Free math worksheets

- Consider what happens when a positive number is multiplied by a factor greater than \(1\). - Find the factor that would produce exactly \(\frac{1}{8}\). - Choose a smaller positive factor for part b).

1. For a), multiplying a positive number by a factor greater than \(1\) increases it. For example, \(\frac{3}{2}\times\frac{3}{8}=\frac{9}{16}>\frac{3}{8}\). 2. For b), a factor smaller than \(\frac{1}{3}\) makes the product less than \(\frac{1}{8}\), because \(\frac{1}{3}\times\frac{3}{8}=\frac{1}{8}\). For example, \(\frac{1}{4}\times\frac{3}{8}=\frac{3}{32}<\frac{1}{8}\).

a) For example, \(\frac{3}{2}\) b) For example, \(\frac{1}{4}\)

- Compare the fractional factor with \(1\). - Think about whether taking only part of a positive quantity makes it larger or smaller.

1. The factor \(\frac{5}{6}\) is greater than \(0\) and less than \(1\). 2. Multiplying a positive number by a factor less than \(1\) makes the product smaller than the original number.

The product is less than \(3\frac{1}{2}\).

- Compare each multiplying factor with \(1\). - Think about whether taking less than one whole, exactly one whole, or more than one whole of \(9\) changes its size.

1. \(\frac{5}{6}<1\), so \(\frac{5}{6}\times9<9\). 2. \(1\times9=9\). 3. \(\frac{7}{6}>1\), so \(\frac{7}{6}\times9>9\).

\(\frac{5}{6}\times9<1\times9<\frac{7}{6}\times9\).

- Compare the scale factor with \(1\) and \(2\). - Think about what multiplying a positive length by more than \(1\), but less than \(2\), does to its size.

1. The scale factor \(\frac{7}{5}\) is greater than \(1\) but less than \(2\). 2. Multiplying a positive length by a factor between \(1\) and \(2\) makes the result greater than the original length but less than twice the original length. 3. Therefore the new route is between \(10\) and \(20\) miles long.

The new route is between \(10\) and \(20\) miles long.

- Compare the batch-size factor with \(1\). - Think about whether making only part of a full batch uses more or less than the full-batch amount.

1. The scale factor \(\frac{7}{9}\) is between \(0\) and \(1\). 2. Multiplying the positive amount \(\frac{2}{3}\) cup by a factor less than \(1\) makes the amount smaller.

Maya uses less than \(\frac{2}{3}\) cup of oats.

- Compare the numerator and denominator of the multiplier. - Decide what value a fraction with equal numerator and denominator represents. - Recall what happens when any number is multiplied by \(1\).

1. The numerator and denominator are equal, so \(\frac{5}{5}=1\). 2. Multiplying a quantity by \(1\) leaves its value unchanged. 3. Therefore \(12\times\frac{5}{5}=12\) miles. 4. Fatima is not correct; the factor changes the written form of the multiplier but not its value.

Fatima is not correct. Since \(\frac{5}{5}=1\), \(12\times\frac{5}{5}=12\) miles, so the route length is unchanged.

- Compare \(\frac{2}{3}\) with \(\frac{1}{2}\) and \(1\). - Use the fraction as a scale factor for the route's length.

1. The scale factor \(\frac{2}{3}\) is greater than \(\frac{1}{2}\) and less than \(1\). 2. Therefore the shortcut is more than half as long as the route but shorter than the full route.

The shortcut is between half as long as the route and equally long.

- Compare each scale factor with \(1\). - Decide which factor shortens the route and which lengthens it.

1. The scale factor \(\frac{4}{5}\) is less than \(1\), so \(\frac{4}{5}\times10<10\). 2. The scale factor \(\frac{6}{5}\) is greater than \(1\), so \(\frac{6}{5}\times10>10\). 3. Therefore \(\frac{4}{5}\times10<10<\frac{6}{5}\times10\).

\(\frac{4}{5}\times10<10<\frac{6}{5}\times10\).

- Compare each multiplier with \(1\). - Decide which multiplier shrinks \(6\) and which enlarges it.

1. Since \(\frac{5}{4}=1+\frac{1}{4}\), multiplying \(6\) by \(\frac{5}{4}\) gives one full copy of \(6\) plus an additional positive part, so \(\frac{5}{4}\times6>6\). 2. Since \(\frac{4}{5}=1-\frac{1}{5}\), multiplying \(6\) by \(\frac{4}{5}\) gives one full copy of \(6\) with a positive part removed, so \(\frac{4}{5}\times6<6\). 3. Thus \(\frac{4}{5}\times6<6<\frac{5}{4}\times6\), and \(\frac{5}{4}\times6\) is the greater product.

\(\frac{4}{5}\times6<6<\frac{5}{4}\times6\).

- Notice that every expression has the same positive whole-number factor. - Compare the fraction multipliers with one another instead of multiplying first. - Fractions below \(1\) shrink the fixed factor, while fractions above \(1\) enlarge it.

1. The fixed factor \(12\) is positive, so the products have the same order as the multipliers. 2. Order the multipliers: \(\frac{2}{5}<\frac{3}{4}<\frac{5}{4}<\frac{8}{5}\). 3. Therefore the products occur in that same order.

\(12\times\frac{2}{5}<12\times\frac{3}{4}<12\times\frac{5}{4}<12\times\frac{8}{5}\).

- Think about which multiplier changes \(8\) the least. - Compare each multiplier with \(1\) instead of multiplying.

1. A product stays closest to \(8\) when the other factor is closest to \(1\). 2. \(\frac{9}{10}\) is \(\frac{1}{10}\) from \(1\), \(\frac{3}{4}\) is \(\frac{1}{4}\) from \(1\), and \(\frac{11}{8}\) is \(\frac{3}{8}\) from \(1\). 3. Therefore \(8\times\frac{9}{10}\) is closest to \(8\).

\(8\times\frac{9}{10}\) is closest to \(8\).

- Compare the multiplier with \(1\) to find the lower bound. - Find a simple factor that would make the product exactly \(6\), then compare the given multiplier with it.

1. \(\frac{11}{10}>1\), so the product is greater than \(5\). 2. \(\frac{11}{10}<\frac{6}{5}\), and \(5\times\frac{6}{5}=6\). 3. Therefore the product is between \(5\) and \(6\).

The product is between \(5\) and \(6\).

- Compare each multiplier's distance from \(1\). - The same base number is being scaled in both expressions.

1. \(\frac{7}{8}\) is \(\frac{1}{8}\) below \(1\), while \(\frac{9}{8}\) is \(\frac{1}{8}\) above \(1\). 2. The multipliers are equally far from \(1\), so scaling the same positive number \(8\) changes it by equal amounts in opposite directions. 3. The products are equally close to \(8\).

The two products are equally close to \(8\).

- Use \(1\) as the benchmark scale factor. - Ask which kind of positive factor makes an amount smaller.

1. Multiplying \(6\) by \(1\) would leave the amount at \(6\) cups. 2. Multiplying by a positive factor greater than \(1\) would make the amount greater than \(6\) cups. 3. Because the scaled amount is less than \(6\) cups, the scale factor must be less than \(1\).

The scale factor must be less than \(1\).

- First compare each multiplier with \(1\). - For a multiplier greater than \(1\), think about the extra fractional part beyond one whole copy of \(10\). - Compare that extra amount with the \(2\)-unit width of the target interval. - Use exact multiplication only after you have made your size prediction.

1. Multiplying by \(\frac{3}{4}<1\) makes the product less than \(10\), so the first expression does not fit. 2. \(\frac{9}{8}=1+\frac{1}{8}\). The extra \(\frac{1}{8}\) of \(10\) is less than \(2\), so this product is greater than \(10\) but less than \(12\). 3. \(\frac{7}{5}=1+\frac{2}{5}\). The extra \(\frac{2}{5}\) of \(10\) is \(4\), so this product is \(14\), which is greater than \(12\). 4. Check the middle expression: \(10\times\frac{9}{8}=\frac{90}{8}=\frac{45}{4}=11\frac{1}{4}\).

\(10\times\frac{9}{8}\), and its value is \(11\frac{1}{4}\).

- Compare the new route with the full \(18\)-mile route to bound the scale factor by \(1\). - Use the fact that half of \(18\) is \(9\) to find the other benchmark.

1. A scale factor less than \(1\) is required because the new route is shorter than \(18\) miles. 2. Multiplying \(18\) by \(\frac{1}{2}\) gives \(9\). Because the new route is longer than \(9\) miles, the scale factor must be greater than \(\frac{1}{2}\). 3. Therefore the scale factor is between \(\frac{1}{2}\) and \(1\).

The scale factor is between \(\frac{1}{2}\) and \(1\).

- Translate the target interval for the product into an interval for the multiplier. - Think about which familiar fractions of \(4\) give the endpoints \(2\) and \(3\). - Use exact multiplication only after making the comparison.

1. A product between \(2\) and \(3\) requires a multiplier between \(\frac{1}{2}\) and \(\frac{3}{4}\), because \(4\times\frac{1}{2}=2\) and \(4\times\frac{3}{4}=3\). 2. Only \(\frac{3}{5}\) lies between \(\frac{1}{2}\) and \(\frac{3}{4}\). 3. Check: \(4\times\frac{3}{5}=\frac{12}{5}=2\frac{2}{5}\), which is between \(2\) and \(3\).

\(4\times\frac{3}{5}\), with value \(2\frac{2}{5}\).

- Use the lower bound to decide whether the scale factor is below or above \(1\). - Find the scale factor that would make exactly \(12\) feet. - Eliminate choices using those benchmarks before checking exactly.

1. Because the second ribbon is longer than \(8\) feet, the scale factor must be greater than \(1\), so \(\frac{3}{4}\) cannot work. 2. Since \(12=8\times\frac{3}{2}\), the scale factor must also be less than \(\frac{3}{2}\). 3. Of the remaining choices, only \(\frac{5}{4}<\frac{3}{2}\). 4. Check: \(8\times\frac{5}{4}=10\), which is between \(8\) and \(12\).

The scale factor is \(\frac{5}{4}\), giving a ribbon length of \(10\) feet.

- Find the benchmark scale factor that makes exactly \(4\frac{1}{2}\) from \(9\). - With the original route fixed, relate a shorter product to a smaller positive scale factor.

1. Multiplying \(9\) by \(\frac{1}{2}\) gives \(4\frac{1}{2}\). 2. The actual route is shorter than \(4\frac{1}{2}\) miles. 3. Because the positive route length \(9\) is fixed, the scale factor must be less than \(\frac{1}{2}\).

The scale factor must be less than \(\frac{1}{2}\).

- Interpret \(\frac{6}{5}\) as one whole plus one fifth of the starting amount. - Bound one fifth of \(23\) between nearby whole numbers before choosing an estimate. - Use the exact multiplication only after the scaling estimate has been justified.

1. Since \(4<\frac{23}{5}<5\), multiplying by \(1+\frac{1}{5}\) adds between \(4\) and \(5\) to \(23\). 2. Therefore the product lies between \(27\) and \(28\), so \(28\) is the most reasonable estimate. 3. The exact product is \(23\times\frac{6}{5}=\frac{138}{5}=27\frac{3}{5}\), which confirms the estimate.

Using scaling, the product is between \(27\) and \(28\), so \(28\) is the best estimate. The exact product is \(27\frac{3}{5}\).

- Compare the fraction multiplier with \(1\) to decide whether the product grows or shrinks. - Estimate the extra amount represented by the part of the multiplier above \(1\). - Use exact multiplication only after choosing an interval.

1. The multiplier \(\frac{7}{6}=1\frac{1}{6}\) means one full copy of \(5\frac{1}{2}\) plus one sixth of it. 2. Since \(3<5\frac{1}{2}<6\), one sixth of \(5\frac{1}{2}\) is between \(\frac{1}{2}\) and \(1\). Adding that increase to \(5\frac{1}{2}\) puts the product between \(6\) and \(6\frac{1}{2}\), so it lies between \(6\) and \(7\). 3. Exact check: \(\frac{11}{2}\times\frac{7}{6}=\frac{77}{12}=6\frac{5}{12}\). 4. The exact product confirms the interval.

The product is between \(6\) and \(7\); exactly, it is \(6\frac{5}{12}\).

- Compare the multiplier with \(1\) to decide whether the product should be above or below \(17\). - Estimate the missing one-eighth part of \(17\). - Use the exact product only to confirm your estimate.

1. The multiplier \(\frac{7}{8}\) is slightly less than \(1\), so the product should be slightly less than \(17\). 2. One eighth of \(17\) is a little more than \(2\), so reducing \(17\) by about \(2\) gives about \(15\). 3. Exact check: \(17\times\frac{7}{8}=\frac{119}{8}=14\frac{7}{8}\). 4. The exact value confirms that \(15\) is the best estimate.

\(15\) is the best estimate; the exact product is \(14\frac{7}{8}\).

- Use scaling only to place both products near \(20\), not to declare a winner. - Calculate each product exactly and compare each result with \(20\). - Notice what the equality shows about the limits of a rough estimate in a near-tie.

1. The scaling observation places both products near \(20\), but it does not quantify the two changes closely enough to order the results. 2. \(18\times\frac{11}{10}=\frac{198}{10}=\frac{99}{5}=19\frac{4}{5}\). 3. \(22\times\frac{9}{10}=\frac{198}{10}=\frac{99}{5}=19\frac{4}{5}\). 4. The products are equal, so each is \(\frac{1}{5}\) below \(20\). Exact calculation was needed to settle the near-tie.

Both products equal \(19\frac{4}{5}\), so each is \(\frac{1}{5}\) below \(20\). The rough estimates alone could not decide the comparison.

- Compare what is known about the multipliers with what is different about the starting numbers. - Find each exact product before deciding a close comparison. - Compare absolute distances from \(30\), not just which side of \(30\) each product lies on.

1. The multipliers are equally far from \(1\), but they act on different starting numbers. Also, \(28\) and \(32\) are on opposite sides of \(30\). Therefore the multiplier comparison alone cannot settle which product is closer to \(30\). 2. \(28\times\frac{9}{8}=31\frac{1}{2}\), which is \(1\frac{1}{2}\) from \(30\). 3. \(32\times\frac{7}{8}=28\), which is \(2\) from \(30\). 4. Therefore \(28\times\frac{9}{8}\) is closer to \(30\).

The equal distances of the multipliers from \(1\) do not decide the comparison because the starting numbers differ. \(28\times\frac{9}{8}=31\frac{1}{2}\) is \(1\frac{1}{2}\) from \(30\), while \(32\times\frac{7}{8}=28\) is \(2\) from \(30\). The first product is closer.

- Compare each multiplier with \(1\). - Describe each alternate route as the original length plus or minus a fractional part of that same original length. - Equal changes in opposite directions place the original length at the midpoint.

1. A factor of \(\frac{11}{12}\) makes the first alternate route one twelfth of the original route shorter than \(24\) miles. 2. A factor of \(\frac{13}{12}\) makes the second alternate route one twelfth of the original route longer than \(24\) miles. 3. The two changes have the same size and go in opposite directions. 4. Therefore the original \(24\)-mile route is exactly halfway between the two alternate route lengths.

Yes. The first alternate route is one twelfth of the original length shorter, and the second is one twelfth of the original length longer, so \(24\) miles is exactly halfway between them.

- Compare the multiplier with \(1\) and identify the small extra fractional part. - Think about what that extra fraction of \(14\) equals. - Combine one full copy of \(14\) with the extra amount.

1. Rewrite \(\frac{15}{14}=1+\frac{1}{14}\). 2. Multiplying \(14\) by \(1\) gives \(14\), and multiplying \(14\) by \(\frac{1}{14}\) gives \(1\). 3. Therefore the product is \(14+1=15\).

\(14\times\frac{15}{14}=15\).

- Compare the multiplier with \(1\) and identify the small missing fraction. - Find that missing fraction of \(25\). - Subtract the missing amount from one full copy of \(25\).

1. Rewrite \(\frac{24}{25}=1-\frac{1}{25}\). 2. One twenty-fifth of \(25\) is \(1\). 3. Multiplying by \(\frac{24}{25}\) keeps one whole copy of \(25\) except for that \(1\), so the product is \(25-1=24\).

\(25\times\frac{24}{25}=24\).

- Rewrite the multiplier as one whole plus an extra fraction. - Estimate the extra fraction of the starting mixed number. - Use exact multiplication only after deciding which side of \(8\) the product should lie on.

1. The multiplier \(\frac{5}{4}=1+\frac{1}{4}\), so the product is \(6\frac{1}{2}\) plus one fourth of \(6\frac{1}{2}\). 2. One fourth of \(6\frac{1}{2}\) is greater than \(1\frac{1}{2}\), so the product is greater than \(8\). 3. Exact check: \(\frac{13}{2}\times\frac{5}{4}=\frac{65}{8}=8\frac{1}{8}\). 4. The product is indeed greater than \(8\).

The product is greater than \(8\); exactly, it is \(8\frac{1}{8}\).

- Rewrite \(\frac{3}{2}\) as one whole plus one half. - Compare the increase from the original path with the remaining increase to twice the original length. - Equal increases on both sides place the second length at the midpoint.

1. A factor of \(\frac{3}{2}=1+\frac{1}{2}\) makes the second path one half of the original length longer than \(16\) feet. 2. The third path is two copies of the original, so it is one full original length longer than the first path. 3. From the second path to the third path, one more half of the original length remains. 4. The second path is therefore exactly halfway between the other two lengths. Numerically, the lengths are \(16\), \(24\), and \(32\) feet.

Yes. The second path is \(24\) feet long, exactly halfway between \(16\) feet and \(32\) feet.

- Separate the scale factor into one whole plus an extra fraction. - The extra part of the multiplier describes the increase over the original length. - Compare that fractional increase with \(\frac{1}{2}\), then use the route length as a check.

1. Rewrite \(\frac{7}{5}=1+\frac{2}{5}\). 2. The new route contains one original route plus an increase equal to \(\frac{2}{5}\) of the original length. 3. Since \(\frac{2}{5}<\frac{1}{2}\), the increase is less than one half of the original route. 4. For the \(20\)-mile route, the increase is \(\frac{2}{5}\times20=8\) miles, while half the original length is \(10\) miles.

The increase is \(\frac{2}{5}\) of the original route, or \(8\) miles. It is less than one half of the original length.

- Compare each multiplier with \(1\). - The amount below or above \(1\) gives the fractional decrease or increase. - Compare the two change fractions, then use the original length to find each change in miles.

1. A factor of \(\frac{4}{5}=1-\frac{1}{5}\) decreases the trail by \(\frac{1}{5}\) of its original length. 2. A factor of \(\frac{5}{4}=1+\frac{1}{4}\) increases the trail by \(\frac{1}{4}\) of its original length. 3. Since \(\frac{1}{4}>\frac{1}{5}\), the increase is the larger fractional change. 4. The decrease is \(\frac{1}{5}\times20=4\) miles, and the increase is \(\frac{1}{4}\times20=5\) miles.

The increase is larger. The decrease is \(\frac{1}{5}\) of the original trail, or \(4\) miles; the increase is \(\frac{1}{4}\), or \(5\) miles.

- When the same positive number is multiplied by two fractions, compare the fractions first. - Use equivalent fractions to see which multiplier is larger. - Find the products only after making the size prediction.

1. Compare the multipliers using denominator \(120\): \(\frac{13}{15}=\frac{104}{120}\) and \(\frac{7}{8}=\frac{105}{120}\). Therefore, \(\frac{13}{15}\) is smaller. 2. Compute \(30\times\frac{13}{15}=26\). 3. Compute \(30\times\frac{7}{8}=\frac{105}{4}=26\frac{1}{4}\). 4. The first product is smaller by \(26\frac{1}{4}-26=\frac{1}{4}\).

\(30\times\frac{13}{15}\) is smaller. The products are \(26\) and \(26\frac{1}{4}\), a difference of \(\frac{1}{4}\).

- Rewrite the two scale factors with a common denominator. - The difference between the factors gives the difference as a fraction of the original route. - Find that fraction of \(40\) miles to express the difference in miles.

1. Rewrite \(\frac{3}{4}=\frac{6}{8}\). 2. The difference between the scale factors is \(\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\). 3. Therefore the first trail is \(\frac{1}{8}\) of the original route shorter. 4. One eighth of \(40\) miles is \(\frac{1}{8}\times40=5\) miles.

The first planned trail is \(\frac{1}{8}\) of the original route shorter, which is \(5\) miles.

- Find a fraction that would scale \(2\frac{1}{2}\) to exactly \(2\). - Compare the given multiplier with that benchmark fraction.

1. For \(2\frac{1}{2}\), multiplying by \(\frac{4}{5}\) would give exactly \(2\). 2. The given multiplier \(\frac{3}{4}\) is less than \(\frac{4}{5}\). 3. Therefore \(2\frac{1}{2}\times\frac{3}{4}<2\).

The product is less than \(2\).

- Find the fraction of \(4\frac{1}{2}\) that would make exactly \(4\). - Rewrite the benchmark fraction and \(\frac{5}{6}\) with a common denominator. - With the trail length fixed, a smaller positive fraction gives a shorter distance.

1. Since \(4\frac{1}{2}=\frac{9}{2}\), the fraction of the trail that equals \(4\) miles is \(\frac{8}{9}\), because \(\frac{8}{9}\times\frac{9}{2}=4\). 2. Compare \(\frac{5}{6}=\frac{15}{18}\) and \(\frac{8}{9}=\frac{16}{18}\), so \(\frac{5}{6}<\frac{8}{9}\). 3. Therefore \(\frac{5}{6}\) of the trail is less than \(4\) miles.

Exactly \(4\) miles is \(\frac{8}{9}\) of the trail. Because \(\frac{5}{6}<\frac{8}{9}\), \(\frac{5}{6}\) of the trail is below \(4\) miles.

- Rewrite \(\frac{4}{3}\) as one whole plus an extra fractional part. - Match the stated \(6\)-mile increase to that extra part of the original route. - Rebuild the whole original length from one of its equal thirds.

1. The factor \(\frac{4}{3}=1+\frac{1}{3}\), so the detour is the original route plus one third of the original route. 2. The extra one-third length is \(6\) miles. 3. Three equal thirds make the original route, so its length is \(3\times6=18\) miles. 4. Check: \(18\times\frac{4}{3}=24\), which is \(6\) miles longer than \(18\).

The original route is \(18\) miles long.