pleasr explain this concept.
Two consecutive no.s are removed from a list of first n natuaral no.s . The average of the remaining no .s is 64/3. What is the product of the two no. that have been removed?
Karuni, I guess the product will be 210 (as Viren has already said).
Suppose there are n numbers initially. That means after removing 2 of them the total is n-2. Also if the average of a set of first n (mostly) consecutive numbers is 64/3 (which is 21.33..) the value of n should be near double this i.e around 42 to 43. Also, n-2 should be a multiple of 3 (since the average has a 3 left in the denominator). So let’s try n-2 = 42 and 39.
Trying n-2 = 42, n becomes 44. Now the 42 numbers sum up to 42 * 64/3 = 896 while the sum of the first 44 numbers is 44 * 45/2 = 990. The two missing numbers would then add to 94 (which cannot be the total for two consecutive numbers and is also too high)
Trying n-2 = 39, n becomes 41. So the 39 numbers add to 39 * 64/3 = 832 while the 41 numbers add to 41 * 42/2 = 861. So the two missing numbers add to 961 – 832 = 29 and so must be 14 and 15. The product will therefore be 210…
Hey J,
“Also if the average of a set of first n (mostly) consecutive numbers is 64/3 (which is 21.33..) the value of n should be near double this i.e around 42 to 43.”
I did not understand this. Could you please explain?
Thanks 🙂
Ratish, try it for yourself. First, take a set of “n” natural numbers starting from 1. Find the average; this will turn out to be the middle number (or the average of the middle two, for even “n”). So it will be near n/2.
Then remove one of the numbers. Since this removed number will be at most “n”, the average will change by less than n/n i.e. 1. Try it for yourself and see.
By this kind of experimentation, you will develop the “feel” necessary for that statement of mine to make immediate sense.
Ok, I understood what you are trying to say. That when 2 numbers are removed from a set of numbers, the range of the new average varies between (a+ or – 1), where a is the original average of all numbers.
Now, you’ve stated above that “if the average of a set of first n (mostly) consecutive numbers is a, the value of n should be near double this” (This is after we have removed two numbers)
So, suppose I take the numbers 1,2,3,4,5. Here, n is 5.
When I remove 1 and 2, I am left with an average of 4. So, according to your logic above, n should be nearly double of 4, i.e., 8. But 5 and 8 arent close enough.
Please correct me. Not able to wrap my head around this. 😦
Too small a set Ratish. What we are doing is approximation, it will give us a reasonable answer when we take n = 20 or 100, since removing 1 or 2 numbers is a tiny fraction, but if you take say n = 5 you are removing 40% of the numbers at one fell swoop!
Sir, this is not related to your above post, but will there be apost on question like ‘There are 200 boys.. 150 Got grade A in Maths, 170 got grade A in Science …. what is the minimum number of boys who got Grade A in all subjects? ‘ I just can’t seem to grasp this concept.
This kind of question is based on the Pigeonhole Principle (though one can also use the Venn diagram logic – Inclusion-Exclusion principle). I am hoping to do a post or three on the PP sometime but so far haven’t managed….
In simple terms, if the above question has 3 subjects and asks 150 in Maths, 170 in Science and 180 in History for example, then say 150 + 170 + 180 – (200*2) = 100. (Idea being we are making all 200 pass in 2 subjects and seeing how many extra “passes” are left over.
CAT-holics 
pleasr explain this concept.
Two consecutive no.s are removed from a list of first n natuaral no.s . The average of the remaining no .s is 64/3. What is the product of the two no. that have been removed?
Averge of consecutive numbers.
Product is 210.
Karuni, I guess the product will be 210 (as Viren has already said).
Suppose there are n numbers initially. That means after removing 2 of them the total is n-2. Also if the average of a set of first n (mostly) consecutive numbers is 64/3 (which is 21.33..) the value of n should be near double this i.e around 42 to 43. Also, n-2 should be a multiple of 3 (since the average has a 3 left in the denominator). So let’s try n-2 = 42 and 39.
Trying n-2 = 42, n becomes 44. Now the 42 numbers sum up to 42 * 64/3 = 896 while the sum of the first 44 numbers is 44 * 45/2 = 990. The two missing numbers would then add to 94 (which cannot be the total for two consecutive numbers and is also too high)
Trying n-2 = 39, n becomes 41. So the 39 numbers add to 39 * 64/3 = 832 while the 41 numbers add to 41 * 42/2 = 861. So the two missing numbers add to 961 – 832 = 29 and so must be 14 and 15. The product will therefore be 210…
Hope this helps!
regards
J
Hey J,
“Also if the average of a set of first n (mostly) consecutive numbers is 64/3 (which is 21.33..) the value of n should be near double this i.e around 42 to 43.”
I did not understand this. Could you please explain?
Thanks 🙂
Ratish, try it for yourself. First, take a set of “n” natural numbers starting from 1. Find the average; this will turn out to be the middle number (or the average of the middle two, for even “n”). So it will be near n/2.
Then remove one of the numbers. Since this removed number will be at most “n”, the average will change by less than n/n i.e. 1. Try it for yourself and see.
By this kind of experimentation, you will develop the “feel” necessary for that statement of mine to make immediate sense.
regards
J
Ok, I understood what you are trying to say. That when 2 numbers are removed from a set of numbers, the range of the new average varies between (a+ or – 1), where a is the original average of all numbers.
Now, you’ve stated above that “if the average of a set of first n (mostly) consecutive numbers is a, the value of n should be near double this” (This is after we have removed two numbers)
So, suppose I take the numbers 1,2,3,4,5. Here, n is 5.
When I remove 1 and 2, I am left with an average of 4. So, according to your logic above, n should be nearly double of 4, i.e., 8. But 5 and 8 arent close enough.
Please correct me. Not able to wrap my head around this. 😦
Too small a set Ratish. What we are doing is approximation, it will give us a reasonable answer when we take n = 20 or 100, since removing 1 or 2 numbers is a tiny fraction, but if you take say n = 5 you are removing 40% of the numbers at one fell swoop!
regards
J
Sir, this is not related to your above post, but will there be apost on question like ‘There are 200 boys.. 150 Got grade A in Maths, 170 got grade A in Science …. what is the minimum number of boys who got Grade A in all subjects? ‘ I just can’t seem to grasp this concept.
This kind of question is based on the Pigeonhole Principle (though one can also use the Venn diagram logic – Inclusion-Exclusion principle). I am hoping to do a post or three on the PP sometime but so far haven’t managed….
In simple terms, if the above question has 3 subjects and asks 150 in Maths, 170 in Science and 180 in History for example, then say 150 + 170 + 180 – (200*2) = 100. (Idea being we are making all 200 pass in 2 subjects and seeing how many extra “passes” are left over.
regards
J
Eagerly waiting for your post(s) on PP 🙂
Would you please elaborate on the 4th question? The cat and tunnel one…
I didn’t understand the explanation provided in the answers.
Also, your posts are wonderful! Thank you, keep posting! 🙂
Mihir, see this earlier post https://crackthecat.wordpress.com/2013/07/08/singles-1/ – chances are it will be clear after that.
regards
J
Thanks J,
I get it now. 🙂
Also, sorry; should have checked “singles – 1” first; the answer mentioned it. :p
Regards,
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