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Byju's Answer
Standard VIII
Mathematics
Solving Linear Equations Using Transpose of Terms
Ajay daily jo...
Question
Ajay daily jogs
2
5
k
m
.
and Sheelu jogs
1
3
k
m
.
How many km. more than Ajay does Sheelu jog?
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Solution
Ajay jogs
2
5
t
h
of a kilometre
=
2
5
×
1000
m
=
2
×
200
=
400
m
Sheelu jogs
1
3
r
d
of a kilometre
=
1
3
×
1000
m
=
1000
3
=
333.33
m
∴
Ajay jogs more distance than Sheelu by
400
−
333.33
=
66.67
m
We know that
1000
m
=
1
k
m
⇒
66.67
m
=
66.67
1000
k
m
=
0.06667
k
m
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