Difference between revisions of "1992 AJHSME Problems/Problem 12"

m (Solution)
(Solution)
Line 8: Line 8:
  
 
In the <math> 30,000 </math> miles, four tires were always used at one time, so the amount of miles the five tires were used in total is <math> 30,000 \times 4=120,000 </math>. Five tires were used and each was used equally, so each tire was used for <math> \frac{120,000}{5}=\boxed{\text{(C)}\ 24,000} </math>.
 
In the <math> 30,000 </math> miles, four tires were always used at one time, so the amount of miles the five tires were used in total is <math> 30,000 \times 4=120,000 </math>. Five tires were used and each was used equally, so each tire was used for <math> \frac{120,000}{5}=\boxed{\text{(C)}\ 24,000} </math>.
 +
 +
<math> 30000 \times 10 </math>
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1992|num-b=11|num-a=13}}
 
{{AJHSME box|year=1992|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:21, 27 November 2022

Problem

The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first $30,000$ miles the car traveled. For how many miles was each tire used?

$\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad \text{(E)}\ 37,500$

Solution

In the $30,000$ miles, four tires were always used at one time, so the amount of miles the five tires were used in total is $30,000 \times 4=120,000$. Five tires were used and each was used equally, so each tire was used for $\frac{120,000}{5}=\boxed{\text{(C)}\ 24,000}$.

$30000 \times 10$

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png