A student starting from his house walks at a
speed of 2 1/2 km/hour and reaches his school 6
minutes late. Next day starting at the same time
he increases his speed by 1 km/hour and reaches
6 minutes early. The distance between the school
and his house is ?
➦4 km
➦3 1/2 km
➦ 1 3/4 km
➦6 km
Answers
Answered by
1
Explanation:
Let the distance between the home & the school of the boy = x km.
In the first case, the speed is 2
21
km/hr =
2
5
km/hr.
∴ Time taken = x÷
2
5
hr.
=
5
2x
hr.
In the second case his speed= 1 +
2
3
km/hr. =
2
7
km/hr.
∴ Time taken=x÷
2
7
hr.=
7
2x
hr.
∴ The difference in time = (
5
2x
−
7
2x
)hr.=
35
4x
hr.=
35
4x
×60min.=
7
48x
×min.
Now , in the first case he is 6 min late = −6 min. and
In the second case, he is 6 min early = +6 min.
∴ The difference in time=[6−(−6)]min.=12 min.
∴
7
48x
min.=12 min.
Or 4x=7
Or x=
4
7
=1.75.
∴ The distance between the home & the school of the boy = 1.75 km.
Answered by
3
Answer:
Oye ldke....vas' happenin'??
um... unfortunately mai on nhi hoti jb tum on aate ho┐(´ー｀)┌
Similar questions