CAT 2002 Question Paper

If $$f(x) = \log \frac{(1+x)}{(1-x)}$$ then $$f(y) = \log \frac{(1+y)}{(1-y)}$$

Also Log (A*B)= Log A + Log B 

f(x)+f(y) = $$ \log \frac{(1+x)(1+y)}{(1-x)(1-y)}$$ 

=$$\log\frac{\left(1+xy\ +x\ +y\right)}{\left(1+xy-x-y\right)}$$ 

Dividing numberator and denominator by (1+xy)

$$\log\frac{\frac{\left(1+xy\ +x\ +y\right)}{1+xy}}{\frac{\left(1+xy-x-y\right)}{1+xy}}$$

 =$$\log\frac{\frac{1+xy\ }{1+xy}+\frac{\left(x+y\right)}{1+xy}}{\frac{1+xy\ }{1+xy}-\frac{\left(x+y\right)}{1+xy}}$$

= $$\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$$

Hence option B.

Length of the rope tied to each horse = 7 m.
Total area of the portion that the horses can graze = 4* $$\pi 7^2$$/4 = 49$$\pi$$
Area of the circular pond = 20 $$m^2$$
So, area left ungrazed = $$14^2 - 20 - 49\pi m^2$$ = 22 $$m^2$$ (approx)

The weights are at distances of 0 m, 2 m, 4 m, 6 m and 8 m from X.

Let us first keep all the weights at a distance of 8 m from X. This would be 8 + 6*2 + 4*2 + 2*2 = 32 m.

Now from the point where all the weights are kept is at a distance of 92 m from Y. So total distance required = 184+184+184+184+92 = 828 m.

So in all 860 m.

Let AB = BE = x

Area of triangle ABE = $$\dfrac{x^2}{2}$$ = 7; we get x = $$\sqrt{14}$$

So we have side BC = 4*$$\sqrt{14}$$

Now area is AB*BC = 14 *4 = 56 $$cm^2$$

The triangle we have is :

The length of three sides is $$\sqrt 2, \sqrt 2$$ and $$2$$.
This is a right-angled triangle.
Hence, it's area equals $$1/2 * \sqrt 2 * \sqrt 2 = 1$$
So, the correct answer is b)
Alternate Approach :
Area of triangle = $$\frac{1}{2}\times\ base\times\ height$$
So we get $$\frac{1}{2}\times2\times\ 1$$=1 square units

Let x be the shorter side and y be the longer one. The shortcut route would be of length $$\sqrt{x^2+y^2}$$. According to given condition we know that (x+y)- $$\sqrt{x^2+y^2}$$ = y/2 . Solving, we get 1=(x/y)+(1/4) 

=> x : y = 3 : 4. Hence option D is the correct answer.

Let the speed of an express train be 4x, normal train be 2x and passenger train be x.
Let the distance between the 2 stations be D. 
Since there is only 1 railway track, train N must reach station B before train S leaves. 
Therefore, D/4x + D/x = 60 
5D/4x = 60
D/x = 48

Train N leaves 20 minutes late. Therefore, the 2 trains must have covered the distance within 40 minutes on this particular day. 
Train N doubles its speed. Therefore, speed of train N will be 8x. Let the new speed of the passenger train be y.

D/8x  + D/y = 40
48/8 + D/y = 40
D/y = 34.

Speed of super fast train = D/8x = 6
Speed of passenger train = D/y = 34

Ratio of the speeds = 6/34 = 3/17.
The ratio is approximately equal to 1:6. Therefore, option D is the right answer.

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5

It takes 1 min to load and unload the patient.

Hence, total time = 5 + 32.5 + 1 = 38.5 mins

So, the doctor would get 1.5 min to attend to the patient.

Consider the options:

24: (Square of sum of digits - the number) = 36 - 24 = 12

54: (Square of sum of digits - the number) = 81 - 54 = 27

34: (Square of sum of digits - the number) = 49 - 34 = 15

45: (Square of sum of digits - the number) = 81 - 45 = 36

So, option b) is the correct answer.

Let x = 1, so, $$X_0$$ = 1
$$X_1$$ = -1
$$X_2$$ = -1
$$X_3$$ = 1
$$X_4$$ = 1
$$X_5$$ = -1
$$X_6$$ = -1
So, $$X_n$$ need not be positive when n is even, $$X_n$$ need not be positive when n is odd, $$X_n$$ need not be negative when n is even. So, none of the first three options are correct.