1. \u00a0\u00a0 [pmath]12\/1[\/pmath] x \u00a0 [pmath]1\/4[\/pmath] =\u00a0 [pmath]12\/4[\/pmath] = 3<\/p>\n
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2.\u00a0\u00a0\u00a0\u00a0 [pmath]3\/3[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]2\/3[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]1\/3[\/pmath]<\/p>\n
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3.\u00a0\u00a0\u00a0\u00a0 [pmath]6\/6[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]1\/6[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]2\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/6[\/pmath]<\/p>\n
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4.\u00a0\u00a0\u00a0\u00a0 [pmath]2\/3[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]2\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]4\/18[\/pmath]<\/p>\n
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5.\u00a0\u00a0\u00a0\u00a0 [pmath]3\/4[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]1\/12[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/48[\/pmath]<\/p>\n
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6.\u00a0\u00a0\u00a0\u00a0 Since you want to make twice as many cookies, you need twice as much sugar: \u00a0 [pmath]2\/1[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]2\/3[\/pmath] cup\u00a0 =\u00a0\u00a0 [pmath]4\/3[\/pmath]\u00a0 cup<\/p>\n
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7.\u00a0\u00a0\u00a0\u00a0 Since you want to make three times as many cookies, you need three times the amount of vanilla:\u00a0\u00a0\u00a0 [pmath]3\/1[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]1\/2[\/pmath]\u00a0 teaspoon\u00a0 =\u00a0\u00a0 [pmath]3\/2[\/pmath]\u00a0 teaspoon<\/p>\n
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8.\u00a0\u00a0\u00a0\u00a0 [pmath]2\/5[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]3\/5[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]6\/25[\/pmath]<\/p>\n
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9.\u00a0\u00a0\u00a0\u00a0 First we have to find out what fraction of the m&m’s in the bowl are peanut m&m’s.\u00a0 This is a subtraction problem because we were told that two-fifths of all the m&m’s are chocolate:\u00a0 The fraction that are peanut m&m’s must be \u00a0\u00a0 [pmath]5\/5[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]2\/5[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/5[\/pmath].\u00a0 The fraction of these m&m’s that are yellow is one-fifth.\u00a0 So the fraction of all the m&m’s in the bowl that are yellow peanut m&m’s is\u00a0\u00a0\u00a0 [pmath]1\/5[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]3\/5[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/25[\/pmath]<\/p>\n
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10.\u00a0\u00a0\u00a0 This is another subtraction problem. \u00a0 The fraction representing all the m&m’s in the bowl is\u00a0\u00a0\u00a0 [pmath]25\/25[\/pmath].\u00a0 The fraction that is red chocolate is\u00a0\u00a0 [pmath]6\/25[\/pmath]\u00a0 (the answer to problem 8), and the fraction that is yellow peanut is\u00a0\u00a0 [pmath]3\/25[\/pmath]\u00a0 (the answer to Problem 9).\u00a0 The remainder must therefore be:<\/p>\n
[pmath]25\/25[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]6\/25[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]3\/25[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]16\/25[\/pmath]<\/p>\n","protected":false},"excerpt":{"rendered":"
1. \u00a0\u00a0 [pmath]12\/1[\/pmath] x \u00a0 [pmath]1\/4[\/pmath] =\u00a0 [pmath]12\/4[\/pmath] = 3 2.\u00a0\u00a0\u00a0\u00a0 [pmath]3\/3[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]2\/3[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]1\/3[\/pmath] 3.\u00a0\u00a0\u00a0\u00a0 [pmath]6\/6[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]1\/6[\/pmath]\u00a0 –\u00a0\u00a0 [pmath]2\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/6[\/pmath] 4.\u00a0\u00a0\u00a0\u00a0 [pmath]2\/3[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]2\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]4\/18[\/pmath] 5.\u00a0\u00a0\u00a0\u00a0 [pmath]3\/4[\/pmath]\u00a0 x\u00a0\u00a0 [pmath]1\/12[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]3\/48[\/pmath] 6.\u00a0\u00a0\u00a0\u00a0 Since you want to make twice as many cookies, you need twice as … Continue reading Answers to Book 2 Extra Problems<\/span>