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Happy weekend questions

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CEO
CEO
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
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Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Happy weekend questions  [#permalink]

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New post 03 Sep 2010, 20:39
1
Hi all,

I will post weekly question on Saturday's and the Solution on Monday's 22:00 IST. Pm me if I forget . :P

Lets not discuss whether this type of question can come on Gmat or not rather learn something new from the question. Even if you are not able to solve it in 2mins just try to crack some logic. I m sure you guys will learn something new every week.

Strictly No replies here. This thread will be updated with question and its solution separately so that it can be revised on one page.


All the replies on this thread will be deleted and warning will be issued. Kindly be supportive.



-----------------------NO Discussion------------------------

Disclaimer : Questions not created by me , but are from multiple mixed sources. The credit goes to the creator.

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CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2532
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Re: Happy weekend questions  [#permalink]

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New post 03 Sep 2010, 20:52
04 Sept 2010

Question : 1
Q1. Find the remainder when \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\) is divided by 101.

A) 0
B) 1
C) 100
D) 99
E) 2

Discussion: happy-weekend-questions-04-sept-100375.html



Spoiler: :: solution: 1
Sn = \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\)

= summation of n*(n+1) where n varies from 1 to 99

= summation of n^2 + n = \(n*(n+1)*\frac{(2n+1)}{6}\) \(+ n*\frac{(n+1)}{2}\)

= \(n*(n+1)*\frac{(n+2)}{3}\)
Now put n =99 we get sum = \(99*100*\frac{101}{3}\)

This is divisible by 101, hence remainder is 0.
Answer is A.

_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
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CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2532
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Re: Happy weekend questions  [#permalink]

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New post 11 Sep 2010, 11:45
11 Sept 2010

Question : 2
Q2. Find the remainder when \(1^{39} +2^{39} + 3^{39}..........12^{39}\) is divided by 39.

A) 0
B) 1
C) 3
D) 2
E) 12

Discussion: happy-weekend-question-11-sept-100862.html

Spoiler: :: solution: 2
Using the property : \(a^n + b^n\) is always divisible by \(a+b\) if n is odd

\(1^{39} +2^{39} + 3^{39}..........12^{39}\)

=> \(1^{39} +12^{39} + 2^{39} + 11^{39} +3^{39} + 10^{39} + ......6^{39} + 7^{39}\) is divisible by

1+12 + 2+11 + 3+10 + 4+9 + 5+8 + 7+6 = 6*13 = 39*2

=> it is divisible by 13 and the remainder is 0.
Answer is A.

In general, \(a^n + b^n +c^n........ z^n\) is divisible by \(a+b+c......z\) if n is odd

_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2532
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: Happy weekend questions  [#permalink]

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New post 18 Sep 2010, 23:06
18 Sept 2010

Question : 3
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same previous speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Discussion: http://gmatclub.com/forum/happy-weekend ... 01295.html


Spoiler: :: solution: 3
Distance traveled is proportional to speed if time taken is same.
\(\frac{D_{X}}{D_{Y}} = \frac{V_{X}}{V_{Y}}\)

\(\frac{100}{90}= \frac{D_{X}}{D_{Y}} = \frac{10}{9}\)

In the second case

\(\frac{D_{X}}{D_{Y}} =\frac{V_{X}}{V_{Y}} = \frac{10}{9}\)

Since \(D_{X} > D_{Y}\) => X beats Y by some distance => X must have traveled 110 m

\(\frac{10}{9} =\frac{D_{X}}{D_{Y}}=\frac{110}{D_{Y}}\)

\(D_{Y} = 99\), Hence X beats Y by 1 m

Answer is B


--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
GMAT Club Bot
Re: Happy weekend questions   [#permalink] 18 Sep 2010, 23:06
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