Preeti, think of an AP with, say, 7 terms a, b, c, d, e, f, g. Now these 7 terms can be thought of as d-3x, d-2x, d-x, d, d+x, d+2x and d+3x. (Try cross-checking with numerical values, say 5, 7, 9, 11, 13, 15, 17)
The average of the 3rd, 4th and 5th terms (c, d, e) will be the same as the 4th term (d). So will the average of the 2nd till 6th terms (b to f) and even of all 7 terms (a to g) – check it and see. Now try and apply the learning from this to the above question! You should be able to work out what I did…
Hey J,
In the third question, aren’t you working under the assumption that there are an even number of terms (2n)?
What if there are odd number of terms? Won’t there be an extra even number at the end then?
Since the set goes till 2n+1 and it is not mentioned whether n specifically takes even or odd values…say for example 2,3,4,5,6…here S-Y is 2
Ratish, the set starts from 2 (which is even) and goes till 2n+1 (which is always odd) and hence will always contain 2n numbers (i.e an even number of numbers). The question is deliberately worded in such a way as to ensure an even total.
mr j
In the last question it is mentioned p ,q,r all are distinct but you have solve the problem by taking 2 terms similar .how ? can we take like that ?
It will give us a limiting case. If you want to split hairs, you could take 1/(3+delta), 1/3 and 1/(3-delta) which would give you the same answer in three times the time and effort. The key is to know when to find the limiting case!
Yes, but if x > y is true, then x >= y is also true 🙂 I am assuming that was what was intended at any rate (this is a CAT question, not my own, so the options are retained as they were in the paper I have)
Hello J,
I have been following your blog for sometime now and would like to thank you for all the quality material! Thumbs up for that 😀 also are you still interested in posting some new stuff? I would love to see some quadratics and higher degree polynomial questions, especially the ones which involve common roots ( just can’t get the hang of those ( Non-math kid after all)) Care to add some of those too?
Well, I don’t really have that much more to add; for example, whatever I would have to say about most of Arithmetic and Algebra could be found in any standard textbook. Besides, health and work have been conspiring to keep me occupied, precluding the possibility of any new posts!
How did you just replace 2008 with 2 , 2007 with 1 and 2009 with 3 directly in the options ?
And that makes 1st option 3/2 same as only the first term. I think I’m missing something from here but I’m sure it’s interesting.
Would you Please explain the solution provided ?
I am trying to identify a pattern. I see that for 2008 terms, the answer choices were in terms of 2008 and nearby numbers 2007 (2008 – 1) and 2009 (2008 + 1). This can’t be a coincidence! I therefore take a simpler analogy, asking myself what would happen if I had just 2 terms, and expressing the answers in terms of 2, (2-1) and (2+1). If more than one answer choice matches, I would try for 3 terms as well.
CAT-holics 
I didnt understand how is the average of terms 11 till 20 is 0 and so on. Can you please elaborate a bit more on that ?
Preeti, think of an AP with, say, 7 terms a, b, c, d, e, f, g. Now these 7 terms can be thought of as d-3x, d-2x, d-x, d, d+x, d+2x and d+3x. (Try cross-checking with numerical values, say 5, 7, 9, 11, 13, 15, 17)
The average of the 3rd, 4th and 5th terms (c, d, e) will be the same as the 4th term (d). So will the average of the 2nd till 6th terms (b to f) and even of all 7 terms (a to g) – check it and see. Now try and apply the learning from this to the above question! You should be able to work out what I did…
regards
J
Hey J,
In the third question, aren’t you working under the assumption that there are an even number of terms (2n)?
What if there are odd number of terms? Won’t there be an extra even number at the end then?
Since the set goes till 2n+1 and it is not mentioned whether n specifically takes even or odd values…say for example 2,3,4,5,6…here S-Y is 2
Ratish, the set starts from 2 (which is even) and goes till 2n+1 (which is always odd) and hence will always contain 2n numbers (i.e an even number of numbers). The question is deliberately worded in such a way as to ensure an even total.
regards
J
Ya, right. My bad.
Thanks!
mr j
In the last question it is mentioned p ,q,r all are distinct but you have solve the problem by taking 2 terms similar .how ? can we take like that ?
It will give us a limiting case. If you want to split hairs, you could take 1/(3+delta), 1/3 and 1/(3-delta) which would give you the same answer in three times the time and effort. The key is to know when to find the limiting case!
regards
J
In Q4 option 4, shouldn’t equality be removed, as they are distinct.
Yes, but if x > y is true, then x >= y is also true 🙂 I am assuming that was what was intended at any rate (this is a CAT question, not my own, so the options are retained as they were in the paper I have)
regards
J
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Hello J,
I have been following your blog for sometime now and would like to thank you for all the quality material! Thumbs up for that 😀 also are you still interested in posting some new stuff? I would love to see some quadratics and higher degree polynomial questions, especially the ones which involve common roots ( just can’t get the hang of those ( Non-math kid after all)) Care to add some of those too?
Well, I don’t really have that much more to add; for example, whatever I would have to say about most of Arithmetic and Algebra could be found in any standard textbook. Besides, health and work have been conspiring to keep me occupied, precluding the possibility of any new posts!
regards
J
How did you just replace 2008 with 2 , 2007 with 1 and 2009 with 3 directly in the options ?
And that makes 1st option 3/2 same as only the first term. I think I’m missing something from here but I’m sure it’s interesting.
Would you Please explain the solution provided ?
I am trying to identify a pattern. I see that for 2008 terms, the answer choices were in terms of 2008 and nearby numbers 2007 (2008 – 1) and 2009 (2008 + 1). This can’t be a coincidence! I therefore take a simpler analogy, asking myself what would happen if I had just 2 terms, and expressing the answers in terms of 2, (2-1) and (2+1). If more than one answer choice matches, I would try for 3 terms as well.
regards
J
Got it ! Thanks !