CAT 2008: QQAD Practice Test Discussions

[Aarav]

Originally Posted by umeshlives

I request AARAV to attach(upload) a copy of every coming test so that ppl like me who r nt able 2 get mail .....able 2 solve it...

This will be done in future.

@All -> (5) and (15) are easy. Then the two sets from 13-14 and 22-25 -> please write the solutions for these as well.

[thebornattitude]

Question 15

b= 10^(ab^2)

now CASE1-- take a as positive..
so 10^(ab^2) >1 always,,,hence b>1 always....
but,,if b>1...and a is positive integer,,,then for any case 10^ (ab^2) will be too larger than b...
ie 10^(ab^2)>>b......so not possible...

CASE 2 -- take a as negative.

so let a= -c( a negative number)..

so b= 1/10^(cb^2)....
as 10^(cb^2)>1

if we take b<1...then RHS of the above term is <1...hence LHS <1...hence b<1....so we can interpret from the above equation, that it can be possible for some set of a,b,c.....

so now the order of the set becomes....

{ a,0,b,1,(1/b)}..

hence median= b....answer

[Aarav]

Originally Posted by thebornattitude

Question 15

b= 10^(ab^2)

now CASE1-- take a as positive..
so 10^(ab^2) >1 always,,,hence b>1 always....
but,,if b>1...and a is positive integer,,,then for any case 10^ (ab^2) will be too larger than b...
ie 10^(ab^2)>>b......so not possible...

CASE 2 -- take a as negative.

so let a= -c( a negative number)..

so b= 1/10^(cb^2)....
as 10^(cb^2)>1

if we take b<1...then RHS of the above term is <1...hence LHS <1...hence b<1....so we can interpret from the above equation, that it can be possible for some set of a,b,c.....

so now the order of the set becomes....

{ a,0,b,1,(1/b)}..

hence median= b....answer

Ek example lekar kahani khatm karo -> b = 1/10 as b can only be less than 1 => a = -100. For our choice of b all the 5 numbers are distinct and hence no trick of having a clash. This was a 1 minute question.

[diptichinu]

First of all I would like to thank AARAV for starting such a thread....Really acknowledge that...

Had taken the test today morning...and quickly realized that need to brush up the concepts again...attempted 11 and got 7 correct....need to improve.....

[dewan_iitr]

[FONT='Georgia','serif'] (5)[/font][FONT='Georgia','serif'] What is the length of the side AB of triangle ABC?[/font]

[FONT='Georgia','serif'](X) AB <= AC = 2, and area of triangle ABC is 2[/font]
[FONT='Georgia','serif'](Y) Exactly two sides have integer length [/font]


(X) Ar if ABC = 1/2* b*c*sinA
AC= b and AB= c

Area = c*sinA =2
c min value = 2
and c<=2 (given)
thus c=2

thus AB = 2 root(2)

[implex]

question 21
Attendance in delhi 50000+-5000

1.10x=60000
or 0.90y=60000 is greater than x
so for max
y=60000/90=66667
now 66667 can be round off to 67000
so max difference is 67000-45000=22000

[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?

[srikar2097]

Problem 9

Well this was simple!
Total revolutions = 33*30 =990
if 'x' min rev. at 45rpm and 'y' min rev at 33rpm
then x+y=26
and 45x + 33y=990

Solve to get x=11min. Hence (C)

[implex]

question 23)
Very simple we can find only 1 such square ! chk for urself

Question 24
one out of 0x or or is 00 but
r can't be 0 as there is already a 0 in its column
so x is 0
option A
QUestion 25
using the latin squares given we can easily check for n =3 and n=4
and so it should be true for all n (>2)

[srikar2097]

Originally Posted by Aarav

This will be done in future.

@All -> (5) and (15) are easy. Then the two sets from 13-14 and 22-25 -> please write the solutions for these as well.

Problem 23


Works on the concept of Sudoku! Just fill in the squares with the remaining digits without repeating same digit any row or column.

Here since for each row/column at least one digit or the other is fixed, there can be only one possible structure. Hence (B)

Just a thought, if there were no fixed digits, we could move a row or a column. So there could be more than one structure. I guess there will be 12?