Question

The speed of a stream is 9
km./hr. A boat goes 21 km
and comes back to the start-
ing point in 8 hours. What is
the speed (in km/hr.) of the
boat in still water?​

Answer 1

Basic Concept Used :-

Writing System of Equation from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

$\large\underline{\sf{Solution-}}$

↝ Let speed of the boat in still water be 'x' km per hour.

It is given that,

• ↝ Speed of stream is 9 km per hour.

So,

• ↝ Speed of upstream = x - 9 km per hour

and

• ↝ Speed of downstream = x + 9 km per hour

Now,

• ↝ Distance covered in upstream = 21 km

So,

• ↝ Time taken to travel 21 km in upstream is

$\rm :\longmapsto\:t_1 = \dfrac{21}{x - 9} \: hrs - - - (1)$

Also,

• ↝ Distance covered in downstream = 21 km

So,

• ↝ Time taken to cover 21 km in downstream is

$\rm :\longmapsto\:t_2 = \dfrac{21}{9 + x} \: hrs - - - (2)$

Now,

According to statement,

• ↝ Total time taken to travel 21 km and to come back is 8 hours

$\rm :\implies\:t_1 + t_2 = 8$

$\rm :\longmapsto\:\dfrac{21}{x - 9} + \dfrac{21}{x + 9} = 8$

$\rm :\longmapsto\:\dfrac{21(x + 9) + 21(x - 9)}{(x + 9)(x - 9)} = 8$

$\rm :\longmapsto\:\dfrac{21x + 189 + 21x - 189}{ {x}^{2} - 81} = 8$

$\rm :\longmapsto\:\dfrac{42x}{ {x}^{2} - 81} = 8$

$\rm :\longmapsto\:\dfrac{21x}{ {x}^{2} - 81} = 4$

$\rm :\longmapsto\:4 {x}^{2} - 324 = 21x$

$\rm :\longmapsto\:4 {x}^{2} - 21x - 324 = 0$

$\rm :\longmapsto\:4 {x}^{2} - 48x + 27x - 324 = 0$

$\rm :\longmapsto\:4x(x - 12) + 27(x - 12) = 0$

$\rm :\longmapsto\:(x - 12)(4x + 27) = 0$

$\bf\implies \:x \: = \: 12 \: \: \: or \: \: \: x = - \dfrac{27}{4} \: \{ \: rejected \: \}$

$\boxed{ \red{ \rm \:Hence, \: speed \: of \: boat \: in \: still \: water \: is \: 12 km \: per \: hour}}$