Question

During a race on a circular race track, a race car burns fuel at a constant rate. After lap 4, the race car has 22 gallons left in its tank. After lap 7, the race car has 18 gallons left in its tank. Assuming the race car does not refuel, after which lap will the race car have 6 gallons left in its tank?
A. Lap 13
B. Lap 15
C. Lap 16
D. Lap 19

Answer 1

Hint: In order to solve the question, we first need to find the slope the slope by using the formula \[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\] and then we will find the equation by using point or slope formula. After that we will find the value of x i.e. lap number by putting the value of y=6 and then we get the required solution of the given question.

Complete step by step solution:
Assuming that the race car does not need to refuel, we have been asked to find after which lap the race car will have 6 gallons left in its tank.
We can write the standard form of linear equation
\[y=ax+b\] Where,
\[x=lap\,number\]and
\[y=no.\ of\ fuel\ gallons\ left\]
Then,
According to question,

\[x=lap\,number\]	\[{{x}_{1}}=4\]	\[{{y}_{1}}=22\]
\[y=no.\ of\ fuel\ gallons\ left\]	\[{{x}_{2}}=7\]	\[{{y}_{2}}=18\]

Use the slope formula, \[slope=m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\], we obtain
\[\Rightarrow slope=m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{18-22}{7-4}=\dfrac{-4}{3}\]
\[\therefore m=-\dfrac{4}{3}\]
Now, we will find the equation using the slope or point formula:
\[\Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\]
Put the value of \[{{y}_{1}}=22\] and \[{{x}_{1}}=4\] in the above equation, we obtain
\[\Rightarrow y-22=m\left( x-4 \right)\]
Putting the value of \[m=-\dfrac{4}{3}\] in the above equation, we obtain
\[\Rightarrow y-22=-\dfrac{4}{3}\left( x-4 \right)\]
Adding 22 to both the sides of the equation, we get
\[\Rightarrow y-22+22=-\dfrac{4}{3}\left( x-4 \right)+22\]
Simplifying the above equation, we get
\[\Rightarrow y=-\dfrac{4}{3}\left( x-4 \right)+22\]
Simplifying the above, we obtain
\[\Rightarrow y=-\dfrac{4x}{3}+\dfrac{16}{3}+22\]
By taking LCM of the denominator on the right side of the equation, we get
\[\Rightarrow y=-\dfrac{4x}{3}+\dfrac{16}{3}+\dfrac{66}{3}\]
\[\Rightarrow y=-\dfrac{4x}{3}+\dfrac{82}{3}\]
Now, according to the question,
We need to find after which lap, car has left with 6 gallons i.e. y=6
Putting y=6 in plug-in solved equation, we obtain
\[\Rightarrow y=-\dfrac{4x}{3}+\dfrac{82}{3}\]
\[\Rightarrow 6=-\dfrac{4x}{3}+\dfrac{82}{3}\]
Simplifying the above equation, we obtain
\[\Rightarrow 6=\dfrac{-4x+82}{3}\]
Multiplying both the sides of the equation by 3, we get
\[\Rightarrow 6\times 3=\dfrac{-4x+82}{3}\times 3\]
Simplifying the above, we get
\[\Rightarrow 18=-4x+82\]
Subtracting 82 from both the sides of the equation, we get
\[\Rightarrow 18-82=-4x+82-82\]
\[\Rightarrow -64=-4x\]
Dividing both the sides of the equation by -4, we get
\[\Rightarrow 16=x\]
\[\therefore x=16\]
Thus, after 16 laps the car has only 6 gallons of fuel left.

Note: In solving these types of questions, most of the time it may have seen that students made common mistakes while choosing and putting the value of x1, x2 and y1, y2 while calculating the slope and it will result in the wrong answer. For that you should choose and put all the values of the variables explicitly and hence avoid making errors.