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Chapter 4 Class 10 Quadratic Equations
Last updated at May 29, 2018 by Teachoo
Ex 4.3 ,9 Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Let the time taken by smaller tap to fill tank completely = x hours So, Volume of tank filled by smaller tap in 1 hour = 1/π₯ Also , it is given that time take by larger tap is 10 hour less Time taken by larger tap to fill tank completely = Time taken by smaller tap β 10 = x β 10 hours Volume of Tank filled by larger tap in 1 hour = 1/(π₯ β10) Now, it is given that Time taken by both taps to fill = 9 3/8 hours = (9(8)+ 3)/8 hours = 75/8 hours Also, Tank filled by smaller tap in 1 hrs. = 1/π₯ Tank filled by smaller tap in 75/8 hrs. = 1/π₯Γ75/8 = 75/8π₯ Also, Tank filled by larger tap in 1 hrs. = 1/(π₯ β 10) Tank filled by larger tap in 75/8 hours = 1/(π₯ β10)Γ75/8=75/(8(π₯β10)) Tank filled by smaller tap + tank filled by larger tap = 1 75/8π₯+75/(8(π₯ β 10))=1 75/8 (1/π₯+1/(π₯ β 10))=1 1/π₯+1/(π₯ β 10)=8/75 (π₯ β10 + π₯)/(π₯(π₯ β 10))=8/75 (2π₯ β10)/(π₯2 β10π₯)=8/75 (2x β 10) 75 = 8 (x2 β 10x) 150x β 750 = 8x2 β 80x 150x β 750 β 8x2 + 80x = 0 β 8x2 + 150x + 80x β 750 = 0 β 8x2 + 230x β 750 = 0 0 = 8x2 β 230x + 750 8x2 β 230x + 750 = 0 We solve this by quadratic method Comparing equation with ax2 + bx + c = 0, Here a = 8, b = β230, c = 750 We know that D = b2 β 4ac D = (β230)2 β 4 Γ 8 Γ ( 750) D = 52900 β 24000 D = 28900 So, the roots to equation are x = (βπ Β± βπ·)/2π Putting values x = (β(β230) Β± β28900)/(2 Γ 8) x = (230 Β± β28900)/16 x = (230 Β± β(289 Γ 100))/16 x = (230 Β± β289 Γ β100)/16 x = (230 Β± β(γ17γ^2 ) Γ β(γ10γ^2 ))/16 x = (230 Β± 17 Γ 10)/16 x = (230 Β± 170)/16 Solving