Chapter 4 Class 10 Quadratic Equations

Class 10
Important Questions for Exam - Class 10

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Example 15 A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Given that speed of the boat = 18 km/ hr. Let the speed of the stream = x km / hr. Given that Time taken upstream is 1 hour more than time taken downstream Time upstream = Time downstream + 1 24/((18 β π₯)) = 24/((18 + π₯)) + 1 24/((18 β π₯)) β 24/((18 + π₯)) = 1 (24(18 + π₯) β 24(18 β π₯))/((18 β π₯)(18 + π₯)) = 1 24((18 + π₯) β (18 β π₯))/((18 β π₯)(18 + π₯)) = 1 24(18 + π₯ β 18 + π₯)/((18 β π₯)(18 + π₯)) = 1 24(2π₯)/((18 β π₯)(18 + π₯)) = 1 48π₯/((18 β π₯)(18 + π₯)) = 1 48x = (18 β x) (18 + x) 48x = 182 β x2 48x = 324 β x2 x2 + 48x β 324 = 0 Comparing equation with ax2 + bx + c = 0, Here a = 1, b = 48, c = β324 We know that D = b2 β 4ac D = (48)2 β 4 Γ 1 Γ (β324) D = 2304 + 4 Γ 324 D = 2304 + 1296 D = 3600 So, the roots to equation are x = (βπ Β± βπ·)/2π Putting values x = (β(48) Β± β3600)/(2 Γ 1) x = (β 48 Β± β(60 Γ 60))/(2 Γ 1) x = (β 48 Β± 60)/2 Solving So, x = 6 & x = β 54 Since, x is the speed , so it cannot be negative So, x = 6 is the solution of the equation Therefore, speed of the stream (x) = 6 km /hr.