Question 1139833
r * t = q


r is the rate
t is the time
q is the quantity.


the quantity in this case is one ploughed field.


the carabao can do it in 6 hours, therefore r * t = q becomes r * 6 = 1.
solve for r to get r = 1/6.


the tractor can do it in 4 hours, therefore r * t = q becomes r * 4 = 1.
solve for r to get r = 1/4.


the rate of the carabao is 1/6 of the field in 1 hour.
the rate of the tractor is 1/4 of the field in 1 hour.


the farmer uses the carabao to plough the field for 2 hours.


r * t = q becomes 1/6 * 2 = q.
solve for q to get q = 1/6 * 2 = 1/6 = 1/3.


in 2 hours, the farmer has ploughed 1/3 of the field.


that leave 2/3 of the field to still be ploughed.


another farmer decides to help him using the tractor.


the last 2/3 of the field is then ploughed using the carabao and the tractor.


when both are used, their rates are additive.


the formula of r * t = q becomes (1/6 + 1/4) * t = 2/3


simplify to get 5/12 * t = 2/3


solve for t to get t = 2/3 * 12/5 = 8/5.


the remaining 2/3 of the field is ploughed in 8/5 = 1 and 3/5 hours.


that's your solution.


r * t = q becomes 1/6 * 2 + 5/12 * 8/5 = q
1/6 * 2 is with the carabao only.
5/12 * 8/5 is with both the carabao and the tractor.
solve for q to get q = 1
the whole field is ploughed in 2 hours and 8/5 hours = 3 and 3/5 hours.
that was 2 hours using the carabao and 1 and 3/5 hours using both the tractor and the carabao.


any questions, write to dtheophilis@gmail.com