CAT 1999 Question Paper

y = $$\frac{4x-1}{17}$$

The integral values of x for which y is an integer are 13, 30, 47,......

The values are in the form 17n + 13, where $$ n \geq 0$$

17n + 13 < 1000

=> 17n < 987

=> n < 58.05

=> n can take values from 0 to 58 => Number of values = 59

$$(ab)^2$$ = ccb

ccb > 300

The last digit of the number ab must be same as that of the square of ab.

So, b can be 0, 1, 5 or 6.

$$20^2$$=400 and $$30^2$$=900 are three digit numbers and greater than 300. But the first 2 digits are not same. Hence, b is not 0.

If b is 5, then the ten's digit of ab's square will be 2 => c = 2. But if c is 2, then ccb is not greater than 300. Hence, b is not 5.

If b is 6, then $$26^2$$ = 576 is the only three digit number that is greater than 300. But, it is not in the form of ccb => b is not 6.

If b is 1, then $$21^2$$=441 satisfies all the given conditions => b is 1.

$$7^3$$ = 343

$$7^{84}$$ = $$(7^3)^{28}$$ = $$343^{28}$$

$$343^{28}$$ mod 342 = $$1^{28}$$ mod 342 = 1

For a triangle to be formed, we need three points.

Case 1: Select 2 points on the line that has 10 points and 1 point on the line that ha 11 points.

This can be done in $$^{10}C_2$$*$$^{11}C_1$$ ways = 495 ways.

Case 2: Select 2 points on the line that has 11 points and 1 point on the line that ha 10 points.

This can be done in $$^{11}C_2$$*$$^{10}C_1$$ ways = 550 ways.

495 + 550 = 1045 ways.

At least one candidate and at most n candidates among 2n+1 candidates => 

^{2n+1}C_1Total number of ways alteast one candidate can be selected = Total number of ways of selecting a candidate - total number of ways of selecting no candidates

$$\Rightarrow ^{2n+1}C_1+^{2n+1}C_2+^{2n+2}C_3+....+^{2n+1}C_{n-1}+^{2n+1}C_n = 63 $$

We know that $$^{2n+1} C_0$$ and $$^{2n+1} C_{2n+1}$$ are equal to 1.

By Binomial Expansion 

$$ ^{2n+1}C_0+^{2n+1}C_1$$+... + $$^{2n+1}C_{2n}+^{2n+1}C_{2n+1} = (1+1)^{2n+1} $$ ---- Eq 1

Also $$^{2n+1}C_1$$ = $$^{2n+1}C_{2n}$$ by symmetry

and $$^{2n+1}C_2$$ = $$^{2n+1}C_{2n-1}$$ and so on

So $$\Rightarrow ^{2n+1}C_{n+1}+^{2n+1}C_{n+2}+....+^{2n+1}C_{2n-1}+^{2n+1}C_{2n} = 63 $$

Therefore, on substituting these values in Eq 1 we get

1 + 63 + 63 +1 = $$2^{2n+1}$$

$$2^{2n+1}$$ = 128

2n+1 = 7

Therefore, n=3

As at most n students can be selected, the correct answer is 3.

The function of the speed of the train = 42 - k$$\sqrt{n}$$ where n is the number of compartments and k is a constant.

42 - k$$\sqrt{9}$$ = 24

=> 3k = 18 => k = 6

=> Function of speed = 42 - 6$$\sqrt{n}$$

Speed is 0 when 42 - 6$$\sqrt{n}$$ = 0

=> 42 = 6$$\sqrt{n}$$

=> n = 49

=> So, with a positive speed, the train can carry 48 compartments.

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $$\frac{55000}{100}$$ = Rs.550.

Let the number of employees be 100.

=> 40 men and 60 women.

Number of men getting more than 25000 = 30

Number of people getting more than 25000 = 45

Number of women getting more than 25000 = 45 - 30 = 15

Fraction of women getting less than 25000 = $$\frac{45}{60}$$ = $$\frac{3}{4}$$

| r-6 | = 11 => r = -5 or 17

| 2q - 12 | = 8 => q = 10 or 2

So, the minimum possible value of q/r = 10/(-5) = -2

Let the four consecutive positive integers be $$a,a+1,a+2$$ and $$a+3$$.
Therefore, $$n=1+a(a+1)(a+2)(a+3)$$
Or, $$n = 1+(a^2+3a)*(a^2+3a+2)$$
Or, $$n = (a^2+3a)^2 + 2*(a^2+3a)+1 = (a^2+3a+1)^2$$
Hence, n is a perfect square and therefore not a prime.

The product of four consecutive positive integers is always even. Hence, n is always odd.
Therefore, from the given statements, only A and C are true.