The Official CAT 2007 Quant Thread

[warrior]

Originally Posted by vineet.nitd

Do this ..

Q> 3 identical blue balls , 4 identical red balls and 5 identical white balls are to be arranged in a row , find the number of ways that this can be done if all the balls with the same colour do not form a single block .

total ways t = 12!/3!4!5!

but as blockas are not required

blocks of blue b = 10!/4!5!
blocks of red r = 9!/3!5!

blocks of white w = 8!/3!4!

but the above we have counted blocks whch occur together twice

so block of blue and red br = 7!/5!

block of blue and white bw = 61/4!

block of red and white rw = 5!/3!

also block of all three bwr = 3!

so ans is t- (b+r+w- br-bw-rw+bwr)

[vineet.nitd]

Originally Posted by doomsayer

3 balls arranged as a block = 10!/(5!* 4!) =a
4"" = 9!/ 5!*3! =b
5 " " = 8!/ 3!*4! =c

3,4 balls = 7!/5!=d
3,5 balls= 6!/4!=e
4,5 = 5!/3! =f

3,4,5 = 3! =g

total ways in which thwy can arrage as single block=z= a+b+c -d-e-f +g
total ways = 12! / 3! * 4! *5!


total ways not as single block =12! / 3! * 4! *5! - z
little lazy to do the calculation

Excellent .. ...This is the right answer and utilises the inclusion-exclusion principle

[doomsayer]

Originally Posted by doomsayer

Quote:
Originally Posted by vineet.nitd (The Official CAT 2007 Quant Thread)
ok ..lemmee explain the concept behind this ..
ac+bc+ad+bd+ae+be/6 >= (a^3b^3c^2d^2e^2)^1/6

So basically u ve got to maximise Z=a^3b^3c^2d^2e^2.
Z=3^3.3^3.2^2.2^2.2^2(a/3)^3(b/3)^3(c/2)^2(d/2)^2(e/2)^2


(a/3)^3(b/3)^3(c/2)^2(d/2)^2(e/2)^2 will be maximum when all the factors are equal ...

Hope u got the funda

here we have to find max value of ac+bc+ad+bd+ae+be/6 ..
but in this method ur finding out the min value...right????

i didnt get the concept...plz elaborate

krish i got ur concept....but didnt understand vineet's aswer completly..
it will be helpful if somebody throw light on the concept..

[vineet.nitd]

Originally Posted by warrior

total ways t = 12!/3!4!5!

but as blockas are not required

blocks of blue b = 10!/4!5!
blocks of red r = 9!/3!5!

blocks of white w = 8!/3!4!

but the above we have counted blocks whch occur together twice

so block of blue and red br = 7!/5!

block of blue and white bw = 61/4!

block of red and white rw = 5!/3!

also block of all three bwr = 3!

so ans is t- (b+r+w- br-bw-rw+bwr)

Good warrior ..now try this one ...

#How many arrangements of the digits 0,1,2,...9 are there that do not end with 8 and do not begin with 3 ?

[vineet.nitd]

Originally Posted by doomsayer

krish i got ur concept....but didnt understand vineet's aswer completly..
it will be helpful if somebody throw light on the concept..

What i meant is that for
ac+bc+ad+bd+ae+be/6 >= (a^3b^3c^2d^2e^2)^1/6
and to have max ac+bc+ad+bd+ae+be , we are trying to maximise a^3b^3c^2d^2e^2 also coz the larger this value is the larger is our original expr..

and that will be max only when each of a/3,b/3,c/2,d/2and e/2 are equal and then i have used componendo to get the values of a,b,c,d,e which ensure max value of the original expression

[doomsayer]

Originally Posted by vineet.nitd

Good warrior ..now try this one ...

#How many arrangements of the digits 0,1,2,...9 are there that do not end with 8 and do not begin with 3 ?

for a ten digit number it's 8*9*10^8
similarly we have to find it for 9 digit number, 8digit etc...

[CAT 2007]

Here goes set 2:

1)Given f(x) is a function satisfying f(x + y) = f(x) X f(y) for every real value of x and y. If f(1) = 3 and f(1) + f(2) + f(3) + …….+ f(n) = 1092, then find the value of n.


a) 5 b) 6 c) 7 d) 8

2)P denotes the sum of two three digit numbers such that P is 606. How many of the 10 digits from 0 to 9 cannot appear as the last digit of the product of these two three digit numbers?


a) 4 b) 3 c) 5 d) 2

3)ABCD is a parallelogram. A point E is selected on AD such that BE is 2(square root of)6. Also 2AE = 2AB = BD = 8 units and 5CF = BC where F is a point on BC. Find the ratio of the quadrilateral DEFC to the area of parallelogram ABCD.


a) 7/20 b) 5/16 c) 2/5 d) 3/10

4)Three persons A, B and C start running clockwise and simultaneously from a point on the circular track of radius (105/2pi) km. The ratio of the speeds of A, B and C is 3 : 5 : 7. What is the difference between the distance traveled by A and that of C when all the three meet for the first time at a point on the track?


a) 105 km b) 131.25 km c) 262.5 km d) None of these

5) If x + y +z = 1 and x, y and z are non-negative numbers, what is the minimum value of the expression (2 – x + yz)^(-1) + (2 – y +xz)^(-1) + (2 – z +xy)^(-1)


a) 27/16 b) 25/9 c) 38/25 d) 3/2

[vineet.nitd]

Originally Posted by CAT 2007

Here goes set 2:

1)Given f(x) is a function satisfying f(x + y) = f(x) X f(y) for every real value of x and y. If f(1) = 3 and f(1) + f(2) + f(3) + …….+ f(n) = 1092, then find the value of n.


a) 5 b) 6 c) 7 d) 8

3/2(3^n -1)=1092
3^n=729 =>n=6

[krsh.vik]

Originally Posted by CAT 2007

Here goes set 2:

1)Given f(x) is a function satisfying f(x + y) = f(x) X f(y) for every real value of x and y. If f(1) = 3 and f(1) + f(2) + f(3) + …….+ f(n) = 1092, then find the value of n.


a) 5 b) 6 c) 7 d) 8
4)Three persons A, B and C start running clockwise and simultaneously from a point on the circular track of radius (105/2pi) km. The ratio of the speeds of A, B and C is 3 : 5 : 7. What is the difference between the distance traveled by A and that of C when all the three meet for the first time at a point on the track?

a) 105 km b) 131.25 km c) 262.5 km d) None of these

1) B 3+3^2+....3^6
4) let the speeds be 3 5 7
A(n rounds) and B(n-1 rounds) will meet after A makes 2.5 rounds
A(n rounds) and C(n-1 rounds) will meet after A makes 1.75 rounds
all of them will meet after A makes 17.5 rounds
hence C will make 0.75*10 7.5 rounds
difference 10 rounds distance 10*105 not in the options

[vineet.nitd]

Originally Posted by CAT 2007

Here goes set 2:


2)P denotes the sum of two three digit numbers such that P is 606. How many of the 10 digits from 0 to 9 cannot appear as the last digit of the product of these two three digit numbers?


a) 4 b) 3 c) 5 d) 2

1,2,6,7 wont be there ..hence ans is 4 option (a)