A dog takes 4 leaps for every 5 leaps of a hare, but 3 leaps of the dog cover…
2025
A dog takes 4 leaps for every 5 leaps of a hare, but 3 leaps of the dog cover the same distance as 4 leaps of the hare. Compare the speed of the dog to that of the hare.
- A.
25:21
- B.
7:5
- C.
16:15
- D.
5:3
Attempted by 2 students.
Show answer & explanation
Correct answer: C
Concept: When two bodies move by taking leaps, each one's speed equals its leap length multiplied by how many leaps it takes per unit of time. To compare two such speeds, first convert the given leap-count information into a leap-length ratio, then multiply that ratio by the leap-frequency ratio for the same time interval.
Let the length of one dog leap be D and the length of one hare leap be H.
Since 3 dog leaps cover the same distance as 4 hare leaps: 3D = 4H, so D = (4/3)H.
In a fixed time interval the dog takes 4 leaps while the hare takes 5 leaps, so the leap-frequency ratio (dog : hare) is 4 : 5.
Distance covered by the dog in that time is proportional to D x 4 = (4/3)H x 4 = (16/3)H.
Distance covered by the hare in that time is proportional to H x 5 = 5H.
Speed ratio (dog : hare) = (16/3)H : 5H = 16 : 15.
Cross-check: Take H = 3 units, so D = 4 units. In one time unit the dog covers 4 x 4 = 16 units while the hare covers 5 x 3 = 15 units -- the same 16 : 15 ratio, confirming the result independently.