A boat running upstream takes 8 hours 48 min to cover a certain distance, while it takes 4 hours to cover the same distance downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?
(a) 2:1
(b) 3:1
(c) 8:3
(d) 4:3
Answer
293.7k+ views
Hint: First, before proceeding for this, we need to suppose the speed of the boat in upstream be x km/hr and speed for the downstream as y km/hr. Then, the condition given in the question clearly says that the distance covered for upstream in 8 hrs 48 min is equal to the distance covered for downstream in 4 hrs. Then, now we should be aware of the formulas for the speed of the boat and speed of the water stream as $\dfrac{x+y}{2}$and $\dfrac{y-x}{2}$respectively. Then, by using the ratio calculated above, we get the ratio between the speed of the boat and the speed of the water current.
Complete step-by-step answer:
In this question, we are supposed to find the ratio between the speed of the boat and the speed of the water current respectively when a boat running upstream takes 8 hours 48 min to cover a certain distance, while it take 4 hours to cover the same distance downstream.
So, before proceeding for this, we need to suppose the speed of the boat in upstream be x km/hr and speed for the downstream as y km/hr.
Now, the condition given in the question clearly says that the distance covered for upstream in 8 hrs 48 min is equal to the distance covered for downstream in 4 hrs.
Now, by using the above condition we get the equation as multiplication of speed for downstream to its time and multiplication of upstream speed with its time as:
$x\times 8\dfrac{48}{60}=y\times 4$
Then, we will solve the above expression to get the value of the ratio x:y as:
$\begin{align}
& 8\dfrac{4}{5}x=4y \\
& \Rightarrow \dfrac{44}{5}x=4y \\
& \Rightarrow \dfrac{x}{y}=\dfrac{4\times 5}{44} \\
& \Rightarrow \dfrac{x}{y}=\dfrac{5}{11} \\
\end{align}$
So, now we should be aware of the formulas for the speed of the boat and speed of the water stream as:
Then, speed of the boat is given by $\dfrac{x+y}{2}$.
Then, speed of the water stream is given by $\dfrac{y-x}{2}$.
So, now by dividing the speed of the boat with the speed of the water current to get the ratio as:
$\dfrac{\dfrac{x+y}{2}}{\dfrac{y-x}{2}}=\dfrac{x+y}{y-x}$
Now, divide the numerator and denominator by y to get the required ratio as:
$\dfrac{\dfrac{x}{y}+1}{1-\dfrac{x}{y}}$
Then, by substituting the value of $\dfrac{x}{y}$as $\dfrac{5}{11}$ calculated above to get the ratio as:
\[\begin{align}
& \dfrac{\dfrac{5}{11}+1}{1-\dfrac{5}{11}}=\dfrac{\dfrac{5+11}{11}}{\dfrac{11-5}{11}} \\
& \Rightarrow \dfrac{16}{6} \\
& \Rightarrow \dfrac{8}{3} \\
\end{align}\]
So, we get the ratio between the speed of the boat and the speed of the water current respectively as 8:3.
Hence, option (c) is correct.
Note: Now, to solve these types of the questions we need to know some of the basic formulas for the time, speed and distance. So, we must know the relation between speed(s), time (t) and distance (d) as:
$s=\dfrac{d}{t}$
Complete step-by-step answer:
In this question, we are supposed to find the ratio between the speed of the boat and the speed of the water current respectively when a boat running upstream takes 8 hours 48 min to cover a certain distance, while it take 4 hours to cover the same distance downstream.
So, before proceeding for this, we need to suppose the speed of the boat in upstream be x km/hr and speed for the downstream as y km/hr.
Now, the condition given in the question clearly says that the distance covered for upstream in 8 hrs 48 min is equal to the distance covered for downstream in 4 hrs.
Now, by using the above condition we get the equation as multiplication of speed for downstream to its time and multiplication of upstream speed with its time as:
$x\times 8\dfrac{48}{60}=y\times 4$
Then, we will solve the above expression to get the value of the ratio x:y as:
$\begin{align}
& 8\dfrac{4}{5}x=4y \\
& \Rightarrow \dfrac{44}{5}x=4y \\
& \Rightarrow \dfrac{x}{y}=\dfrac{4\times 5}{44} \\
& \Rightarrow \dfrac{x}{y}=\dfrac{5}{11} \\
\end{align}$
So, now we should be aware of the formulas for the speed of the boat and speed of the water stream as:
Then, speed of the boat is given by $\dfrac{x+y}{2}$.
Then, speed of the water stream is given by $\dfrac{y-x}{2}$.
So, now by dividing the speed of the boat with the speed of the water current to get the ratio as:
$\dfrac{\dfrac{x+y}{2}}{\dfrac{y-x}{2}}=\dfrac{x+y}{y-x}$
Now, divide the numerator and denominator by y to get the required ratio as:
$\dfrac{\dfrac{x}{y}+1}{1-\dfrac{x}{y}}$
Then, by substituting the value of $\dfrac{x}{y}$as $\dfrac{5}{11}$ calculated above to get the ratio as:
\[\begin{align}
& \dfrac{\dfrac{5}{11}+1}{1-\dfrac{5}{11}}=\dfrac{\dfrac{5+11}{11}}{\dfrac{11-5}{11}} \\
& \Rightarrow \dfrac{16}{6} \\
& \Rightarrow \dfrac{8}{3} \\
\end{align}\]
So, we get the ratio between the speed of the boat and the speed of the water current respectively as 8:3.
Hence, option (c) is correct.
Note: Now, to solve these types of the questions we need to know some of the basic formulas for the time, speed and distance. So, we must know the relation between speed(s), time (t) and distance (d) as:
$s=\dfrac{d}{t}$
Recently Updated Pages
The average and RMS value of voltage for square waves class 12 physics JEE_Main

The force between two short electric dipoles placed class 12 physics JEE_Main

Number of sigma and pi bonds in C2 molecule isare A class 11 chemistry JEE_Main

The dimensional formula of k Coulombs Constant is Take class 11 physics JEE_Main

The force of interaction of two dipoles if the two class 12 physics JEE_Main

The value of current through 2Omega resistor is A 10A class 12 physics JEE_MAin

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

