Official Quant thread for CAT 2013

Find the number of ternary sequences of length 4 where 0 is not followed by 1 and 1 is not followed by 2. (Ternary sequence of length n is a sequence having n terms and each term is either 0,1 or 2)
a) 47 b)37 c) 72 d)54 e)44

Total ternary sequences possible = 3^4 = 81 {There are 4 positions and 3 choices for each position}

Invalid sequences:
Sequences which contain 01.
The location for 01 can be selected in 3C2 or 3 ways. {01xx, x01x, xx01}
The other two digits can be selected in 3*3 or 9 ways. {Filling up the x}
No. of invalid sequences = 3*9 = 27

Similarly, no. of invalid sequences which contain 12 = 27

Some of the invalid sequences have been counted twice. They are:
Type 1 - 0101, 1212, 0112, 1201 - There will be 4 of these
Type 2 - 012x & x120 - There will be 6 of these.

Total invalid sequences = 27 + 27 - (4 + 6) = 44

Total valid sequences = 81 - 44 = 37. Option B

cc: @sujamait
@sujamait - I don't think I have seen this question before. Where did you get this from?
y=rt(1+2 rt(1+3 rt(1+ ....... ))) ?
@sujamait said:
y=rt(1+2 rt(1+3 rt(1+ ....... ))) ?
Oh God ! This is Ramanujan's problem from his notebook ! :splat: ..

I will not even try :splat:
@YouMadFellow said:
Oh God ! This is Ramanujan's problem from his notebook ! .. I will not even try
But the answer is 3 :D

@ashishpai2001 said:
But the answer is 3
The simplicity of the answer doesn't make this an easy question :splat:

Share the solution please 😃
@YouMadFellow said:
The simplicity of the answer doesn't make this an easy question Share the solution please
Very true.

1^1 * 2^2 * 3*3 * 4^4......... 100^100. Find the total numbers of ZERO at the end ? Kindy share the approach as well.. Thanks!


@gupanki2 said:
1^1 * 2^2 * 3*3 * 4^4......... 100^100. Find the total numbers of ZERO at the end ? Kindy share the approach as well.. Thanks!
count the number of 5's in expression.

5^5*5^10*5^15.....
= 5^(5+10+15+20......100)
= 5^{105*20/2}=5^1050

but in 25, 50, 75, 100 5 comes 2 times, so we have count those powers twice.
hence, No. of zeros at the end=number of 5's=1050+25+50+75+100=1300
@YouMadFellow said:
Oh God ! This is Ramanujan's problem from his notebook ! .. I will not even try

root x + y = 7
x + root y = 11
find x,y...straight from google doodle...


@chillfactor
y=rt(1+2 rt(1+3 rt(1+ ....... ))) ?
@sujamait said:
root x + y = 7x + root y = 11find x,y...straight from google doodle...


x + y + rt x + rt y = 18

now only two numbers satisfy x = 9; y = 4 / x = 4 ; y = 9
from first equation y = 4 x = 9
so x = 9 y = 4

@YouMadFellow

vinay fires two bullets from the same place at an interval of 12 minutes but raju sitting in a train approaching the place hears the second report 11 minutes 30 seconds after the first.What is the approximate speed of the train ?(is sound travels at the speed of 330 metres per second)

I am sorry for again troubling u for the same question i absolutely got the left hand side i.e 3960 as a distance.now on the right hand side why it is (330+v)*11.5 and not v*0.5 ?? I know this is relative velocity question,but after i have got the distance travelled by train between the first and second sound.

now to travel this distance(3960) train required 30 seconds,coz he heard the second sound 30 seconds after the first. the distance 3960 was travelled by train alone right,so why are we adding velocities?

why m i not doing just 3960=v*0.5 ? I said that i understood that day but when i tried again,was not able to convince myself
y=rt(1+2 rt(1+3 rt(1+ ....... ))) ?

Then y = 3
@floamiya said:
y=rt(1+2 rt(1+3 rt(1+ ....... ))) ?Then y = 3
approach ?
@sujamait




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@sujamait said:
root x + y = 7x + root y = 11find x,y...straight from google doodle...
This one doesn't seem so tough

_/x =a, _/y = b

a + b^2 = 7
a^2 + b = 11

substract => (a -b) *( a + b - 1) = 4 => (1*4, 2*2, and negatives too)

So, (a - b) = 1, (a + b - 1) = 4 => a = 3,b = 2

=> x = 9, y= 4 ?

Rest all either gives negative solutions, or the solutions don't satisfy the original equation
@nole said:
@YouMadFellowvinay fires two bullets from the same place at an interval of 12 minutes but raju sitting in a train approaching the place hears the second report 11 minutes 30 seconds after the first.What is the approximate speed of the train ?(is sound travels at the speed of 330 metres per second)I am sorry for again troubling u for the same question i absolutely got the left hand side i.e 3960 as a distance.now on the right hand side why it is (330+v)*11.5 and not v*0.5 ?? I know this is relative velocity question,but after i have got the distance travelled by train between the first and second sound.now to travel this distance(3960) train required 30 seconds,coz he heard the second sound 30 seconds after the first. the distance 3960 was travelled by train alone right,so why are we adding velocities?why m i not doing just 3960=v*0.5 ? I said that i understood that day but when i tried again,was not able to convince myself.

The bold statement statement is wrong. He heard it after 11 minutes 30 seconds

If the person on train was not moving with respect to ground, it would have taken 12 minutes.
Find the coefficient of x^7 in the expansion of (1+3x-2x^3)^10.
PS:As it is not only the matter of this question,I want to learn the right way to approach these type of questions
Except basic expansion formula for binomial expressions I am unable to recall anything,so please share your approach .
@19rsb said:
Find the coefficient of x^7 in the expansion of (1+3x-2x^3)^10.PS:As it is not only the matter of this question,I want to learn the right way to approach these type of questionsExcept basic expansion formula for binomial expressions I am unable to recall anything,so please share your approach .
Although there may be many approaches, but it surely varies. Here I will try to use the general term formula of the binomial expression. The basic idea is to limit the expression in such a way so that we reach a point where the coefficients of x become higher than what is asked in the question.

[(1 + 3x) - (2x^3) ]^10

T(r) = 10C(r) * (1+ 3x)^(10 - r) * (-2x^3)^r

Now, to get power of x as 7, only possible value of r = 0,1,2 ..

r= 0 => 10C0*(1 + 3x)^10 => (x^7 in this) = 10C7*(3)^7
r= 1 => 10C1*(1 + 3x)^9 *(-2) => (x^4 in this) = (10C1)*(-2)*(9C4)*(3^4)
r= 2 => 10C2*(1 + 3x)^8 *( 4) => (x in this) = (10C2)*(4)*(8C1)*(3)

The answer should be the sum of the above three. :splat: ..