CAT 2003 Question Paper (Leaked)

Consider the first statement:
When the common ratio is less than 1 we can apply the formula of sum of infinite terms.
So, LHS = $$1/(a^2 - 1)$$
RHS = $$a/(a^2 - 1)$$
If a<1 then LHS<RHS

If a = 1,then LHS = RHS

So, we cannot answer the question using statement 1 alone

Using statement 2 alone, we know that a = 1/2. So, RHS > LHS
Hence, option a)

From statement 1 alone, we can infer that the triangle ABC is an equilateral triangle with side = 2 cm
Similarly, from statement 2 alone, we can infer that the triangle ABC is an equilateral of side 2 cm
So, the question can be answered using either statement alone

By the end of the year 2002, Shepard bought 4 times and sold 4 times. He is left with the initial number of goats that he had in 1998. If the percentage of goats bought is equal to or lesser than the percentage of goats sold, then there would be a net decrease in the total number of goats. For the number of goats to remain the same, p has to be greater than q, because p% is being calculated in a lesser number and q% is being calculated on a greater number. Hence, p > q.

Let the total number of sides be x.

Sum of internal angles in a polygon = (x-2)*180 where x is the number of sides.

It is given that the polygon has 25 convex sides, then the number of concave sides = x-25

(25*90)+(x-25)*270 = (x-2)180

x = 46

Number concave corners = x-25 = 46-25 = 21

1, 2, 3, 4,....n such that the sum is greater than 288
If n = 24, n(n+1)/2 = 12*25 = 300
So, n = 24, i.e. the 24th letter in the alphabet is the letter at position 288 in the series
​So, answer = x

Let $$\alpha $$ be equal to k.

=> f(x) = $$x^2-(k-2)x-(k+1) = 0$$

p and q are the roots

=> p+q = k-2 and pq = -1-k

We know that $$(p+q)^2 = p^2 + q^2 + 2pq$$

=> $$ (k-2)^2 = p^2 + q^2 + 2(-1-k)$$

=> $$p^2 + q^2 = k^2 + 4 - 4k + 2 + 2k$$

=> $$p^2 + q^2 = k^2 - 2k + 6$$

This is in the form of a quadratic equation.

The coefficient of $$k^2$$ is positive. Therefore this equation has a minimum value.

We know that the minimum value occurs at x = $$\frac{-b}{2a}$$

Here a = 1, b = -2 and c = 6

=> Minimum value occurs at k = $$\frac{2}{2}$$ = 1

If we substitute k = 1 in $$k^2-2k+6$$, we get 1 -2 + 6 = 5.

Hence 5 is the minimum value that $$p^2+q^2$$ can attain.

Let R ,r be radius of big and small circles respectively. We know that R=2r.

And since area = 12 ;

$$R^2 = \frac{12}{\pi}$$.
$$r^2$$ = $$\frac{3}{\pi}$$

By pythagoras theorem in the small triangle with side 'x' we have x = $$\sqrt{\ 3}r$$.

Angle OAB = $$\tan\ \theta\ \ =\ \ \frac{\ r}{\sqrt{\ 3}r}\ =\ 30$$
Hence Angle CAB = 60.

So area = $$\ \frac{\ 1}{2}\ \times\ 2\sqrt{\ 3}r\times\ 2\sqrt{\ 3}r\times\ \sin\ 60$$.
On substituting the value of sin 60 and $$r^2$$, we get

Area = $$9\sqrt3/\pi$$.

Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,

take a=b=c=1 and d=2

we get $$a^2 + b^2 + c^2 + d^2 = 7$$ which is equal to $$4m^2+2m+1$$ for other values it is greater than $$4m^2+2m+1$$ . so option B

Let R be radius of big circle and r be radius of circle with centre S. Radius of 2 semicircles is R/2.

 From Right angled triangle OPS, using pythagoras theorem we get

$$(r+0.5R)^2 = (0.5R)^2 + (R-r)^2$$ . We get R=3r .

Now the area of big semicircle that cannot be grazed is Area of big S.C - area of 2 semicircle - area of small circle = $$\pi*R^2$$/2 - 2*$$\pi*(0.5R)^2$$/2-$$\pi*r^2$$ = $$\pi*R^2$$/2 - 2*$$\pi*(0.5R)^2$$/2-$$\pi*(R/3)^2$$= $$\pi*R^2$$/2 - $$\pi*(R)^2$$/4-$$\pi*(R)^2$$/9 = 5*$$\pi*R^2$$/36.  this is about 28 % of the area $$\pi*R^2$$/2 . Hence option B.

When the hexagon is divided into number of similar triangle AOF we get 12 such triangles . Hence required ratio of area is 1/12.
Alternate Approach :

Let us take side of the hexagon as a
Now We get AF = a
Now As OF || ED
Angle OFE +Angle E = 180
We know Angle E =120
So angle OFE=60 degrees.
Now therefore Angle AFO=120-60=60 degrees
Now in triangle AFO
Cos60=$$\frac{OF}{AF}$$
$$\frac{1}{2}=\frac{OF}{AF}$$
We get OF = $$\frac{a}{2}$$
Now therefore area of triangle AOF =$$\frac{1}{2}\times\ AF\times\ OF\times\ \sin60$$
So we get $$\frac{1}{2}\times\ a\times\ \frac{a}{2}\times\ \frac{\sqrt{\ 3}}{2}$$
=$$\frac{\sqrt{\ 3}}{8}a^2$$     (1)
Now area of hexagon = $$6\times\ \frac{\sqrt{\ 3}}{4}a^2$$      (2)
Dividing (1) and (2) we get ratio as $$\frac{1}{12}$$
Hence option A is the correct answer.