Official Quant thread for CAT 2013

@ScareCrow28 said:
Just check those values and see whether a and b are coming out to be integers and satisfying n^2 = a + b and n^3 = a^2 + b^2 or not.Ans is 3 values
I can not find any method to proceed further
Just by hit and trial
n=a=b=2
n=1 a=0 b=1
n=1 a=1 b=0
@gnehagarg said:
I can not find any method to proceed furtherJust by hit and trialn=a=b=2n=1 a=0 b=1n=1 a=1 b=0
Though underestimated; Hit & Trial is, in fact, a legitimate method
You are correct!

@ScareCrow28 said:
According to your solution, the hands of the clock meet 24 times in a day. But, that is not the case. The hands , in fact, meet 22 times in a day. So , I think that is the flaw in your solution.
@grkkrg said:
the minute hand and the hour hand don't meet every 60 minutes.. they meet every 65 5/11 minutes.. So it will be (60*24)/(65 5/11 * 60/70 * 24) = 25.667
No no... actually I was trying to calculate relative speed and then time taken for first meeting considering 60 points on dial 😁

Finally got the mistake. What I did was that I calculated wrong relative speed of MH wrt HH. It should have been 70-5*70/60
and now for first meeting time taken= 60/(70-5*70/60) =3600/3850
hence in 24 hrs, number of meetings=24*3850/3600=25.67

Thanks btw. I will learn by your help. I am novice
@sparklingaubade said:
@TootaHuaDilNopesThere are three watches €“ W1, W2 and W3. Once an alarm goes off in W1, it rings continuously for20 seconds, then pauses, then starts ringing again after 2 minutes, and so on. The respectivevalues for W2 are 50 seconds and 5 minutes, and for W3 are 1 minute and 6 minutes. An alarm isset in each of the three watches for 06:00 am. At what time after 06:00 am will the three alarms gooff simultaneously for the first time?(a) 06:35 am (b) 06:42 am (c) 06:30 am (d) None of these
W1=20+120=140 sec
W2=50+300=350 sec
W3=60+360=420 sec

LCM(140,350,420)=2100sec(35min)

6:35 am
The domain of the function f(x) = log7{log3(log5(20x– x^2 – 91))} ?
@gnehagarg said:
The domain of the function f(x) = log7{log3(log5(20x €“ x^2 €“ 91))} ?
Domain : (8,12) ??
@gnehagarg said:
The domain of the function f(x) = log7{log3(log5(20x– x^2 – 91))} ?
log3(log5(20x– x^2 – 91)) > 0

log5(20x– x^2 – 91) > 1

20x– x^2 – 91 > 5

=> x^2 - 20x + 96
(x - 12)*(x-8)

Domain is (8,12) ?
Three people A, B and C were all born on 1st February but in different years. How old was A when B was twice as old as C?

When A was four years old, B was three times as old as C.

When C was twice as old as A, B was five times as old as A.
@gnehagarg said:
The domain of the function f(x) = log7{log3(log5(20x– x^2 – 91))} ?
8
log3(log5(20x– x^2 – 91))
> 0
log5(20x– x^2 – 91) > 1
20x– x^2 – 91
> 5
x^2 - 20x + 96 (x - 12)(x - 8)
x 8
OR
x 12
@gnehagarg said:
The domain of the function f(x) = log7{log3(log5(20x €“ x^2 €“ 91))} ?
(8,12)??

log3(log5(-x^2+20x-91))>0
x^2-20x+96
x=8,12

@grkkrg

alarm is set at 6.00 am that means it will not ring for 120, 300, 360 sec respectively .
then it will ring for 20 sec, 50 sec, 60 sec.. continuously. we have to find the first sec when all of three are buzzing. right ???

120____20_________120________20
300_______50__________300__________50

....................................

Why you apply LCM here.

@chillfactor @ScareCrow28 @grkkrg sir help


@shashankbapat23 said:
Three people A, B and C were all born on 1st February but in different years. How old was A when B was twice as old as C?When A was four years old, B was three times as old as C.When C was twice as old as A, B was five times as old as A.
16 years old?
How many distinct regular polygons can be constructed by using 24 sticks, each of length 4.75 cm,
such that all 24 sticks are used together and no two sticks overlap?
(a) 8 (b) 6 (c) 7 (d) 12

@shashankbapat23 said:
Three people A, B and C were all born on 1st February but in different years. How old was A when B was twice as old as C?When A was four years old, B was three times as old as C.When C was twice as old as A, B was five times as old as A.
16 years?
@ScareCrow28 You are on fire today!!!
@sparklingaubade said:
@grkkrgalarm is set at 6.00 am that means it will not ring for 120, 300, 360 sec respectively . then it will ring for 20 sec, 50 sec, 60 sec.. continuously. we have to find the first sec when all of three are buzzing. right ??? 120____20_________120________20300_______50__________300__________50....................................Why you apply LCM here. @chillfactor@ScareCrow28@grkkrgsir help
It's similar to 3 people running on a circular track at different speeds.
If they start at a common point then they meet at the same point after the LCM of their times.

take examples of (1,1) (1,2) and (1,3)
W1 _ W1 _ W1 _ W1 _ W1 _ W1 _ W1
W2 _ _ W2 _ _ W2 _ _ W2 _ _ W2
W3 _ _ _ W3 _ _ _ W3 _ _ _ W3
As you can see, they buzz together after LCM(1+1, 1+2, 1+3) = LCM(2,3,4) = 12


PS: This is how LCM is defined :)

@sparklingaubade said:
@grkkrgalarm is set at 6.00 am that means it will not ring for 120, 300, 360 sec respectively . then it will ring for 20 sec, 50 sec, 60 sec.. continuously. we have to find the first sec when all of three are buzzing. right ??? 120____20_________120________20300_______50__________300__________50....................................Why you apply LCM here. @chillfactor@ScareCrow28@grkkrgsir help
Suppose a person A comes only on every second day of a month and a person B comes every 3rd day of a month and a person C comes evry 5th day of a month to the Office.

When will A and B meet ??
A comes every second day, B comes every 3rd day, hence they will meet every day which is a multiple of both 3 and 2; which is nothing but LCM of 3 and 2.

Similarly when will they all meet??
They all meet on days which is a multiple of 2, 3 and 5; i.e nothing but LCM of (2,3,5)

Similarly here also, The alarms will ring simultaneously for the first time after LCM(T1, T2, T3) minutes/seconds.
Hope you understand now/
@sparklingaubade said:
How many distinct regular polygons can be constructed by using 24 sticks, each of length 4.75 cm,such that all 24 sticks are used together and no two sticks overlap?(a) 8 (b) 6 (c) 7 (d) 12
factors of 24= 1,2,3,4,6,8,12,24

but we can not use 12,24 no of sticks we will not get polygon..

so 6 ans
@sparklingaubade said:
How many distinct regular polygons can be constructed by using 24 sticks, each of length 4.75 cm,such that all 24 sticks are used together and no two sticks overlap?(a) 8 (b) 6 (c) 7 (d) 12
Since regular polygons are to be made, hence we consider factors of 24
24 = 2^3 * 3
No of factors = 4*2 = 8 (including 12 and 24)
no polygon of 12 or 24 length of a side can be made
Hence total polygons that can be formed are : 8-2 = 6 ..??
@soumitrabengeri said:
16 years?@ScareCrow28 You are on fire today!!!
Are sir kripa hai apki
The number 165 is split into three parts €“ two perfect squares and one perfect cube. If all the three parts are positive and distinct,
find the three parts.
A. One of the perfect squares is 36.
B. The perfect cube is the greatest of the three parts.

Each question below is followed by two statements, A and B. Answer each question using the
following instructions:
Choose 1 if the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone.
Choose 2 if the question can be answered by using either statement alone.
Choose 3 if the question can be answered by using both the statements together, but cannot be answered by using either statement alone.
Choose 4 if the question cannot be answered even by using both the statements together.