Question

Anita goes to school at 15 km/h and she is late by 4 minutes. The next
time she goes to school at 20 km/h and she reaches school a minute
early. What is the distance to her school?​

Answer 1

Answer:

let tie be x

1st case,

$speed(v_{1}) = 15km{h}^{-1} \\= \frac{25}{6}{ms}^{-1}\\</p><p>time(t_{1}) = x + 4 minutes = 60(x + 4) seconds \\</p><p> = 60x + 240 seconds\\</p><p>distance(s_{1}) = ?$

$speed(v) = \frac{distance(s)}{time(t)} \\ s = v \times t \\ s= \frac{25}{6} \times (60x + 240) \\ s = \frac{25}{6} \times 6(10x + 40) \\ s = 25(10x + 40) \\ s = 250x + 1000...(1)$

2nd case,

$speed(v_{2}) = 20km{h}^{-1}\\ = \frac{50}{9}{ms}^{-1}\\</p><p>time(t_{2}) = x - 1 minutes = 60(x - 1) seconds\\</p><p> =60x-60 seconds\\</p><p>distance(s_{2}) = ?$

$speed(v) = \frac{distance(s)}{time(t)} \\ s = v \times t \\ s= \frac{50}{9} \times (60x + 240) \\ s = \frac{50}{3 \times 3} \times 3(20x + 80) \\s = \frac{50}{3} (20x + 80) \\s = \frac{1000x + 4000}{3} ...(2)$

As the distance to school will remain same,

Thus the distances of case 1 and 2 will be equal

.°. $s_{1} = s_{2} = s$

$250x + 1000 = \frac{1000x + 4000}{3} \\ 3(250x + 1000) = 1000x + 4000 \\ 750x + 3000 = 1000x + 4000 \\ 3000 - 4000 = 1000x - 750x \\ - 1000 = 250x \\ x = \frac{ - 1000}{250} \\ x = \frac{ - 100}{25} \\ x = \frac{25( - 4)}{25(1)} \\ x = - 4$

Substituting x = -4 in (1)

$s = 250x + 1000 \\ s = 250( - 4) + 1000 \\s = - 1000 + 1000 \\s = 0$

Thus the distance will be 0 m

Answer 2

Given :- Anita goes to school at 15 km/h and she is late by 4 minutes. The next time she goes to school at 20 km/h and she reaches school a minute early. What is the distance to her school ?

Solution :-

Let us assume that, the distance of her school is x km.

Case 1) :-

→ Time taken to go to school at 15 km/h = D/S = (x/15) hours.

and she reaches 4 minutes late.

so, we can say that,

→ Actual time to reach school = {(x/15) - (4/60)} hours.

Case 2) :-

→ Time taken to go to school at 20 km/h = D/S = (x/20) hours.

and she reaches 1 minutes earlier.

so, we can say that,

→ Actual time to reach school = {(x/20) + (1/60)} hours.

therefore,

→ {(x/15) - (4/60)} = {(x/20) + (1/60)}

→ (x/15) - (x/20) = (1/60) + (4/60)

→ (4x - 3x)/60 = (5/60)

→ (x/60) = (5/60)

→ x = 5 km/h. (Ans.)

Hence, the distance of her school is 5 km/h .

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