Question

Which situation can be represented by 20s >30 + 14s?

A. Jeannie makes and sells personalized picture frames. She sells the frames for $20 each. She spends $30 for an electric drill and the materials for each frame costs $14. How many frames does she have to sell to make a profit?
B. Sheila buys books for resale. She spent less on 20 books than she spent on 14 books and $30 worth of shipping material. How much did she pay for each book?
C. Twenty times a number is equal to 14 times the number increased by 30. What is the number?
D. A number increased by 20 is 30 more than the number increased by 14. What is the number?

Answer 1

The situation that can be represented by the inequality 20s > 30 + 14s is option C: Twenty times a number is equal to 14 times the number increased by 30.

Step-by-step explanation:

The situation that can be represented by the inequality 20s > 30 + 14s is option C. Twenty times a number is equal to 14 times the number increased by 30. In other words, the situation represents an equation where 20 times the number is greater than 30 plus 14 times the number. The situation that can be represented by the inequality 20s > 30 + 14s is A. Jeannie makes and sells personalized picture frames. She sells the frames for $20 each. In order for Jeannie to make a profit, she needs to earn more money from selling the frames than she spends on the materials and the drill. The inequality shows that the amount she earns from selling frames (20 times the number of frames sold, or 20s) must be greater than the total cost of $30 for the electric drill plus $14 for materials for each frame (14 times the number of frames made, or 14s, plus $30). When we solve the inequality 20s > 30 + 14s for 's', which represents the number of frames, we find out how many frames Jeannie needs to sell to make a profit. By subtracting 14s from both sides of the inequality, we simplify it to 6s > 30. Dividing both sides by 6, we get s > 5. This means Jeannie must sell more than 5 frames to start making a profit.