Official Quant thread for CAT 2013

MBA CAT 2024
28505
@ThankYou said:
All those guys having Arun Sharma Quant book 5th edition, I have doubts in geometry. Kindly solve these questions and pls tag me with the answers:1. Page no=384 , Q.no=142. Page no=385 , Q.no=223. Page no=385 , Q.no=24
Kindly post the questions

Not every1 has the book
28506
@ThankYou Post a couple of lines from each question. People have different editions. And may be google those couple of lines - you might find the questions as well.
28507
@albiesriram said:
OA for Last problem set is C n D . Last one of the day....
1) (x-1)^2 x^2 -3x+2 (x-2)(x-1)
2) sin(cosx) = cos(sinx) --> cosx = pi/2 - sinx --> cosx + sinx --> pi/2 max. value of sinx + cosx is root 2
28508
1. abc is an equilateral triangle. point d is on ac and point e is on bc, such that ad=2cd, and ce=eb.. If we draw perpendiculars from d and e to other two sides and find the sum of the length of two perpendiculars for each set, that is , for d and e individuallly and denote them as per (D) and per(E) respectively, then
1.per(D) > per(E) 2.per(D)
3.per(D) = per(E) 4.none of these ...give with explanation.
28509
A, B, C started a business with their investments in the ratio 1:3:5 after 4 months A invested thesame amount as before and B as well as C withdrew half of their investments. The ratio of their profitsat the end of the year is
a)4:3:5 b)5:6:10
c)6:5:10 d)10:5:6
28510
@sumit99 said:
A, B, C started a business with their investments in the ratio 1:3:5 after 4 months A invested thesame amount as before and B as well as C withdrew half of their investments. The ratio of their profitsat the end of the year isa)4:3:5 b)5:6:10 c)6:5:10 d)10:5:6
A=x*12 +x*8=20x
B=3x*4 +3x/2 *8 =24x
C=5x*4 +5x/2 *8 =40x
5:6:10
28511
@ThankYou
Page 384 is not available in Google preview.
Page 385. Q - 22. Assume the circles to be touching and having the same radii
Then, PP' = RS = 2r and AB = 0. Option B
Page 385. Q - 24. The triangles PAB and PDC will be similar and so PA/PB = PD/PC. Option B
28512
@ThankYou said:
1. abc is an equilateral triangle. point d is on ac and point e is on bc, such that ad=2cd, and ce=eb.. If we draw perpendiculars from d and e to other two sides and find the sum of the length of two perpendiculars for each set, that is , for d and e individuallly and denote them as per (D) and per(E) respectively, then 1.per(D) > per(E) 2.per(D) 3.per(D) = per(E) 4.none of these ...give with explanation.
Both should be equal....I guess from any point inside the triangle equal...

regards
scrabbler
28513
@scrabbler said:
Both should be equal....I guess from any point inside the triangle equal...regardsscrabbler


@ThankYou
28514
@Subhashdec2 said:
To yeh equal hi hai na? How did you get greater than?

regards
scrabbler
28515
@scrabbler said:
To yeh equal hi hai na? How did you get greater than?regardsscrabbler
sorry i quoted u by mistake
was quoting the person who asked it...
and will edit my previous post..
28516
If f(x) = x^2 + 4x + 4 and g(x) = x^2 + 4x + 3, then find the value of x such that
f(g(x)) = g(f(x)).
28517
@amresh_maverick said:
If f(x) = x^2 + 4x + 4 and g(x) = x^2 + 4x + 3, then find the value of x such that f(g(x)) = g(f(x)).
f(x) = (x + 2)^2
g(x) = (x + 2)^2 - 1

f(g(x)) = [(x + 2)^2 + 1]^2

g(f(x)) = [(x + 2)^2 + 2]^2 - 1

[(x + 2)^2 + 2]^2 - 1 = [(x + 2)^2 + 1]^2

[(x + 2)^2 + 3][(x + 2)^2 + 1] = [(x + 2)^2 + 1]^2

[(x + 2)^2 + 3] = [(x + 2)^2 + 1]

no such value ..??

28518
@amresh_maverick said:
If f(x) = x^2 + 4x + 4 and g(x) = x^2 + 4x + 3, then find the value of x such that f(g(x)) = g(f(x)).
@techgeek2050 said:
f(x) = (x + 2)^2g(x) = (x + 2)^2 - 1f(g(x)) = [(x + 2)^2 + 1]^2g(f(x)) = [(x + 2)^2 + 2]^2 - 1[(x + 2)^2 + 2]^2 - 1 = [(x + 2)^2 + 1]^2[(x + 2)^2 + 3][(x + 2)^2 + 1] = [(x + 2)^2 + 1]^2[(x + 2)^2 + 3] = [(x + 2)^2 + 1]no such value ..??
Correct

sum of roots |x-2|^2 + |x-2| -2=0

options :
8
4
7
3
28519
@amresh_maverick said:
If f(x) = x^2 + 4x + 4 and g(x) = x^2 + 4x + 3, then find the value of x such that f(g(x)) = g(f(x)).
f(x) = g(x) + 1

f(g(x)) = g(g(x)) + 1
g(f(x)) = g(g(x) + 1)

g(x) = a

g(a) + 1 = g(a + 1)

=> g(a+ 1) - g(a) = 1
=> (a+1)^2 - a^2 + 4 = 1
=> 2a = -4

=> a = -2

g(x) = -2

x^2 + 4x + 3 = -2
x^2 + 4x + 5 = 0

No such x ?

PS: Calculation mistake somewhere may be :(
28520
@amresh_maverick said:
Correctsum of roots |x-2|^2 + |x-2| -2=0options :8473
3? Only root I am getting is x-2 = 1...

Edited: |x-2| = 1 hoga

regards
scrabbler
28521
@amresh_maverick said:
Correctsum of roots |x-2|^2 + |x-2| -2=0options :8473
4 ?
28522
@scrabbler said:
3? Only root I am getting is x-2 = 1...regardsscrabbler
shouldn't it be should be |x - 2| = 1 ?
28523
@techgeek2050 said:
shouldn't it be should be |x - 2| = 1 ?
Oops yes. That's what I get for not writing :(

regards
scrabbler
28524
@scrabbler said:
Oops yes. That's what I get for not writing regardsscrabbler
sir, by now you would have saved millions of trees from being felled.