Difference between revisions of "1987 AJHSME Problems/Problem 15"
m (→Solution) |
|||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | The sale ad read: "Buy three tires at the regular price and get the fourth tire for | + | The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire? |
<math>\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}\ 79\text{ dollars} \qquad \text{(E)}\ 80\text{ dollars}</math> | <math>\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}\ 79\text{ dollars} \qquad \text{(E)}\ 80\text{ dollars}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Let the regular price of one tire be <math>x</math>. We | + | Let the regular price of one tire be <math>x</math>. |
| − | + | We are given: | |
| − | |||
| − | |||
| − | |||
| − | <math>\boxed{\ | + | <cmath> |
| − | + | \begin{align*} | |
| + | 3x + 3 &= 240 \\ | ||
| + | 3x &= 237 \\ | ||
| + | x &= \boxed{\textbf{(D) 79}} | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | To get the price, we can also work backwards without introducing a variable. | ||
| + | |||
| + | To get the price of all of the tires outside of the fourth tire, we can subtract <math>3</math> from the total, which gives <math>240 - 3</math> = <math>237</math>. | ||
| + | |||
| + | Dividing this by <math>3</math> to get the price of a single tire without the sale, we get: | ||
| + | |||
| + | <math>\frac{237}{3} = \boxed {\textbf {(D) } 79}</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:26, 18 June 2025
Contents
Problem
The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?
Solution 1
Let the regular price of one tire be 
.
We are given:
Solution 2
To get the price, we can also work backwards without introducing a variable.
To get the price of all of the tires outside of the fourth tire, we can subtract 
from the total, which gives 
= 
.
Dividing this by to get the price of a single tire without the sale, we get:
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
