Last updated at May 29, 2018 by Teachoo

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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 = 1/๐ฅ 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 โ 10 = x โ 10 hours Volume of Tank filled by larger tap in 1 hour = 1/(๐ฅ โ10) 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. = 1/๐ฅ Tank filled by smaller tap in 75/8 hrs. = 1/๐ฅร75/8 = 75/8๐ฅ Also, Tank filled by larger tap in 1 hrs. = 1/(๐ฅ โ 10) Tank filled by larger tap in 75/8 hours = 1/(๐ฅ โ10)ร75/8=75/(8(๐ฅโ10)) Tank filled by smaller tap + tank filled by larger tap = 1 75/8๐ฅ+75/(8(๐ฅ โ 10))=1 75/8 (1/๐ฅ+1/(๐ฅ โ 10))=1 1/๐ฅ+1/(๐ฅ โ 10)=8/75 (๐ฅ โ10 + ๐ฅ)/(๐ฅ(๐ฅ โ 10))=8/75 (2๐ฅ โ10)/(๐ฅ2 โ10๐ฅ)=8/75 (2x โ 10) 75 = 8 (x2 โ 10x) 150x โ 750 = 8x2 โ 80x 150x โ 750 โ 8x2 + 80x = 0 โ 8x2 + 150x + 80x โ 750 = 0 โ 8x2 + 230x โ 750 = 0 0 = 8x2 โ 230x + 750 8x2 โ 230x + 750 = 0 We solve this by quadratic method Comparing equation with ax2 + bx + c = 0, Here a = 8, b = โ230, c = 750 We know that D = b2 โ 4ac D = (โ230)2 โ 4 ร 8 ร ( 750) D = 52900 โ 24000 D = 28900 So, the roots to equation are x = (โ๐ ยฑ โ๐ท)/2๐ Putting values x = (โ(โ230) ยฑ โ28900)/(2 ร 8) x = (230 ยฑ โ28900)/16 x = (230 ยฑ โ(289 ร 100))/16 x = (230 ยฑ โ289 ร โ100)/16 x = (230 ยฑ โ(ใ17ใ^2 ) ร โ(ใ10ใ^2 ))/16 x = (230 ยฑ 17 ร 10)/16 x = (230 ยฑ 170)/16 Solving

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.