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12 + 13 + 14 + ... 51 + 52 + 53 = ?

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12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post Updated on: 22 Jul 2014, 09:28
1
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E

Difficulty:

  5% (low)

Question Stats:

83% (00:56) correct 17% (01:26) wrong based on 180 sessions

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12 + 13 + 14 + ... 51 + 52 + 53 = ?

A. 1361
B. 1362
C. 1363
D. 1364
E. 1365


explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42
and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

Originally posted by sagnik2422 on 22 Jul 2014, 09:26.
Last edited by Bunuel on 22 Jul 2014, 09:28, edited 1 time in total.
Edited the question
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 22 Jul 2014, 09:37
2
The sum is simply no. of terms multiplied by the average.

Number of terms:

53 - 11 = 42.

(This answers your question: No. of pairs = 42/2 = 21)


Average of each pair:

(53+12)/2 = 32.5
(52+13)/2 = 32.5
(51+14)/2 = 32.5
. .
. .

. .
So, total average = 32.5

Ans: 42*32.5 = 21*65
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 22 Jul 2014, 23:53
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2
sagnik2422 wrote:
12 + 13 + 14 + ... 51 + 52 + 53 = ?

A. 1361
B. 1362
C. 1363
D. 1364
E. 1365


explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42
and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?


You can do this question in different ways:

Sum = 12 + 13 + 14 + ... 51 + 52 + 53

Method 1:

Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2

Sum of first 53 positive integers = 53*54/2
Sum of first 11 positive integers = 11*12/2

Sum = 12 + 13 + 14 + ... 51 + 52 + 53 = 53*54/2 - 11*12/2 = 53*27 - 66
Note that 3*7 is 21 so 53*27 will end with 1 and when we subtract 66 from it, last digit will be 5. So answer must be (E)

Method 2:

From 12 to 53, how many numbers are there? 53 - 12 + 1 = 42 numbers
Number of positive integers from A to B are calculated as B - A + 1. You add an extra 1 to make up for the 12 that you subtracted. Since you want to include 12 as well in your counting, you add 1.

The average of these terms is (First term + last term)/2 = (53 + 12)/2 = 32.5
Average of a list is the number which can replace all numbers such that the sum of the list stays the same. So sum of the list can be calculated as Average*Number of elements in the list

Sum = 32.5*42 = 65*21
Ends with 5 so answer (E)
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 23 Jul 2014, 07:57
VeritasPrepKarishma wrote:
sagnik2422 wrote:
12 + 13 + 14 + ... 51 + 52 + 53 = ?

A. 1361
B. 1362
C. 1363
D. 1364
E. 1365


explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42
and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?


You can do this question in different ways:

Sum = 12 + 13 + 14 + ... 51 + 52 + 53

Method 1:

Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2

Sum of first 53 positive integers = 53*54/2
Sum of first 11 positive integers = 11*12/2

Sum = 12 + 13 + 14 + ... 51 + 52 + 53 = 53*54/2 - 11*12/2 = 53*27 - 66
Note that 3*7 is 21 so 53*27 will end with 1 and when we subtract 66 from it, last digit will be 5. So answer must be (E)

Method 2:

From 12 to 53, how many numbers are there? 53 - 12 + 1 = 42 numbers
Number of positive integers from A to B are calculated as B - A + 1. You add an extra 1 to make up for the 12 that you subtracted. Since you want to include 12 as well in your counting, you add 1.

The average of these terms is (First term + last term)/2 = (53 + 12)/2 = 32.5
Average of a list is the number which can replace all numbers such that the sum of the list stays the same. So sum of the list can be calculated as Average*Number of elements in the list

Sum = 32.5*42 = 65*21
Ends with 5 so answer (E)


Where did 65 and 21 come from in this : Sum = 32.5*42 = 65*21
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 23 Jul 2014, 22:42
sagnik2422 wrote:

Where did 65 and 21 come from in this : Sum = 32.5*42 = 65*21


Sum = 32.5 * 42 = 32.5 * 2 * 21 = 65 * 21
We do this to remove the decimal and get integers.
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 27 Feb 2017, 06:41
1
Sum = sum(1 +2 +3…..53) - sum(1 +2+3….11)
= 53(53 +1)/2 - 11(11 +1)/2
= 1431- 66
= 1365. Option E
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ? [#permalink]

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New post 09 Sep 2017, 23:46
Well there are faster methods that have been posted by various members , this method shows one more possibility of how some seq can be solved.

how many terms are there between 12 and 53 ( inclusive) = 42
If we can see that 12 +13= 25 and then 14+15= 29 and 16+17 =33 So each next pair is 4 more than the previous pair
How many pairs can we form then 21.

Sum = N ( FT + LT)/2
21( 25 +105)/2= 21*65
( 20+1)* 65= 1300+65= 1365
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Re: 12 + 13 + 14 + ... 51 + 52 + 53 = ?   [#permalink] 09 Sep 2017, 23:46
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12 + 13 + 14 + ... 51 + 52 + 53 = ?

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