Question

A motorboat heads upstream a distance of 30 km on a river whose current is running at 5km/hr. The trip up and back takes 4 hours. Write a quadratic equation to find the speed of the boat.

Answer 1

Hint: We will assume the speed of the boat to be \[x\]. We will find the time taken to go upstream and the time taken to come downstream in terms of \[x\]. We will add the time and equate it with 4 which is the total time taken. We will simplify the expression to obtain a quadratic equation for the speed of the boat.

Formulas used:
1.\[{\rm{time}} = \dfrac{{{\rm{distance}}}}{{{\rm{speed}}}}\]
2.\[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]

Complete step-by-step answer:
We will assume that the speed of the boat is \[x\] km/hr.
We know that the speed of the boat while going upstream will be \[\left( {x - 5} \right){\rm{km/hr}}\]
We will assume that the boat took \[{t_1}\] hours to travel 30 Km upstream.
We will find \[{t_1}\] by substituting \[\left( {x - 5} \right)\] for speed and 30 for distance in the formula for time:
\[{t_1} = \dfrac{{30}}{{x - 5}}\]
We know that the speed of the boat while going downstream will be \[\left( {x + 5} \right){\rm{km/hr}}\]
We will assume that the boat took \[{t_2}\] hours to travel 30 Km downstream.
We will find \[{t_2}\] by substituting \[\left( {x + 5} \right)\] for speed and 30 for distance in the formula for time:
\[{t_2} = \dfrac{{30}}{{x + 5}}\]
We know that the total time taken is 4 hours:
\[{t_1} + {t_2} = 4\]
We will substitute \[\dfrac{{30}}{{x - 5}}\] for \[{t_1}\] and \[\dfrac{{30}}{{x + 5}}\] for \[{t_2}\] in the above equation:
\[\Rightarrow \dfrac{{30}}{{x - 5}} + \dfrac{{30}}{{x + 5}} = 4\]
We will take L.C.M of the denominators of the fractions on the left-hand side:
\[\dfrac{{30\left( {x + 5} \right) + 30\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = 4\]
We will substitute \[x\] for \[a\] and 5 for \[b\] in the 2nd formula. We will simplify the numerator and denominator:
\[\Rightarrow \dfrac{{30 \cdot x + 30 \cdot 5 + 30 \cdot x - 30 \cdot 5}}{{{x^2} - {5^2}}} = 4\\
\Rightarrow \dfrac{{60x}}{{{x^2} - {5^2}}} = 4\]
We will cross multiply the denominator on the left-hand side with the numerator on the right-hand side:
\[\Rightarrow 60x = 4\left( {{x^2} - 25} \right)\\
\Rightarrow 60x = 4{x^2} - 100\]
We will subtract \[60x\] from both sides of the equation:
\[\Rightarrow 60x - 60x = 4{x^2} - 60x - 100\\
\Rightarrow 0 = 4{x^2} - 60x - 100\]
We will divide the whole equation by 4:
\[ \Rightarrow \dfrac{0}{4} = \dfrac{{4{x^2} - 60x - 100}}{4}\\
\Rightarrow 0 = {x^2} - 15x - 25\]

Note: We know that a quadratic equation is an equation in which the power of the variable is 2. We know that the standard form of a quadratic equation is \[a{x^2} + bx + c = 0\] where \[a \ne 0\]. Here, the value of \[a\] is 1, the value of \[b\] is \[ - 15\] and the value of \[c\] is \[ - 25\].