CAT 2003 Question Paper (Leaked)

 x, y and z in the hundred's, ten's and unit's place. So y should start from 2

If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.

Also if y=3 , possible values of x=1,2 and z=0,1,2.

Here 6 three digit nos. possible .

Similarly for next cases would be 3*4=12,4*5=20,5*6=30,.....,8*9=72 . Adding all we get 240 cases. 

Consider the triangle APB.  $$\angle$$P = 60 and AP = BP => APB is an equilateral triangle.

Hence AP = $$b$$    ...(1)

$$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$$

$$\text{AC}^2 = b^2 + b^2$$  =>  $$\text{AC} = \sqrt{2}\times b$$  =>  $$\text{AO} = \dfrac{\text{AC}}{2} = \dfrac{b}{\sqrt{2}}$$

$$\text{AP}^2 = \text{AO}^2 + \text{OP}^2$$

$$b^2 = \dfrac{b^2}{2} + h^2$$    ...From (1)

$$2h^2 = b^2$$

Hence, option B is the correct answer.

Let BD = x .Semi-perimeter of triangle ABC = 12. Now by herons formula area of ABC is 24. Also Area = 0.5*x*10 . We get x = 24/5 .  AP = 6/5  and CQ = 16/5 . Hence the required ratio is 3:8. 

Let BQ = z , QD = y , PQ = x.

From similar triangles PQD and ABD we have

(y/x) = (z+y)/3 .

Also from  similar triangles PQB and CBD we have

(z/x) = z+y .

Solving we get z = 3*y.

Now required ratio is (z+y)/z.

We get eual to 4/3 which is equal to 1:0.75. 

For the given problem ,

$$\sum {n(n+1)/2} = 8436 $$ which is 

$$\sum {n^2/2} + \sum{n/2} = 8436 $$ which is equal to

n*(n+1)(2n+1)/12 + n*(n+1)/4 = 8436 , solving we get n=36.

Solving the equation might be lengthy. you can substitute the values in the options to arrive at the answer. 

Let the numbers be $$a_1,a_2....a_n.$$

Since the numbers are positive,

$$AM\geq GM$$

$$\frac{a_1+a_2+a_3....+a_n}{n}\geq (a_1*a_2....*a_n)^{1/n}$$

$$a_1+a_2+a_3....+a_n \geq n$$

$$2 log (2^x - 5) = log 2 + log (2^x - 7/2)$$
Let $$2^x = t$$
=> $$(t-5)^2 = 2(t-7/2)$$
=> $$t^2 + 25 - 10t = 2t - 7$$
=> $$t^2 - 12t + 32 = 0$$
=> t = 8, 4
Therefore, x = 2 or 3, but $$2^x$$ > 5, so x = 3

Since Angle BOC = Angle BCO = y.

Angle OBC = 180-2y .

Hence Angle ABO =  z = 2y = Angle OAB.

Now since x is exterior angle of triangle AOC .

We have x = z + y = 3y.

Hence option A. 

As seen in the fig. we have a right angled triangle with sides  r ,r-10 , r-20.

Using pythagoras we have $$r^2 = (r-10)^2 + (r-20)^2$$.

Solving the equation, we get r = 10 or 50.

But 10 is not possible , so r = 50.

Hence radius is 50.

We know $$w = vz /u$$ so taking max value of u and min value of v and z to get min value of w which is -4.

Similarly taking min value of u and max value of v and z to get max value of w which is 4

Take v = 1, z = -2 and u = -0.5, we get w = 4

Take v = -1, z = -2 and  u = -0.5, we get w = -4