1. [pmath]12/1[/pmath] x [pmath]1/4[/pmath] = [pmath]12/4[/pmath] = 3
2. [pmath]3/3[/pmath] – [pmath]2/3[/pmath] = [pmath]1/3[/pmath]
3. [pmath]6/6[/pmath] – [pmath]1/6[/pmath] – [pmath]2/6[/pmath] = [pmath]3/6[/pmath]
4. [pmath]2/3[/pmath] x [pmath]2/6[/pmath] = [pmath]4/18[/pmath]
5. [pmath]3/4[/pmath] x [pmath]1/12[/pmath] = [pmath]3/48[/pmath]
6. Since you want to make twice as many cookies, you need twice as much sugar: [pmath]2/1[/pmath] x [pmath]2/3[/pmath] cup = [pmath]4/3[/pmath] cup
7. Since you want to make three times as many cookies, you need three times the amount of vanilla: [pmath]3/1[/pmath] x [pmath]1/2[/pmath] teaspoon = [pmath]3/2[/pmath] teaspoon
8. [pmath]2/5[/pmath] x [pmath]3/5[/pmath] = [pmath]6/25[/pmath]
9. First we have to find out what fraction of the m&m’s in the bowl are peanut m&m’s. This is a subtraction problem because we were told that two-fifths of all the m&m’s are chocolate: The fraction that are peanut m&m’s must be [pmath]5/5[/pmath] – [pmath]2/5[/pmath] = [pmath]3/5[/pmath]. The fraction of these m&m’s that are yellow is one-fifth. So the fraction of all the m&m’s in the bowl that are yellow peanut m&m’s is [pmath]1/5[/pmath] x [pmath]3/5[/pmath] = [pmath]3/25[/pmath]
10. This is another subtraction problem. The fraction representing all the m&m’s in the bowl is [pmath]25/25[/pmath]. The fraction that is red chocolate is [pmath]6/25[/pmath] (the answer to problem 8), and the fraction that is yellow peanut is [pmath]3/25[/pmath] (the answer to Problem 9). The remainder must therefore be:
[pmath]25/25[/pmath] – [pmath]6/25[/pmath] – [pmath]3/25[/pmath] = [pmath]16/25[/pmath]