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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