Free Math Worksheets

- Think of a division expression as a fraction. - Compare how the dividend and divisor are scaled in each choice. - The quotient stays the same when both are scaled by the same nonzero factor.

1. In expression 1, both dividend and divisor are multiplied by \(10\), so the quotient stays the same. 2. In expression 2, both are multiplied by \(100\), so the quotient stays the same. 3. In expression 3, both are divided by \(10\), so the quotient stays the same. 4. In expression 4, the dividend is multiplied by \(100\) but the divisor by only \(10\), so the quotient changes. 5. Expressions 1, 2, and 3 are equivalent to the original division.

Expressions \(1\), \(2\), and \(3\) have the same quotient as \(2.4\div0.08\).

- Create an equivalent division with a whole-number divisor. - Scale the dividend and divisor by the same power of \(10\). - Use familiar whole-number division facts after scaling.

1. For a), \(15\div0.3=150\div3=50\). 2. For b), \(0.8\div0.04=80\div4=20\). 3. For c), \(1.2\div0.06=120\div6=20\). 4. For d), \(0.45\div0.15=45\div15=3\). 5. For e), \(10\div0.005=10{,}000\div5=2000\).

a) \(50\) b) \(20\) c) \(20\) d) \(3\) e) \(2000\)

- Create an equivalent quotient with a whole-number divisor. - Scale the dividend and divisor by the same power of \(10\). - Check that the size of each quotient is reasonable.

1. \(7.2\div0.8=72\div8=9\). 2. \(1.26\div0.3=12.6\div3=4.2\). 3. \(0.048\div0.006=48\div6=8\). 4. \(3.45\div0.15=345\div15=23\).

a) \(9\) b) \(4.2\) c) \(8\) d) \(23\)

- Ask how many times each piece length fits into the total length. - Create an equivalent division with a whole-number divisor if helpful. - Check that the number of pieces times the piece length gives the full strip length.

1. For a), \(3.5\div0.25=14\), so \(14\) pieces can be cut. 2. For b), \(3.5\div0.14=25\), so \(25\) pieces can be cut.

a) \(14\) pieces b) \(25\) pieces

- Divide the total stack height by the number of identical sheets. - Pay attention to the requested unit in part a). - For part b), determine how many sheet thicknesses fit into the target height.

1. Divide the stack height by the number of sheets: \(13.75\div25=0.55\,\text{cm}\) per sheet. 2. Convert the thickness: \(0.55\,\text{cm}=5.5\,\text{mm}\). 3. For b), \(44\div0.55=80\), so \(80\) sheets are needed.

a) \(5.5\,\text{mm}\) b) \(80\) sheets

- Compare each expression with the original pair of numbers. - Scale the dividend and divisor by the same factor. - Use the simplest equivalent division to find the final quotient. - Check that every completed expression gives the same value.

1. From \(1.2\) to \(12\), multiply by \(10\), so also multiply \(0.3\) by \(10\): the first blank is \(3\). 2. From \(0.3\) to \(30\), multiply by \(100\), so also multiply \(1.2\) by \(100\): the second blank is \(120\). 3. From \(1.2\) to \(0.12\), divide by \(10\), so also divide \(0.3\) by \(10\): the third blank is \(0.03\). 4. Use the easiest equivalent expression: \(12\div3=4\).

\(1.2\div0.3=12\div3=120\div30=0.12\div0.03\). The common quotient is \(4\).

- Think of the division as a fraction and recall how equivalent fractions are formed. - Compare the factor used to scale the dividend with the factor used to scale the divisor. - Make the divisor a whole number while keeping the quotient unchanged.

1. To keep a quotient unchanged, the dividend and divisor must be multiplied by the same nonzero number. 2. Making \(0.5\) into \(5\) multiplies the divisor by \(10\), so the dividend must also be multiplied by \(10\): \(0.45\) becomes \(4.5\), not \(45\). 3. The equivalent division is \(4.5\div5=0.9\). 4. The student's quotient \(9\) is ten times too large because the dividend was scaled more than the divisor.

The method is incorrect because the dividend and divisor were scaled by different factors. Correctly, \(0.45\div0.5=4.5\div5=0.9\).

- Compare how the decimal place changes from one expression to another. - Compare a) with b) to isolate the change in the divisor. - Compare a) with d) to see what happens when both numbers are scaled equally.

1. For a), \(2.4\div0.6=4\). 2. For b), \(2.4\div0.06=40\). 3. For c), \(0.24\div0.6=0.4\). 4. For d), \(0.24\div0.06=4\). 5. Making only the divisor ten times smaller makes the quotient ten times larger. Making only the dividend ten times smaller makes the quotient ten times smaller. Scaling both by the same factor leaves the quotient unchanged.

a) \(4\) b) \(40\) c) \(0.4\) d) \(4\) Scaling the dividend and divisor by the same nonzero factor leaves the quotient unchanged.

- Rewrite each division with a whole-number divisor when useful. - Compare how the dividend and divisor are scaled across the expressions. - Calculate each quotient and compare the results.

1. A: \(0.45\div0.09=5\). 2. B: \(4.5\div0.9=5\). 3. C: \(45\div9=5\). 4. D: \(0.045\div0.9=0.05\). 5. Expression D is different because the other three all equal \(5\).

D has the different value: \(0.05\). A, B, and C each equal \(5\).

- Compare each dividend and divisor with \(156\) and \(5\). - Scaling both numbers by the same nonzero factor leaves the quotient unchanged. - If only one number changes by a power of \(10\), determine how the quotient changes.

1. For a), the dividend is one tenth as large, so the quotient is one tenth as large: \(3.12\). 2. For b), multiply dividend and divisor by \(10\): \(1.56\div0.5=15.6\div5=3.12\). 3. For c), the divisor is one hundredth as large as \(5\), so the quotient is \(100\) times as large: \(31.2\times100=3120\). 4. For d), the dividend is one thousandth as large as \(156\), so the quotient is one thousandth as large: \(0.0312\).

a) \(3.12\) b) \(3.12\) c) \(3120\) d) \(0.0312\)

- How much do the glitter stickers cost altogether? - How much money remains for regular stickers? - How many regular stickers can be bought with that amount? - Remember to add both kinds of stickers at the end.

1. Find the cost of the glitter stickers: \(3 \times \$0.70 = \$2.10\). 2. Find the amount spent on regular stickers: \(\$3.60 - \$2.10 = \$1.50\). 3. Find the number of regular stickers: \(\$1.50 \div \$0.30 = 5\). 4. Find the total number of stickers: \(3 + 5 = 8\).

Jonas buys \(8\) stickers altogether.

- First find the cost of one sticker. - How many times does one sticker's price fit into the total amount of money?

1. Find the cost of one sticker: \(\$0.80 \div 4 = \$0.20\). 2. Divide the available money by the price of one sticker: \(\$2.00 \div \$0.20 = 10\).

\(10\) stickers can be bought.

- Find how much money was handed to the cashier. - Subtract the change to find the purchase cost. - Divide by the price per kilogram, then subtract the purchased mass from the container’s capacity.

1. The beekeeper paid \(2 \times \$50=\$100\). 2. The honey cost \(\$100-\$14.50=\$85.50\). 3. The purchased mass was \(\$85.50 \div \$4.50=19\,\text{kg}\). 4. The starting mass was \(25\,\text{kg}-19\,\text{kg}=6\,\text{kg}\).

\(6\,\text{kg}\)