CAT 2008 Question Paper

The circumferences of the two circle pass through each other's centers. Hence, O1A = O1B=O1O2 = 1cm

By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P.

O1P = Half of O1O2 = 1/2 cm
So, the angle AO1P = 60 degrees as cos 60 = 1/2
By symmetry, BO1P = 60 degrees.
So, angle AO1B= 120 degrees

In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) - A(triangleO1AB )(red)

Area of sector = 120°/360° * $$\pi * 1^2 $$ = $$\pi/3$$

Area of triangle = 1/2 * b * h = 1/2 * (2* 1 cos 30°) * (1/2) = √3/4

Hence, required area = $$\frac{\pi}{3}-\frac{\sqrt 3}{4}$$ . Hence so the required area is 2 times the above value which is $$\frac{2\pi}{3}-\frac{\sqrt 3}{2}$$

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle. 

Hence, $$\angle\ ABC=60^{\circ\ }$$, $$\angle\ CAB=30^{\circ\ }$$, which implies $$\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$$

Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.

We know that $$\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$$

Similarly, $$\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$$

So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around $$\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.

Let us assume that three positive consecutive integers are x, x+1, x+2. They are raised to first, second and third powers respectively. 

$$x^{1} + (x+1)^{2} + (x+2)^{3} = (x + (x+1) +(x+2))^{2}$$

$$x^{1} + (x+1)^{2} + (x+2)^{3}$$ = $$(3x + 3)^{2}$$

$$x^{3} + 7x^{2} + 15x + 9$$ = $$9x^{2} + 9 + 18x$$

After simplifying you get,

$$x^{3} - 2x^{2} - 3x = 0$$

=> x=0,3,-1

Since x is a positive integer, it can only be 3.

So, the minimum of the three integers is 3. Option a) is the correct answer.

Consider the first term:
$$\sqrt{1+1/1^2+1/2^2}$$ = $$\sqrt{9/4}$$ = 3/2
Second term: $$\sqrt{1+1/2^2+1/3^2}$$ = $$\sqrt{49/36}$$ = 7/6
First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 - 1/3
.
.
​Required sum = 2008 - 1/2008

Trangle AGF and triangle FHC are similar.
So, $$\frac{10-h}{x}=\frac{h}{4-x}$$
=> $$\frac{10-h}{h}=\frac{x}{4-x}$$
=> $$\frac{10}{h}-1=\frac{x}{4-x}$$
=> $$\frac{10}{h}=\frac{4}{4-x}$$
=> $$h=\frac{10}{4}(4-x)$$

Total surface area= $$2 \pi x^2 + 2 \pi x h= 2 \pi (x^2 + x \frac{10}{4}(4-x)=2 \pi (10x - \frac{3}{2} (x^2))$$

=$$2\pi\frac{-3}{2}[x^{2}-\frac{20}{3}x] $$. We can add and subtract 100/9 to get a perfect square.

=$$2\pi\frac{-3}{2}[x^{2}-\frac{20}{3}x+\frac{100}{9}-\frac{100}{9}]$$

=$$2 \pi (-\frac{3}{2}(x-\frac{10}{3})^{2}+ \frac{50}{3})$$

This value will be maximum when the negative factor is minimum i.e. 0 because it is multiplied by a square. Thus, when x=10/3 the surface area is max.

So, the surface area is $$\frac{100}{3} \pi$$

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We know that P is opposite to R and Q is opposite to U => S is opposite to T

It is given that T is opposite to red house => S is the red house and hence T is the blue house.

So, we know the colours of all houses and heights of P, R, T and U.

In this question, we are asked to find the house that is opposite to yellow house. R is the yellow house, P is opposite to R and S is on the other corner in P’s row. Hence S is the house that is diagonally opposite to yellow house and the colour of S is Red.

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We only know that the second tallest house is either Q or S.  Hence the answer is cannot be determined.

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We know that P is opposite to R and Q is opposite to U => S is opposite to T

It is given that T is opposite to red house => S is the red house and hence T is the blue house.

T is the tallest house and hence the colour of the tallest house is blue.

If the volume of data transfer in India is the same as that of Singapore then we have ($$R_i$$ * 7.5 / 0.9 )= (20*$$R_s$$ / 9.5) ; we get $$R_s$$ nearly equal to 4*$$R_i$$.
Hence  Total revenue in Singapore is about 4 time that of India.

Revenue from data transfer as a percentage of total revenue will triple for India and double for Sweden.
Also the total revenue in India is twice that of Sweden.
Taking volume constant we have fir India  $$3*0.9*R_i = ARDT * vol$$
For Sweden  $$0.18*2*R_s = 6 * Vol$$
Since vol is constant and $$R_i = 2* R_s$$ we have ARDT = 9 .
Thus % increase in ARDT = 800%.