A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is:

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

As per the given question ,
Food of 150 men for 45 days = 150 × 45 = 6750 unit
After 10 days , Food of 150 men for 150 × 10 = 1500 unit
And after 10 days remaining men = 150 - 25 = 125
and remaining food = 6750 - 1500 = 5250 unit
Let D be the number of days for which the remaining food .
So, 125 × D = 5250
? D = 5250 /125 = 42 days

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.