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M24-18

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Bunuel
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M24-18 [#permalink] New post   16 Sep 2014, 01:21
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00:00
A
B
C
D
E

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5% (low)

Question Stats:

93% (00:51) correct 7% (00:41) wrong based on 76 sessions
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What is the value of \(1 + 2 + 3 + ... + 47 + 48 + 49\) ?

A. 1168
B. 1176
C. 1195
D. 1215
E. 1225
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E
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Bunuel
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Re M24-18 [#permalink] New post   16 Sep 2014, 01:21
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Official Solution:

What is the value of \(1 + 2 + 3 + ... + 47 + 48 + 49\) ?

A. 1168
B. 1176
C. 1195
D. 1215
E. 1225


To find the sum of an evenly spaced set, we multiply the average (arithmetic mean) by the number of terms.

The average of an evenly spaced set is the average of the first and the last terms, hence it's \(\frac{1 +49}{2}=25\). Therefore, the sum is \(25*49 = 1225\).


Answer: E
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ss3617
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Re: M24-18 [#permalink] New post   06 Feb 2018, 10:31
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Hi,

1,2,3..49 is an AP series.

So for that we can take the sum as averageXno of integers

Here average = (first term+last term)/2--> 1+49/2=25

Sum =25x49 = 1225
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tightrope
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Re: M24-18 [#permalink] New post   03 Mar 2018, 02:00
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Agree with previous calculations.

You also don't really need to count how many multiples of 50 there are (e.g. (1+49) + (2+48) etc).

You can realise very quickly that 25 will be "on its own" (as 25+25 would equal 50 but obviously we can't have 2 copies of the same number).

Therefore the final answer will either end with 25 (if there are an even number of multiples of 50) or 75 (if there are an odd number of multiples of 50).

Only 1225 satisfies this requirement so you can almost do this one "at a glance".
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Bunuel
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Re: M24-18 [#permalink] New post   20 Oct 2023, 00:41
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M24-18 [#permalink]
20 Oct 2023, 00:41
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