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Smart Shopping - Reason Proportionally Using Least Common Multiples

Teacher Notes

Smart Shopping shows prices for the same item from two different stores. To determine the better buy, students equate the number of items in order to compare prices. To accomplish this, they 1) identify the least common multiple of number of items, 2) the prices for those numbers of items, and 3) and compare prices to determine the better buy.

Solutions

1. \quad \text{Least common multiple of} \ 3 \ \text{and} \ 4 \ \text{is} \ 12.\!\\{\;} \quad \ \text{Brown's Books has the better buy.}\!\\\\{\;} \quad \ \text{Brown:} \ 3 \ \text{for} \ \$5 \ \text{is}\!\\ {\;} \quad \quad \ 12 \ \text{for} \ 4 \times \$5 \ \text{or} \ \$20\!\\{\;} \quad \ \text{Noble:} \ 4 \ \text{for} \ \$7 \ \text{is}\!\\ {\;} \quad \quad \ 12 \ \text{for} \ 3 \times \$7, \ \text{or} \ \$21\!\\{\;} \quad \quad \ \$20 < \$21

2. \quad \text{Least common multiple of} \ 2 \ \text{and} \ 5 \ \text{is} \ 10.\!\\{\;} \quad \ \text{Brown's Books has the better buy.}\!\\\\{\;} \quad \ \text{Brown:} \ 2 \ \text{for} \ \$11 \ \text{is}\!\\ {\;} \quad \quad \ 10 \ \text{for} \ 5 \times \$11, \ \text{or} \ \$55\!\\{\;} \quad \ \text{Noble:} \ 5 \ \text{for} \ \$31 \ \text{is}\!\\ {\;} \quad \quad \ 10 \ \text{for} \ 2 \times \$31, \ \text{or} \ \$62\!\\{\;} \quad \quad \ \$55 < \$62

3. \quad \text{Least common multiple of} \ 5 \ \text{and} \ 3 \ \text{is} \ 15.\!\\{\;} \quad \ \text{Noble Books has the better buy.}\!\\\\{\;} \quad \ \text{Brown:} \ 5 \ \text{for} \ \$59 \ \text{is}\!\\ {\;} \quad \quad \ 15 \ \text{for} \ 3 \times \$59, \ \text{or} \ \$177\!\\{\;} \quad \ \text{Noble:} \ 3 \ \text{for} \ \$34 \ \text{is}\!\\ {\;} \quad \quad \ 15 \ \text{for} \ 5 \times \$34, \ \text{or} \ \$170\!\\{\;} \quad \quad \ \$177 > \$170

Which store has the better buy for calendars?

& \mathbf{Describe:} && \text{At Brown's Books,} \ 2 \ \text{calendars are} \ \$17. \ \text{At Noble Books,} \ 3 \ \text{calendars are} \ \$23.\\  & \mathbf{Plan:} && \text{First, figure out the least common multiple of the number of calendars,} \ 2 \ \text{and} \ 3.\\&&& \text{Second, figure out the price for that number of calendars at each store. Last,}\\&&& \text{compare prices.}\\& \mathbf{Solve:} && \text{The least common multiple of} \ 2 \ \text{and} \ 3 \ \text{is} \ 6.\\&&& \text{Brown's:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \text{Noble:}\\&&& 2 \ \text{for} \ \$17 \ \text{is} \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3 \ \text{for} \ \$23 \ \text{is}\\&&& (3 \times 2) \ \text{for} \ (3 \times \$17), \ \text{or} \qquad \qquad \qquad \qquad \ (2 \times 3) \ \text{for} \ (2 \times \$23), \ \text{or}\\&&& 6 \ \text{for} \ \$51 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 6 \ \text{for} \ \$46\\&&& \$46 < \$51\\&&& \text{Noble Books has the better buy.}\\& \mathbf{Check:} && \text{Brown:} \ 2 \ \text{for} \ \$17 \ \text{is} \ \$17 + \$17 + \$17 \ \text{or} \ \$51 \ \text{for} \ 6 \ \text{calendars}\\&&& \text{Noble:} \ 3 \ \text{for} \ \$23 \ \text{is} \ \$23 + \$23 \ \text{or} \ \$46 \ \text{for} \ 6 \ \text{calendars}\\&&& \$46 < \$51

What is the least common multiple of 3 and 4?

Which store has the better buy? How did you figure it out?

What is the least common multiple of 2 and 5?

Which store has the better buy? How did you figure it out?

What is the least common multiple of 5 and 3?

Which store has the better buy? How did you figure it out?

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