CAT 2005 Question Paper

If the moduli are removed, the equations formed are

x+y+x-y = 4 => x=2

x+y-x+y = 4 => y =2

-x-y+x-y = 4 => y=-2

-x-y-x+y = 4 => x=-2

The area enclosed by these equations is a square with vertices at (2,2), (-2,2), (-2,-2), (2,-2) as shown in figure.

The required area = 4*4 = 16

Let EO = x, So, AE = 1.5 - x
AE : EB = 1:2 => x = 1/2

(1.5-x):(1.5+x) = 1:2.

x=0.5.

So, EO = 0.5
Similarly, OL = 0.5
Now, EOLH is a parallelogram and EO = OL = 0.5
In triangle DOL, DO = radius = 1.5 and OL = 0.5
So, DL = $$\sqrt2$$
=> DH = $$(2\sqrt2 - 1)/2$$

Consider triangles ABC and BDC
Angle B is common in both triangles. Also $$\angle A = \angle C$$
Since 2 angles are equal, the third angles $$\angle{ACB} = \angle{BDC}$$
Hence, triangle BCA $$\sim$$ BDC (BCA and BDC are similar)
AC/DC = BC/BD = AB/BC 
BC = 12 cm, DB = 9 cm, CD = 6 cm
=> AC = 12/9 * 6 = 8 cm
AB = 8/6 * 12 = 16 cm
So, AD = 16 - 9 = 7 cm
Perimeter of ADC = 7+8+6 = 21 cm
Perimeter of BCD = 27 cm
Ratio = 21/27 = 7/9

Let PQR be an equilateral triangle with side equal to x and let the intersection point of PS and QR be M.

Clearly, the circle is the circumcircle of the triangle PQR.

QR = x => QM = $$\frac{x}{2}$$ because a perpendicular from the centre to any chord bisects the chord.

Angle OQM = 30 degrees and QM is equal to $$\frac{x}{2}$$ => OQ = $$\frac{\frac{x}{2}}{cos(30)}$$ = $$\frac{x}{\sqrt{3}}$$

Hence the radius of the circumcircle of an equilateral triangle is equal to $$\frac{x}{\sqrt{3}}$$.

Angle PQS = 90 degrees as it is an angle in a semicircle. PS bisects angle QPR => angle QPS is 30 degrees. Hence QS subtends an angle of 30 degrees in the major arc => QS subtends an angle of 60 degrees at the centre because angle subtended by a chord at the centre is twice the angle subtended by the chord in the major arc.

Angle QOS = 60 degrees => Triangle QOS is equilateral and hence QS is equal to radius of the circle => QS = $$\frac{x}{\sqrt{3}}$$

Given that radius is r => r = $$\frac{x}{\sqrt{3}}$$ => x = $$r\sqrt{3}$$

=> Perimeter of PQRS = PQ+QS+SR+RP= $$ r\sqrt{3} + r + r + r\sqrt{3}$$ = $$2r(1+\sqrt{3})$$

The no. has all the digits as odd no. and is divisible by 3. So the possibilities are 

1113
1119
1131
1137
1155
1173
1179
1191
1197
Hence 9 possibilities . 

$$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$$

=> $$x = \sqrt{4+\sqrt{4-x}}$$

=> $$x^2 = 4 + \sqrt{4-x}$$

=>$$x^4 + 16 - 8x^2 = 4 - x$$

=> $$x^4 - 8x^2 + x +12 = 0$$

On substituting options, we can see that option C satisfies the equation.

According to given condition we have , g(x+1) = -g(x-1) + g(x)

Putting x=x+1 we get g(x+2) = g(x+1) - g(x) = -g(x-1)

Putting x=x+2 we get g(x+3)=-g(x)

Similarly g(x+4)=-g(x+1), g(x+5)=-g(x+2)=-g(x+1) + g(x)  and g(x+6) = g(x+1)-g(x+2)=g(x).

So p=6.

Let x be no. of male and y be no. of female operators.
We have 40x+50y=1000 .
So x = 25-(5*y/4) also 7<=y<=12.
So y can be 8 or 12.
If y=8 then x=15 and y=12 then x=10 .
Then we have to find total cost incurred in both the cases.
We find that cost is minimum in 2nd case when no. of males are 10.

Consider there are 6 people numbered 1-3 englishmen and 3-6 frenchmen, let 3 know both english and french.
First call would be between 1-3 then 2-3 such that 3 know secret of all 3 englishmen.
Let 3 call 4 .
Similarly there would be call between 4-5 then 4-6 such that 4 know secret of all 3 frenchmen.
Now 3 would call 4 . Such that 3 and 4 would know secret of all 6 members.
Now to let this know to 1,2,5,6 more 4 calls would be required.

Hence, minimum calls required would be 9.

Let C and R be no. of columns and rows respectively.

The number of red coloured tiles would be given by (R-2)(C-2). This is because two outer rows made of white tiles and the two outer columns made up of outer columns are removed.

Similarly the number of white tiles would be given by R*2 + (C-2)*2. Two tiles are removed from columns because the corner tiles would have already been included while considering the rows.

So according to given condition we have (C-2)*2 + 2*R = (C-2)(R-2).
Now start putting value of c from options into this equation. Only for one option B we get an integer value of R .