CAT 2006 Question Paper

The area of triangle ABC is 1/2 * $$\sqrt2$$ * $$\sqrt2$$ = 1

Area of semi-circle ABC = $$\pi/2$$

So, area of circle outside the square = $$\pi/2$$ - 1 = ($$\pi$$ -2)/2

So, area of circle inside the sheet = $$\pi$$ - ($$\pi/2$$ - 1) = 1 + $$\pi/2$$

Area of original square = 2*2 = 4

So, area of the sheet after punching = 4 - 1 - $$\pi/2$$ = 3 - $$\pi/2$$

So, proportion of sheet that remains after punching = (3 - $$\pi/2$$)/4 = (6 - $$\pi$$)/8

The area of triangle ABC is 1/2 * $$\sqrt2$$ * $$\sqrt2$$ = 1

Area of semi-circle ABC = $$\pi/2$$

So, area of circle outside the square = $$\pi/2$$ - 1 = ($$\pi$$ -2)/2

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
So the charge for the excess of (20-$$\frac{2x}{3}$$) = k(20-$$\frac{2x}{3}$$) = 1200  ....(1)
Also, the charge for the excess of 60-x = k(60-x) = 5400 .....(2)
Dividing (1) by (2), we get
=>$$\frac{\left(60-2x\right)}{3\times\ \left(60-x\right)}=\frac{1200}{5400}$$ 
Solving this, x = 15 kg
So, Praja's luggage = 35 kg

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
=> (60-2x)/3*(60-x) = 1200/5400
Solving this, x = 15 kg

$$2^p$$ is always positive

$$x^2$$ is always non negative.

$$1/\sqrt{-x}$$ is always positive.

$$\frac{1}{x}$$ is negative when x is negative.

In this case, x is negative => $$\frac{1}{x}$$ is smallest.

Make the power equal and compare the denominators.

$$2^{1/2}$$ can be written as $$64^{1/12}$$

$$3^{1/3}$$ can be written as $$81^{1/12}$$

$$4^{1/4}$$ can be written as $$64^{1/12}$$

$$6^{1/6}$$ can be written as $$36^{1/12}$$

Among these, $$81^{1/12}$$ is the greatest => $$3^{1/3}$$ is the greatest.

a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3

Similarly, b/e and c/f are 3 and 3/8 respectively.

b/e = b/c*c/d*d/e = 3

c/f = c/d*d/e*e/f = 3/8

=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8

The area of the four walls is length*height*2 + breadth*height*2
Initial area = 3*1*2 + 2*1*2 = 10
Final area = 6*1/2*2 + 1*1/2*2 = 7
So, the area decreased by 30%

substituting 3,4...53 in the given function, we get
$$t_3 = \frac{3}{5}$$
$$t_4 = \frac{4}{6}$$
$$t_5 = \frac{5}{7}$$
$$t_6 = \frac{6}{8}$$

Multiplying the values, we get $$\frac{3}{5}*\frac{4}{6}*\frac{5}{7}*....\frac{52}{54}*\frac{53}{55} $$ which ultimately after cancellations give $$\frac{3*4}{54*55}=\frac{2}{495}$$

Let x be in the front row.

So no. of children in next rows will be x-3,x-6,x-9,x-12,x-15,x-18,x-21....

Suppose there are 6 rows, then the sum is equal to x + x-3 + x-6 + x-9 + x-12 + x-15 = 6x - 45

This sum is equal to 630.

=> 6x - 45 = 630 => 6x = 585

Here, x is not an integer.

Hence, there cannot be 6 rows.