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A French club collected the same amount from each student going on a trip to Le Cercle Molière in Winnipeg. When six students could not go, each of the remaining students was charged an extra $3. If the total cost was $540, how many students went on the trip?

A French club collected the same amount from each student going on a trip to Le Cercle Molière in Winnipeg. When six students could not go, each of the remaining students was charged an extra $3 \$ 3 . If the total cost was $540 \$ 540 , how many students went on the trip?

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Q. A French club collected the same amount from each student going on a trip to Le Cercle Molière in Winnipeg. When six students could not go, each of the remaining students was charged an extra $3 \$ 3 . If the total cost was $540 \$ 540 , how many students went on the trip?
  1. Denote Original Number of Students: Let's denote the original number of students as xx. Each student was supposed to pay an equal amount, so the total cost divided by the number of students would give us the amount each student pays. When six students dropped out, the remaining students (x−6)(x - 6) had to pay an extra $3\$3 each. We can set up an equation to represent this situation.
  2. Set Up Equation: The total amount collected would be the same in both cases. Initially, it would be the number of students times the amount each student pays. After six students dropped out, it would be the remaining number of students times the increased amount each pays. We can express this as:\newlinex×amount_per_student=(x−6)×(amount_per_student+$(3))x \times \text{amount\_per\_student} = (x - 6) \times (\text{amount\_per\_student} + \$(3))
  3. Substitute Total Amount: We know the total amount collected was $540\$540. So we can substitute this into our equation:\newlinex×amount_per_student=$540x \times \text{amount\_per\_student} = \$540\newline(x−6)×(amount_per_student+$3)=$540(x - 6) \times (\text{amount\_per\_student} + \$3) = \$540
  4. Distribute and Simplify: Now we have two equations with two variables. However, we can simplify this by realizing that the amount each student pays initially is $540\$540 divided by xx. We can substitute this into the second equation:\newline(x−6)×($540x+$3)=$540(x - 6) \times (\frac{\$540}{x} + \$3) = \$540
  5. Clear Fraction: To solve for xx, we need to distribute the (x−6)(x - 6) across the terms inside the parentheses: (x−6)×540x+(x−6)×3=540(x - 6) \times \frac{540}{x} + (x - 6) \times 3 = 540
  6. Simplify Further: This simplifies to: $540−6×$540x+3x−18=$540\$540 - \frac{6 \times \$540}{x} + 3x - 18 = \$540
  7. Quadratic Equation: We can subtract $540\$540 from both sides to get rid of the $540\$540 on the right side of the equation:\newline−6×$540x+3x−18=0-\frac{6 \times \$540}{x} + 3x - 18 = 0
  8. Calculate Discriminant: Multiplying through by xx to clear the fraction gives us: −6×$540-6 \times \$540 + 33x^22 - 1818x = 00\)
  9. Find Solutions for x: This simplifies to: 3x2−18x−6×540=03x^2 - 18x - 6 \times 540 = 0
  10. Discard Negative Solution: Now we can divide the entire equation by 33 to simplify it further: x2−6x−2×($540)=0x^2 - 6x - 2 \times (\$540) = 0
  11. Final Answer: Plugging in the value of $540\$540, we get:\newlinex2−6x−1080=0x^2 - 6x - 1080 = 0
  12. Final Answer: Plugging in the value of $540\$540, we get:\newlinex2−6x−1080=0x^2 - 6x - 1080 = 0 This is a quadratic equation, and we can solve for x using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−6b = -6, and c=−1080c = -1080.
  13. Final Answer: Plugging in the value of $540\$540, we get: x2−6x−1080=0x^2 - 6x - 1080 = 0 This is a quadratic equation, and we can solve for x using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−6b = -6, and c=−1080c = -1080. First, we calculate the discriminant, b2−4ac\sqrt{b^2 - 4ac}: (−6)2−4⋅1⋅−1080=36+4320=4356\sqrt{(-6)^2 - 4 \cdot 1 \cdot -1080} = \sqrt{36 + 4320} = \sqrt{4356}
  14. Final Answer: Plugging in the value of $540\$540, we get: x2−6x−1080=0x^2 - 6x - 1080 = 0 This is a quadratic equation, and we can solve for x using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−6b = -6, and c=−1080c = -1080. First, we calculate the discriminant, b2−4ac\sqrt{b^2 - 4ac}: (−6)2−4⋅1⋅−1080=36+4320=4356\sqrt{(-6)^2 - 4 \cdot 1 \cdot -1080} = \sqrt{36 + 4320} = \sqrt{4356} The square root of 43564356 is 6666, so the discriminant is 6666. Now we can find the two possible solutions for x: x=6±662x = \frac{6 \pm 66}{2}
  15. Final Answer: Plugging in the value of $540\$540, we get: x2−6x−1080=0x^2 - 6x - 1080 = 0 This is a quadratic equation, and we can solve for x using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−6b = -6, and c=−1080c = -1080. First, we calculate the discriminant, b2−4ac\sqrt{b^2 - 4ac}: (−6)2−4⋅1⋅−1080=36+4320=4356\sqrt{(-6)^2 - 4 \cdot 1 \cdot -1080} = \sqrt{36 + 4320} = \sqrt{4356} The square root of 43564356 is 6666, so the discriminant is 6666. Now we can find the two possible solutions for x: x=6±662x = \frac{6 \pm 66}{2} This gives us two possible solutions for x: x=6+662=722=36x = \frac{6 + 66}{2} = \frac{72}{2} = 36 x2−6x−1080=0x^2 - 6x - 1080 = 000
  16. Final Answer: Plugging in the value of $540\$540, we get: x2−6x−1080=0x^2 - 6x - 1080 = 0 This is a quadratic equation, and we can solve for xx using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−6b = -6, and c=−1080c = -1080. First, we calculate the discriminant, b2−4ac\sqrt{b^2 - 4ac}: (−6)2−4⋅1⋅−1080=36+4320=4356\sqrt{(-6)^2 - 4 \cdot 1 \cdot -1080} = \sqrt{36 + 4320} = \sqrt{4356} The square root of 43564356 is 6666, so the discriminant is 6666. Now we can find the two possible solutions for xx: x2−6x−1080=0x^2 - 6x - 1080 = 000 This gives us two possible solutions for xx: x2−6x−1080=0x^2 - 6x - 1080 = 022 x2−6x−1080=0x^2 - 6x - 1080 = 033 Since the number of students cannot be negative, we discard the negative solution. Therefore, the number of students who went on the trip is x2−6x−1080=0x^2 - 6x - 1080 = 044.

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