Question

Sangeetha travels 14 Km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 Km by rickshaw, and the remaining distance by bus. On the other hand, if she travels 4 Km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

Answer 1

Hint: We assume the speed of the rickshaw and the speed of the bus as different variables. Using the formula of speed we write the formula of time in terms of speed and distance. Form two equations using time as a factor. We will solve for the values of variables using the required method.

* If ‘d’ distance is covered in time ‘t’, then speed ‘s’ is given by \[s = \dfrac{d}{t}\]

Complete step by step answer:
We are given the total distance covered by Sangeetha on her way home is 14 Km.
Since, she travels some distance by rickshaw and some distance by bus, we form two cases which depict two situations in the question.
Let us denote the speed of the rickshaw by ‘x’ and the speed of the bus by ‘y’.
Since, we know speed is given by the formula \[s = \dfrac{d}{t}\]
Then by cross multiplying the values
\[ \Rightarrow t = \dfrac{d}{s}\] …………….… (1)
Let time taken by rickshaw be \[{t_1}\] and time taken by bus be \[{t_2}\]; Then \[{t_1} + {t_2} = T\] ………….… (2)
Where, T is the total time taken to travel.
Case 1: 2 Km by rickshaw and rest by bus
We are given that Sangeetha travels 2 Km by rickshaw
Since, total distance \[ = 14\]Km
\[ \Rightarrow \]Distance travelled by bus \[ = 14 - 2\]
\[ \Rightarrow \]Distance travelled by bus \[ = 12\]Km
Also, we are given the total time taken in this case as half hour.
\[\therefore T = \dfrac{1}{2}\]hour
We find the value of \[{t_1}\]and \[{t_2}\]
From equation (1): \[t = \dfrac{d}{s}\]
\[ \Rightarrow \]Time taken by rickshaw, \[{t_1} = \dfrac{2}{x}\]
\[ \Rightarrow \]Time taken by bus, \[{t_2} = \dfrac{{12}}{y}\]
Substitute values of \[{t_1}\]and \[{t_2}\]in equation (2)
\[ \Rightarrow \dfrac{2}{x} + \dfrac{{12}}{y} = \dfrac{1}{2}\] ……………..… (3)
Case 2: 4 Km by rickshaw and rest by bus
We are given that Sangeetha travels 4 Km by rickshaw
Since, total distance \[ = 14\]Km
\[ \Rightarrow \]Distance travelled by bus \[ = 14 - 4\]
\[ \Rightarrow \]Distance travelled by bus \[ = 10\]Km
Also, we are given the total time taken in this case as half hour plus 9 minutes
\[ \Rightarrow T = (30 + 9)\]minutes
\[ \Rightarrow T = 39\]minutes
Since, 1 hour has 60 minutes
\[ \Rightarrow T = \dfrac{{39}}{{60}}\]hour
Cancel same factors from numerator and denominator
\[\therefore T = \dfrac{{13}}{{20}}\]hour
We find the value of \[{t_1}\]and \[{t_2}\]
From equation (1): \[t = \dfrac{d}{s}\]
\[ \Rightarrow \]Time taken by rickshaw, \[{t_1} = \dfrac{4}{x}\]
\[ \Rightarrow \]Time taken by bus, \[{t_2} = \dfrac{{10}}{y}\]
Substitute values of \[{t_1}\]and \[{t_2}\]in equation (2)
\[ \Rightarrow \dfrac{4}{x} + \dfrac{{10}}{y} = \dfrac{{13}}{{20}}\] …………..… (4)
Now we solve the two equations, equation (3) and (4).
Let us assume \[\dfrac{1}{x} = a,\dfrac{1}{y} = b\]
Then, from equation (3); \[\dfrac{2}{x} + \dfrac{{12}}{y} = \dfrac{1}{2}\]
\[ \Rightarrow 2a + 12b = \dfrac{1}{2}\] ……………...… (5)
Also, from equation (4); \[\dfrac{4}{x} + \dfrac{{10}}{y} = \dfrac{{13}}{{20}}\]
\[ \Rightarrow 4a + 10b = \dfrac{{13}}{{20}}\] …………….… (6)
From equation (5):\[2a + 12b = \dfrac{1}{2}\]
Shift all the values except one with variable ‘a’ to one side of the equation
\[ \Rightarrow 2a = \dfrac{1}{2} - 12b\]
Take LCM on RHS of equation
\[ \Rightarrow 2a = \dfrac{{1 - 24b}}{2}\]
Multiply both sides by 2
\[ \Rightarrow 2 \times 2a = 2 \times \dfrac{{1 - 24b}}{2}\]
\[ \Rightarrow 4a = 1 - 24b\] ……………….… (7)
Substitute the value of 4a from equation (7) in equation (6)
\[ \Rightarrow (1 - 24b) + 10b = \dfrac{{13}}{{20}}\]
\[ \Rightarrow 1 - 24b + 10b = \dfrac{{13}}{{20}}\]
\[ \Rightarrow 1 - 14b = \dfrac{{13}}{{20}}\]
Shift all constants to one side of the equation
\[ \Rightarrow 1 - \dfrac{{13}}{{20}} = 14b\]
Take LCM on LHS of the equation
\[ \Rightarrow \dfrac{{20 - 13}}{{20}} = 14b\]
\[ \Rightarrow \dfrac{7}{{20}} = 14b\]
Divide both sides of the equation by 14
\[ \Rightarrow \dfrac{7}{{20 \times 14}} = \dfrac{{14b}}{{14}}\]
Cancel same factors on both sides of the equation
\[ \Rightarrow \dfrac{1}{{40}} = b\]
Since, we know \[\dfrac{1}{y} = b\]
\[ \Rightarrow \dfrac{1}{{40}} = \dfrac{1}{y}\]
\[ \Rightarrow y = 40\]
\[\therefore \] The speed of the bus is 40 Km per hour.
Substitute the value of b in equation (7)
\[ \Rightarrow 4a = 1 - 24 \times \dfrac{1}{{40}}\]
Cancel same factors from numerator and denominator in RHS
\[ \Rightarrow 4a = 1 - \dfrac{3}{5}\]
Take LCM on RHS of the equation
\[ \Rightarrow 4a = \dfrac{{5 - 3}}{5}\]
\[ \Rightarrow 4a = \dfrac{2}{5}\]
Divide both sides by 4
\[ \Rightarrow \dfrac{{4a}}{4} = \dfrac{2}{{4 \times 5}}\]
Cancel same factors from numerator and denominator on both sides of the equation
\[ \Rightarrow a = \dfrac{1}{{10}}\]
Since, we know \[\dfrac{1}{x} = a\]
\[ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{10}}\]
\[ \Rightarrow x = 10\]
\[\therefore \]Speed of rickshaw is 10 Km per hour.

Thus, the speed of the rickshaw is 10 Kmph and the speed of the bus is 40Kmph.

Note:
Students are likely to make mistakes when they try to form equations with respect to speed, they tend to think we have to find speed in the question so equations formed will be of speed. But we are given the time clearly in which she covers the total distance in two cases so we form equations on the basis of time. Also, don’t solve the equations having the variables in the denominator using LCM as that will only cause more confusion in the solution.