Official Quant thread for CAT 2013

MBA CAT 2024
15243
@raopradeep
how many non real roots does the eqn have if it has 1 negative and 2 positive roots
x^5-9x^4+13x^3+57x^2-86x-120=0

this equation will have either 3 or 0 positive real roots and either 2 or 0 negative real roots.
as per give condition in the problem...there will be zero non real roots.
15244
@pankaj1988 said:
d)4&5f(4)=-5 and f(5)=35Since there is a sign change between 4 and 5 so the graph would have crossed at least once between 4 & 5. So there is at least one root between 4 and 5.
but OA is B

15245
@vbhvgupta said:
Q2
Say the series is 1,r,r^2,r^3,...
Now lets take the first term 1
So 1 = x (r+r^2+...) = x r/(1-r) => r=1/(1+x)
15246
@Tiws said:
@raopradeephow many non real roots does the eqn have if it has 1 negative and 2 positive rootsx^5-9x^4+13x^3+57x^2-86x-120=0this equation will have either 3 or 0 positive real roots and either 2 or 0 negative real roots.as per give condition in the problem...there will be zero non real roots.
PLZ ELABORATE MORE
I THINK IT HAS 3 OR 1 POSITIVE ROOTS AND 2 OR 0 NEGATIVE ROOT
15247
@chandrakant.k 'c' the other two options donot go with the question
15248
if f(x)= x^4 - 9x^3 + 26x^2 - 24x + 5 then one root of f(x) lies btwn
a)-1 &0
b)0&2
c)2&3
d)4&5
@raopradeep

f(0) = 5 ; F(1) = -1 ;

hence thr is a root between x=0 and x=1
thus option B is correct
15249

Q3

15250
@vbhvgupta said:
Q3
1/64

a = 3( ar+ ar^2 +.....)
r= 1/4
a*r^4 = 4/(4^4) = 1/4^3 = 1/64
15251
@Tiws said:
1/64
explain.
15252
@vbhvgupta said:
Q3
Let the series be => a,ar,ar^2,.........

Acc to Q , a = 3ar / 1-r
Hence, r = 1/4

5th Term => ar^4 => 4*(1/4)^4 => 1/64 ?
15253
@vbhvgupta said:
Q3
1/64 is it ??
15254

Q4

15255

find remainder when S is devided by 13.

S =[ 5+ 5^2 +......+ 5^ 55]
a.0
b.1
c.7
d.12

15256
@Tiws said:
find remainder when S is devided by 13.

S =[ 5+ 5^2 +......+ 5^ 55]
a.0
b.1
c.7
d.12
d 12...
15257
@vbhvgupta 3:2

15258
@Tiws said:
find remainder when S is devided by 13.S =[ 5+ 5^2 +......+ 5^ 55]a.0b.1c.7d.12
5^2 mod 13= -1

so 5+(-1)+(-5)+1+..................+5+(-1)+(-5)

pair of four terms will get canceled
so just the last three terms
=> -1 or 12

option D
15259
@amitranjan2311 said:
@vbhvgupta 3:2
explain kar do.
15260
@vbhvgupta said:
Q4
3:2

[ n*d1+ (2*a1-d1)] / [n*d2+ (2*a2-d2) ] = 7n-17 / 4n +16

so, d1 = 7 , d2 = 4
2*a1- d1 =-17
2*a2 -d2 = 16

thus a1 = -5
a2 = 10

T21 (1 ) / T21(2) = 135/90

15261
@Tiws said:
find remainder when S is devided by 13.S =[ 5+ 5^2 +......+ 5^ 55]a.0b.1c.7d.12
12
15262

@vbhvgupta said:
explain kar do.

ratio of the sum of n terms is 7n-17 : 4n+16 --->(2a1+(n-1)d1)/(2a2+(n-1)d2)=7n-17/4n+16
now we have to find the ratio of 21 terms then ratio will be (a1+20d1)/(a2+20d2)
if we put n=41 in the previous expression we will get (2a1+40d1)/(2a2+40d2) =our requirement
so put n-41 in the 7n-17/4n+16 you will get 270/180