Official Quant thread for CAT 2013

@pavimai said:
A welder can finish job A and job C in 10 days and 20 days respectively.A Blacksmith can finish job B and job c in 15 days and 10 days respectively.On the first day ,two welders begin work on job A and a day later three blacksmiths begin work on job B .what is the least time required to complete all the three jobs???1.33/42.37/43.31/44.none of these
33/4 ?
@bullseyes said:
Log(1/2+1/5+1/9+1/14+........)^2 = 2 base of log is (x) find x
X^2=(1/2+1/5+1/9+1/14+---------)^2
X=(1/2+1/5+1/9+1/14+--------)

General term =1/(ax^2+bx+c)
a=1/2
b=3/2
c=0

General term =1/(1/2*a^2+3/2x)
=2/(x)(x+3)
Summation of 2/(x)(x+3)
2[1/(1*4)+1/(2*5)+1/(3*6)+--------]
2/3[1-1/4+1/2-1/5+1/3-1/6+--------]
2/3[1+1/2+1/3+1/4+1/5+1/6+1/7+--------]-[1/4+1/5+1/6+1/7+1/8+------]
2/3[1+1/2+1/3]
2/3[11/6]
11/9

@pavimai said:
A welder can finish job A and job C in 10 days and 20 days respectively.A Blacksmith can finish job B and job c in 15 days and 10 days respectively.On the first day ,two welders begin work on job A and a day later three blacksmiths begin work on job B .what is the least time required to complete all the three jobs???1.33/42.37/43.31/44.none of these
A - 5 days
B - 5 days + 1

so 2 welders worked on C for 1 extra day - 2/20

work left = 1-2/20 = 9/10

work rate for C = 2/20+3/10=4/10=2/5 / day

for 9/10 C - 9/10*5/2 = 9/4

so time required = 6+9/4 = 33/4
@chillfactor said:
Read the question carefully !a, b, c are natural numbers such that a²+ b² = c², what can be said about a*b*ci) It will always be a multiple of 30ii) It will always be a multiple of 60iii) It will always be a multiple of 120iv) None of the above@anytomdickandhary Sir, its good to see that you are active here
ii) It will always be a multiple of 60 ?
@gnehagarg said:
X^2=(1/2+1/5+1/9+1/14+---------)^2X=(1/2+1/5+1/9+1/14+--------)General term =1/(ax^2+bx+c)a=1/2b=3/2c=0General term =1/(1/2*a^2+3/2x) =2/(x)(x+3) Summation of 2/(x)(x+3)2[1/(1*4)+1/(2*5)+1/(3*6)+--------]2/3[1-1/4+1/2-1/5+1/3-1/6+--------]2/3[1+1/2+1/3+1/4+1/5+1/6+1/7+--------]-[1/4+1/5+1/6+1/7+1/8+------]2/3[1+1/2+1/3]2/3[11/6]11/9
explain why u think its a quadratic and not cubic equation
@chillfactor said:
m and n are natural numbers such that hcf(m, n) + lcm(m, n) = m + n, then which of the following is definitely correct?a) m = nb) One of m or n is multiple of the other onec) Both are correctd) None of these
c?
@chillfactor said:
m and n are natural numbers such that hcf(m, n) + lcm(m, n) = m + n, then which of the following is definitely correct?a) m = nb) One of m or n is multiple of the other onec) Both are correctd) None of these
HCF=k
LCM=mn/k

k+mn/k=m+n
k^2-k(m+n)+mn=0
k=m or n

HCF= m or n
LCM=n or m

number m=mx
number n=my
x and y are coprime to each other

number m=nx
number n=ny
x and y are coprime to each other
m/n=x/y
When x and y are odd prime numbers then m and n are not multiple of each other
So, b is not definitely true
m=kn
k can be any value
So,a is not definitely true

I think d is answer.


@chillfactor said:
Read the question carefully !a, b, c are natural numbers such that + b² = c², what can be said about a*b*ci) It will always be a multiple of 30 (but not always a multiple of 60)ii) It will always be a multiple of 60 (but not always a multiple of 120)iii) It will always be a multiple of 120iv) None of the above@anytomdickandhary Sir, its good to see that you are active here


shld be option 2

a pythogorean triplet has following property.

a= x² - y²
b= 2*x*y
c = x² + y²

so a*b*c = (x² - y²)(2*x*y)(x² + y²)

=> 2(x-y)(x+y)(x*y)(x² + y²)

will always be divisible by 60.
@chillfactor said:
Read the question carefully !a, b, c are natural numbers such that a²+ b² = c², what can be said about a*b*ci) It will always be a multiple of 30 (but not always a multiple of 60)ii) It will always be a multiple of 60 (but not always a multiple of 120)iii) It will always be a multiple of 120iv) None of the above@anytomdickandhary Sir, its good to see that you are active here
a,b,c can be 3,4,5
multiple of 60k

2nd 180,300-- are not multiple of 120
@chillfactor c is the right answere lets take 7,7
then hcf 7 and lcm also 7
so hcf +lcm =14 and n+m=14
now if m=7,n=14
then hcf 7 and lcm 14 and hcf+lcm=21
and m+n=21........similarly everything goes
@chillfactor smalest pythogorean triplet of natural number ia
3,4,5
and 3*4*5=60
next 6,8,10
and 6*8*10=480
then
8,15,17
and 8*15*17=2040
all dividsible by 60 :P
@bullseyes said:
explain why u think its a quadratic and not cubic equation
2,5,9,14---
first take common difference 3,4,5
Step 2-> again take common diff 1,1
In step 2 we get same common difference as 1.Now,process is completed.
We reach step two so we take quadratic equation.
In case,process is completed at step 3 we take cubic equation
@chillfactor said:
Read the question carefully !a, b, c are natural numbers such that a²+ b² = c², what can be said about a*b*ci) It will always be a multiple of 30 (but not always a multiple of 60)ii) It will always be a multiple of 60 (but not always a multiple of 120)iii) It will always be a multiple of 120iv) None of the above@anytomdickandhary Sir, its good to see that you are active here
2?
Sanjay Mallaya daily removes 6 litres of Kingfisher(KF) from 60 litres cask of only KF and replaces it with 4 litres of Blender's Pride(BP) and 2 litres of Jack Daniels(JD).He continues doing this for 2 weeks.What is the final quantity of KF,BP and JD in the mixture at the end of two weeks?
@19rsb said:
Sanjay Mallaya daily removes 6 litres of Kingfisher(KF) from 60 litres cask of only KF and replaces it with 4 litres of Blender's Pride(BP) and 2 litres of Jack Daniels(JD).He continues doing this for 2 weeks.What is the final quantity of KF,BP and JD in the mixture at the end of two weeks?
KF => 60 => 54 => 54 * 54/60 => ... =>60 * 0.9^14
BP => 0 => 4 => 4*54/60 + 4 = 4(1 + 0.9) => ... => 4 (1 - 0.9^14)/0.1 = 40 * (1 - 0.9^14)
JD => 0 => 2 => 2*54/60 + 2 = 2(1 + 0.9) => ... => 20 * (1 - 0.9^14)

So final quantity
KF = 60 * 0.9^14
BP = 40 * (1 - 0.9^14)
JD = 20 * (1 - 0.9^14)
@bullseyes said:
u explained it so eloquently.. answer is 11/9 Four angles of trapezium are in AP. if one of the angles is 65, find the largest possible angle that trapezium can have ?
a-2d,a-d,a+d,a+2d=360
4a=360
a=90
a-d=65
d=25
40,65,115,140

@bullseyes said:
u explained it so eloquently.. answer is 11/9 Four angles of trapezium are in AP. if one of the angles is 65, find the largest possible angle that trapezium can have ?
a-3d+a-d+a+d+a+3d=360
a=90
a-d=65
d=25
a+3d=90+3*25=165??
@gnehagarg nahhh
360-65=295
now 295 in pattern
15,65,115,165

@gautam22 said:
4(1 + 0.9) => ... => 4 (1 - 0.9^14)/0.1....sir ye kaise kara hai?
ye ek GP banta hai
After n replacements the quantity will be 4(1 + 0.9 + 0.9^2 + 0.9^3 + ... + 0.9^(n-1))
a = 1
r = 0.9
sum = 4 * 1 * (1 - 0.9^n)/(1 - 0.9) = 40 * (1 - 0.9^n)
Here n = 14

PS : no sir
Carl walks over a railway-bridge. At the moment that he is just ten meters away from the middle of the bridge, he hears a train coming from behind. At that moment, the train, which travels at a speed of 90 km/h, is exactly as far away from the bridge as the bridge measures in length. Without hesitation, Carl rushes straight towards the train to get off the bridge. In this way, he misses the train by just four meters! If Carl would, however, have rushed exactly as fast in the other direction, the train would have hit him eight meters before the end of the bridge. What is the length of the railway-bridge?
In a class are twenty-six children. None of the children was born on February 29th. What is the probability that at least two children have their birthdays on the same day?