Question

X goes to the office by driving at 5/7 times his usual speed. If he normally takes half an
hour to reach his office by how many minutes will he be late?
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Answer 1

Answer:

12 minutes

Step-by-step explanation:

time = distance ÷ speed

let t (=30minutes) be time x normally takes & d be the distance & v be speed

t= d/v......1

according to question, x drive at (5/7)v

=> t' = d÷(5/7)v

=> t' = (7/5) (d/v)

putting value of (d/v) form equation 1

we get,

t' = 7/5t = (7/5) × 30 minutes = 42minutes

t' - t = 42-30 = 12 minutes

so, x will be 12minutes late

Answer 2

Answer:

12 min

Step-by-step explanation:

let his actual speed is x km/hr

distance to office is d km

As given in the ques he drives 5/7 of his actual speed i:e 5/7x.

We know that

speed=dist/time

so 5/7x=d/t (where t is the time taken with speed 5/7x)----------- 1

In normal case

x=d/0.5 ----> x=2d-------------------2

becz distance is same

so we can equate 1 and 2

----> 5/7x * t=x/2

-----> t=7/10 hrs

Now convert this into min

t=7/10 * 60  ----->42min

Normal time is 30min and time cover with new speed is 42 min

Therefore he will be late by 12 min