Question
The time taken by Boat 'X' to travel a distance of '2a'
km upstream is the same as the time it takes to travel '3a' km downstream. Similarly, the time taken by Boat 'Y' to cover '3b' km upstream equals the time to cover '5b' km downstream, and the time taken by Boat 'Z' to cover '5c' km upstream matches the time to cover '7c' km downstream. Quantity I: Boat 'X' travels 144 km downstream and then returns to the starting point in a total of 20 hours. Find the speed of Boat 'X' in still water. Quantity II: Boat 'Y' travels 150 km downstream and then returns to the starting point in a total of 16 hours. Determine the speed of Boat 'Y' in still water. Quantity III: Boat Z travels 140 km downstream and then returns to the starting point in a total of 12 hours. Calculate the speed of Boat 'Z' in still water. In the question, three quantities i.e. Quantity I, Quantity II and Quantity III are given. Solve the given quantities to establish the correct relation between them and choose the correct option.Solution
ATQ, For Boat X, Downstream Speed : Upstream Speed = 3a : 2a = 3 : 2 For Boat Y, Downstream Speed : Upstream Speed = 5b : 3b = 5 : 3 For Boat Z, Downstream Speed : Upstream Speed = 7c : 5c = 7 : 5 Quantity I, For Boat 'X', Downstream Speed = 3p km/hr Upstream Speed = 2p km/hr 144/3p + 144/2p = 20 48/p + 72/p = 20 120/p = 20 p = 6 Speed of Boat 'X' in still water = (3p + 2p)/2 = 5p/2 = 2.5 × 6 = 15 km/hr Quantity II, For Boat 'Y', Downstream Speed = 5q km/hr Upstream Speed = 3q km/hr 150/5q + 150/3q = 16 30/q + 50/q = 16 80/q = 16 q = 5 Speed of Boat 'Y' in still water = (5q + 3q)/2 = 8q/2 = 4 × 5 = 20 km/hr Quantity III, For Boat Z, Downstream Speed = 7r km/hr Upstream Speed = 5r km/hr 140/7r + 140/5r = 12 20/r + 28/r = 12 48/r = 12 r = 4 Speed of Boat Z in still water = (7r + 5r)/2 = 12r/2 = 6 × 4 = 24 km/hr Quantity I < Quantity II < Quantity III
The LCM of six consecutive numbers is 420. The sum of the first and fourth numbers is equal to the sixth number. What is the product of six numbe...
The ratio of two numbers is 9:12 and their LCM is 324. The numbers are:
Let N be the greatest number that will divide 85, 116, 147 leaving the same remainder in each case. Then sum of the digits in N is:
- Determine the difference between the LCM and HCF of the numbers 15, 25, and 50.
What are the Highest Common Factor and Lowest Common Multiple of 6, 72, 120?
Find the LCM of x² – 8x + 15 and x² – 5x + 6 is.Â
The smallest number which when divided by 25 and 30 leaves remainder 10 and 15 respectively is:
If total number of factors of 1,260 is 'x', then find the value of (x - 4) (x + 6).
Calculate the Least Common Multiple (L.C.M) of 144 and 180.
The LCM of two numbers is 1991 and their HCF is 23. If one of the numbers is 181, the other number is