Difference between revisions of "2006 AMC 12B Problems/Problem 6"
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== Problem == | == Problem == | ||
| + | Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? | ||
| − | == Solution == | + | <math> |
| + | \text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411 | ||
| + | </math> | ||
| + | |||
| + | ==Solution== | ||
| + | Francesca makes a total of <math>100+100+400=600</math> grams of lemonade, and in those <math>600</math> grams, there are <math>25</math> calories from the lemon juice and <math>386</math> calories from the sugar, for a total of <math>25+386=411</math> calories per <math>600</math> grams. We want to know how many calories there are in <math>200=600/3</math> grams, so we just divide <math>411</math> by <math>3</math> to get <math>137\implies\boxed{(\text{B})}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=B|num-b=5|num-a=7}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 12:57, 4 December 2015
Problem
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?
Solution
Francesca makes a total of 
grams of lemonade, and in those 
grams, there are 
calories from the lemon juice and 
calories from the sugar, for a total of 
calories per grams. We want to know how many calories there are in 
grams, so we just divide 
by 
to get 
.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
