Question

Point A and B are \[90km\] apart from each other on a highway. A car starts from A and another from B at the same time. If they go in the same direction, they will meet in \[9{\text{ }}hrs\] and if they go in opposite directions, they meet in \[\dfrac{9}{7}\]hrs. Find their speeds.
A. The speed of car starting from point A =\[30km/h\]and speed of car starting from point B \[ = 550km/h\].
B. The speed of car starting from point A = 60km/h and speed of car starting from point B \[ = 40km/h\].
C. The speed of car starting from point A = 40km/h and speed of car starting from point B \[ = 30km/h\].
D. The speed of car starting from point A = 25km/h and speed of car starting from point B \[ = 20km/h\]

Answer 1

Hint: To solve these types of questions, we will form two equations with two variables and solve them algebraically. We also use speed, distance and time relationship to form the equations.

Formula used: We use basic formulas, which describe the speed of moving objects.
$speed = \dfrac{{distance}}{{time}}$
Or ${\text{ }}Distance = speed \times time$

Complete step by step solution: Let us assume that speed of the car starting from point A = \[x{\text{ }}km/h\]
And speed of the car starting from point B=\[{\text{y }}km/h\]
Part 1: When both move in the same direction and meet in \[9{\text{ }}hrs\].
Distance travelled by car A in i.e. \[AP = 9x\] \[[as{\text{ }}Distance = speed \times time]\]
Distance travelled by car B in i.e. \[BP = 9y\]

Clearly from figure:
$AP - BP = AB$
$9x - 9y = 90................(1)$
Part 2: When both are moving in opposite direction:

Distance travelled by car A in \[\dfrac{9}{7}hour = \dfrac{9}{7}x\]
Distance travelled by car B in \[\dfrac{9}{7}hour = \dfrac{9}{7}y\]
Clearly from figure :
$ {AP + BP = AB}$
${\dfrac{9}{7}x{\text{ }} + \dfrac{9}{7}y{\text{ }} = 90} $
Taking\[\dfrac{1}{7}\]common and shift it to RHS,
${9x{\text{ }} + {\text{ }}9y{\text{ }} = {\text{ }}90 \times 7}$
${Or{\text{ }}9x{\text{ }} + {\text{ }}9y{\text{ }} = {\text{ }}630 \ldots \ldots \ldots \ldots \ldots ..\left( 2 \right)}$
Adding equation (1) and (2),
${18x = 90+630}$
$ \Rightarrow 18x=720$
$\therefore x= \dfrac{{720}}{{18}}=40$
Put this value of x in equation (2),
${9 \times 40{\text{ }} + {\text{ }}9y{\text{ }} = {\text{ }}630}$
${ \Rightarrow 360{\text{ }} + 9y{\text{ }} = {\text{ }}630}$
${Or{\text{ }}9y{\text{ }} = {\text{ }}630{\text{ }}-{\text{ }}360}$
${ \Rightarrow {\text{ }}9y{\text{ }} = {\text{ }}270}$
${y{\text{ }} = {\text{ }}\dfrac{{270}}{9} = {\text{ }}30{\text{ }}km/h}$
Hence speed of car starting from A is 40km/h and speed of car starting from point B is 30km/h.

Option C is our correct option.

Additional information: we know that linear equations in one variable have one solution. If any equation has two unknown values then it is known as a linear equation in two variables. Hence we get that the number of solutions varies as per the count of variables an equation contains. The equation of a straight line is an example of linear equation.

Note: Students generally make mistakes while solving algebraic equations. If the side of variable changes, the sign and operator must be changed accordingly. Avoid confusion of the same and opposite direction by making a simple diagram. We get two equations having velocities of two cars as variable or unknown quantities.