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30-04-2008, 02:18 PM
Hello All,
We will start QQAD from tomorrow. Since the software takes approximately 30 minutes to deliver the NL - the time by which you will receive the NL daily would be between 0630 and 0700 Hours daily.
Whoever is earliest on the thread - please post the problem of the day first before writing the solution.
Hope each one of you learn as much from one another and the discussions turn out to be very fruitful and enriching.
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30-04-2008, 02:45 PM
yupp phplist sucks  if you got loads of consumers, its hell of a task. I guess same is the case here, quite many registered users | | | | | The Following 3 Users Say NO Thank You to implex For This Un-useful Post: | | | The Following 3 Users Say Thank You to implex For This Useful Post: | | | | | |
Persevering to be the best
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01-05-2008, 06:38 AM
Expecting many to come out with the right answer  let's see who writes the solution with correct reasoning. ------------------------------------------------------
Quantitative Question # 001
------------------------------------------------------ Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ? (1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing What lies in front of you or behind you is nothing compared to what lies within you - T.M.W.S.H.F The greatest events in the life aren't the loudest, but the quietest hours - Anonymous Subscribe to QQAD: http://www.pagalguy.com/index.php?categoryid=65 | | | | | The Following User Says NO Thank You to Aarav For This Un-useful Post: | | | The Following 2 Users Say Thank You to Aarav For This Useful Post: | | | | | |
is Bak
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01-05-2008, 06:44 AM
Quote:
Originally Posted by implex yupp phplist sucks  if you got loads of consumers, its hell of a task. I guess same is the case here, quite many registered users | Quantitative Question # 001
------------------------------------------------------
Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ?
(1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing
clealry if a=b suppose
then
2a(a+1)+ c(c+1) +d(d+1)=312
now as all are natural nos
c(c+1) +d(d+1) >=4
308-2a(a+1)>=0
154-a(a+1)>=0
(a-11)(a+14)<=0
clearly 1<=a<=9
a^2+b^2+c^+d^2=292
the only solution comes 17,1,1,1
if a=!b we dont know which one is a which one is b but we know c,d=1 option b is right
if a>b
a=17 ,b,c,d=1 so option c is right
so overall option e is right
Last edited by implex; 01-05-2008 at 06:47 AM.
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01-05-2008, 06:52 AM
Quote:
Originally Posted by implex Quantitative Question # 001
------------------------------------------------------
Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ?
(1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing
so overall option e is right | That was fast  I was curious to see the reasoning out there - and must say it wasn't bad - though if possible try finding the perfect one What lies in front of you or behind you is nothing compared to what lies within you - T.M.W.S.H.F The greatest events in the life aren't the loudest, but the quietest hours - Anonymous Subscribe to QQAD: http://www.pagalguy.com/index.php?categoryid=65 | | | | | The Following User Says NO Thank You to Aarav For This Un-useful Post: | | | The Following 3 Users Say Thank You to Aarav For This Useful Post: | | | | | |
is Bak
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01-05-2008, 06:54 AM
Quote:
Originally Posted by Aarav That was fast  I was curious to see the reasoning out there - and must say it wasn't bad - though if possible try finding the perfect one  | perfection and "alcohol+me" don't go together, but will try!!
edit:
a^2+b^2+ c^2+ d^2=292
using am >=gm
and assuming a=b
2a^2+ 2cd<=292
146-a^2>=cd=1
a^2<=145
clearly a=b=1 if a=b
option a is wrong
the rest logic remains the same!!
Last edited by implex; 01-05-2008 at 07:02 AM.
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01-05-2008, 08:31 AM
Quote:
Originally Posted by implex perfection and "alcohol+me" don't go together, but will try!!
edit:
a^2+b^2+ c^2+ d^2=292
using am >=gm
and assuming a=b
2a^2+ 2cd<=292
146-a^2>=cd=1
a^2<=145
clearly a=b=1 if a=b
option a is wrong
the rest logic remains the same!! | Ok - same problem but we now have six natural numbers whose sum is 100 such that a(a+1) + b(b+1) + c(c+1) + d(d+1) + e(e+1) + f(f+1) = 9130.
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Join Date: Feb 2006 Location: Pune Age: 22 | Re: CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions -
01-05-2008, 08:36 AM
Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ? (1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing
Convert the given equation to: Code: a^2 + b^2 + c^2 + d^2= 292
And, Take option 1. If a=b, then it becomes 2(a^2) +c^2 + d^2= 292 and, 2a+c+d=20. a has to be at most 9. If a=9, then c=d=1. Putting various values, you'll find that you don't get a solution set. So option 1 is right
From option 2, you get a solution set of (17,1,1,1). Just ,maximise one value and go about it. You'll see that it is the only solution set, when you take lesser values like 16,15.. for one of the unknowns.
But, still we don't have a unique answer because we are not sure which is which, i.e., is a= 17, or b=17, and so on. Now option 3 comes to our help, which says a>b. So our problem is solved.
So, the answer is option 4) All of the foregoing | | | | | The Following User Says NO Thank You to comradeharsh For This Un-useful Post: | | | The Following 4 Users Say Thank You to comradeharsh For This Useful Post: | | | | | |
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01-05-2008, 08:42 AM
Quote:
Originally Posted by Aarav Ok - same problem but we now have six natural numbers whose sum is 100 such that a(a+1) + b(b+1) + c(c+1) + d(d+1) + e(e+1) + f(f+1) = 9130.
Find the numbers. | again we see that a^2+b^2+.. +f^2=9030
and clealry 1<=a,b,..f<=95
(95)^2=9025
one of the solutions comes to 95,1,1,1,1,1.
if we go lower than 95 we are going to lose much more than we can gain by adding one to one of the smaller nos
the fact being x^2 increases very fast for a higher x
clearly this is the only solution possible
now i think i am closer to a better logic | | | | | The Following User Says Thank You to implex For This Useful Post: | | | | | |
is Bak
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01-05-2008, 08:45 AM
Quote:
Originally Posted by comradeharsh Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ? (1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing
Convert the given equation to: Code: a^2 + b^2 + c^2 + d^2= 292
And, Take option 1. If a=b, then it becomes 2(a^2) +c^2 + d^2= 292 and, 2a+c+d=20. a has to be at most 9. If a=9, then c=d=1. Putting various values, you'll find that you don't get a solution set. So option 1 is right
From option 2, you get a solution set of (17,1,1,1). Just ,maximise one value and go about it. You'll see that it is the only solution set, when you take lesser values like 16,15.. for one of the unknowns.
But, still we don't have a unique answer because we are not sure which is which, i.e., is a= 17, or b=17, and so on. Now option 3 comes to our help, which says a>b. So our problem is solved.
So, the answer is option 4) All of the foregoing | what if a=b=1
then (c,d)=(17,1) or (1,17)
so option a is wrong for sure, we can find a solution set but not a unique one | | | | | The Following User Says NO Thank You to implex For This Un-useful Post: | | | The Following 2 Users Say Thank You to implex For This Useful Post: | | | |