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Tricky problem that seems easy

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Tricky problem that seems easy

by PN123 » Fri Mar 21, 2008 5:11 am
I have done the GMAT twice and both times encountered a problem like this which I could not answer. (Please note this may not be the exact problem but very similar)

If today is Friday the 5th of November 1995, which day would be today + 1000 days?

Now you have to take into account leap years and the whole thing. Perhaps the question was a bit simpler than this. But the idea is that you take a certain day and try to approximate what the future day would be ( ex. Sunday, Monday, etc.) based on many parameters like leap year.

Both times, I could not answer this question on time. This means, this type of question should be fairly common in GMAT.

Can some one please help or point me to a question/answer that is similar in this forum?

Thanks a lot.

Pete
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Re: Tricky problem that seems easy

by HarvardDreamin » Fri Mar 21, 2008 5:54 am
PN123 wrote:I have done the GMAT twice and both times encountered a problem like this which I could not answer. (Please note this may not be the exact problem but very similar)

If today is Friday the 5th of November 1995, which day would be today + 1000 days?

Now you have to take into account leap years and the whole thing. Perhaps the question was a bit simpler than this. But the idea is that you take a certain day and try to approximate what the future day would be ( ex. Sunday, Monday, etc.) based on many parameters like leap year.

Both times, I could not answer this question on time. This means, this type of question should be fairly common in GMAT.

Can some one please help or point me to a question/answer that is similar in this forum?

Thanks a lot.

Pete


Would definitely like to see an answer to this as well - using logic, got to 1 Aug 2008 however not sure how to determine the exact day

1000 - (366(leap year) - 365) = 279 279/30= 9.3 months approx hence Nov + 9 months is approx August.

However im sure there must be a more elegant solution.
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by PN123 » Fri Mar 21, 2008 6:01 am
The question asked for the exact day - i.e. Monday vs. a Friday

Also there are some months with 31 days while others with 30 days. Then non-leap year has February with 28 days? How do you calculate some thing over a multi year period, taking into account all those variations?

I got beaten with this question twice before and it can happen to any of you, unless we can find an easy way to tackle this.

Thanks.

Pete
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simple approach

by ritz » Fri Mar 21, 2008 6:34 am
Hi,

here is what you can do...
Q. today is wednesday Nov 5th, 1980. What would be the day Nov5th 2099.
Ans -
Step 1:- Always find out the diff bet today's year & the future year
diff bet 2099 & 1980 = 119 (keep that aside as A)
Step 2:- Find out number of leap year bet given year & future year
1980 & 2099 = 119/4 = 29 (lets say this is B)
Step 3:- Add A & B
(119 + 29 = 148)
Step 4:- Divide that by 7 & find out the remainder
(148/7)
that is 1. Add this 1 to Wednesday (today's day as given in problem) & here is your answer THURSDAY.

Any day can be found out using this.

regards
Ritz
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by PN123 » Fri Mar 21, 2008 6:57 am
Ritz,

Thanks. But I don't follow step 3 and 4. What is the the logic behind that?

Let us go from 1980 to 1981. Can you please explain how this works for one year?

Pete
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Re: simple approach

by HarvardDreamin » Fri Mar 21, 2008 7:23 am
ritz wrote:Hi,

here is what you can do...
Q. today is wednesday Nov 5th, 1980. What would be the day Nov5th 2099.
Ans -
Step 1:- Always find out the diff bet today's year & the future year
diff bet 2099 & 1980 = 119 (keep that aside as A)
Step 2:- Find out number of leap year bet given year & future year
1980 & 2099 = 119/4 = 29 (lets say this is B)
Step 3:- Add A & B
(119 + 29 = 148)
Step 4:- Divide that by 7 & find out the remainder
(148/7)
that is 1. Add this 1 to Wednesday (today's day as given in problem) & here is your answer THURSDAY.

Any day can be found out using this.

regards
Ritz
I disagree with the part in BOLD because its a 1000 DAYS not YEARS ...
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by ritz » Fri Mar 21, 2008 7:26 am
Hey Pete,

First of all, FORGET THERE IS SUCH A THING AS A LEAP YEAR.NOW THINK
If Feb 1, 1980 was friday, next year (1981) the same day it will be Saturday & the year after that (1982) it will be Sunday & year after that (1983) it will be Monday & so on..
In short, with every passing year, we need to advance the DAY for the same date. That is because if today is friday, 7 days later it will be friday & 14 later it will be friday & 21 later it will be friday & so on until 364 days later it will be friday. But becuase there are 365 days in the year, we are left with 1 more day. So we need to advance the day by 1 DAY for every passing year.
Now bring the leap year knowledge into consideration. becasue 1980 was a LEAP year, it will add 1 additional day and hence 1981 feb 1 will not be Saturday, we will need to add 1 day to that. So 1 feb 1981 will become Sunday.

If you have understood that, you will realize that
Feb 1
1981 Sunday
1982 Monday
1983 Tuesday
1984 Wedneday
1985 Friday (because 1984 was a LEAP year, we need to add 1 for the next year)

I hope this is clear. let me know if it is not.

regards
Ritz
PS: for any dates after the month of February, we need to add 1 day to that year itself & not the next year. Think why & you should be able to get that.
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Re: simple approach

by ritz » Fri Mar 21, 2008 7:34 am
HarvardDreamin wrote:
ritz wrote:Hi,

here is what you can do...
Q. today is wednesday Nov 5th, 1980. What would be the day Nov5th 2099.
Ans -
Step 1:- Always find out the diff bet today's year & the future year
diff bet 2099 & 1980 = 119 (keep that aside as A)
Step 2:- Find out number of leap year bet given year & future year
1980 & 2099 = 119/4 = 29 (lets say this is B)
Step 3:- Add A & B
(119 + 29 = 148)
Step 4:- Divide that by 7 & find out the remainder
(148/7)
that is 1. Add this 1 to Wednesday (today's day as given in problem) & here is your answer THURSDAY.

Any day can be found out using this.

regards
Ritz
I disagree with the part in BOLD because its a 1000 DAYS not YEARS ...
If it is only days, then simply divide that by 7 & add the remainder (if any ) to arrive at the day.
for example, if today is friday & we need to find the day after 1200 days,
1200/7 (remainder is 3). so it will be Monday (3 days after friday)
hope this is clear now.
Regards
Ritz
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Thanks

by PN123 » Fri Mar 21, 2008 8:30 am
Hey Ritz,

Thanks a lot for that detailed explanation. By and large I feel comfortable with your step by step by approach. While I don't remember the exact question, with the knowledge given by you, I feel I am ready to go.

How ever, I still have one question. You wrote,

> PS: for any dates after the month of February, we need to add 1 day to that year itself & not the next year. Think why & you should be able to get that.

I am probably not that smart but I could not figure this out. I simply thought the step by step approach you gave earlier works for any date problem.

Can you please clarify?

Thanks and regards,

Pete
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Re: simple approach

by HarvardDreamin » Fri Mar 21, 2008 8:34 am
If it is only days, then simply divide that by 7 & add the remainder (if any ) to arrive at the day. for example, if today is friday & we need to find the day after 1200 days, 1200/7 (remainder is 3). so it will be Monday (3 days after friday)
hope this is clear now.
Regards
Ritz


Good stuff! :!:

It worked for this particular example - Im guessing its a general principle right?
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Re: simple approach

by ritz » Fri Mar 21, 2008 10:07 am
HarvardDreamin wrote:If it is only days, then simply divide that by 7 & add the remainder (if any ) to arrive at the day. for example, if today is friday & we need to find the day after 1200 days, 1200/7 (remainder is 3). so it will be Monday (3 days after friday)
hope this is clear now.
Regards
Ritz


Good stuff! :!:

It worked for this particular example - Im guessing its a general principle right?
Yes
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by ritz » Fri Mar 21, 2008 10:13 am
The approach i gave will work for every problem. I just suggested that for the days prior to feb last day, we need to be careful.

Here is an example.
Feb 1 is monday for 1979
so going by my approach
Feb 1 in 1980 will be monday but wednesday in 1982 as we will add a day for 1980 is a LEAP year.
However,
if 1 mar was Monday in 1979, we will see that Mar 1 in 1980 is on wednesday & not tuesday. That is because, 1 day delay becasue of the leap year has already happened on Feb 29 1980.
So watch out for that & dont make that mistake.

I hope it is clear.
Regards
Ritz
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by PN123 » Fri Mar 21, 2008 11:04 am
Ritz,

Thank you for the message. But I wonder if there is a small typo here.
Let me repeat and try to analyze what you said.

> Here is an example.
> Feb 1 is monday for 1979
> so going by my approach
> Feb 1 in 1980 will be monday

I don't get quite this. Going by your approach, I thought Feb 1 in 1980 should be Tuesday and not Monday becuase we have passed 365 days. Each year, the day gets advanced by 1 day.

> but wednesday in 1982 as we will add a day for 1980 is a LEAP year.

Now if we go to 1981, we hit the leap year in 1980 itself and 1 more day added because of 365 days, so Feb 1, 1981 should be Thursday.
Now if we analyze 1982, Feb 1, 1982 should be Friday? Did you make a typo when when you said Wednesday in 1982?

>However,
>if 1 mar was Monday in 1979, we will see that Mar 1 in 1980 is on >wednesday & not tuesday. That is because, 1 day delay becasue of the >leap year has already happened on Feb 29 1980.

This I follow. But please see if I analyzed it correctly above.

Thanks and regards,

Prasath
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by ritz » Fri Mar 21, 2008 12:28 pm
I am sorry, my bad..
it should read
Feb 1 is monday for 1979
so going by my approach
Feb 1 in 1980 will be Tuesday but Thursday in 1982 as we will add a day for 1980 is a LEAP year.

regards
Ritz
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by ritz » Fri Mar 21, 2008 3:22 pm
typo again :(
I am sorry, my bad..
it should read
Feb 1 is monday for 1979
so going by my approach
Feb 1 in 1980 will be Tuesday but Thursday in 1981 as we will add a day for 1980 is a LEAP year.

regards
Ritz
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