Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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<math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10</math> | ||
− | ==Solution | + | ==Solution 1== |
We know that there sandwiches cost <math>4.50</math> dollars. We can multiply <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 = 9 }</math> | We know that there sandwiches cost <math>4.50</math> dollars. We can multiply <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 = 9 }</math> | ||
- SBose | - SBose |
Revision as of 20:23, 11 October 2020
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 there sandwiches cost dollars. We can multiply 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