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19 Sep 2023, 08:25
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
62% (02:33) correct
38%
(02:37)
wrong
based on 172
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When Friedrich was adding the first n positive integers, he mistakenly added one of the numbers twice, resulting in a total of 640. What was the number he added twice?
A. 5
B. 10
C. 12
D. 35
E. 45
A. 5
B. 10
C. 12
D. 35
E. 45
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19 Sep 2023, 11:03
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Bunuel
A shortcut that can come handy to solve this question -
The sum of the first 10 positive integers = 55
Sum of (1 to 10) = 55
Sum of (11 to 20) = 155
Sum of (21 to 30) = 255
Sum of (31 to 40) = 355
.
.
So on
We know that the resultant sum that Friedrich calculated is \(640\)
So let's add first few terms = \(55 + 155 + 255 = 465\)
If we add the sum of the next 10 numbers (sum from 1 to 40) = 465 + 355 ⇒ the Value is greater than 640. So we need to do a bit of approximation post 30.
Let's take the sum of 1 to 35 ( I have taken 35 as it is the midpoint of 30 and 40)
Sum (1 to 35) = \(\frac{35 * 36}{2}\) = 630
If we add the next value i.e. if we compute the sum from 1 to 36
Sum of (1 to 36) = \(630 + 36 = 666\) ⇒ the value is greater than 640
Hence, the intended sum was 630, but Friedrich calculated 640. Hence, 10 was counted twice.
Option B
20 Dec 2024, 20:42
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Sum of first n integers can be written as Sn= n/2(1+n)
Let us consider the no. added 2 times as x,
n/2(1+n) + x =640
n(1+n) + 2x =1280
we can find that 35x36= 1260
so,
1260 + 2x= 1280
2x=20
x=10
Ans= 10 (B)
Let us consider the no. added 2 times as x,
n/2(1+n) + x =640
n(1+n) + 2x =1280
we can find that 35x36= 1260
so,
1260 + 2x= 1280
2x=20
x=10
Ans= 10 (B)
