QUESTION:
\(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}=\Box\text{ hour}\)

SOLUTION:
We need to equalize the denominator first before subtracting.
     \(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}\)

\(=\frac{3}{4}\text{ jam}-\frac{1\color{orange}{\times2}}{2\color{orange}{\times2}}\text{ jam}\)

\(=\frac{3}{4}\text{ jam}-\frac{2}{4}\text{ jam}\)

\(=\frac{1}{4}\text{ jam}\)

ANSWER:
\(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}=\frac{1}{4}\text{ hour}\)

QUESTION:
\(1\text{ hour }40\text{ minutes}-0.95\text{ hour}=\Box\text{ minutes}\)

SOLUTION:
We can convert everything to minute units for ease of operation.
     \(1\text{ hour }40\text{ minutes}-0.95\text{ hour}\)
\(=(1\times60)\text{ minutes} + 40\text{ minutes}-(0.95\times60)\text{ minutes}\)
\(=60\text{ minutes} + 40\text{ minutes}-57\text{ minutes}\)
\(=43\text{ minit}\)

ANSWER:
\(1\text{ hour }40\text{ minutes}-0.95\text{ hour}=43\text{ minutes}\)

QUESTION:
\(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}=\Box\text{ hours}\)

SOLUTION:
Equalize the denominators.
     \(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}\)

\(=\frac{5\color{red}{\times3}}{8\color{red}{\times3}}\text{ day}-\frac{1\color{red}{\times8}}{3\color{red}{\times8}}\text{ day}\)

\(=\frac{15-8}{24}\text{ day}\)

\(=\frac{7}{24}\text{ day}\to\frac{7}{24}\times24\text{ hours}\)

\(=7\text{ hours}\)

ANSWER:
\(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}=7\text{ hours}\)

QUESTION:
\(0.25\text{ day}-0.125\text{ day}=\Box\text{ day}\)

SOLUTION:

\(\frac{ \begin{array}{lr} &\color{red}{^{4}}\,\color{red}{^{10}}\,\,\,\,\,\,\,\,\,\,\\&0.\,2\,{\not}5\,{\not}0\text { day}\\ -&0.\,1\,2\,5\text{ day} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,0.\,1\,2\,5\text{ day}\\\hline \end{array} }\)

ANSWER:
\(0.25\text{ day}-0.125\text{ day}=0.125\text{ day}\)

QUESTION:
\(4\frac{1}{2}\text{ years} - 1\frac{5}{12}\text{ years}=\Box\text{ years }\Box\text{ months }\)

SOLUTION:
We need to equalize the denominator of the fraction first.

     \(4\frac{1\color{orange}\times6}{2\color{orange}\times6}\text{ years} - 1\frac{5}{12}\text{ years}\)

\(=4\frac{6}{12}\text{ years} - 1\frac{5}{12}\text{ years}\)

\(=3\frac{1}{12}\text{ years}\)

We need to convert fractions to units of months.

     \(3\frac{1}{12}\text{ years}\)

\(=3\text{ years }+(\frac{1}{12}\times24)\text{ months}\)

\(=3\text{ years }2\text{ months}\)

ANSWER:
\(4\frac{1}{2}\text{ years} - 1\frac{5}{12}\text{ years}=3\text{ years }2\text{ months }\)

QUESTION:
\(7.75\text{ years}-4.5\text{ years} = \Box\text{ years }\Box\text{ months}\)

SOLUTION:
We can convert decimals to monthly units first.
\(0.75\times12\text{ months}=9\text{ months}\)
\(0.5\times12\text{ months}=6\text{ months}\)

     \(7.75\text{ years}-4.5\text{ years}\)
\(=7\text{ years }9\text{ months}-4\text{ years }6\text{ months}\)
\(=3\text{ years }3\text{ months}\)

ANSWER:
\(7.75\text{ years}-4.5\text{ years} = 3\text{ years }3\text{ months}\)

QUESTION:
\(9\frac{7}{10}\text{ decades}-2\text{ decades }3\text{ years}=\Box\text{ years}\)

SOLUTION:
Convert all to units of years.
     \(9\frac{7}{10}\text{ decades}\)
\(=(9\times10)+(\frac{7}{10}\times10)\text{ years}\)
\(=(90 +7) \text{ years}\\ =97\text{ years}\)

     \(2\text{ decades }3\text{ years}\)
\(=(2\times10)+3\text{ years}\\\)
\(=(20+3)\text{ years}\\ =23\text{ years}\)

\(97\text{ years}-23\text{ years}=74\text{ years}\)

ANSWER:
\(9\frac{7}{10}\text{ decades}-2\text{ decades }3\text{ years}=74\text{ years}\)

QUESTION:
\(8.1\text{ decades} - 27\text{ years}=\Box\text{ decades }\Box\text{ years}\)

SOLUTION:
Convert 8.1 decades to decades and years.
     \(8.1\text{ decades}\)
\(=8\text{ decades}+(0.1\times10)\text{ years}\\=8\text{ decades }1\text{ year}\)

    \(27\text{ years}\)
\(=27\div10\text{ decades}\\=2.7\text{ decades}\\=2\text{ decades}+(0.7\times10)\text{ years}\\=2\text{ decades }7\text{ years}\)

Perform a subtraction operation.
    \(8.1\text{ decades} - 27\text{ years}\)
\(=8\text{ decades }1\text{ year}-2\text{ decades } 7\text{ years}\)
\(=7\text{ decades }{11}\text{ years}-2\text{ decades } 7\text{ years}\)
\(=5\text{ decades } 4\text{ years}\)

ANSWER:
\(8.1\text{ decades} - 27\text{ years}=5\text{ decades }4\text{ years}\)

QUESTION:
\(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}=\Box\text{ decades}\)

SOLUTION:
Convert to decade units.
      \(2\text{ centuries } 4\text{ decades}\)    
\(=(2\times10)\text{ decades}+4 \text{ decades}\\\)
\(=20\text{ decades}+4\text{ decades}\\=24\text{ decades}\)

     \(1\frac{1}{5}\text{ centuries}\)    
\(=\frac{6}{5}\times10\text{ decades}\\\)
\(=12\text{ decades}\)

    \(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}\\\)
\(=24\text{ decades}-12\text{ decades}\\ =12\text{ decades}\)

ANSWER:
\(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}=12\text{ decades}\)

QUESTION:
\(8.1\text{ centuries}-6.7\text{ centuries}=\Box\text{ centuries }\Box\text{ decades}\)

SOLUTION:

\(\frac{ \begin{array}{lr}&\color{red}{^{7}}\color{red}{^{11}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\&{\not}8.\,{\not}1\,\text { centuries}\\ -&6.\,7\,\text{ centuries} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,1.\,4\,\text{ centuries}\\\hline \end{array} }\)

    \(1.4\text{ centuries}\\\)
\(= 1 \text{ century} +(0.4\times10)\text{ decades}\\ = 1 \text{ century } 4\text{ decades}\)

ANSWER:
\(8.1\text{ centuries}-6.7\text{ centuries}=1\text{ centuries }4\text{ decades}\)

QUESTION:
\(2\frac{1}{2}\text{ centuries} -\frac{3}{4}\text{ century}=\Box\text{ years}\)

SOLUTION:
    \(2\frac{1}{2}\text{ centuries}\\\)
\(=\frac{5}{2}\times100\\=250\text{ tahun}\)

    \(\frac{3}{4}\text{ century}\\\)
\(=\frac{3}{4}\times100\\=75\text{ years}\)

\(\frac{ \begin{array}{lr} &\color{red}{^{4}}\,\color{red}{^{10}}\,\,\,\,\,\,\,\,\,\,\,\\&2\,{\not}5\,{\not}0\text { years}\\ -&\,7\,5\text{ years} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,1\,7\,5\text{ years}\\\hline \end{array} }\)

ANSWER:
\(2\frac{1}{2}\text{ centuries} -\frac{3}{4}\text{ century}=175\text{ years}\)

QUESTION:
\(9.1\text{ decades}-6.3\text{ decades}=\Box\text{ decades }\Box\text{ years }\)

SOLUTION:

\(\frac{ \begin{array}{lr} &\color{red}{^{8}}\,\color{red}{^{11}}\,\,\,\,\,\,\,\,\,\,\,\\&{\not}9.\,{\not}1\text { decades}\\ -&6.\,3\text{ decades} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,2.\,8\text{ decades}\\\hline \end{array} }\)

    \(2.8\text{ decades}\\\)
\(=2\text{ decades}+(0.8\times10)\text{ years}\\ =2\text{ decades } 8\text{ years}\)

ANSWER:
\(9.1\text{ decades}-6.3\text{ decades}=2\text{ decades }8\text{ years }\)

QUESTION:
\(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}=\Box\text{ hour}\)

SOLUTION:
We need to equalize the denominator first before subtracting.
     \(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}\)

\(=\frac{3}{4}\text{ jam}-\frac{1\color{orange}{\times2}}{2\color{orange}{\times2}}\text{ jam}\)

\(=\frac{3}{4}\text{ jam}-\frac{2}{4}\text{ jam}\)

\(=\frac{1}{4}\text{ jam}\)

ANSWER:
\(\frac{3}{4}\text{ hour}-\frac{1}{2}\text{ hour}=\frac{1}{4}\text{ hour}\)

QUESTION:
\(1\text{ hour }40\text{ minutes}-0.95\text{ hour}=\Box\text{ minutes}\)

SOLUTION:
We can convert everything to minute units for ease of operation.
     \(1\text{ hour }40\text{ minutes}-0.95\text{ hour}\)
\(=(1\times60)\text{ minutes} + 40\text{ minutes}-(0.95\times60)\text{ minutes}\)
\(=60\text{ minutes} + 40\text{ minutes}-57\text{ minutes}\)
\(=43\text{ minit}\)

ANSWER:
\(1\text{ hour }40\text{ minutes}-0.95\text{ hour}=43\text{ minutes}\)

QUESTION:
\(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}=\Box\text{ hours}\)

SOLUTION:
Equalize the denominators.
     \(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}\)

\(=\frac{5\color{red}{\times3}}{8\color{red}{\times3}}\text{ day}-\frac{1\color{red}{\times8}}{3\color{red}{\times8}}\text{ day}\)

\(=\frac{15-8}{24}\text{ day}\)

\(=\frac{7}{24}\text{ day}\to\frac{7}{24}\times24\text{ hours}\)

\(=7\text{ hours}\)

ANSWER:
\(\frac{5}{8}\text{ day}-\frac{1}{3}\text{ day}=7\text{ hours}\)

QUESTION:
\(0.25\text{ day}-0.125\text{ day}=\Box\text{ day}\)

SOLUTION:

\(\frac{ \begin{array}{lr} &\color{red}{^{4}}\,\color{red}{^{10}}\,\,\,\,\,\,\,\,\,\,\\&0.\,2\,{\not}5\,{\not}0\text { day}\\ -&0.\,1\,2\,5\text{ day} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,0.\,1\,2\,5\text{ day}\\\hline \end{array} }\)

ANSWER:
\(0.25\text{ day}-0.125\text{ day}=0.125\text{ day}\)

QUESTION:
\(4\frac{1}{2}\text{ years} - 1\frac{5}{12}\text{ years}=\Box\text{ years }\Box\text{ months }\)

SOLUTION:
We need to equalize the denominator of the fraction first.

     \(4\frac{1\color{orange}\times6}{2\color{orange}\times6}\text{ years} - 1\frac{5}{12}\text{ years}\)

\(=4\frac{6}{12}\text{ years} - 1\frac{5}{12}\text{ years}\)

\(=3\frac{1}{12}\text{ years}\)

We need to convert fractions to units of months.

     \(3\frac{1}{12}\text{ years}\)

\(=3\text{ years }+(\frac{1}{12}\times24)\text{ months}\)

\(=3\text{ years }2\text{ months}\)

ANSWER:
\(4\frac{1}{2}\text{ years} - 1\frac{5}{12}\text{ years}=3\text{ years }2\text{ months }\)

QUESTION:
\(7.75\text{ years}-4.5\text{ years} = \Box\text{ years }\Box\text{ months}\)

SOLUTION:
We can convert decimals to monthly units first.
\(0.75\times12\text{ months}=9\text{ months}\)
\(0.5\times12\text{ months}=6\text{ months}\)

     \(7.75\text{ years}-4.5\text{ years}\)
\(=7\text{ years }9\text{ months}-4\text{ years }6\text{ months}\)
\(=3\text{ years }3\text{ months}\)

ANSWER:
\(7.75\text{ years}-4.5\text{ years} = 3\text{ years }3\text{ months}\)

QUESTION:
\(9\frac{7}{10}\text{ decades}-2\text{ decades }3\text{ years}=\Box\text{ years}\)

SOLUTION:
Convert all to units of years.
     \(9\frac{7}{10}\text{ decades}\)
\(=(9\times10)+(\frac{7}{10}\times10)\text{ years}\)
\(=(90 +7) \text{ years}\\ =97\text{ years}\)

     \(2\text{ decades }3\text{ years}\)
\(=(2\times10)+3\text{ years}\\\)
\(=(20+3)\text{ years}\\ =23\text{ years}\)

\(97\text{ years}-23\text{ years}=74\text{ years}\)

ANSWER:
\(9\frac{7}{10}\text{ decades}-2\text{ decades }3\text{ years}=74\text{ years}\)

QUESTION:
\(8.1\text{ decades} - 27\text{ years}=\Box\text{ decades }\Box\text{ years}\)

SOLUTION:
Convert 8.1 decades to decades and years.
     \(8.1\text{ decades}\)
\(=8\text{ decades}+(0.1\times10)\text{ years}\\=8\text{ decades }1\text{ year}\)

    \(27\text{ years}\)
\(=27\div10\text{ decades}\\=2.7\text{ decades}\\=2\text{ decades}+(0.7\times10)\text{ years}\\=2\text{ decades }7\text{ years}\)

Perform a subtraction operation.
    \(8.1\text{ decades} - 27\text{ years}\)
\(=8\text{ decades }1\text{ year}-2\text{ decades } 7\text{ years}\)
\(=7\text{ decades }{11}\text{ years}-2\text{ decades } 7\text{ years}\)
\(=5\text{ decades } 4\text{ years}\)

ANSWER:
\(8.1\text{ decades} - 27\text{ years}=5\text{ decades }4\text{ years}\)

QUESTION:
\(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}=\Box\text{ decades}\)

SOLUTION:
Convert to decade units.
      \(2\text{ centuries } 4\text{ decades}\)    
\(=(2\times10)\text{ decades}+4 \text{ decades}\\\)
\(=20\text{ decades}+4\text{ decades}\\=24\text{ decades}\)

     \(1\frac{1}{5}\text{ centuries}\)    
\(=\frac{6}{5}\times10\text{ decades}\\\)
\(=12\text{ decades}\)

    \(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}\\\)
\(=24\text{ decades}-12\text{ decades}\\ =12\text{ decades}\)

ANSWER:
\(2\text{ centuries } 4\text{ decades}-1\frac{1}{5}\text{ centuries}=12\text{ decades}\)

QUESTION:
\(8.1\text{ centuries}-6.7\text{ centuries}=\Box\text{ centuries }\Box\text{ decades}\)

SOLUTION:

\(\frac{ \begin{array}{lr}&\color{red}{^{7}}\color{red}{^{11}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\&{\not}8.\,{\not}1\,\text { centuries}\\ -&6.\,7\,\text{ centuries} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,1.\,4\,\text{ centuries}\\\hline \end{array} }\)

    \(1.4\text{ centuries}\\\)
\(= 1 \text{ century} +(0.4\times10)\text{ decades}\\ = 1 \text{ century } 4\text{ decades}\)

ANSWER:
\(8.1\text{ centuries}-6.7\text{ centuries}=1\text{ centuries }4\text{ decades}\)

QUESTION:
\(2\frac{1}{2}\text{ centuries} -\frac{3}{4}\text{ century}=\Box\text{ years}\)

SOLUTION:
    \(2\frac{1}{2}\text{ centuries}\\\)
\(=\frac{5}{2}\times100\\=250\text{ tahun}\)

    \(\frac{3}{4}\text{ century}\\\)
\(=\frac{3}{4}\times100\\=75\text{ years}\)

\(\frac{ \begin{array}{lr} &\color{red}{^{4}}\,\color{red}{^{10}}\,\,\,\,\,\,\,\,\,\,\,\\&2\,{\not}5\,{\not}0\text { years}\\ -&\,7\,5\text{ years} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,1\,7\,5\text{ years}\\\hline \end{array} }\)

ANSWER:
\(2\frac{1}{2}\text{ centuries} -\frac{3}{4}\text{ century}=175\text{ years}\)

QUESTION:
\(9.1\text{ decades}-6.3\text{ decades}=\Box\text{ decades }\Box\text{ years }\)

SOLUTION:

\(\frac{ \begin{array}{lr} &\color{red}{^{8}}\,\color{red}{^{11}}\,\,\,\,\,\,\,\,\,\,\,\\&{\not}9.\,{\not}1\text { decades}\\ -&6.\,3\text{ decades} \end{array} }{ \begin{array}{lr} &\,\,\,\,\,2.\,8\text{ decades}\\\hline \end{array} }\)

    \(2.8\text{ decades}\\\)
\(=2\text{ decades}+(0.8\times10)\text{ years}\\ =2\text{ decades } 8\text{ years}\)

ANSWER:
\(9.1\text{ decades}-6.3\text{ decades}=2\text{ decades }8\text{ years }\)