Question

A person walking at 4 kmph reaches his office 8 minutes late. If he walks at 6 kmph, he reaches there 8 minutes earlier. How far is the office from his house? A. 3 1/5 km. B. 2 1/5 km. C. 4 1/5 km. D. 5 1/5 km.

Answer 1

ANSWER:

Distance = 3.2 km = 3 1/5 km

CONVERSIONS UESD:

1 kmph = 1000/3600 m/s = 5/18 m/s

1 minute = 60 s

1 km = 1000 m

GIVEN:

$v_1 = 4 \:kmph = 4(5/18) \\v_1 = 20/18 = 10/9\: m/s\\ t_1 = x + 480\\ \\ v_2 = 6\: kmph = 6(5/18) \\v_2= 30/18 = 5/3 \:m/s\\ t_2 = x - 480$

FORMULA:

DISTANCE = SPEED × TIME

EXPLANATION:

$d = \frac{10}{9} (x + 480) \\ \\d= \frac{5}{3} (x - 480) \\ \\ Equate \: d \: values \\ \\ \frac{10}{9} (x + 480) = \frac{5}{3} (x - 480) \\ \\ \frac{10}{5} (x + 480) = \frac{9}{3} (x - 480) \\ \\ 2(x + 480) = 3(x - 480) \\ \\ 2x + 960 = 3x -1440 \\ \\ x = 960 + 1440 \\ \\ x = 2400 \: s \\ \\ Substitute \: x = 2400 \: in \: d \\ \\ d = \frac{10}{9} (2400 + 480) \\ \\ d = 10( \frac{2880}{9} ) \\ \\ d = 10 \times 320 \\ \\ d = 3200 \: m \: (or) \: 3.2 \: km$

Distance = 3.2 km = 3 1/5 km

Answer 2

$\large\bf\underline{Given:-}$

• A person walking at 4 kmph reaches his office 8 minutes late. If he walks at 6 kmph, he reaches there 8 minutes earlier.

$\large\bf\underline {To \: find:-}$

• Distance of office from house

$\huge\bf\underline{Solution:-}$

Let the distance from house to office be x km.

Let the actual time be t.

$\large \star{ \underline{\rm \:Condition \: 1st : - }}$

If the person walks with the speed of 4km/hr then he reaches his office 8 min late.

• ✥ Time = Distance/speed

↣x/4 = t + 8/60

↣ x/4 = (60t + 8)/60

↣60x = 240t + 32

Divide both side by 4.

≫ 15x = 60t + 8

≫ 15x - 60t = 8.....(1)

$\large \star{\underline{\rm \:Condition \: 2nd : - }}$

If the person walks at 6 km/hr, he reaches there 8 minutes earlier.

• ✥ Time = Distance/speed

↣x/6 = t - 8/60

↣x/6 = (60t - 8)/60

↣60x = 360t - 48

Divide both side by 6

↣10x = 60t - 8

≫ 10x - 60t = -8....(2)

• ▶️Solving 1) and 2)

$\tt 15x - 60t = \: \: \: 8 \\ \tt 10x - 60t =- 8 \\ \rm \underline{(- ) \: \: \: \: \: ( + ) \: \: \: \: \: \: \: \: ( + ) \: \: } \\ \tt \underline{ 5x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \: \: 16 \: \: \: \: \: \: }$

• ≫ x = 16/5

Hence,

Distance from house to office is 16/5

16/5 can be written as = $\sf \: 3 \frac{1}{5}$

So,

Option A) $\sf \: 3 \frac{1}{5}$ ✔is correct

$\rule{200}3$