Decimal Arithmetic

Stack the numbers with the points aligned and pad \(1.7\) to \(1.70\) so both show a hundredths digit. $$\begin{array}{r} 1.70 \\ +\,0.45 \\ \hline 2.15 \end{array}$$ Adding right to left, hundredths give \(0 + 5 = 5\), tenths give \(7 + 4 = 11\) so write \(1\) and carry \(1\), and ones give \(1 + 0 + 1 = 2\). The point drops straight down. The runner covers \(\boxed{2.15}\) km.

Pad \(12.5\) to \(12.50\) so both numbers show two decimal places, then stack them with the points in one vertical line. $$\begin{array}{r}12.50\\ +\;0.45\\ \hline 12.95\end{array}$$ The hundredths give \(0 + 5 = 5\), the tenths give \(5 + 4 = 9\), and the \(12\) comes straight down. The total is \(\boxed{12.95}\) dollars. Line up the right edges instead and the \(5\) tenths sits above the \(5\) hundredths, adding digits of different sizes and inflating a few cents into a few dollars.

Pad \(5.2\) to \(5.20\) and stack it over \(3.65\) with the points aligned. In the hundredths column, borrow \(1\) tenth to get \(10 - 5 = 5\). That leaves \(1\) in the tenths, so borrow \(1\) one to get \(11 - 6 = 5\). The ones column is then \(4 - 3 = 1\). Bringing the point straight down gives \(\boxed{1.55}\) litres. Checking it, \(3.65 + 1.55 = 5.20\), the full bottle again.

Write each decimal as a fraction over ten and multiply across, \(\tfrac{3}{10} \times \tfrac{2}{10} = \tfrac{3 \times 2}{10 \times 10} = \tfrac{6}{100}\). A denominator of \(100\) puts the digits two places past the point, so the answer is \(0.06\) and not \(0.6\). So \(0.3 \times 0.2 = \boxed{0.06}\). Taking a fraction of a fraction always lands below both factors, which is why the product is smaller than \(0.3\) and \(0.2\).

Write the decimals as fractions and multiply across, \(\tfrac{6}{10} \times \tfrac{7}{10} = \tfrac{6 \times 7}{10 \times 10} = \tfrac{42}{100}\). A denominator of \(100\) is the hundredths place, so that reads as forty-two hundredths, and the bed is \(\boxed{0.42}\) square metres. There is a shortcut hiding in that. The digits gave \(6 \times 7 = 42\), and each factor had one decimal place, so the product carries two.

Strip the point and multiply the digits like ordinary whole numbers, \(7 \times 25 = 175\). Now total the decimal places in the two factors. The whole number \(7\) carries zero places and \(0.25\) carries two, so the product keeps \(0 + 2 = 2\) places. Placing the point two digits from the right turns \(175\) into \(1.75\). A quick gut check, \(0.25\) is a quarter of a dollar, so seven quarters is \(1.75\) dollars, exactly as expected. So \(7 \times 0.25 = \boxed{1.75}\) dollars.

Ignore the points and multiply the digits, \(4 \times 2 = 8\). Then count places. \(0.04\) has two and \(0.2\) has one, so the product needs \(2 + 1 = 3\). Two zeros push the \(8\) out to the thousandths spot, so the recipe uses \(\boxed{0.008}\) grams. The fractions agree, since \(\tfrac{4}{100} \times \tfrac{2}{10} = \tfrac{8}{1{,}000}\) and thousandths is exactly that third place.

Write the division as \(\tfrac{4.5}{0.5}\) and multiply the top and bottom by \(10\), which slides each point one place right and gives \(\tfrac{45}{5}\). Since \(5 \times 9 = 45\), the rope makes \(\boxed{9}\) pieces. Multiplying top and bottom by the same number is multiplying by \(1\), so the slide cannot change the value.

Multiply the top and bottom of \(\tfrac{1.44}{1.2}\) by \(10\) to clear the divisor, which turns the problem into \(14.4 \div 12\). Twelve fits into \(14\) once, leaving \(2.4\), and twelve fits into \(2.4\) exactly \(0.2\) times, so \(1.44 \div 1.2 = \boxed{1.2}\). Check it by multiplying back, since \(1.2 \times 1.2 = 1.44\).

Three spools of \(1.2\) m each is \(3 \times 1.2\). The digits give \(3 \times 12 = 36\), and \(1.2\) carries one decimal place, so there is \(3.6\) m on hand. Pad that to \(3.60\), line up the points, and subtract what the build uses. $$3.60 - 2.75 = 0.85$$ That leaves \(\boxed{0.85}\) m of wire.

Both numbers already show one decimal place, so stack them with the points aligned and add. The tenths give \(3 + 6 = 9\) and the ones give \(4 + 5 = 9\), with the point dropping straight down. The two stretches total \(\boxed{9.9}\) meters.

Pad \(7.2\) to \(7.20\) so both numbers show hundredths, then line up the points and subtract from the right. The hundredths borrow to give \(10 - 5 = 5\). The tenths are now \(1\), so borrow again for \(11 - 8 = 3\). The ones give \(6 - 3 = 3\). Bringing the point down leaves \(\boxed{3.35}\) liters.

Write \(6\) as \(6.00\), the same value with two decimal places, and stack it over \(2.35\) with the points aligned. $$\begin{array}{r} 6.00 \\ -\,2.35 \\ \hline 3.65 \end{array}$$ The hundredths borrow to give \(10 - 5 = 5\), the tenths become \(9 - 3 = 6\) as the borrow cascades, and the ones give \(5 - 2 = 3\). So \(\boxed{3.65}\) units of charge are left. Checking it, \(2.35 + 3.65 = 6.00\), a full battery again.

Ignore the decimal points and multiply the digits like whole numbers, \(8 \times 4 = 32\). Then count the decimal places in the factors, one in \(0.8\) and one in \(0.4\), which makes two in all. Giving \(32\) two decimal places turns it into \(\boxed{0.32}\).

Half of the fogged area means \(0.5 \times 0.6\). The digits give \(5 \times 6 = 30\), and each factor has one decimal place, so the product needs two, giving \(\boxed{0.30}\) of the window. It landed below both \(0.5\) and \(0.6\), which is what taking a part of a part always does.

Set the points aside and multiply like whole numbers, \(25 \times 4 = 100\). There is one decimal place in \(2.5\) and one in \(0.4\), so the product needs two, and counting two spots from the right of \(100\) gives \(1.00\). Trailing zeros after the point add nothing, so the recipe uses \(\boxed{1}\) gram. Two numbers that both carry decimals can still multiply to a clean whole number, when the places fill in exactly.

Start with the digits and forget the decimals. Multiplying \(3 \times 3\) gives \(9\). Now count how many decimal places the factors have. The number \(0.03\) has two places and \(0.3\) has one place, for a total of three. The product must therefore have three decimal places. Writing \(9\) with three places means filling in zeros, so \(9\) sits in the thousandths spot as \(0.009\). \(\boxed{0.009}\)

Sliding both numbers one place to the right means multiplying each by \(10\), and doing the same thing to both keeps the quotient the same. So \(7.2 \div 0.9\) becomes \(72 \div 9\). Now the divisor is a whole number and the work is easy. Since \(9 \times 8 = 72\), we get \(72 \div 9 = 8\). Therefore \(7.2 \div 0.9 = \boxed{8}\).

Start by making the divisor whole. Slide both points one place to the right, so \(2.56\) becomes \(25.6\) and \(1.6\) becomes \(16\). The quotient does not change, so \(2.56 \div 1.6 = 25.6 \div 16\). Now divide. \(16\) goes into \(25\) once, leaving \(9\). Bring down the \(6\) to make \(96\), and \(16 \times 6 = 96\), so the next digit is \(6\). Keeping the decimal point in place gives \(1.6\). Check by multiplying back, \(1.6 \times 1.6 = 2.56\), which matches the ribbon. So the ribbon is \(\boxed{1.6}\) times as long as one strip.

Pad \(0.9\) to \(0.90\) and line up the points. The hundredths give \(8 + 0 = 8\) and the tenths give \(0 + 9 = 9\). $$0.08 + 0.90 = 0.98$$ The total is \(\boxed{0.98}\) mm. The inner zero in \(0.08\) is doing real work, since it holds the \(8\) in the hundredths place instead of letting it drift up to the tenths.

Ignore the points and multiply the digits, \(14 \times 5 = 70\). There is one decimal place in \(1.4\) and one in \(0.5\), so the answer needs two, giving \(1.4 \times 0.5 = \boxed{0.70}\) m. That matches taking half of \(1.4\).

Start with the cost of the jam. Multiply \(4 \times 0.85\). Ignoring the point for a moment, \(4 \times 85 = 340\). The factor \(0.85\) has two digits after the point, so the product does too, giving \(4 \times 0.85 = 3.40\). Now find the change. Write the \(5\) dollar bill as \(5.00\) so the points line up, then subtract. \(5.00 - 3.40 = 1.60\). So your change is \(\boxed{1.60}\) dollars.