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Happy weekend questions

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gurpreetsingh
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Happy weekend questions [#permalink] New post   03 Sep 2010, 20:39
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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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gurpreetsingh
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Re: Happy weekend questions [#permalink] 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



Show Spoilersolution: 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.
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Re: Happy weekend questions [#permalink] 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

Show Spoilersolution: 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
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gurpreetsingh
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Joined: 12 Oct 2009
Status:<strong>Nothing comes easy: neither do I want.</strong>
Posts: 2279
Own Kudos [?]: 3595 [0]
Given Kudos: 235
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] 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: happy-weekend-question-18-sept-101295.html


Show Spoilersolution: 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




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Re: Happy weekend questions [#permalink]
18 Sep 2010, 23:06
Moderator:
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Senior Moderator - Masters Forum
3137 posts

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