medication. If the only way for him to get more medication is to return to his starting point, how much longer can he bike before he has to turn around and ride back to his starting point along the same path?

A. 1/36 day B. 1/18 day C. 1/9 day D. 1/4 day E. 1/2 day

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Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 29 May 2019, 04:35
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dhritidutta wrote:
Mike has been biking through the Rockies for 2 days and has used up 4/9 of his medication. If the only way for him to get more medication is to return to his starting point, how much longer can he bike before he has to turn around and ride back to his starting point along the same path?

A. 1/36 day B. 1/18 day C. 1/9 day D. 1/4 day E. 1/2 day

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In 2 days, Mike uses = 4/9 of his meds

To return, he needs 1/2 of his meds left on him.
Therefore, he needs = 1/2 * 9/9 = 4.5/9 to make the return trip

Thus, he can use another 0.5/9 unit (i.e. 4.5/9 - 4/9)

Since he uses 4/9 in 2 days
he'll use 0.5/9 in = 2/(4/9) * 0.5/9 = 2/4 * 0.5 = 1/4 day

He has 1/4 day left before he has to turn around.
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Re: Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 17 Jul 2019, 04:27
chetan2u VeritasKarishma Bunuel

Quote:
Mike has been biking through the Rockies for 2 days and has used up 4/9 of his medication. If the only way for him to get more medication is to return to his starting point, how much longer can he bike before he has to turn around and ride back to his starting point along the same path?


Can you elaborate the text in blue? I unable to infer solutions from above steps?
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Re: Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 17 Jul 2019, 11:06
2 days of medication = 4/9 so 1 day medication=2/9
remaining medication =1-(4/9)= 5/9
so total number of days remaining= (5/9)/(2/9)= 2.5 days this includes further journey(lets x) and return journey(2 days +x days)
So remaining days 2+x+x=2.5 so x=0.25 days=1/4 days
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Re: Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 18 Jul 2019, 04:26
adkikani wrote:
chetan2u VeritasKarishma Bunuel

Quote:
Mike has been biking through the Rockies for 2 days and has used up 4/9 of his medication. If the only way for him to get more medication is to return to his starting point, how much longer can he bike before he has to turn around and ride back to his starting point along the same path?


Can you elaborate the text in blue? I unable to infer solutions from above steps?


The highlighted part means that he cannot stock up on medicine on the way. So whatever distance he travels, he must travel the same distance back and then he will get more medicine. So he cannot finish more than 1/2 his medicine on one way route. He will need the other half for the return journey.

Say Mike has medicine for 10 days. He can go one way for only 5 days. He will need other 5 days to go back to his starting point and will need to conserve half his medicine for that.

We are given that he used (4/9)th of medicine in 2 days so (2/9)th of his medicine every day.
He can consume only upto 1/2 of his medicine so he can go further only on 1/2 - 4/9 = 1/18 of the medicine.

He consumes (2/9)th of his medicine in 1 day
He will consume (1/18)th of his medicine in [1/(2/9)] * 1/18 = (1/4)th of a day

So he can bike for only 1/4th of a day and then he would have consumed 1/2 the medicine. The other half is needed for the return trip along the same route.

Answer (D)
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Re: Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 18 Jul 2019, 05:00
Let the total medication of Mike be 90 (as we have given in fraction that 4/9 he has used)
Used medication 4/9 * 90 = 40
Remaining medication = 90 - 40 = 50
To get more medication he has to travel back and medication required for both trips should be same that means Mike can use 45 of his medication for single journey
40 points he has already used in 2 days he has left with 5 more points after which he has to return

so days left will be 2/40 * 5 = 1/4 days
Correct Option D
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Re: Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 19 Jul 2019, 23:51
Let's say he is at Point P of the trail. He needs to conserve (4/9)th of the remaining (5/9)th of his medicine for the two-day return trip from P to his starting point. So he can bike onward and return to P only as long as (1/9)th of his medicine will last. (4/9)th of medicine lasts for 2 days so (1/9)th will last 1/2 day. So he can bike further for (1/4)th day (half of 1/2 day) before he has to return.

ANS: D
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Mike has been biking through the Rockies for 2 days  [#permalink]

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New post 20 Jul 2019, 08:32
DiyaDutta wrote:
Mike has been biking through the Rockies for 2 days and has used up 4/9 of his medication. If the only way for him to get more medication is to return to his starting point, how much longer can he bike before he has to turn around and ride back to his starting point along the same path?

A. 1/36 day B. 1/18 day C. 1/9 day D. 1/4 day E. 1/2 day

Posted from my mobile device


let t=time allowed before turning around
2/(4/9)=t/(1-4/9)→
t=5/2 day
if M turns around right now, and returns in 2 days,
he will have 5/2-2=1/2 day to spare
so he can bike (1/2)/2=1/4 day longer
D
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Mike has been biking through the Rockies for 2 days   [#permalink] 20 Jul 2019, 08:32
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