Question

Xi rode her cycle at 2/3 m/s from her hostel to reach LHC 5 minutes late. Had she driven 25% faster than average she would have reached LHC 4 minutes earlier. How far is her LHC from her hostel?

Answer 1

Answer:

Distance = 6.3 m

Step-by-step explanation:

Suppose Xi was supposed to reach the hostel in t time, but he took 5 minutes more. So,

Time take by Xi to reach hostel = t +5

Speed of Xi Initially = V₁ = $\frac{2}{3}$ m/s

As we know

Distance  = Speed x Time

d = $\frac{2}{3} (t + 5)$ ....(1)

Now if Xi travels 25 percent faster his new speed can be calculated as

V₂ = V₁ + 25 % of V₁

V₂ = $\frac{2}{3}$ + $\frac{25}{100} . \frac{2}{3}$

V₂ = $\frac{2}{3}$ + $\frac{1}{4} . \frac{2}{3}$

V₂ = $\frac{2}{3}$ + $\frac{1}{6}$

V₂ = $\frac{4 + 1}{6}$

V₂ = $\frac{5}{6}$

V₂ = .84 m/s

Also

Time take by Xi to reach hostel = t - 4

So,

Distance = Speed x Time

d = $\frac{5}{6} (t - 4)$  .....(2)

Comparing equation (1) and (2)

$\frac{2}{3} (t + 5)$ =  $\frac{5}{6} (t - 4)$

$2(t + 5)$ =  $\frac{5}{2} (t - 4)$

using cross multiplication

4 (t + 5) = 5 (t - 4)

4t + 20 = 5t - 20

20 + 20 = 5t + 4t

40 = 9t

t = $\frac{40}{9}$

Now substitute value of t into equation (1)

d = $\frac{2}{3} (\frac{40}{9} + 5)$

d = $\frac{2}{3} (\frac{40 + 45}{9})$

d = $\frac{170}{27}$

d = 6.3 meters

So, our desired distance is 6.3 m

Answer 2

Answer:

1.8 km

Step-by-step explanation:

Define x:

let the distance be x

Convert 2/3 m/s to m/min

1 sec = 2/3 m

60 sec = 2/3 x 60 = 40 m

The speed is 40m/min

25% faster than the average speed:

25% of the speed = 0.25 x 40 = 10 m

The faster speed = 40 + 10 = 50 m/min

Find the time taken when she is 5 mins late:

Time = Distance ÷ Speed

Time = x/40 min

Find the time taken when she is 4 mins earlier:

Time = Distance ÷ Speed

Time = x/50 min

Solve x:

The difference in time = 4 - (- 5) = 9 mins

x/40 - x/50 = 9

50x - 40x = 9(40)(50)

10x = 18000

x = 1800 m = 1.8 km

Answer: The distance is 1.8 km