CAT 1998 Question Paper

The first wheel completes a revolution in $$\frac{60}{60}=1$$ second
The second wheel completes a revolution in $$\frac{60}{36}=1\frac{2}{3}$$ second
The third wheel completes a revolution in $$\frac{60}{24}=2\frac{1}{2}$$ second

The three wheels touch the ground simultaneously at time which are multiples of the above three times.
Hence, the required number is $$LCM(1,\frac{5}{3},\frac{5}{2}) = 5$$ seconds.

So, the correct option is option (c)

Let's say N is our number
N = (899K + 63) or N = ($$29 \times 31$$K) + 63
So when it is divided by 29, remainder will be $$\frac{63}{29}$$ = 5

Let the number 'n' belong to the set A.
Hence, the remainder when n is divided by 2 is 1
The remainder when n is divided by 3 is 2
The remainder when n is divided by 4 is 3
The remainder when n is divided by 5 is 4 and 
The remainder when n is divided by 6 is 5

So, when (n+1) is divisible by 2,3,4,5 and 6.
Hence, (n+1) is of the form 60k for some natural number k.
And n is of the form 60k-1

Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1

As least number of coin, one can have is 10
And no two of them possess same number of coins, hence for having maximum number of coins to one person, the distribution of coins will be = 10,12,14,64
maximum no. of coins = 64

If A collected 54 coins, remaining will be 46.
Now B,C,D must have atleast 10,12,14 coins 
So remaining coins will be 10
So difference between maximum and second maximum should be least when second highest will be as maximum as possible.
I.e. difference = 54- (14+10) = 30

Number of coins with A = 54

Number of coins with B = 2x+2

Number of coins with C = x

Number of coins with D = y

54+3x+2+y = 100

3x+y = 44

If 2x+2 = 28, then x = 13 and y = 5 which is not possible.

If 2x+2 = 20, then x = 9 which is not possible.

If 2x+2 = 26, then x = 12 and y = 8 which is not possible

If 2x+2 = 22, then x = 10 and y = 14 which is feasible

Akbar can wear either red or blue.
Anthony can wear either red or green.
So when both of them wear same colour that has to be red.
And Amar can wear either blue or green.
Hence our answer will be A.

The different possible cases are:

From the table we can see among all of the statements, only fourth statement is false, as if amar wears green, anthony has to wear green.

As we can see here that total number of students are = 60+84+108 = 252

Now given condition is that in one room only the students of the same subject can be there and the number of rooms should be minimum that means the number of students in a particular room will be maximum.

This Maximum number of students will be HCF (Highest common factor) of 60, 84 and 108 and that will be 12

Hence, number of rooms will be = 252/12 = 21

As we know for a number to be divisible by 125, its last three digits should be divisible by 125
So for a five digit number, with digits 2,3,8,7,5 its last three digits should be 875 and 375
Hence only 4 numbers are possible with its three digits as 875 and 375
I.e. 23875, 32875, 28375, 82375