CAT 2002 Question Paper

If the child adds all the numbers from 1 to 34, the sum of the numbers would be 1+2+3+...+34 = 34*35/2 = 595
Since the child got the sum as 575, he would have missed the number 20.

Consider different values of x and y:

x = -1.5 and y = -1.5; x = 1.5 and y = -1.5; x = -1.5 and y = 1.5; x = 1.5 and y = 1.5.

For these possibilities, options A,B and C gets satisfied , but it is impossible to find any two positive real numbers x and y for which L(x, y) > R(x, y).

If there are 'm' non-parallel lines, then the maximum number of regions into which the plane is divided is given by

[m(m+1)/2]+1

In this case, 'm' = 10

So, the number of regions into which the plane is divided is (10*11/2) + 1 = 56

$$2^4 = 16 = -1$$ (mod $$17$$)
So, $$2^{256} = (-1)^{64} $$(mod $$17$$)
$$= 1$$ (mod $$17$$)
Hence, the answer is 1. Option a).

The given equation can be written as : $$A^2 * (x-1) + B^2 * x = x^2 - x$$
=> $$x^2 + x(-1 - A^2 - B^2) + A^2 = 0$$
Discriminant of the equation = $$ (-1 - A^2 - B^2) ^2 - 4A^2$$
=$$ A^4 + B^4 + 1 - 2A^2 + 2B^2 + 2A^2B^2$$
= $$ A^4 + B^4 + 1 - 2A^2 - 2B^2 + 2A^2B^2 + 4B^2$$
= $$ (A^2 + B^2 - 1)^2 + 4B^2$$
>= 0, 0 when B =0 and A =1
Hence, the number of roots can be 1 or 2.
Option d) is the correct answer.

In this problem, the lights are flashed at the intervals 2.5, 4.25 and 5.125 seconds and put off after one second each.

The total duration of intervals of these lights are (2.5+1) = 3.5 s, (4.25+1) = 5.25 s and (5.125+1) = 6.125 s.

We have to find the minimum duration. It would be the LCM of thes three numbers.
Since each word is put after a second. So LCM [$$(\frac{5}{2}+1 )(\frac{17}{4}+1)(\frac{41}{8}+1)$$] = LCM of numerator / HCF of denominator = 49*3/2 = 73.5. Hence they will glow for full one second after 73.5-1 =72.5 sec.

HCF of [(9/2), (27/4), (36/5)] = HCF of numerators / LCM of denominators = 9/20

Total weight = 18.45 lb

So no. of parts = 18.45 / (9/20) = 18.45*20/9 = 41 

Hence option d) is the correct answer.

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K 

Let k = 1; the number becomes 53

If it is divided by 84, the remainder is 53.

Option d) is the correct answer.

Alternative Solution.

Consider only for 3 and 4 and the remainders are 2 and 1 respectively.

So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48x4+5= 53

The correct original circular arrangement sequence in clockwise manner is Suresh, Dhiraj, Ajay, Raghuveer, Pramod, Yogendra.

So after the changes, Suresh is to the left of Dhiraj.

Let the length of the tunnel be x and distance of the train to entrance A be y. Let the speeds of train and cat be t and c respectively.

Hence, when the cat runs 3x/8, the train covers y.

=> (3x/8)/c = y/t --- (1)

When the cat runs 5x/8 to the other end, the train covers x+y

=>(5x/8)/c = (x+y)/t ---(2)

Taking ratio of (1) to (2)

3/5 = y/(x+y) => 3x = 2y ---(3)

Substituting (3) in (1)

(2y/8)/c = y/t

=> t = 4c

Hence the ratio t:c is 4:1.