Tricky problem that seems easy

Expert replies

PN123: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Mar 21, 2008 5:08 am

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

Like this post

Source: Beat The GMAT — Problem Solving | Fri Mar 21, 2008 5:11 am

HarvardDreamin: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Sun Jan 27, 2008 1:35 am; Thanked: 2 times

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.

ON MY WAY TO HBS......

Like this post

PN123: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Mar 21, 2008 5:08 am

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

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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

Like this post

PN123: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Mar 21, 2008 5:08 am

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

Like this post

HarvardDreamin: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Sun Jan 27, 2008 1:35 am; Thanked: 2 times

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 ...

ON MY WAY TO HBS......

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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.

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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

Like this post

PN123: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Mar 21, 2008 5:08 am

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

Like this post

HarvardDreamin: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Sun Jan 27, 2008 1:35 am; Thanked: 2 times

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?

ON MY WAY TO HBS......

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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
[/i]

Like this post

PN123: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Mar 21, 2008 5:08 am

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

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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

Like this post

ritz: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Sat Mar 08, 2008 5:15 pm; Location: Cincinnati; Thanked: 3 times

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

Like this post

Post new topic Post Reply

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review