Posted by Aldwin on Thursday, November 10, 2011 at 8:25am.
mr kumar lives in the eastern part of spore . he visits his aged parents , who lives 36 km away , every weekend he finds that is he increases the average speed of his vehicle by 12km/h, he could save 9 minutes of his travelling time. find the speed at which he travels before the increase in speed

MAth grade 8  Reiny, Thursday, November 10, 2011 at 9:23am
speed before increase : x km/h
time for trip with old speed = 36/x hrs
speed after increase : x+12 km/h
time for trip after increase = 36/(x+12)
36/x  36/(x+12) = 9/60 = 3/20
divide by 3
12/x  12/(x+12) = 1/20
multiply be 20x(x+12)
240(x+12)  240x = x(x+12)
240x + 2880  240x = x^2 + 12x
x^2 + 12x  2880 = 0
x^2 + 12 + 36 = 2880 + 36
(x+6)^2 = 2916
x+6 = ± √2916 = ±54
x = 6 ± 54
x = 48 or a negative
So he originally travelled at 48 km/h
check:
first time = 36/48= .75 hrs = 45 minutes
second time = 36/60 = .6 hrs = 36 minutes,
sure enough, the difference he gains is 9 minutes.
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