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

[anuragagrawal]

Originally Posted by shanky_2k8

dude did u try back substituting ur answers... i m afraid none of them satisfy the equation

Hey this is Anurag,
I m new to this site.
U shud have substituted (root)x=a(root)5

[raghavsahu]

i haven't recieved d daily question , plz. help m out so tat i can
get it . plz..........

[rsriram84]

Originally Posted by raghavsahu

i haven't recieved d daily question , plz. help m out so tat i can
get it . plz..........

Check the following post :
http://www.pagalguy.com/forum/quanti...ml#post1145802 (CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions)

[sushantgupta]

This is for AARAV
hi... sorry for asking this query late..but i thought i wud be able to work it out myself..but now need some help...
last year QQAD'S ques no 33...
1/3+2/21+3/91+4/273... the sum of this series..
i just wanted to know how to proceed to find the nth term of the series..please try to explain step wise..i am messing it up everytime..going by conventional approach of S(N) - S(N-1)...

[nbangalorekar]

Originally Posted by Aarav

Ok -> this is the warm-up problem since I want Implex also to have a go at this

This is last year's QQAD problem that took Vineet, me and a friend of mine in Intel research team to come to any conclusion in 4 hours. Treat this as a warm up and real contest (easy + difficult problems) with points will be held tomorrow.

Allwin likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example, 89.34 is 89 and 99.54 is 100, and 115.5 is 116. Allwin always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true?

(a) The minimum possible number of courses is less than 15
(b) The maximum possible number of courses is not more than 50
(c) The minimum possible current rounded average is 95
(d) Either of 126 or 127 can be the current rounded average
(e) none of the foregoing

I guess the best way to approach this qustion wud be to assume one of them as correct n then go on to show which one is wrong. if none of the other conditions is satisfied then our assumed option is wrong.
Serious number crunching is required for this problem..
will attempy iy later... too tied up with work now...
but a terrific question.... very rarely do we come across such mind blowing questions as this...
tried a bit now bt not reached a definite conclusion.... so will try again l8r in the night when i hv time...

[Aarav]

Originally Posted by sushantgupta

This is for AARAV
hi... sorry for asking this query late..but i thought i wud be able to work it out myself..but now need some help...
last year QQAD'S ques no 33...
1/3+2/21+3/91+4/273... the sum of this series..
i just wanted to know how to proceed to find the nth term of the series..please try to explain step wise..i am messing it up everytime..going by conventional approach of S(N) - S(N-1)...

A little observation would help here -> 3 = 1.3, 21 = 3.7, 91 = 7.13, ...
thus our series is 1/2* (1-1/3 + 1/3 - 1/7 + 1/7 - 1/13 + ...) = 1/2.

[Aarav]

Hello All,

Due to NL issues, we will take a small break for few days so that when we resume the recurring problem of variable delivery of NL is resolved for the good.

However tomorrow, you may participate in the discussions here on problems that will be posted in the form of a contest. The problems will be a mix of objective and subjective Quant/Logical questions. Set of 4 problems each will be posted on this thread at 7 am, 11 am, and 3 pm tomorrow. Each problem would be based on graded system (for calculating scores e.g. A+, A, B+ etcetera). But wait, if you do very well in the 1st set of 7 am then life will get tough for you as you will have lesser problems in set 2 (at 11 am) for answering (must do types) correctly and so for the 3rd set, but this doesn't mean we nullify your early mover advantage. It should be FUN, how this will turn out tomorrow, even I have no idea about

[rajeev_hts]

Hi Arav,

It is going to be Big fun. A BIG thank to you and Implex to keep this thread amzingly intrested.

[sanchit_states]

The number of ordered (x, y) such that 1/√x + 1/√y = 1/√20 is
(1) 1(2) 3(3) 5(4)7(5) none of these
Sol:
1/√x + 1/√y = 1/√20
1/√y = 1/√20-1/√x
For (x,y) cannot be perfect square but they can be of form
A^2*5 for having √20.
A=1,x=5-->y=not integer.
A=2,x=20--->y--cannot be zero
A=3,x=45--y=180
A=4,x=80-->y=80
A=6,x=180-->y=45
for further A no value will lead y as integer..thus only
three sol
Option 3

But is there any Generic answert for it?Do let me know.
"Challenges Accompany Opportunities"

[ankaj]

Originally Posted by sanchit_states

The number of ordered (x, y) such that
1/√x + 1/√y = 1/√20 is
(1) 1(2) 3(3) 5(4)7(5) none of these

the equation can be modifed as
√(y/x) + 1= √(y/20)

now substitute the values of y/x, such that y is an integer and x is integer

y/x = 1----y = 80 x=80

y/x = 4----y=180 x=45

y/x = 9----y= 320 x=320/9 (not an integer) similarly for other y/x sqaure values...we will get Y as integer but not x

y/x = 1/4--Y=45 x=180

So the answer is B.... 3


But, yes this is might not be a smart approach...any way shorter and better than this will help.........