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