Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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| − | + | 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{ | ||
- SBose | - SBose | ||
Revision as of 12:19, 11 September 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
