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Question

Difficulty: HardSystems of Linear Equations

A software company offers a Basic subscription for $15\$15 per month and a Premium subscription for $30\$30 per month. In April, the company had a total of 1,2001,200 active subscriptions. In May, the number of Basic subscriptions increased by 20%20\%, and the number of Premium subscriptions increased by 5%5\% compared to April. If the company's total monthly revenue increased by $2,700\$2,700 from April to May, how many Premium subscriptions did the company have in April?

Answer: 600

Answer

600
To find the number of Premium subscriptions in April, we can set up a system of linear equations. Let BB represent the number of Basic subscriptions in April and PP represent the number of Premium subscriptions in April. Since the total number of subscriptions is 1,2001,200, we have B+P=1,200B + P = 1,200. The revenue increase from April to May is $2,700\$2,700. The increase in Basic subscriptions is 20%20\%, which contributes 15×0.20B=3B15 \times 0.20B = 3B dollars to the revenue increase. The increase in Premium subscriptions is 5%5\%, which contributes 30×0.05P=1.5P30 \times 0.05P = 1.5P dollars to the revenue increase. Thus, 3B+1.5P=2,7003B + 1.5P = 2,700. Substituting B=1,200−PB = 1,200 - P into this equation yields 3(1,200−P)+1.5P=2,7003(1,200 - P) + 1.5P = 2,700. Solving for PP gives 3,600−1.5P=2,7003,600 - 1.5P = 2,700, which simplifies to 1.5P=9001.5P = 900, or P=600P = 600.

Step-by-Step Solution

1
Define variables and write the first equation based on the total number of subscriptions in April.
B+P=1,200B + P = 1,200, where BB is the number of Basic subscriptions and PP is the number of Premium subscriptions.
To represent the relationship between the two types of subscriptions in April.
2
Write the second equation representing the change in monthly revenue from April to May.
3B+1.5P=2,7003B + 1.5P = 2,700
The change in revenue is the sum of the increase in revenue from each subscription type: 15(0.20B)+30(0.05P)=2,70015(0.20B) + 30(0.05P) = 2,700.
3
Solve the system of equations by substituting B=1,200−PB = 1,200 - P into the revenue equation.
3(1,200−P)+1.5P=2,700  ⟹  3,600−1.5P=2,7003(1,200 - P) + 1.5P = 2,700 \implies 3,600 - 1.5P = 2,700
Substitution eliminates the variable BB, allowing us to solve for PP directly.
4
Isolate the variable PP to find the number of Premium subscriptions in April.
−1.5P=−900  ⟹  P=600-1.5P = -900 \implies P = 600
Dividing the revenue difference by the coefficient solves for the value of PP.

Key Concept

Solving systems of linear equations in real-life contexts involving percentage changes.
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Topics

Systems of Linear Equations

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