CAT 2008 Question Paper

The shortest route from A to B is via the diagonal OQ in the square P. One can travel from A to O in 4!/2!*2! ways. The shortest way from O to Q is through the diagonal only.From Q to B can be travelled in 6!/4!*2! ways.

The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) = 6*15 = 90

The shortest route from A to B is via the diagonal in the square P. A to the north-west corner of square P can be travelled in 4!/2!*2! ways. Number of ways to travel from the south-east corner of square P to B is 6!/4!*2!.
B to the north-east corner of D can be travelled in 6!/5! ways. From there to C can be travelled in 2 ways. There is 1 other way of travelling from B to C (along the perimeter of the field).
The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) * (6!/5! * 2 + 1)
= 6*15*13 = 1170

$$f(1)^2$$ = f(1) => f(1) = 1

f(2)*(f(1/2) = f(1) => 4x = 1

So, f(1/2) = 1/4

For seed (n) = 9, all the numbers below 500 must have a digit sum of 9.

These numbers are all divisible by 9.

So total number of numbers below 500 and divisible by 9 is 55.

Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x/(4*R), where R is circumradius.

Also Area = 0.5*x*3 .

Equating both, we have 3 = 17.5*9 / (2*R)

=> R = 26.25.

For obtuse-angles triangle, $$c^2 > a^2 + b^2$$ and c < a+b
If 15 is the greatest side, 8+x > 15 => x > 7 and $$225 > 64 + x^2$$ => $$x^2$$ < 161 => x <= 12
So, x = 8, 9, 10, 11, 12
If x is the greatest side, then 8 + 15 > x => x < 23
$$x^2 > 225 + 64 = 289$$ => x > 17
So, x = 18, 19, 20, 21, 22
So, the number of possibilities is 10

Consider side of square as 10 units. So HD=5 and HP=$$\frac{5}{\sqrt3}$$ . So now area of triangle HPD=$$\frac{12.5}{\sqrt3}$$. Also Area APD=Area BQC=2*AreaHPD=$$\frac{25}{\sqrt3}$$. So numerator of required answer is $$100-\frac{50}{\sqrt3}$$ and denominator as $$\frac{50}{\sqrt3}$$. Solving we get answer as $$2\sqrt{3}-1$$.

The power is 20.

20 has to be divided among a, b and c. This can be done in $$^{20+3-1}C_{3-1}$$ = $$^{22}C_2$$ = 231

Option a) is the correct answer.

In total Raju bets 6000Rs and ends up with no profit - no loss. So there are 3 possibilities.
1) White comes 2nd, Black comes 4th and Red comes 5th.
2) Black comes 1st, White comes 3rd and Red comes 4th or 5th.
3) Black comes 2nd, Red comes 3rd and White comes 4th or 5th.

So there can never be 3 horses between white and red according to above to possibilities. Hence option D cannot be true.

There are total 3 cases which satisfies the condition "no profit and no loss."

Case 1: White comes 2nd.(remaining two horses(red/black) come 4th/5th)

Profit from white horse = Final Amount - Initial Amount = 2000*3 - 2000 = 4000

Loss from Red and Black horse = 3000+1000 = 4000

Net profit = 4000-4000 = 0

Case 2: Black, Red come second, third respectively.(remaining one horse(white) comes 4th/5th)

Profit from Black = 1000*3-1000 = 2000

Profit from Red = 3000 - 3000 = 0

Loss from white = 2000

Net profit = 2000-2000 = 0

Case 3: black, white come first, third respectively.(remaining one horse(red) comes 4th/5th)

Profit from Black = 1000*4-1000 = 3000

Profit from White = 2000 - 2000 = 0

Loss from Red = 3000

Net Profit = 3000-3000=0

And it is mentioned that grey case 4th. ==> case 1 is wrong.(because, in that case red, black should come 4th,5th)

So option C cannot be true.