consider the number 4n,where n is a natural number . Check whether there is any value of n for which 4n ends with the digit zero.
Answers
Given:−
\sf{\implies Total\: distance=400km}⟹Totaldistance=400km
\sf\small\underline\green{To\: Find:-}
ToFind:−
\sf{\implies speed\:of\: train=?}⟹speedoftrain=?
\sf\small\underline\green{Given:-}
Given:−
we will solve this question by applying formula of time = Distance/speed.
\sf\small\underline\green{Case-(i):-}
Case−(i):−
\sf{\implies Speed=40km/h\:\:\: Distance=250km}⟹Speed=40km/hDistance=250km
\tt{\implies Time\:_{(1)}=\dfrac{Distance}{Speed}}⟹Time
(1)
=
Speed
Distance
\tt{\implies Time\:_{(1)}=\dfrac{250}{40}}⟹Time
(1)
=
40
250
\tt{\implies Time\:_{(1)}=\dfrac{25}{4}}⟹Time
(1)
=
4
25
\sf{\implies Speed=x\:km/h\:\:\: Distance=150km}⟹Speed=xkm/hDistance=150km
\tt{\implies Time\:_{(2)}=\dfrac{Distance}{Speed}}⟹Time
(2)
=
Speed
Distance
\tt{\implies Time\:_{(2)}=\dfrac{150}{x}}⟹Time
(2)
=
x
150
\sf\small\underline\green{Case-(ii):-}
Case−(ii):−
\tt{\implies Total\:_{(Distance)} = 400km}⟹Total
(Distance)
=400km
\tt{\implies Total\:_{(Time)} =T\:_{(1)}+T\:_{(2)}}⟹Total
(Time)
=T
(1)
+T
(2)
Now speed of train in next case:-
\tt{\implies Average\:_{(speed)}=\dfrac{Total\: distance}{Total\:time}}⟹Average
(speed)
=
Totaltime
Totaldistance
\tt{\implies 60=\dfrac{400}{25/4+150/x}}⟹60=
25/4+150/x
400
\tt{\implies 60=\dfrac{400}{(25x+600)/4x}}⟹60=
(25x+600)/4x
400
\tt{\implies 60=\dfrac{1600x}{25x+600}}⟹60=
25x+600
1600x
\tt{\implies 60(25x+600)=1600x}⟹60(25x+600)=1600x
\tt{\implies 1500x + 36000=1600x}⟹1500x+36000=1600x
\tt{\implies 1600x-1500x=36000}⟹1600x−1500x=36000
\tt{\implies 100x=36000}⟹100x=36000
\tt{\implies x=360km/h}⟹x=360km/h
\sf\large{Hence}Hence
\bf{\implies Speed\:of\: train\:_{(next\:150km)}=360km/h}⟹Speedoftrain
(next150km)
=360km/h