Question

If an employee walks at 2/3 of his usual speed, he gets late to the office by 15 mins. What time does he usually take to reach the office? A. 10 min. B. 30 min C. 35 min, D. 25 min

Answer 1

Answer is 30 minutes....



Solution...
 
Let original speed be x and original time taken be t
let d be the distance 
 
Speed = d/t
x=d/t

if speed be 2x/3 time will be t+15


2x/3 = d/(t+15)
2(d/t)/3 = d/(t-+15)
2/3t = 1/(t+15)

Cross multiply

2(t+15) = 3t
2t+30 = 3t
3t-2t = 30
 
t = 30..............(answer)

Hope it is correct

Answer 2

B. 30 min

Step-by-step explanation:

Let the actual time to reach office by an employee be 't'.

Also Let the speed be denoted by 's'.

Now Let distance traveled be denoted by 'd'.

Now we know that;

Speed is equal to distance upon time.

So,

$s= \frac{d}{t} \ \ \ \ equation \ 1$

Now Given:

2/3 of his usual speed, he gets late to the office by 15 mins.

When speed is  $\frac{2}{3} \times s$ then time will be equal to $t+15$

Speed is equal to distance upon time.

$\frac{2}{3}\times s = \frac{d}{t+15}$

Substituting the value of s from equation 1 in above equation we get;

$\frac{2}{3}\times \frac{d}{t} = \frac{d}{t+15}$

By Using Cross Multiplication we get;

$2(t+15)=\frac{3td}{d}\\\\2t+30 = 3t\\\\3t-2t=30\\\\t =30\ mins$

Hence the Actual time required to reach the office by an employee is in 30 mins.