CAT 2004 Question Paper

Radius of the circles $$C_1, C_2, C_3,...$$ would be in GP with (R/4),(R/8),(R/16) and so on. Radii of circles are in the ratio 1:4. 

Ratio of unshaded region to the ratio of original circle = 1-$$\frac{Ratio\ of\ shaded\ region}{Ratio\ of\ original\ circle}$$

= 1-$$\frac{\pi r^2/16 + \pi r^2/64+.....}{\pi r^2}$$ = 1-$$\frac{1/16}{(1-1/4)}$$ = 1- 1/12 = $$\frac{11}{12}$$ = 11:12

According to given conditions the terms are 81.33, -19, -100.33, -81.33, 19, 100.33, 81.33,-19,.. Hence the series repeats after every 6 terms . Also summation of these 6 terms is 0. Hence summation is 60002 terms will we sum of first 2 terms which is 62.33.

Let radius be 1 units and p = 3.14 or $$\pi$$ . So circumference is $$2*\pi$$.

According to given condition distance covered in first 1/2 mins = $$\pi$$/2 km, distance covered in next 1 min = $$\pi$$/2 km, distance covered in next 2 mins = $$\pi/2$$ km and finally distance covered in next 4 minutes = $$\pi/2$$ km.

Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes.

Now time taken to cover $$\pi/2$$ is in GP.

For the second round the time taken is = 8+16+32+64 = 120

Ratio = 120/7.5 = 16

$$x^u = 256$$

Taking log to the base 2 on both the sides, 

$$u * \log_{2}{x} = \log_{2}{256}$$

=>$$[({\log_2 x})^2 - 6 {\log_2 x} + 12] * \log_{2}{x} = 8$$

$$(log_2 x)^3 - 6(log_2 x)^2 + 12log_2 x = 8$$

Let $$log_2 x = t$$

$$t^3 - 6t^2 +12t - 8 = 0$$

$$(t-2)^3 = 0$$

Therefore, $$log_2 x = 2$$ 

=> $$x = 4$$ is the only solution

Hence, option B is the correct answer.

Consider side of the cube as x.

So diagonal will be of length $$\sqrt{3}$$ * x.

Now if diagonals are side of equilateral triangle we get area = 3*$$\sqrt{3}*x^2$$ /4 .

Also in a triangle

4 * Area * R = Product of sides

4* 3*$$\sqrt{3}*x^2$$ /4 * R = .3*$$\sqrt{3}*x^3$$

R = x

The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35

In the triangle BEO, BE = $$\sqrt{4^2-2^2}$$ = $$2\sqrt3$$

==> BC = $$4\sqrt3$$

Using pythagoras we can find diagonal of small square shown in fig. = 2*$$\sqrt2$$ . Now this is equal to 2 + r +  $$\sqrt2$$ *r. Equating we get r = $$6- 4 \sqrt{2}$$

If EBC = 65 then EOC = 130 then OEC = OCE = 25. NOw since OC and ED are parallel we have OCE = OED = 25. Hence option D. 

The remainder when $$15^{23}$$ is divided by 19 equals $$(-4)^{23}$$
The remainder when $$23^{23}$$ is divided by 19 equals $$4^{23}$$
So, the sum of the two equals$$(-4)^{23}+(4)^{23}=0$$