Question

How to calculate the speed of the slower horse in the following question?

Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km, what is the speed of the slower horse?
Options are as follows:
1) 30 km/h
2) 15 km/h
3) 20 km/h
4) 25 km/h

Answer 1

3) 20 km/h

Explanation:

We have two different times and distances related by the same rate. The basic equation is `D = RxxT`. The first set is the meeting point, which must be at a point `50 - x` from one side and `x` from the other. Thus, the two rate equations are:
`D_1 = R_1xx1` and `D_2 = R_2xx1`
`50 - x = R_1` and `x = R_2`

The second set are the time equations related to the total distance:
`50 = R_1xxT_1` and `50 = R_2xxT_2`
We also know that `T_1 = T_2 - 5/6` or `T_2 = T_1 + 5/6`

`50 = R_2xx (T_1 + 5/6)` ; `R_2 = 50/(T_1 + 5/6)`
`50 = R_1xxT_1` ; `R_1 = 50/(T_1)`
Pick a number from the options to avoid further math manipulations.
For:
`R_2 = 20`
`20 = 50/(T_1 + 5/6)` ; `20T_1 + 100/6 = 50`
`T_1 = (50 - 100/6)/20 = 1.67`
`50 = R_1xxT_1` ; `50 = R_1xx1.6667`
`R_1 = 30`
Check:
`50 - x = R_1` and `x = R_2`
`50 - x = 30` and `x = 20`
`50 - 20 = 30` CORRECT!

For:
`R_2 = 30`
`30 = 50/(T_1 + 5/6)` ; `30T_1 + 100/6 = 50`
`T_1 = (50 - 100/6)/30 = 1.11`
`50 = R_1xxT_1` ; `50 = R_1xx1.11`
`R_1 = 45`
Check:
`50 - x = R_1` and `x = R_2`
`50 - x = 45` and `x = 30`
`50 - 30 != 45` INcorrect!
You can try out the others.