Together two computers cost $2700 per year to rent. If one costs twice other, what is the monthly cost of the most expensive one? OA. $900 OB. $1,800 O C. $225 O D. $150 O E. $75
Together two computers cost $2700 per year to rent. If one costs twice other, what is the monthly cost of the most expensive one? OA. $900 OB. $1,800 O C. $225 O D. $150 O E. $75
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Mathematics Final Exam Question
**Question 102 of 190:**
Together, two computers cost $2700 per year to rent. If one costs twice as much as the other, what is the monthly cost of the most expensive one?
**Options:**
- A. $900
- B. $1,800
- C. $225
- D. $150
- E. $75
### Explanation:
To find the monthly cost of the most expensive computer, follow these steps:
1. Let \( x \) be the yearly rental cost of the less expensive computer.
2. Therefore, the yearly rental cost of the more expensive computer would be \( 2x \).
3. Together, their total yearly rental cost is \( x + 2x = 3x \).
4. Given that \( 3x = $2700 \), solve for \( x \):
\[ 3x = 2700 \]
\[ x = \frac{2700}{3} \]
\[ x = 900 \]
So, the yearly rental cost of the less expensive computer is $900, and the more expensive one is \( 2 \times 900 = $1800 \).
5. To find the monthly cost of the most expensive computer:
\[ \frac{1800}{12} = 150 \]
Thus, the monthly cost of the most expensive computer is $150.
**Correct Answer:** D. $150
---
This explanation enriches the student's understanding of how to break down the problem into manageable steps, use simple algebra to solve for the unknown variables, and convert from annual costs to monthly costs.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc22a9166-79ef-4489-a230-81be74daa87f%2F7c564b38-1260-4a0d-808e-d4940be4c815%2Fvjt1pod_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Mathematics Final Exam Question
**Question 102 of 190:**
Together, two computers cost $2700 per year to rent. If one costs twice as much as the other, what is the monthly cost of the most expensive one?
**Options:**
- A. $900
- B. $1,800
- C. $225
- D. $150
- E. $75
### Explanation:
To find the monthly cost of the most expensive computer, follow these steps:
1. Let \( x \) be the yearly rental cost of the less expensive computer.
2. Therefore, the yearly rental cost of the more expensive computer would be \( 2x \).
3. Together, their total yearly rental cost is \( x + 2x = 3x \).
4. Given that \( 3x = $2700 \), solve for \( x \):
\[ 3x = 2700 \]
\[ x = \frac{2700}{3} \]
\[ x = 900 \]
So, the yearly rental cost of the less expensive computer is $900, and the more expensive one is \( 2 \times 900 = $1800 \).
5. To find the monthly cost of the most expensive computer:
\[ \frac{1800}{12} = 150 \]
Thus, the monthly cost of the most expensive computer is $150.
**Correct Answer:** D. $150
---
This explanation enriches the student's understanding of how to break down the problem into manageable steps, use simple algebra to solve for the unknown variables, and convert from annual costs to monthly costs.
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