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

[imj0ker]

Originally Posted by abhilash108

Sorry for late reply.....
60 stdnts may have cleared cutoff in DI.......but all of thm may not clear cutoff in othr subjects....any doubts on that proc, u can msg again.
I will explain.
But most of the junta has come up with 40 as answr......
let us wait fr official solution.

The question was at max, how many students can fail in all 4 subjects. So that is why I said, that since 60 have already passed in DI, so we can not have more than 40 ppl failing in all 4.

[Varun Khullar]

if 100 students are there and 60 students pass in DI.. and etc..
then shouldn't 40 be the max person that can fail..

about the other question..
..use pythagoras in both
x^2 =(b-x)^2 + a^2/4
y^2 = (a-y)^2 + b^2/4
add
x+y =a/2+b/2 +a^2/8b +b^2/8a
x+y = a/2+b/2 + (a+b)/ab +a/b+b/a -1
5/8 is the answer..

aarav i feel jaded.. lack of practice..give some tips. to take off the rust..
hi prade good to see u around

[atombomb_blast]

In a test taken by 100 students, 60 cleared cut-off in DI, 44 cleared cut-off in mathematics, 38 cleared cut-off in English and 27 students cleared cut-off in GK. 20 students cleared cut-off in all 4 sections. How many maximum students could have failed to clear the cut-off in all four sections?


(1) 38 (2) 41 (3) 47 (4) 50 (5) none of these

[imj0ker]

Originally Posted by atombomb_blast

In a test taken by 100 students, 60 cleared cut-off in DI, 44 cleared cut-off in mathematics, 38 cleared cut-off in English and 27 students cleared cut-off in GK. 20 students cleared cut-off in all 4 sections. How many maximum students could have failed to clear the cut-off in all four sections?


(1) 38 (2) 41 (3) 47 (4) 50 (5) none of these

As students passing in DI is tha maximum, i.e. 60 out of 100, we can safely say that maximum students who can fail in all four subjects is (100-60)= 40
. That would be the case when all 40 students failing in DI are also failing in other 3 subjects.

[tnj07iitr]

Hi...

my solution is as follows...

(x-b)^2 + a^2/4 = x^2
(a-y)^2

[atombomb_blast]

Originally Posted by tnj07iitr

Hi...

my solution is as follows...

(x-b)^2 + a^2/4 = x^2
(a-y)^2

could u be more elaborate plz

[atombomb_blast]

Originally Posted by imj0ker

As students passing in DI is tha maximum, i.e. 60 out of 100, we can safely say that maximum students who can fail in all four subjects is (100-60)= 40
. That would be the case when all 40 students failing in DI are also failing in other 3 subjects.

hey pal will this holds ud for all i mean wen the max pass say is 62 will the max fail in all 4 be 38????

[tnj07iitr]

Originally Posted by rajeev_hts

Hi Srikar, Please find the diagram with nomincleture. The case where the radius of circle is half of arm length gives the minimum possible radius by that a circle can passthough AB and touch CD (See small circle), if we increase the radius of circle from here we can also satisfy the conditions given.

so x >= 1/2 b

like above y>= 1/2 a

combining x+y >= 1/2(a+b)

Please ask me if it is not clear.

Hi

just evaluating what the solution actually implies..

The equality holds when x = y

For this to be the case a = b

or ABCD is a square :

in that case:

(a-y)^2 + a^2/4 = y^2
=> y = 5a/8

[thebornattitude]

Quantitative Question # 026
------------------------------------------------------



Let ABCD be a rectangle with AB = a, and BC = b. Suppose x is the length of the radius of the circle passing through A and B and touching CD, and y be the length of the circle passing through B and C and touching AD. If x + y ≥ k.(a+b) for all a and b, then k =


(1) √3/2 (2) 5/8 (3) 1/√2 (4) 1/2 (5) none of these




The diagram roughly looks like ...


Question.bmp


Now in taking pythagorus in traingle O1AN
(a-x)^2 + (b/2)^2 =x^2

In Triangle O2AN,
(y-b)^2 + (a/2)^2= y^2


solving we get x= b^2/8a + a/2
y= a^2/8b +b/2

putting in x+y>= k(a+b) we get
x+y= (a+b)/2 + b^2/8a + a^2/8b >= k(a+b)cancelling (a+B) from both sides , we get
ab(8k-4)<= a^2 +b^2 -ab
subtracting ab from both sides we get
ab(8k-5)<= (a-b)^2
now RHS >=0 always,, so k =5/8
answer - option 2- 5/8

[sudeepdeb]

Originally Posted by imj0ker

As students passing in DI is tha maximum, i.e. 60 out of 100, we can safely say that maximum students who can fail in all four subjects is (100-60)= 40
. That would be the case when all 40 students failing in DI are also failing in other 3 subjects.

hey pal will this holds ud for all i mean wen the max pass say is 62 will the max fail in all 4 be 38????

Well u have to look at other conditions also if they satisfy or not.
Let say students passing in DI =62
and some extra condition is given that, students who only passed in maths = 8 , then the max number of failures can be 100 - 62 - 8.

Solely, the highest number doesnt determine anything. Have to satisfy each condition.
Drawing Venn Diagrams and finding number of elements in each set n intersection of sets will give the actual picture.