CAT 2001 Question Paper

Lets take $$n_1 = n_2 = 10$$ and $$r_1 = r_2 =100$$ .
Using given formula we have $$BA = 20$$ and $$MBA_1 = MBA_2 = 10$$ .
So we have $$BA$$ highest which is not the case in option A,B and C. Hence the correct option is D. 

With no incomplete innings and $$BA$$ of 50 , lets assume $$r_1 = 50$$ and $$n_1 = 1$$.
Now we have $$BA = MBA_2 = 50$$ 
So we have $$r_2 = 45$$ and $$n_2 = 1$$ .
We have $$BA = 95$$ and $$MBA_2 = \frac{95}{2} < 50$$.
Hence $$MBA_2$$ decreases.
So $$BA$$ will increase and $${MBA}_2$$ will decrease. 

At 60 kmph, time taken to reach the destination = 200/60 = 3.33 hours.
Fuel consumed per hour at 60 kmph = 4 liters.
Therefore, total fuel consumed = 3.33*4 = 13.33 liters.

If she drives at 40 kmph, the fuel consumed = 200/40 * 2.5 = 12.5 liters
If she drives at 60 kmph, the fuel consumed = 13.33 liters
If she drives at 80 kmph, the fuel consumed = 200/80 * 7.9 = 19.75 liters
To minimize fuel consumption, Manasa has to decrease the speed.

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

Let the length of each side of the octagon be x.

So, length of the square will be x+2*(x/√2) = 2

=> x(1+√2) = 2 => x = 2/(1+√2)

Take x=3 , z=2 , y=5.

$$y(x-z)^2 = 5(3-2)^2 = 5$$

Option A gives 5 which is odd.

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

-4x = -24, 5y = -30

So none of the options out of a,b or c satisfies .

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min . 

Hence they flash together after every 2 min. So in an hour they flash together 30 times .

Each box contains at least 120 and at most 144 oranges.

So boxes may contain 25 different numbers of oranges among 120, 121, 122, .... 144.

Lets start counting.

1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.

Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.

Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.

Hence the number of boxes containing the same number of oranges is at least 6.