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Tough and tricky: The sum of the even numbers.

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Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 11 Oct 2009, 17:11
2
4
00:00
A
B
C
D
E

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Question Stats:

71% (01:16) correct 29% (00:32) wrong based on 84 sessions

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I found a doc of quant problems, regarded as tough and tricky.

I'm starting posting most interesting of them. Maybe some of them were already posted here but still it would be useful to discuss once more.

They all have the answers but some of them are really doubtful so I'll try to avoid such ones or correct them. Also I'll provide the solutions given, with additional comments when necessary.

First one:

1. The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159
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Re: Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 11 Oct 2009, 18:02
We're looking at the sum of even numbers between 1 and k, with k being an odd number. So essentially, we're looking for the sum of even numbers between 2 and (k-1).

Sum (in consecutive series) = Average * Number of Elements in Subset

Average \(= \frac{2 + (k-1)}{2}\)

Number of Elements in Subset\(= \frac{k-1}{2}\)

Sum \(= [\frac{2 + (k-1)}{2}][\frac{(k-1)}{2}]\)

\(= [\frac{(k+1)}{2}][\frac{(k-1)}{2}]\)

\(= [\frac{(k+1)}{2}][\frac{(k+1)}{2} - 1]\)

It's easy to see that these two terms are consecutive digits. We know that the sum is equal to 79*80, and therefore we can solve for k.

\(80 = \frac{k+1}{2}\)

\(160 = k + 1\)

\(k = 159\)

Therefore, the correct answer is E:159.
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Re: Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 11 Oct 2009, 19:13
\((K-1)*(K+1)/4 = 79*80\)
\((K-1)*(K+1) = 158*160\)
So K = 159.
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Re: Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 12 Oct 2009, 09:38
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Re: Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 04 May 2011, 00:05
(k+1) * (k-1) = 79 *80 * 4

for k = 157, LHS/4 = 78 * 79
for k = 159, LHS/4 = 79 * 80 Hence E.
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Re: Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 06 Nov 2013, 19:41
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The Above are some very good explanations. But I approached the problem in another way.

Notice a general pattern.
2+4 = 6 which is also 2*3=6
2+4+6=12 which is also 3*4=12
2+4+6+8=20 which is also 4*5=20
2+4+6+8+10=30 which is also 5*6=30

(So the sum of a series of positive even integers starting from 2, is equal to the number of integers "n" multiplied by "n+1")

The question tells us that the sum is equal to 79*80 meaning that its the sum of the first 79 even integers.
79*2=158
the only answer choice that encompasses this figure is 159. So the answer is E.
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Tough and tricky: The sum of the even numbers.  [#permalink]

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New post 17 May 2015, 22:08
1
Bunuel wrote:
1. The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159



Ans: E

sum of n consecutive numbers= [n(n+1)]/2 = S
sum of only even or only odd numbers = S/2
now by putting the value we know: n= 158
so the value of k is 159
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Tough and tricky: The sum of the even numbers.   [#permalink] 17 May 2015, 22:08
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