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number properties (600 level question)

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by Night reader » Sat Oct 30, 2010 10:09 am
If the total of 11 consecutive positive integers is x, then the total of the next 11 consecutive integers in terms of x is

(A) x + 11
(B) x + 221
(C) 7x + 11
(D) x + 121
(E) x + 101
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by shovan85 » Sat Oct 30, 2010 10:16 am
Night reader wrote:If the total of 11 consecutive positive integers is x, then the total of the next 11 consecutive integers in terms of x is

(A) x + 11
(B) x + 221
(C) 7x + 11
(D) x + 121
(E) x + 101
imo D

11 consecutive let us say a, a+1, ....., a+10
next 11 consecutive say a+11, a+12, ..... a+21

so each has 11 terms. compare each positioned numbers
I mean a with a+11 (1st of 1st seq to 1st of 2nd seq) diff is 11
similarly, a+1 with a+12: diff is 11

this way 11 times 11 diff will come

Now 1st seq sum is x
then 2nd seq sum is x+(11+11...... 11 times) = x+(11*11) = x+121
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by Night reader » Sat Oct 30, 2010 10:20 am
shovan85 wrote:
Night reader wrote:If the total of 11 consecutive positive integers is x, then the total of the next 11 consecutive integers in terms of x is

(A) x + 11
(B) x + 221
(C) 7x + 11
(D) x + 121
(E) x + 101
imo D


11 consecutive let us say a, a+1, ....., a+10
next 11 consecutive say a+11, a+12, ..... a+21

so each has 11 terms. compare each positioned numbers
I mean a with a+11 (1st of 1st seq to 1st of 2nd seq) diff is 11
similarly, a+1 with a+12: diff is 11

this way 11 times 11 diff will come

Now 1st seq sum is x
then 2nd seq sum is x+(11+11...... 11 times) = x+(11*11) = x+121
Correct!
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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