A person walked at a speed of 9 km/h to taxi
stand and then took taxi, which travelled at
a speed of 12.5 m/sec. If he reached his office
28 km away in 48 minutes, find the distance
between his home and taxi stand and also
the time taken to cover this distance.
Answers
Answer:
The distance from home to taxi stand is 2.07 km in 0.23 hours.
Step-by-step explanation:
Given : A person walked at a speed of 9 km/h to taxi stand and then took taxi, which traveled at a speed of 12.5 m/sec If he reached his office 28 km away in 48 minutes.
To Find : The distance between his home and taxi stand and also the time taken to cover this distance ?
Solution :
Let the distance covered by person in walking is d_1d
1
Speed of the person is s=9\ km/hrs=9 km/hr
Time taken in walking is t_1t
1
According to relationship between them,
d_1=9t_1d
1
=9t
1
.....(1)
Let the distance covered by person in taxi is d_2d
2
Speed of the person is s=12.5\ m/ss=12.5 m/s
Convert km/hr into m/s,
s=12.5\times \frac{18}{5}=45\ km/hrs=12.5×
5
18
=45 km/hr
Time taken in walking is t_2t
2
According to relationship between them,
d_2=45t_2d
2
=45t
2
.....(2)
If he reached his office 28 km away in 48 minutes.
i.e. d_1+d_2=28d
1
+d
2
=28
Substitute the values,
9t_1+45t_2=289t
1
+45t
2
=28 ...(3)
Converting minutes into hours,
48 minutes = 0.8 hours
and t_1+t_2=0.8t
1
+t
2
=0.8 ....(4)
Solving (3) and (4) by multiplying (4) by 9 and subtract from (3),
9t_1+45t_2-9t_1-9t_2=28-7.29t
1
+45t
2
−9t
1
−9t
2
=28−7.2
36t_2=20.836t
2
=20.8
t_2=\frac{20.8}{36}t
2
=
36
20.8
t_2=0.57t
2
=0.57
Substitute in (4),
t_1+0.57=0.8t
1
+0.57=0.8
t_1=0.8-0.57t
1
=0.8−0.57
t_1=0.23t
1
=0.23
Substitute the value of t_1t
1
in (1),
d_1=9(0.23)d
1
=9(0.23)
d_1=2.07d
1
=2.07
The distance from home to taxi stand is 2.07 km in 0.23 hours.
Substitute the value of t_2t
2
in (2),
d_2=45(0.57)d
2
=45(0.57)
d_2=25.65d
2
=25.65
The distance from taxi to office is 25.65 km in 0.57 hours.
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