CAT 1991 Question Paper

$$2^{71} (2^2 - 2^1 - 1)$$
$$2^{71} (4-2-1)$$
$$2^{71}$$

-n+2 >= 0
or n<=2
and 2n>=4
or n>=2
So we can take only one value of n i.e. 2

let's say integers are x and y
so x+y = 10 => y = 10 - x
and $$\frac{1}{x} + \frac{1}{y} = \frac{5}{12}$$
$$\frac{1}{x} + \frac{1}{10-x} = \frac{5}{12}$$

=> (10 - x + x)*12 = 5*x(10-x)

=> $$120 = 50x - 5x^2$$

=> $$24 = 10x - x^2$$

=> x = 4, 6

=> y = 6 or 4

The bigger of the two numbers is 6.

As we know that area of the circle is directly proportional to the square of its radius.
Hence $$\frac{A_{ic}}{A_{cc}} = \frac{\frac{x^2}{4}}{\frac{x^2}{2}}$$
Where $$x$$ is side of square (say), ic is inscribed circle with radius $$\frac{x}{2}$$, cc is circumscribed circle with radius $$\frac{x}{\sqrt{2}}$$
So ratio will be 1:2

Among all options only D satisfies the given equations as follows:
$$f(y) = \frac{1-y}{1+y}$$
and for x:
$$y + xy = 1-x$$
$$x(1+y) = 1-y$$
$$x= \frac{1-y}{1+y}$$
Hence $$x=f(y)$$

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

Here a = b = y = 0

7x/2 = 35

x = 10

Total number of schools having at least one of PG, LIB or LAB = 30+2x+x+x/2 = 30+3x+x/2 = 30+30+5 = 65

Number of schools having neither of the three = 100-65 = 35

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

It has been given that the schools having playground don't have a Library or Laboratory. 

Hence  a = b = y = 0

7x/2 = 35

x = 10

Required ratio = 25:15 = 5:3

The expected money got by the player = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 21/6 = Rs 3.5

So, the player has to pay 3.5 - 1 = Rs 2.5 to get a profit of Re 1 in the long run.

As machine B's efficiency is twice as of A's, Hence, it will complete its work in 30 hours.
And C's efficiency is putting A and B together i.e. = 20 hours $$( (\frac{1}{60} + \frac{1}{30})^{-1})$$
Now if all three work together, then it will be completed in x (say) days.

$$\frac{1}{x} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60}$$

or x = 10 hours

For x<0 ; y=x+2

for 0<x<$$\frac{1}{2}$$ ; $$y=x+2$$

for $$x>\frac{1}{2}$$ ; $$y=3-x$$

Hence, $$y$$ attains its maxima at $$x=\frac{1}{2}$$  i.e. $$y$$ = 2.5