A train travel at the rate of 50 km/h without stoppages and it travels at 40 km/h with stoppages. How many minutes does the train stop on an average per hour?

Since a alone can finish a work in 10 days and B alone can do it in 15 days. So if they work together then the ration of work done by A and B is = (1/10) : (1/15).

? The wages A will get = (1/10)[(1/10) + (1/15)] x 75
= Rs. 45

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

Required number of ways=  8 C 5 × 10 C 6 = ( 8 C 3 × 10 C 4)= 11760.

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4!/2!= 12.

Required number of words = (10080 x 12) = 120960

? Factory A turns out x cars in one hour. Factory B turns out y/2 cars in one hour.
? In one hour both the factories A and B can turn out (x + y/2) cars
? In 8 hours both factories turn out = 8( x + y/2) cars
= 4(2x + y) cars.

Out of 26 alphabets two distinct letters can be chosen in  26 P 2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

Combined with letters there are  6 P 2 X 100 ways = 65000 ways to choose vehicle numbers.

ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

(A's 1 day's work) : (B's 1 day's work) = 2 : 1
Now, ? (A + B)'s day's work = 1/14
? A's 1 day's work = (1/14) x 2/3 = 1/21
? A alone can finish the work in 21 days.
[Dividing 1/14 in the ratio 2 : 1 ]

Choose 2 juniors and 2 seniors.

14 C 2 * 23 C 2 = 23023