A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the begining and 100 more after 35 days and completes the work in stipulated time.
If he had not engaged the additional men, how many days behind schedule would it be finished?

Let required amount of coal be y metric tonnes
9 engines consumes 24 metric tonnes of coal in 8 hours a day.
More engines, more amount of coal (direct proportion)
If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)
More hours, more amount of coal(direct proportion)
Now we have , 9 × ( 1 / 3 ) × 8 × y = 8 × ( 1 / 4 ) × 13 × 24
? 3 × 8 × y = 8 × 6 × 13
? 3 × y = 6 × 13 ? y = 2 x 13 = 26 metric tonnes

Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + ( n - 1 ) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1 ) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + ( n - 1 ) x d = 12 + 37
a + ( 6- 1 ) x d = 39
a + 5d = 39................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 39 - 37
5d - 3d= 2
2d = 2
d = 1
Put the value of d in equation (1), we will get
a + 3 x 1 = 37
a = 37 - 3
a = 34

Second term = a + (n - 1) x d = 34 + (2 - 1) x 1 = 34 + 1 = 35
Six term = a + (n - 1) x d = 34 + (6 - 1) x 1 = 34 + 5 = 39
Sum of Second and Six term = 35 + 39 = 74
Sum of Second and Six term = 74
Answer is 74.

First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
where a = First Number , l = Last Number and d = difference of two consecutive numbers.
Let the number of terms be n. So Last term = t_n
Then t_n = 996
Use the formula for n terms of arithmetic progression.
? a + ( n - 1) x d = 996
? 102 + (n - 1) x 6 = 996
? 6(n - 1) = 894
? (n - 1) = 149
? n = 150
? Numbers of terms = 150