Two bicyclists cover the same path by cycling at the rate of nine km/hr and ten km/hr respectively. Find the length of journey of one takes 32 minutes longer than other.
Answers
Answer:
Two cyclists do the same journey by travelling at 9 km/hr and 10 km/hr respectively. Find the distance travelled when one takes 32 minutes longer than the other. Explanation: Let the distance traveled = x. ∴ x / 9 – x / 10 = 32 / 60 → x = 48 km.
Explanation:
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Given data : Two bicyclists cover the same path by cycling at the rate of 9 km/hr and 10 km/hr respectively.
To Find : The length of journey of one takes 32 minutes longer than other.
Solution : According to given data, both cyclists cover the equal distance.
Let, speed of cyclists A be 11 km/hr and speed of cyclists B be 13 km/hr and assume that cyclist A take 42 minutes more time than cyclist B.
We know,
➜ 42 minute = 42/60 hour = 7/10 hour
Let, time taken by B to cover distance be x
Hence,
➜ Time taken by A = (x + 7/10) hour.
Now,
➜ speed of B = distance/time
➜ 13 = distance/time
➜ 13 = distance/x
➜ distance = 13x ----{1}
Similarly,
➜ speed of A = distance/time
➜ 11 = distance/time
➜ 11 = distance/(x + 7/10)
➜ distance = 11 (x + 7/10) ----{2}
Cyclists A and B cover same distance hence, from eq. {1} and eq. {2}
➜ 13x = 11 (x + 7/10)
➜ 13x = 11x + 77/10
➜ 13x - 11x = 77/10
➜ 2x = 77/10
➜ x = (77/10)/2
➜ x = 77/(10 * 2)
➜ x = 77/20
➜ x = 3.85 hour
Now, put value of x in eq. {1}
➜ distance = 13x
➜ distance = 13 * 3.85
➜ distance = 50.05 km
Answer : The length of the journey is 50.05 km.
{Verification : put distance = 50.05 km in eq. {2}
➜ distance = 11 (x + 7/10)
➜ 50.05 = 11 (x + 7/10)
➜ 50.05 = 11x + (77/10)
➜ 50.05 = 11x + 7.7
➜ 11x = 50.05 - 7.7
➜ 11x = 42.35
➜ x = 42.35/11
➜ x = 3.85
Hence it verified.}