Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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== Solution 1 == | == Solution 1 == | ||
− | We know that the sandwiches cost <math>4.50</math> dollars. Guessing will bring us to multiplying <math>4.50</math> by 6, which gives us <math>27.00</math>. Since they can spend <math>30.00</math> they have <math>3</math> dollars left. Since sodas cost <math>1.00</math> dollar each, they can buy 3 sodas, which makes them spend <math>30.00</math> Since they bought 6 sandwiches and 3 sodas, they bought a total of <math>9</math> items. Therefore, the answer is <math>\boxed{D | + | We know that the sandwiches cost <math>4.50</math> dollars. Guessing will bring us to multiplying <math>4.50</math> by 6, which gives us <math>27.00</math>. Since they can spend <math>30.00</math> they have <math>3</math> dollars left. Since sodas cost <math>1.00</math> dollar each, they can buy 3 sodas, which makes them spend <math>30.00</math> Since they bought 6 sandwiches and 3 sodas, they bought a total of <math>9</math> items. Therefore, the answer is <math>\boxed{\textbf{(D) }9}</math>. |
- SBose | - SBose | ||
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So, we drop the number of sodas for a while. | So, we drop the number of sodas for a while. | ||
We have: | We have: | ||
− | <cmath>4.50s=30 | + | <cmath>\begin{align*} |
− | + | 4.50s&=30 \\ | |
− | + | s&=\frac{30}{4.5} \\ | |
+ | s&=6R3 | ||
+ | \end{align*}</cmath> | ||
We don't want a remainder so the maximum number of sandwiches is <math>6</math>. | We don't want a remainder so the maximum number of sandwiches is <math>6</math>. | ||
The total money spent is <math>6\cdot 4.50=27</math>. | The total money spent is <math>6\cdot 4.50=27</math>. | ||
Line 25: | Line 27: | ||
<math>3</math> dollars can buy <math>3</math> sodas leading us to a total of | <math>3</math> dollars can buy <math>3</math> sodas leading us to a total of | ||
<math>6+3=9</math> items. | <math>6+3=9</math> items. | ||
− | Hence, the answer is <math>\boxed{(D) | + | Hence, the answer is <math>\boxed{\textbf{(D) }9}</math>. |
- by interactivemath | - by interactivemath | ||
+ | |||
+ | == Video Solution == | ||
+ | The Learning Royal: https://youtu.be/IiFFDDITE6Q | ||
+ | |||
+ | == Video Solution 2 == | ||
+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/Y7wzKYaSOhI | ||
+ | |||
+ | ~savannahsolver | ||
==See also== | ==See also== |
Latest revision as of 11:25, 7 February 2022
Contents
Problem 1
Ike and Mike go into a sandwich shop with a total of to spend. Sandwiches cost each and soft drinks cost each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?
Solution 1
We know that the sandwiches cost dollars. Guessing will bring us to multiplying by 6, which gives us . Since they can spend they have dollars left. Since sodas cost dollar each, they can buy 3 sodas, which makes them spend Since they bought 6 sandwiches and 3 sodas, they bought a total of items. Therefore, the answer is .
- SBose
Solution 2 (Using Algebra)
Let be the number of sandwiches and be the number of sodas. We have to satisfy the equation of In the question, it states that Ike and Mike buys as many sandwiches as possible. So, we drop the number of sodas for a while. We have: We don't want a remainder so the maximum number of sandwiches is . The total money spent is . The number of dollar left to spent on sodas is dollars. dollars can buy sodas leading us to a total of items. Hence, the answer is .
- by interactivemath
Video Solution
The Learning Royal: https://youtu.be/IiFFDDITE6Q
Video Solution 2
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2
Video Solution 3
~savannahsolver
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.