The members of a club hire a bus for $2100. Seven members withdraw from the club and the remaining members have to pay $10 more each to cover the cost. How many members originally agreed to go on the bus?
The Correct Answer and Explanation is :
Let’s break down the problem step by step:
- Define variables:
Let ( x ) be the number of members originally in the club. - Original cost per person:
The total cost of the bus is $2100. If there were ( x ) members originally, the cost per member would be:
[
\frac{2100}{x}
] - Cost after 7 members withdraw:
After 7 members withdraw, the remaining number of members is ( x – 7 ). Now, each of the remaining members must pay $10 more than before. The new cost per member is:
[
\frac{2100}{x – 7}
] - Set up the equation:
According to the problem, the new cost per member is $10 more than the original cost per member. Therefore, we can write the equation:
[
\frac{2100}{x – 7} = \frac{2100}{x} + 10
] - Solve the equation:
To solve this, let’s first eliminate the fractions by multiplying through by ( x(x – 7) ):
[
2100x = 2100(x – 7) + 10x(x – 7)
]
Simplify both sides:
[
2100x = 2100x – 14700 + 10x(x – 7)
]
The terms involving ( 2100x ) cancel out:
[
0 = -14700 + 10x(x – 7)
]
Expand the quadratic term:
[
0 = -14700 + 10x^2 – 70x
]
Rearranging the terms:
[
10x^2 – 70x – 14700 = 0
]
Divide through by 10 to simplify:
[
x^2 – 7x – 1470 = 0
] - Solve the quadratic equation:
We can solve this quadratic equation using the quadratic formula:
[
x = \frac{-(-7) \pm \sqrt{(-7)^2 – 4(1)(-1470)}}{2(1)}
]
Simplify:
[
x = \frac{7 \pm \sqrt{49 + 5880}}{2}
]
[
x = \frac{7 \pm \sqrt{5929}}{2}
]
[
x = \frac{7 \pm 77}{2}
]
So, we have two possible solutions:
[
x = \frac{7 + 77}{2} = \frac{84}{2} = 42
]
or
[
x = \frac{7 – 77}{2} = \frac{-70}{2} = -35 \quad (\text{not a valid solution since the number of members can’t be negative})
] - Conclusion:
The original number of members was 42.
Explanation:
The problem involves determining how many members initially agreed to go on the bus, given that when 7 members withdrew, the cost per remaining member increased by $10. We set up an equation to model the relationship between the number of members and the cost per person. Solving the quadratic equation gave us 42 members as the correct answer. This solution ensures that the original members’ cost and the adjusted cost after the withdrawal of 7 members satisfies the conditions given in the problem.