We know that ,
( H1 × D1 )/k1 = ( H2 × D2 )/K2
( 3 × 12 )/2 = ( 4 × x )/10
( 3 × 12 × 10 )/( 2 × 4 ) = x
after cancellation , we get
45 = x
Therefore ,
Number of days = D2 = 45
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2 ) Let the Original average speed of
the train ( s1 ) = x km/hr
distance travelled ( d1 ) = 63 km
time ( t1 ) = d1/s1
t1 = 63/x hr ---( 1 )
3 ) If speed of the train increased,
new speed ( S2 ) = ( x + 6 )km/hr
distance ( d2 ) = 72 km
time ( t2 ) = d2/s2
t2 = 72/( x + 6 ) ---( 2 )
according to the problem given ,
t1 + t2 = 3 hours
63/x + 72/( x + 6 ) = 3
divide each term with 6 , we get
21/x + 24/( x + 6 ) = 1
[ 21( x + 6 ) + 24x ]/[ x( x + 6 ) ] = 1
21x + 126 + 24x = x² + 6x
126 + 45x = x² + 6x
0 = x² + 6x - 45x - 126
x² - 39x - 126 = 0
x² - 42x + 3x - 126 = 0
x( x - 42 ) + 3( x - 42 ) = 0
( x - 42 )( x + 3 ) = 0
x - 42 = 0 or x + 3 = 0
x = 42 or x = -3
but x should not be negative.
Therefore ,
Original average speed of the
train = x = 42km/hr
I hope this helps you.
: )
