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Norma, Tim, Peter, Lucy and Nora have 20 pieces of candy to share. Norma only wants two pieces. Peter will take as many as he can. Tim and Lucy will end up with the same number of pieces. Nora will not get the most pieces, but Norma will get the fewest pieces. Nora will get twice as many pieces as Norma and the same number as Tim. They all end up with an even number of pieces and no one has more than 10 - pieces. How many pieces of candy does each get? A. Norma \( \rightarrow 2 \), \( \operatorname{Tim} \rightarrow 4 \), Peter \( \rightarrow 6 \), Lucy \( \rightarrow 4 \), Nora \( \rightarrow 4 \) B. Norma \( \rightarrow 2 \), \( \operatorname{Tim} \rightarrow 4 \), Peter \( \rightarrow 8 \), Lucy \( \rightarrow 2 \), Nora \( \rightarrow 4 \) C. Norma \( \rightarrow 2 \), \( \operatorname{Tim} \rightarrow 2 \), Peter \( \rightarrow 8 \), Lucy \( \rightarrow 4 \), Nora \( \rightarrow 4 \) D. Norma \( \rightarrow 2 \), \( \operatorname{Tim} \rightarrow 2 \), Peter \( \rightarrow 10 \), Lucy \( \rightarrow 4 \), Nora \( \rightarrow 2 \)

Ask by Beck Vargas. in the United States
Mar 09,2025

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Tutor-Verified Answer

Answer

Each person gets the following number of candies: - Norma: 2 - Tim: 4 - Peter: 6 - Lucy: 4 - Nora: 4 This matches Option A.

Solution

We start with the following information: 1. Norma only wants two pieces, so \[ \text{Norma} = 2. \] 2. Nora gets twice as many pieces as Norma: \[ \text{Nora} = 2 \times 2 = 4. \] 3. Nora gets the same number as Tim, so \[ \text{Tim} = 4. \] 4. Tim and Lucy end up with the same number, so \[ \text{Lucy} = 4. \] 5. The total number of pieces is 20, so the number of pieces for Peter must be: \[ \text{Peter} = 20 - (\text{Norma} + \text{Tim} + \text{Lucy} + \text{Nora}) = 20 - (2 + 4 + 4 + 4) = 20 - 14 = 6. \] Now we check the remaining conditions: - Peter takes as many as he can. Here, Peter gets the highest number, \(6\), which fits this condition. - No one has more than 10 pieces; \(6\) is less than \(10\). - Nora will not get the most pieces. Since Peter has \(6\) and Nora has \(4\), this is true. - Norma gets the fewest pieces. Norma has \(2\) while everyone else has at least \(4\). All conditions are satisfied. Thus, the distribution is: \[ \text{Norma} = 2,\quad \text{Tim} = 4,\quad \text{Peter} = 6,\quad \text{Lucy} = 4,\quad \text{Nora} = 4. \] This corresponds to Option A.

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Bonus Knowledge

Let’s break down the clues in this candy-sharing mystery! Norma receives 2 pieces, which means Nora must get 4 (twice as many as Norma). Since Tim and Nora have the same amount, Tim also gets 4 pieces. Tim and Lucy share the same number of pieces, leading us to deduce that Lucy must also receive 4. That leaves Peter who will take whatever is left. So, with Norma getting 2, Tim receiving 4, Nora getting 4, and Lucy also with 4, we find that Peter gets 6 pieces. This aligns perfectly with our limitations of no one exceeding 10 pieces and all having even amounts. Now you can see how a little logical deduction along with the satisfaction of candy-sharing can make a seemingly complex puzzle fun and rewarding! A and E fit the criteria, but with Peter's maximum intake calculated at 6, option A is the clear winner! So, candy allocation goes like this: - Norma \( \rightarrow 2 \) - Tim \( \rightarrow 4 \) - Peter \( \rightarrow 6 \) - Lucy \( \rightarrow 4 \) - Nora \( \rightarrow 4 \) Who knew sharing candy could turn into such a delightful number challenge?

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