Math, asked by Vivekshukla3181, 1 year ago

Two horses started simultaneously towards each other and meet each other 3 hr 20 mins later. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second 5 hours later than the second arrived at the point of departure of the first?
A.10 hours
B.5 hours
C.15 hours
D.6 hours

Answers

Answered by devansh5390
0
I am sure the answer is d.6hours
Answered by bhavya29007
0

Answer:

Ans: 10 hr

Step-by-step explanation:

Let distance between the two places  =d km

Let total time taken by slower horse  

= ( t + 5 ) hr,

total time taken by faster horse  = t  hr.

Therefore, speed of the slower horse  = d t + 5 km/hr

speed of the faster horse  

= d t  km/hr

The two horses meet each other in 3 hour 20 min.

(i.e., in  

3 1 3  hr  = 10 3 hr)

In this time, total distance travelled by both the horses together is  

d

.

Therefore,

dt + 5 × 10 3 + d t × 10 3 = d 10 3

( t + 5) + 10

3 t = 1 10 t + 10

( t + 5 ) = 3 t ( t + 5 )

20 t + 50 = 3

t 2 + 15 t 3

t 2 − 5 t − 50 = 0

2 + 10 t − 15 t − 50 = 0

t ( 3 t+ 10 ) − 5

( 3 t + 10) = 0

( 3 t + 10 )

( t − 5 ) = 0

t = 5

(ignoring -ve value)

i.e., total time taken by slower horse  = 5 + 5 = 10 hr

Step-by-step explanation:

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