Last updated at May 29, 2018 by Teachoo

Transcript

Example 19 A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water. Let the speed of boat in still water be x km/hr & let the speed of current(stream) be y km/hr Speed downstream = Speed of boat in still water + Speed of stream Speed downstream = x + y Speed upstream = Speed of boat in still water – Speed of stream Speed upstream = x – y Given that A boat goes 30 km upstream and 44 km downstream in 10 hours Time taken to go 30 km upstream + Time taken to go 44 km downstream (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 30 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 44 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 10 30/(𝑥 − 𝑦) + 44/(𝑥 + 𝑦) = 10 Similarly, A boat goes 40 km upstream and 55 km downstream in 13 hours Time taken to go 40 km upstream + Time taken to go 55 km downstream (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 40 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 55 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 13 40/(𝑥 − 𝑦) + 55/(𝑥 + 𝑦) = 13 Our equations are 30 (1/(𝑥 − 𝑦))+44(1/(𝑥 + 𝑦))=10 …(1) 40 (1/(𝑥 − 𝑦))+55(1/(𝑥 + 𝑦))=13 …(2) Solving 30u + 44v = 10 …(3) 40u + 55v = 13 …(4) From (3) 30u + 44v = 10 30u = 10 – 44v u = (10 −44𝑣)/30 Putting value of u in (4) 40u + 55v = 13 40((10−44𝑣)/30)+55𝑣=13 4((10−44𝑣)/3)+55𝑣=13 Multiplying both sides by 3 3 × 4((10−44𝑣)/3)+"3 ×" 55𝑣="3 ×" 13 4(10 – 44v) + 165𝑣= 39 40 – 176v + 165v = 39 – 176v + 165v = 39 – 40 – 11v = -1 v = (−1)/(−11) v = 1/11 Putting v = 1/11 in equation (3) 30u + 44v = 10 30u + 44(1/11) = 10 30u + 4 = 10 30u = 10 – 4 30u = 6 u = 6/30 u = 1/5 So, u = 1/5 & v = 1/11 So, u = 1/5 & v = 1/11 But we need to find x & y We know that So, our equations become x – y = 5 …(6) x + y = 11 …(7) From (6) x – y = 5 x = 5 + y Putting value of x in (7) x + y = 11 (5 + y) + y = 11 5 + 2y = 11 2y = 11 – 5 2y = 6 y =6/2 y = 3 Put y = 3 in (6) x – y = 5 x – 3 = 5 x = 5 + 3 x = 8 So, x = 8, y = 3 is the solution of the given equation Hence Speed of boat in still water = x = 8 km/hr Speed of stream = y = 3 km/hr

Example 6
Important

Example 10 Important

Example 11 Important

Example 13 Important

Example 14 Important

Example 19 Important You are here

Ex 3.2, 1 Important

Ex 3.2, 3 Important

Ex 3.2, 6 Important

Ex 3.2, 7 Important

Ex 3.5, 2 Important

Ex 3.6, 1 (i) and (ii) Important

Ex 3.6, 1 (iii) and (iv) Important

Ex 3.6, 1 (v) and (vi) Important

Ex 3.6, 1 (vii) and (viii) Important

Ex 3.6, 2 Important

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.