CAT 2001 Question Paper

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ =12.5

Approach 2:

$$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$

Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$\sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM).

Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = y +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ = 12.5

We will solve this question by taking the options.

Suppose x = 11

x+4 = 15, x-3 = 8

Hyptenuse = $$\sqrt{225+64} = 17$$

17-11 = 6

$$6^2+(x-3)^2 = y^2$$

x-3 = 8

y = 10 which is similar to what is given in the question.

Hence x = 11

If the width of the walkway is x, then its area = (60+2x)(20+2x) - 1200 = 516
Solving this, we get x = 3m

We know that a=$$b^2-b$$.

So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1)

The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)

In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).

Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24.

Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

The possible arrangements are 1,multiples of 2 ,remaining. So we have 1+2+4+8+16+32+64+31 = 158. Hence minimum no. of bags required is 8. 

We have a +c=e so possible summation 6+4=10 or 4+2 = 6.
Also b=2d so possible values  4=2*2 or 10=5*2.
So considering both we have b=10 , d=5, a=4 ,c=2, e=6.
Hence the correct option is B .

We know that quadratic equation can be written as $$x^2$$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $$x^2$$-(7)*x+(12)=0 where the constant term is wrong.

Keshab got the roots as (3, 2) so the equation is $$x^2$$-(5)*x+(6)=0 where the coefficient of x is wrong .

So the correct equation is $$x^2$$-(7)*x+(6)=0. The roots of above equations are (6,1).

Let the number of coins of the three denominations be x, y and z respectively.
x+y+z = 300  
x+2y+5z = 960
2x+y+5z = 920
=> 3(x+y) + 10z = 1880
=> 3(300 - z) + 10z = 1880
=> 900 + 7z = 1880 => z = 980/7 = 140
So, the number of 5 rupee coins is 140

The distinct paths are:

A -> D -> C -> F
A -> D -> E -> F
A -> D -> C -> E -> F
A -> B -> D -> C -> F
A -> B -> D -> E -> F
A -> B -> D -> C -> E -> F
A -> B -> C -> F
A -> B -> E -> F
A -> B -> F
A -> B -> C -> E -> F

So, the total number of distinct paths is 10

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $$^4C_3 $$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192