CAT 2008: Quantitative Questions a Day 134 -Till end -> The Discussions

IIMA_2009

Can you please explain how C,O and D lie on the cirlce with center Q...

ashishkiran

My Take on:
------------------------------------------------------
Quantitative Question # 153
------------------------------------------------------
f(f(x,y),f(y,x))= 1 -f(x,y)

thus, f(x,y) = 1-x
Thus, f(1,1)= 1-1
=> f(1,1)= 0

Thus the answer is (e) none of these

Just followed gut feel in this one. Correct me if i'm wrong

anupam5138

guys what about the practice test five...it wasnt sent to me and was supposed to come on 14th ...right????
Please upload if it wasnt sent only to me...
thanks...
keep puying...

serena

f(f(x,y),f(y,x))= 1-f(x,y)

if we take that f(x,y)=x and f(y,x)=y then

f(x,y)=1-f(x,y)

2f(x,y)=1

f(x,y)=1/2 .... a

therefore based on a
if x=1 and y=1 we will get a 1/2

therefore answer is option (b)

shivam_01

Quote:

Originally Posted by Aarav

QQAD 153

Let f(f(x, y), f(y, x)) = 1 - f(x, y) for all real x and y. If f(1, 1) ≠ 1, then it is equal to

(a) 1 (b) 1/2 (c) 2 (d) can not be determined (e) none of these

Also,
TQTWLY

The remainder when 100*(99^10) is divided by 100*99 + 1 is

(1) 0 (2) 1 (3) 100 (4) 9900 (5) none of these

My attempt ::::

Take f(1,1) = z
so f(f(x, y), f(y, x)) = 1 - f(x, y)
When x=y=1 then
f[f(1,1),f(1,1)] = 1- f(1,1)
or f[z,z] = 1-z

Taking options into count

f(1,1) = z = 1 => f(1,1) = 1-1 = 0 Not possible
f(1,1) = z = 1/2 => f(1,1) = 1 - 1/2 = 1/2 Possible
f(1,1) = z= 2 =>o f(1,1) = 1-2 = -1 Not possible

Option (b) 1/2

Previous year QQAD
remainder when 100*(99^10) is divided by 100*99 + 1
100*99^10 mod 9900+1
100*99^10 mod 9901
= 100*9801^5 mod 9901
= 100*(-100)^5 mod 9901
= (-100)*100^5 mod 9901
= (-100)*99^2*100 mod 9901
= 100^2*100 mod 9901
(As the 99^2 mod 9901 = -100)

= 9900 is the answer

jayaram85

my take on qqad 152

from the figure we get angle ODP+angle OCP= 180
thus qaud ODPC is cyclic

the circle which circumscribes this quad is tangential to AB at O
< COD=x <AQB=2x

from triangles PDO and PCO OP=OP common side

PD=PC tangents from same external pt to a circle are equal

and OD =OC radii

so they are congruent by SSS

<OPD= < OPC so we can prove that triangle QPD ~= triangle QPC

so we get Q as the centre of the circum circle of cyclic quad PDOC

from tangent chord theorem we get <AOD= <OPD
and <BOC =<OPC

<AOD=<BOC=<COD=60 straight line

so triangle COD is equilateral <CDP IS 30 becos <CDP + <CDO=90 and < CDO =60

so using these results and finding pO we get PO= 2/3^0.5

jayaram85

my soln to qqad 153

let f(1,1)=k
from the eqn we get f(k,k)=1-k

now using teh given eqn we can write the relation
1-f(k,k)= f( f(k,k), f(k,k))
1- f(k,k)= f(1-k,1-k)
substituting f(k,k)=1-k

1-(1-k)= f(1-k,1-k)
k= f(1-k,1-k)

now putting k=0 we get f(1,1)=k=0

so f(1,1)=0
so answer is none of these

garehte

------------------------------------------------------
Answer of Quantitative Question # 152
------------------------------------------------------

C and D are points on the semi-circle with diameter AB and center O such that <(AQB) = 2*(<COD). Q is the point of intersection of AC and DB inside the circle.The tangents at C and D meet at P. The circle has radius 1. The distance of P from its center is

(a) 2 (b) √3 (c) 2/√3 (d) (√5 + 1)/2 (e) none of these

Answer is : (a) 2

jayaram85

Quote:

Originally Posted by garehte

------------------------------------------------------
Answer of Quantitative Question # 152
------------------------------------------------------

C and D are points on the semi-circle with diameter AB and center O such that <(AQB) = 2*(<COD). Q is the point of intersection of AC and DB inside the circle.The tangents at C and D meet at P. The circle has radius 1. The distance of P from its center is

(a) 2 (b) √3 (c) 2/√3 (d) (√5 + 1)/2 (e) none of these

Answer is : (a) 2

provide your solution so that evry one can learn some thing new

Varun Khullar

------------------------------------------------------
Quantitative Question # 154
------------------------------------------------------

I was at a restaurant for lunch the other day with some QQAdites. The bill came and I wanted to give the beautiful waiteress a multiple of 100 number of rupees, with the difference between what I give her and the bill being the tip. I always like to tip between 10 and 15 percent of the bill (ofcourse to a good looking damsel only). But if I give her a certain number of rupees (mutiple of 100), the tip would have been less than 10% of the bill, and if instead I give her 100 rupee more, the tip would have been more than 15% of the bill. The exact bill was a whole number. Let Rs PQRS be the largest possible amount of the bill where P, Q, R and S are each single digit numbers. Then P+Q+R+S is
(1) 14 (2) 15 (3) 17 (4) 22 (5) none of these

The question was that was 154th Last Year
---------------------------------------------
Two kinds of rice are mixed in the ratio 1:2 and 2:1 and then they are sold fetching the profit of 10% and 20% respectively. If they are mixed in equal ratio and the individual profit percents on them are increased 4/3 and 5/3 time respectively, then the profit % would be
(1) 18 (2) 20 (3) 21 (4) 25 (5) none of these