# Question Video: Using Linear Equations to Solve Problems

For journeys between 9 pm and 7 am, a cab company charges a fixed fare of 4.50 dollars plus 40 cents per fifth of a mile. If a passenger takes a cab at midnight and the fare was \$12.10, how far did they travel?

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### Video Transcript

For journeys between 9 pm and 7 am, a cab company charges a fixed fare of four dollars 50 plus 40 cents per fifth of a mile. If a passenger takes a cab at midnight and the fare was 12 dollars and 10, how far did they travel?

The total fare for the journey was 12 dollars and 10 cents. Of this cost, four dollars and 50 cents was made up of a fixed fare. This means that our first step is to subtract these two numbers. One method we could use to do this is column subtraction.

Zero minus zero is equal to zero. We are unable to subtract five from one. Therefore, we need to go next door and borrow one. We now have eleven- tenths minus five-tenths. 11 minus five is equal to six. Finally, 11 minus four is equal to seven. This means that 12 dollars 10 minus four dollars 50 is equal to seven dollars and 60 cents.

We’re also told in the question that the cab company charges 40 cents for every one- fifth of a mile travelled. We need to calculate how far they would travel to be charged seven dollars and 60 cents. 7.6 divided by 0.4 can be rewritten as 76 divided by four by multiplying the numerator and denominator by 10. 76 divided by four is equal to 19.

Therefore, 40 cents multiplied by 19 is equal to seven dollars and 60 cents. As we have multiplied the cost by 19, we need to multiply the distance by 19. One-fifth multiplied by 19 is equal to nineteen-fifths or 19 divided by five. This is equivalent to three and four-fifths or 3.8.

We can therefore say that if the fare was 12 dollars and 10 cents, the passenger travelled 3.8 miles.