CAT 2007 Question Paper

Let w1 and w2 be average of both the groups respectively.

Since average of whole class is 45. $$\frac{50*w1+50*w2}{100} = 45 $$ => w1 + w2 = 90

And if we consider case A we have w2-w1 = 1. From the two equations average weight can be found out.

Further using the equations deduced from given condition which are : $$\frac{50*w1-l+h}{50} = w2$$ and $$\frac{50*w2+l-h}{50} = w1$$ where l and h is weight of poonam and deepak respectively. However, both of the above equations are effectively the same. Hence we have two equations and 3 unknowns, thus we cannot find the weights by using statement I alone.

If we consider statement B, then we can get another equation. So we will have 3 equations and then we can solve them for the 3 variables.
So both the equations are required to answer the question.

Using statement A alone, we can determine the inner volume of the tank to be at least 4/3 $$\pi$$ (125-64) = 256 $$m^3$$.
But we do not know if it is more than 400 $$m^3$$. So, using statement A alone, we cannot determine the answer.
Using statement B alone, volume of tank = 30000 kg / 3gm/cc = $$10^7 * 10^{-6}$$  = 10 $$m^3$$
So, inner volume = 4/3 $$\pi$$ 125 - 10 > 400 $$m^3$$
So, using statement B alone, we can answer the question

We know that the $$x^2 + y^2 + z^2$$ will be minimum when x = y = z = 89/3
Since that is not an integer, we can consider integer values that are closest to 89/3
Let x = y = 30 and z = 29 => $$x^2 + y^2 + z^2$$ = 2641
If x = y = 29 and z = 31,$$x^2 + y^2 + z^2$$ = 2643
So, the minimum is 2641
So, the question can be answered using statment 1 alone
Using statement 2 alone, we cannot answer the question.
Option a)

If the side of the square is x cm, then the maximum length of OM is $$\sqrt2x$$
The minimum length of OL is x.
So, OM can never be 2 times OL
So, using statement A alone, we can conclude that Rahim is unable to draw the square.
Using statement B alone, we cannot answer the question.
So, option a) is the correct answer.

Let the speed of the plane be p Kmph.
So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.
We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.
So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.
We can clearly see that speed of the plane is 550 which satisfies the above equation.
So for the journey of B to A, the plane takes $$\frac{3000}{550-50} = 6$$ hrs.
So time at B when plane reaches at A is 2 pm .
Hence the time difference between A and B is 1 hr.

Alternatively,
Let speed of flight be s,
Since A is to the east of B, A is ahead of be in time
Let A be ahead of B in time by a hours
Departure from A = 4PM, Arrival at B = 8PM
Travel time = 8 - 4 +a = 4 + a
Since City B is behind city A by 'a' hours, the actual travel time is 'a' hours more than the difference of local times.
Similarly when one travels from B to A, since B is ahead of A by 'a' hrs, actual travel time is 'a' hours less than total
i.e. B->A Travel time = (3PM - 8AM) - a = 7 -a
Total distance travelled = Speed $$\times$$ Time taken ....(1)
From A to B, the wind is favourable / in same direction as flight
Hence from (1), we have 
A->B >>> $$3000 = (s+50)(4+a) => 3000(7-a) = (s+50)(4+a)(7-a)$$ ...(2)
B->A >>> $$3000 = (s-50)(7-a) => 3000(4+a) = (s-50)(7-a)(4+a)$$ ...(3)
(2) - (3) => $$3000(3-2a) = 100(7-a)(4+a) => a^2-63a+62=0 => a=1/62$$
Hence the time difference between A and B is 1 hr.

Let the speed of the plane be p Kmph.

So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.

We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.

So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.

On substituting the options, we can clearly see that speed of the plane is 550 which satisfies the above equation.

Let a, b and c be the percentages of amount invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b - 0.025c

Return attained if there is a fall in the stock market => 0.001a - 0.03b + 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b - 0.025c = 0.001a - 0.03b + 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let's put the values for a, b and c that satisfy the above equation.

b = 9, c = 16, a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed return is 0.2%

Let a, b and c be the percentages of amount invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b - 0.025c

Return attained if there is a fall in the stock market => 0.001a - 0.03b + 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b - 0.025c = 0.001a - 0.03b + 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let's put the values for a, b and c that satisfy the above equation.

b = 9, c = 16, a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed return is 0.2% and it is attained when 36% is invested in option B and 64% is invested in option C.

Any ordered pair has 2 elements => There are n-2 elements that are not present in the ordered pair.

The number of enemies of any ordered pair is all the ordered pairs in the set formed using the numbers other than these two elements = $$^{n-2}C_2$$ = $$1/2 * (n^2 - 5n + 6)$$.

For n, the number of elements in set S is $$^nC_2$$.

Lets say the 2 friends are (x,a) and (y,a)

These two friends have 3 numbers in total and 1 common element(say a) (as both elements cannot be exactly same)

They have 2 non common elements(x, y)  

The number of common friends is formed by the non-common elements of the friends (x,y) + the number of elements in the set which have the common element other than the two friends (a,c), (a,d) and so on = 1 + (n-1 - 2) = n-2.

For the example in question, if the friends are (1,2) and (1,3), then common friends are (2,3) and all other elements with 1

All elements with 1 = n-1= 3 which are (1,2) (1,3) (1,4) excluding the friends (1,2) and (1,3) only 1 other friend is common. Hence it is 1+(n-1)-2=n-2