CAT 2008: Quantitative Questions a Day 1 to 50

Aarav

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.

Enjoy the QQAD experience

Good Luck!

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

Aarav

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

implex · 06:47 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

Aarav

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

implex · 07:02 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!!

Aarav

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.
Find the numbers.

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,

Code:

a+b+c+d=20

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

implex

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

implex

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,

Code:

a+b+c+d=20

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