A student starts from his house with a speed 2kmph and reaches his college 3 min late. Next day he increases his speed by 1kmph and reaches college 3 min earlier. Find the distance between his house and the college.
Answers
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Answer:
The distance between his house and college is 0.6 km.
Explanation:
Given : A student starts from his house with a speed 2 kmph and reaches his college 3 min late. Next day he increases his speed by 1 kmph and reaches college 3 min earlier.
To find : The distance between his house and college?
Solution :
Let x be the distance between house and college.
Let t be the time to reach from house to college.
A student starts from his house with a speed 2 kmph and reaches his college 3 min late.
Speed = 2 kmph
Time = t+\frac{3}{60}t+
60
3
Distance = Speed × Time
x=2(t+\frac{3}{60})x=2(t+
60
3
) ....(1)
Next day he increases his speed by 1 kmph and reaches college 3 min earlier.
Speed = 2+1=3 kmph
Time = t-\frac{3}{60}t−
60
3
Distance = Speed × Time
x=3(t-\frac{3}{60})x=3(t−
60
3
) ......(2)
As the distance is same equate (1) and (2),
2(t+\frac{3}{60})=3(t-\frac{3}{60})2(t+
60
3
)=3(t−
60
3
)
2t+\frac{6}{60}=3t-\frac{9}{60}2t+
60
6
=3t−
60
9
3t-2t=\frac{6}{60}+\frac{9}{60}3t−2t=
60
6
+
60
9
t=\frac{6+9}{60}t=
60
6+9
t=\frac{15}{60}t=
60
15
t=\frac{1}{4}t=
4
1
The time taken to reach house to college is \frac{1}{4}
4
1
hours.
Substitute in equation (1),
x=2(\frac{1}{4}+\frac{3}{60})x=2(
4
1
+
60
3
)
x=\frac{1}{2}+\frac{1}{10}x=
2
1
+
10
1
x=\frac{10+2}{20}x=
20
10+2
x=\frac{12}{20}x=
20
12
x=\frac{3}{5}x=
5
3
x=0.6kmx=0.6km
Therefore, The distance between his house and college is 0.6 km.
Hope This may be helpful to you.
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