Official Quant thread for CAT 2013

MBA CAT 2022
761
@grkkrg said:
60
approach ??
762
@Sandy0303 said:
consider a right angle triangle of perimeter 48.inside triangle,there is circle,inside circle there is square,find the area of the square given that circle touches the hypotenuse divide it in ratio 2:3?
32?
763
@maddy2807 plse explain?
764

@grkkrg said:
60
@maddy2807 said:
60?
yes 60... hop down ur approaches
765
@Sandy0303 said:
consider a right angle triangle of perimeter 48.inside triangle,there is circle,inside circle there is square,find the area of the square given that circle touches the hypotenuse divide it in ratio 2:3?
32 ??
766
@Sandy0303 said:
@maddy2807 plse explain?
hypotenuse is a multiple of 5. use triplet (3,4,5)
hypotenuse = 20
other sides 12, 16
Now find radius of incenter using any formula, which will be = 4
so the diagonal wil be= 8 of the square inside the circle
side will be = 4rt(2)

area= 32
767
@bullseyes said:
In a Rectangle, the perpendicular bisector of AC divide the longer side AB in ratio of 2:1. then angle between AC and BD 30,45,60,90
Let say meeting point is O,

length ratio=2k:k

∠AOQ = 90,
Δ AOQ is similar to Δ ABC.
So,AO / AQ = AB / AC
==> y / 2k = 3k / 2y which gives us y^2 = 3k^2 or y = k.

Now applying Pyth Thm ,
OQ = k
==> ∠ OAQ = 30°
==> ∠ BOC = 60
768
@Sandy0303 said:
@maddy2807 plse explain?
take side of hypo =2x,3x

now
we get 5x+ 2x+y+3x+y=48
u get
5x+y=24
y=4 you get a soln
so radius of circle=4
side of sq= 8/rt(2)
area of square=64/2 = 32
769

sorry for wasting your time friends. really. pardon me. i thought it was n*(n+1) . so i framed that question.


Once again sorry, and we can logically pick none of these as answer..
770
@vijay_chandola said:
Let say meeting point is O,length ratio=2k:k∠AOQ = 90,Δ AOQ is similar to Δ ABC. So,AO / AQ = AB / AC ==> y / 2k = 3k / 2y which gives us y^2 = 3k^2 or y = k. Now applying Pyth Thm ,OQ = k ==> ∠ OAQ = 30° ==> ∠ BOC = 60
figure diyo yar 😞 i cant get d figure of the q
771
@Brooklyn said:
figure diyo yar i cant get d figure of the q
PFA 😃 :D
772
@bullseyes said:
In a Rectangle, the perpendicular bisector of AC divide the longer side AB in ratio of 2:1. then angle between AC and BD 30,45,60,90
considering the rectangle ABCD..let the perp bisector of AC be OP..now join OB.

now triangle AOP is similar to ABC..hence we get AO= xsqrt(3)..where AB=3x

now in triangle AOP cos (OAP)= AO/AP
therefore cos(OAP)= sqrt(3)/2=30
hence OBA also equals 30.

therefore BOC= 30+30=60

773
@Sandy0303 said:
consider a right angle triangle of perimeter 48.inside triangle,there is circle,inside circle there is square,find the area of the square given that circle touches the hypotenuse divide it in ratio 2:3?
32
774
@bullseyes said:
yes 60... hop down ur approaches
ABCD is the rectangle.

let O be the meeting point of the diagonal. Use the attached file.

triangle AEO and CEO are similar
AE= EC
AE= 2x= CE
EB=1x
therefore, BC= rt(3)x= AD
Now, we can calculate AC= 2rt(3)
AO=rt(3)

Now in triangle AOE

cos (angle EAO)= AO/AE= rt(3)/2= 30

hence angle OAD= 90-30=60

Now, OAD= ODA= 60= AOD
hence, AOD= 60
775
Let there be two lines AC and BC such that angle ACB=90. We
draw circles C1,C2, ..Cn with radii R1,R2, ..Rn, (Ri > Rj , if i > j). All
these circles are tangent to the lines AC and BC, and any two circles
Cp,Cp+1(1 when n = N where N is very large

a)[(sqrt2 +1)/2]R1 b)(sqrt2 +1)R1 c)[(sqrt2 +1)/2]RN d)(sqrt2 +1)RN


776

if someone has already posted it then please share the link


Consider the following equation; x2 + y2 + z2 = (x €“ y)(y €“ z)(z €“ x)
Which of the following statement is definitely true?

Options:

1) The above equation has no integer solutions for x, y, z.
2) The above equation has finitely many distinct integer solutions for x, y, z.
3) The above equation has 2 distinct integer solutions for x, y, z.
4) The above has infinitely many integer solution for x, y, z.

PS: I was able to eliminate option 1 and 3
(0,0,0)=> 1 solution
(0,1,-1)=> 6 solutions
but I don't know how to find out if it has finite no. of solutions or not

777
@maddy2807
@Brooklyn
@sumeet1489
@vijay_chandola

PFA Cat qstn

Find ratio of - (area under red / area under blue)
778
@nick_baba said:
Let there be two lines AC and BC such that angle ACB=90. We draw circles C1,C2, ..Cn with radii R1,R2, ..Rn, (Ri > Rj , if i > j). Allthese circles are tangent to the lines AC and BC, and any two circlesCp,Cp+1(1
a)[(sqrt2 +1)/2]R1
779
@gs4890 :: wats outer shape??? rectangle or square?? n same for inside
780
@gs4890 said:
@maddy2807@Brooklyn@sumeet1489@vijay_chandolaPFA Cat qstnFind ratio of - (area under red / area under blue)
Oval shape hai ya circle?