CAT 2008 Question Paper

If we consider statement a then 2 possibilities occur , the champion may or may not receive bye and the answer for both possibilities would be different.
If we consider both statements a & b together we can find the exact answer to the question.

Using only statement A, we cannot say anything because the player might have received the bye in any round and the total number of players differs if the round in which the player got a bye changes.

Using only statement B, we cannot saty anything because there might be players who received a bye in other rounds.

Using both the statements, we know that only 1 player got a bye in the whole tournament and that happened in round 3.

If the number of players were between 65 and 128, then the number of rounds is 7.

=> In the fourth round there would be 16 people.

One person got a bye here => 31 people in round 3 => 62 people in round 2 => 124 people in round 1.

Hence, we can ansswer the question using both the statements together.

Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on

So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79

Proceeding this way, the last remaining number will be 781

$$7^4$$ = 2401 = 2400+1
So, any multiple of $$7^4$$ will always end in 01
Since 2008 is a multiple of 4, $$7^{2008}$$ will also end in 01

b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, $$b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 - n + n^2 + n + n^2 - 1$$
$$b = 3n^2 - 1$$. The smallest value is -1.

After the first sale, the remaining quantity would be (x/2)-0.5 and after the second sale, the remaining quantity is 0.25x-0.75

After the last sale, the remaining quantity is 0.125x-(7/8) which will be equal to 0

So 0.125x-(7/8) = 0 => x = 7

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
=> -4 is a root of the equation.

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
a + b + c = a + a - 12a = -10a.
But the value of a is not given. Therefore, the value cannot be determined.

The terms of the first sequence are of the form 4p + 13

The terms of the second sequence are of the form 5q + 11

If a term is common to both the sequences, it is of the form 4p+13 and 5q+11

or 4p = 5q -2. LHS = 4p is always even, so, q is also even.

or 2p = 5r - 1 where q = 2r.

Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.

Hence, p = 5m + 2.

So, the number = 4p+13 = 20m + 21.

Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61,...,401 = 20.

We have to essentially look at numbers between 1000 and 4000 (including both).

The first digit can be either 1 or 2 or 3.

The second digit can be any of the five numbers.

The third digit can be any of the five numbers.

The fourth digit can also be any of the five numbers.

So, total is 3*5*5*5 = 375.

However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376