CAT 2008: QQAD Practice Test Discussions

[dewan_iitr]

19)

See figure.
Consider two triangles -
ACF and ABE

BE = x/2
In trngl ABE (x/2)/AB = tan 30 = 1/root3
AB = root(3)*x/ 2


In trngl ACF
AC^2 + CF^2 = AF^2
AC= AB +1
[root(3)*x/ 2+1]^2 + [x/ 2]^2 = 16

x = 3 root(7) - root(3) /2

[thebornattitude]

Originally Posted by srikar2097

Problem 19

For the problem where 6 rect. tables inside a circle.
This is the approach I took.

Since the inner part of the circle is a hexagon. Each angle is 120 degrees.
=> the triangle between any 2 rect. is an equilateral triangle with side as 1.
=> the height is sqrt(3)/2
=>the radius of the inner circle is 4-sqrt(3)/2 = (8-sqrt(3))/2 = 3.15 (approx).

Since when a hexagon is inscribed in a circle, the radius of the circle equals the side of the hexagon, here x=3.15 i.e. (C)

Although my answer is correct, I did not get it in the form the answers were given. Is there any better approach?

I have solved it in earlier post,, u can go through it

[sanyo]

Answer 24) the two latin squares are orthogonal only for the following values of:

x = 0
y = 1
z = 3
w = 2
q = 3
r = 2
s= 0
t = 1

Just checked in the orthogonal squares, cheked what could be the values for the variables.

have got one more correct answer- forgot to write the answer here in the thread .. Did a blunder in 13 question after finding out the answer marked it wrong(

[Aarav]

Originally Posted by sanyo

Answer 24) the two latin squares are orthogonal only for the following values of:

x = 0
y = 1
z = 3
w = 2
q = 3
r = 2
s= 0
t = 1

Just checked in the orthogonal squares, cheked what could be the values for the variables.

have got one more correct answer- forgot to write the answer here in the thread .. Did a blunder in 13 question after finding out the answer marked it wrong(

Was amused yeserday to note that you marked 13th wrong and 14th correct -> remembered this since only you attempted this set.

@All
I think after reading the answers here, most of you would be wondering as to what stopped you from scoring 50+ in this paper. Find the answer to this yourself.

[deep@k]

Hi any body knows when the solution for the test paper will publish....

I think the time was 1:30 PM today....

but i didnt get ny solution?????

Is there ny change in schedule............????????????????????

[rsriram84]

Originally Posted by deep@k

Hi any body knows when the solution for the test paper will publish....

I think the time was 1:30 PM today....

but i didnt get ny solution?????

Hey deepak, check post 69 in this thread for solutions.. detailed solutions will be mailed to ur inbox by sunday..
Cheers!!

[deep@k]

Thanks sriram,

I just missed that....
my score....

15 attempt 10 correct 5 wrong......

final 35 marks....... very bad......

need to brush up my basics for many topics....

really hard work required.....

Aarav how much percentile for 35 marks......

[Aarav]

Originally Posted by deep@k

Aarav how much percentile for 35 marks......

Too early to think in terms of percentiles -> have a contest with yourself only in the coming days and try to reason out why just 35 in this kinda paper.
It is well known fact that CAT toppers care least about other's score and create benchmarks for themselves and keep raising the bar consistently.
Your benchmark at the end of October should be 60+ in a similar paper.

[thebornattitude]

I am used to scoring less in initial phase .....scored just 23 in test....after the test,,,in 10 mins, did 6 questions...........but most of all I am happy that I could almost do 80% questions, though taking extra time, without seeking help..... Being an experienced CAT taker for 2 yrs now...( CAT 2006- 97.94 %, CAT 2007- 96.7 %....both the times scoring above 99 in quant...),,,

i can just say that,,,, be patient while solving questions,,,, do not carry a constraint of time in ur mind,,,,,don go with preconceived notion that u have to solve x many questions,,,,, if the paper is difficult to u, then it is difficult for all the test takers.... so stay focussed,,, if u are able to justify the cause of the question,,, u r through in quant......
keep rocking...cheerz..

[marijuana_user]

I m not sure if this has been posted b4::

Q.20

kx^3 + 2x^2- 3x + 1 = 0

Let the roots be a,b,c.
then b=2ac / a+c.

then from the equation:: abc= -1/k and a+b+c = -2/k
---> ac= -1/kb and a+c = -2/k - b

putting in b = 2ac / a+c. we will get a quadratic :: b^2 - kb- 2 = 0

if we assume D=0 :: k^2 = -8.

Hence no solution exists!

Regards!