# average of 2 ticket sales price variance

by Lenny
(Memphis, TN, USA)

2 ticket price variance

I calculated the average ticket prices for two different products.

Each product had a variance from the previous price.

One variance was a decline in the price by 1.3 percent.

The other product had an increase of 6.0%.

Combining the two products together I had a price variance of 8.6%.

How can I have an 8.6% increase in both products together while neither product was greater than 6.0%?

### Comments for average of 2 ticket sales price variance

 May 03, 2014 price variance by: Staff Answer Part I The problem statement does not provide any prices, so I can't see what numbers you are working with. The general answer is shown below. Can you have an 8.6% increase in both products together when neither product has an increase of greater than 6.0%? Mathematically, the only way an 8.6% increase for both products can be true is for one of the prices to be positive, and the other price to be a negative price (The purchaser acquiring the ticket does not pay for the ticket. Instead, the purchaser receives a payment equal to the price of the ticket), or both prices must be zero. If I understand your question correctly: x = price of ticket product A y = price of ticket product B A = average ticket price of both products = (x + y) / 2 The initial average ticket prices for the two different products = A₁. x₁ = initial price of ticket product A y₁ = initial price of ticket product B (x₁ + y₁) / 2 = A₁ The price of each product changed in relation to its initial price. The new average ticket prices for the two different products = A₂. x₂ = new price of ticket product A y₂ = new price of ticket product B (x₂ + y₂) / 2 = A₂ The new price of product A is 1.3% lower than its initial price. In other words, the new price of product A is 98.7 % of its initial price. x₂ = (.987) * x₁ The new price of product B is 6.0% higher than its initial price. The means the new price of product B is 106 % of its initial price. y₂ = (1.06) * y₁ When the new prices of the two products were averaged, the new average was 8.6% higher than the initial average. A₂ = (1.086) * A₁ If that's a correct statement of the problem, the initial price of product B (the variable y₁) can be determined. A₂ = (1.086) * A₁ A₂ = (x₂ + y₂) / 2 = ((.987) * x₁ + (1.06) * y₁)/2 A₁ = (x₁ + y₁) / 2 ((.987) * x₁ + (1.06) * y₁)/2 = 1.086(x₁ + y₁)/2 ((.987) * x₁ + (1.06) * y₁)/2 = 1.086(x₁ + y₁)/2 ((.987) * x₁ + (1.06) * y₁) = 1.086(x₁ + y₁) ((.987) * x₁ + (1.06) * y₁) = 1.086 * x₁ + 1.086 * y₁ 0 = (1.086 - .987) * x₁ + (1.086-1.06) * y₁ 0 = (0.099) * x₁ + (0.026) * y₁ 0.099 x₁ + 0.026 y₁ = 0 x₁ = -(26/99) y₁ or y₁ = -(99/26) x₁ ------------------------------------------------

 May 03, 2014 price variance by: Staff ------------------------------------------------ Part II This relationship can be graphed: How can the average increase of both products be 8.6% while the increase in the individual prices of each product is less than 8.6%? The graph shows that this can only occur when both prices are 0, or when one of the prices to is positive, and the other price is a negative price (The purchaser acquiring the ticket does not pay for the ticket. Instead, the purchaser receives a payment equal to the price of the ticket). ------------------------------------------- check the answer select any value for x₁, and then substitute -(99/26) x₁ for y₁ in the original equation A₂ = (1.086) * A₁ ((.987) * x₁ + (1.06) * y₁)/2 = 1.086(x₁ + y₁)/2 ((.987) * x₁ + (1.06) * (-(99/26) x₁))/2 = 1.086(x₁ + (-(99/26) x₁))/2 x₁ can be any value. As an example, let x₁ = 20. ((.987) * 20 + (1.06) * (-(99/26) 20))/2 = 1.086(20 + (-(99/26) 20))/2 -30.4915384615385 = -30.4915384615385 Since both the left hand side and the right hand side of the equation are equivalent, the solution is valid. Thanks for writing. Staff www.solving-math-problems.com