Official Quant thread for CAT 2013

@skt. said:
why is 11 not getting counted ?
Typo! 😃 Edited! :banghead:
@albiesriram said:
option A..
@albiesriram said:
is it 27?

13 to 39...
@albiesriram said:
13^3 = 2197
40^3 = 64000
39^3

So, 27 ?
@albiesriram said:
13 to 39

27 ?
@ all oa.

Next one..


I know ii's smaller .. cant maximize. adjust guys.. do a ctrl +
@albiesriram said:
4/15 ?
@albiesriram said:
Cube root of 2011 = 12.X
Cube root of 63000 = 39.X
Hence 27

@albiesriram said:
@ all oa. Next one..I know ii's smaller .. cant maximize adjust guys.. do a ctrl +
is it 3120??

#RIP_EYES! :P
@abhishek.2011 yes 9/8

Couldn't improve the resolution . the image hosting site trolls me by returning the same image for different resolutions.


Anyway,
@ChirpiBird said:
is it 3120??

OA..

Solution:-
@rnishant231 said:
in ques , it is given that a and b are digits..for denom to be 9 , we should have a multiple of 11 in numerator..but by multiplying any digits , we cant get multiple of 11..(we need to have 11)bhai pls clarify
11/99 = 1/9 :)
@ScareCrow28 said:
11/99 = 1/9
i took ab as a*b..but actually that is a number ab..
anyways thanks bro
@albiesriram said:
13 to 39 = 27 A)
@ChirpiBird said:
is it 3120??#RIP_EYES!
kaise kiya yaar....solution nahi dikh raha dhang se..
@albiesriram said:
it will be 4/15

Solution. for that penatgon wala sum. .

@albiesriram said:
Suppose that a andb are digits, not both nine and not both zero, and the repeating decimal 0.abababababab...... is expressed as a fraction in lowest terms. How many different denominators are possible?a)3 b)4 c)5 d)8 e)9
99=3*3*11
now make pairs
3 9 11 33 99
5 values of denominator possible
@nole said:
A circle circumscribes an equilateral triangle and is inscribed inside another equilateral triangle.Fnd the ratio of area of bigger triangle to the smaller one.I don't have the OA for this question.how to solve this ?
2:1
side of inside triangle a
radius of circle a/_/3
a/_/3 = x/2_/3
x=2
ratio 2:1
@albiesriram said:
kisi se ho jae to tell me i cant solve it ever :P