CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[rajeev_hts]

a+b+c+d=20 –(1)
a(a+1)+b(b+1)+c(c+1)+d(d+1)=312 -----(2)
From 1 & 2
a^2+b^2+c^2+d^2=292 ------------(3)
(a+b+c+d)^2 = 400
a^2+b^2+c^2+d^2+((a+d)*(b+c))+(a*d)+(b*c)=400
((a+d)*(b+c))+(a*d)+(b*c) = 54 --------(4)
Now (a+d)*(b+c) < 54
This could be only with combination 2*18
So a+d =2 or 18
Also b+d=2 or 18
i.e. 3 numbers are 1,1,1 and one is 17
so if a>b means
a=17, b=1, c=1, d=1
My answer is (3)

[rajeev_hts]

My mistake 2 and 3 are correct so 5 shoud be answer.

[mohdfazal]

i understood the logic used by implex(even i used the same logic) but not Aarav... is it tht the logic used cant b used in any particular situation??

please explain...

[Aarav]

Originally Posted by mohdfazal

i understood the logic used by implex(even i used the same logic) but not Aarav... is it tht the logic used cant b used in any particular situation??

please explain...

The thing of importance in this problem was - if a1, a2, ... an are non-negative then (a1)^2 (a2)^2 + ... + (an)^2 <= (a1+a2+...+an)^2 that can easily be proved and the equality occurs iff all but one term is zero.

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[resurgentnaveen]

Hey Monk4lif,

Please checkout this book "Quantitative Aptitude For Cat" By Nishit K Sinha,Pearson Publication

He is an alumni of IIML and has exp in training students for CAT

[ravijanjanam]

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
ans:
a+b+c+d=20(given)
then from eq 2
(a^2+a)+(b^2+b)+(c^2+c)+(d^2+d)=312(given)
then
a^2+b^2+c^2+(a+b+c)=312
=>a^2+b^2+c^2+(20)=312

=>a^2+b^2+c^2=292
from option 1:
if a=b
then
assume a=11
then
11^2+11^2+7^2+1^2=312
therefore
option 1 is wrong
from option 2:
a not equal to b
then
13^2+11^2+1^2+1^2=312
therefore option 2 is correct
from option 3:

a>b
from above see that a is greater then b
so this also satisfying
there option e is the correct answer

[implex]

Originally Posted by ravijanjanam

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
ans:
a+b+c+d=20(given)
then from eq 2
(a^2+a)+(b^2+b)+(c^2+c)+(d^2+d)=312(given)
then
a^2+b^2+c^2+(a+b+c)=312
=>a^2+b^2+c^2+(20)=312

=>a^2+b^2+c^2=292
from option 1:
if a=b
then
assume a=11
then
11^2+11^2+7^2+1^2=312
therefore
option 1 is wrong
from option 2:
a not equal to b
then
13^2+11^2+1^2+1^2=312
therefore option 2 is correct
from option 3:

a>b
from above see that a is greater then b
so this also satisfying
there option e is the correct answer

total wrong!!
how can u take 13,11,1,1??
and 11,11,7,1??
a+b+c+d=20...?? this is given to be used !!!

[nidhisoni24]

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




a^2+b^2+c^2+d^2+a+b+c+d=312
a^2+b^2+c^2+d^2=292
17^2=289

289+1+1+1=392


a&b=1,
c=17
d=1
option 1 is nt correct
option 2 n 3 are correct so ans should be 5

[scribbles]

he he...
I came, i saw, i solved. I had trouble understanding the answer options. I interpretted the 5th option as follows : 'out of a,b, and d. the distinct value of only two can be found out.' since i found out the combination, i striked off 5th as possible answer(really fast, the same way i striked off the 1st option). And then I striked off the 4th option as I inferred the wrong meaning of foregoing. so i was left with 2 and 3 which were hard to eliminate. 2 was starring at my face as the right answer. and so was 3.

If only i knew that the 5th option(the first option that i striked out, meant that i could choose both 2 and 3)...

nice question. happy that i would have almost solved it if it werent for my tepid verbal recognition of the answer options. Cant wait for tomorrows QQ.

Cheers plus Smiles
scribbles

Originally Posted by Aarav

The problem has been solved correctly by most of you in the 1st attempt. Very well done

Here is the official solution to 001 - we might change this if by end of the day someone comes with a better approach

Solution:

Given a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312 => (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 312 - 3S + 4 = 256. Let a-1 = A, b-1 = B, c-1 = C, d-1 = D => we have non-negative integers A, B, C, D such that A^2 + B^2 + C^2 + D^2 = (A+B+C+D)^2 (the LHS <= RHS always) and can only be true if all but one number is zero => three among a, b, c, d are 1 and fourth number is 17.

Thus (3) holds true - (2) can guarantee the values for c and d. For a=b we have exactly 2 determinable values - thus (1) is not true.

=> choice (5) is the right answer

[mohdfazal]

Originally Posted by Aarav

The thing of importance in this problem was - if a1, a2, ... an are non-negative then (a1)^2 (a2)^2 + ... + (an)^2 <= (a1+a2+...+an)^2 that can easily be proved and the equality occurs iff all but one term is zero.

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I understand that would be of importance if we were to think from that point of view but is that really required... Is it not enough to use the other logic?? or will the logic that we use not be useful in certain situations?



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