A group of friends decided to split the cost of renting a car for a trip among themselves equally. The total cost turned out to be 240 dollars. When two of the friends decided not to go on the trip, the ones who were still going divided the 240 dollars equally. But each friend’s share of the cost increased by 20 dollars. How many friends were originally in the group?
We are told that the total cost of renting the car was 240 dollars. Our first step is to let 𝑛 be the number of friends that were originally in the group. We will also let 𝑝 be the original cost per person. To work out this cost per person, we would divide 240 by 𝑛 as they were splitting the cost equally. This gives us an equation 240 divided by 𝑛 is equal to 𝑝. We will call this equation one. Two of the friends decide to not go on the trip. This means that the cost of 240 dollars needs to be split between two less people. 240 needs to be divided by 𝑛 minus two. This is because the number of friends has decreased by two. We’re also told that the cost per person has now increased by 20 dollars. As the original cost per person was 𝑝 dollars, the new cost will be 𝑝 plus 20. We will call this equation two.
As we’re trying to work out the number of friends that were originally in the group, we need to work out the value of 𝑛. We can do this by eliminating 𝑝 from one of the equations. We can substitute 240 divided by 𝑛 into equation two. Rewriting the equation gives us 240 divided by 𝑛 minus two is equal to 240 divided by 𝑛 plus 20. 20 is the same as 20 divided by one. In order to solve this equation, we firstly need to eliminate the fractions. We do this by multiplying by the lowest common denominator. In this case, this will be 𝑛 multiplied by 𝑛 minus two. We need to multiply all three fractions by this.
At this point, we notice that some things will cancel in all three of the fractions. On the left-hand side, 𝑛 minus two cancels, leaving us with 240𝑛. In the second fraction, the 𝑛s cancel, leaving us with 240 multiplied by 𝑛 minus two. We can then distribute the 240 over the bracket or parentheses. We need to multiply 240 by 𝑛 and 240 by negative two. This gives us 240𝑛 minus 480. 20 divided by one is just 20. So for the third term, we need to multiply 20𝑛 by 𝑛 and 20𝑛 by negative two. Once again distributing the parenthesis. 20𝑛 multiplied by 𝑛 is equal to 20𝑛 squared. And 20𝑛 multiplied by negative two is equal to negative 40𝑛. We have 240𝑛 on both sides of the equation. Therefore, these will cancel. We are therefore left with the quadratic equation. 20𝑛 squared minus 40𝑛 minus 480 is equal to zero.
All three of the terms on the right-hand side are divisible by 10. This gives us two 𝑛 squared minus four 𝑛 minus 48. We can divide this by two, giving us 𝑛 squared minus two 𝑛 minus 24. We can solve this quadratic equation by factorizing into two parentheses. The first term in both of them will be 𝑛. The second terms must have a product of negative 24 and a sum of negative two. Negative six multiplied by four is equal to negative 24. And negative six plus four is equal to negative two. Therefore, 𝑛 squared minus two 𝑛 minus 24 can be rewritten as 𝑛 minus six multiplied by 𝑛 plus four. As this needs to be equal to zero, we have two possible solutions. 𝑛 equals six and 𝑛 equals negative four. As 𝑛 was the number of friends, this must be a positive answer. We can therefore conclude that there were six friends originally in the group.
We can now check this answer by substituting this value back into the equations. As the number of friends was equal to six, we need to divide 240 by six to find the cost per person. 240 divided by six is equal to 40. This means that the original cost was 40 dollars per person. In our second equation, the number of friends had decreased by two. This means that we’re now dividing 240 by four. We were told that the cost has increased by 20 dollars. Therefore, we need to add 20 to 40. 240 divided by four is equal to 60. 40 plus 20 is also equal to 60. As both sides of this equation are equal, our answer is correct.
There were six friends originally in the group.