# Answers to Book 2 Extra Problems

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]