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# Estimating Products of Decimals

Alignments to Content Standards: 6.NS.B

2. A school bought 8.5 kilograms of apples. If apples cost $2.45 per kilogram, about how much did they pay (in dollars)?  0.2 2 20 200 3. A laptop contains about 0.0006 kilograms of gold. If a manufacturer plans to produce 10,100 of these laptops, about how many kilograms of gold are needed?  0.6 6 60 600 ## IM Commentary The purpose of this task is to help students estimate products of decimal numbers. Estimation is a valuable strategy when deciding where to put the decimal point when multiplying decimals. Although students will learn an algorithmic approach to multiplying decimals in grade 6, making sense of the context is a powerful tool that can support students' developing fluency. In order to promote sense-making, this task might work best before a teacher explains an algorithmic approach, so that students don't just carry out computations and look to see which choice is closest. ## Solution Answers will vary. Sample responses: 1. There were about 10 liters of water all together. I know this because$8 \times 1$is 8. Since we are multiplying 8 by 1.2, the answer will be greater than 8, but not much greater. 2. They paid about \$20. $8\times2=16$. The numbers we are multiplying are a little larger than 8 and 2, so I'd expect the product to be a little larger than 16.
3. There would be about 6 kilograms of gold in 10,001 laptops. I know that $10\times0.0006$ is 0.006, $100\times0.0006$ is 0.06, $1,000\times0.0006$ is 0.6. Therefore $10,000\times0.0006$ is 6. 10,001 is very close to 10,000, so the answer is very close to 6. We could also approach this estimation verbally: Since 0.0006 is six ten-thousandths or $6 \div 10,000$, multiplying by 10,000 results in 6.