CAT 2007 Question Paper

The number of players in all the teams put together = k * n

The number of players that are common is 1*n = n

So, the number of players in the tournament = kn - n = n(k-1)

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $$(50 \pm a)$$ and $$100 - a$$ where $$0 \leq a \leq 100$$
So, the square would be of form $$(50 \pm a)^2 = 2500 + a^2 \pm 100a$$ or $$ (100 - a)^2$$ i.e. $$10000 + a^2 + 200 a$$

In both cases, only $$a^2$$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

Price of Darjeeling tea on 100th day= 100+(0.1*100)=110
Price of Ooty tea on nth day= 89+0.15n
Let us assume that the price of both varieties of tea would become equal on nth day where n<=100
So
89+0.15n=100+0.1n
n=220 which does not satisfy the condition of n<=100
So the price of two varieties would become equal after 100th day.
89+0.15n=110
n=140
140th day of 2007 is May 20 (Jan=31,Feb=28,March=31,April=30,May=20)

To know the range, we have to take the limiting case.

The limiting case in this case is when the circles pass through each other's centers. 

In this case, PQ = AP = AQ => They form an equilateral triangle => angle AQP = 60 degrees.

So, the maximum possible angle is 60 degrees.

Another limiting case is when the circles touch each other externally.

In this case, angle AQP = 0 degrees.

Hence, the range is 0 to 60.

Let the function be $$ax^2 + bx + c$$.

We know that x=0 value is 1 so c=1.

So equation is $$ax^2 + bx + 1$$.

Now max value is 3 at x = 1.

So after substituting we get a + b = 2.

If f(x) attains a maximum at 'a' then the differential of f(x) at x=a, that is, f'(a)=0.

So in this question f'(1)=0

=> 2*(1)*a+b = 0

=> 2a+b = 0.

Solving the equations we get a=-2 and b=4.

$$ -2x^2 + 4x + 1$$ is the equation and on substituting x=10, we get -159.

We want a diet with 10% minerals and at least 30% protien. The mixture of O and Q is suitable because both contain 10% minerals.In required proportion we can get 10% minerals and at least 30% proteins.
Solution P and R contains minerals which are less than 10% and maximum mineral %age in any component is 10. So if we include P or R in any mixture,we can't get 10% mineral %age.

So only 1 solution of O and Q is possible.

Option A: P and Q in 4:1 ratio gives fat 10% but the protein is 22%

Option B: P and S do not contain fat. Hence we wont analyse the option.

Option C: P and R in 3:1 ratio gives fat 10% but the protein is 27.5%

Option D: Q and S in 1:4 gives fat and protein as 10% and 46 %.

Cost of the mixture= [(1*200)+(4*100)]/5=120

Option E: R and S in 1:3 gives 10% fat and 50%protein.
Cost of the mixture= [(1*500)+(3*100)]/4=200

P,Q,S contain 80%,10% and 45% carbohydrates respectively.
Option A,C and D does not contain 60% carbohydrates.

Option E: Carbohydrate %=[(4*80)+(1*10)+(1*45)]/6= 62.5%
Cost per unit=[(4*50)+(200)+100]/6=83.3

Option B: Carbohydrate %=[(4*80)+(1*20)+(2*45)]/7= 61.4%
Cost per unit=[(4*50)+(200)+(2*100)]/7=600/7=85.7

Two ingredients are mixed in equal proportion. So, the minimum amount of protein and carbohydrate should be 60 each, the minimum amount of minerals is 10 and the maximum amount of fat is 50. The combination of O and S satisfies this requirement.

No. of athletes which are top academic performers: 10
According to statement a) Sixty per cent of the top academic performers were not athletes  so 40% of top academic performers are athletes which we know is 10. So total no. of top academic performers are 25. 
Hence statement a is alone to answer the question.
while we cant deduce anything from statement b .