Tarmizi drew \(3\) straight lines with lengths of \(12\text{ cm }4\text{ mm}\) , \(8\text{ cm }6\text{ mm}\) and \(15\text{ cm }9\text{ mm}\). What is the sum of length for the \(3\) straight lines?

SOLUTION:

\(\begin{aligned} 12\,\text{cm}\;4\,\text{mm}\\8\,\text{cm}\;6\,\text{mm}\\ \underline{+\;15\,\text{cm}\;9\,\text{mm}}\\ 35\,\text{cm}\;19\,\text{mm} \\\hline\end{aligned}\)

\(\color{orange}1\text{ cm}=10\text{ mm}\)
\(35\text{ cm }19\text{ mm}\rightarrow\color{blue}{36}\text{ cm }\color{red}9\text{ mm}\)

In conclusion, sum of the length for the \(3\) lines are \({36}\text{ cm }9\text{ mm}\).

Distance between Arvin's house and Bunga's house is \(12\text{ km }18\text{ m}\). Distance between Arvin's house and Chong's house is \(5\text{ km }30\text{ m}\) shorter than the distance between Arvin's house and Bunga's house. Find the distance between Arvin's house and Chong's house.

SOLUTION:

Distance between Arvin's house and Bunga's house: \(12\text{ km }18\text{ m}\)
Distance between Arvin's house and Chong's house: \(12\text{ km }18\text{ m}-5\text{ km }30\text{ m}\)

\(\color{orange}1\text{ km}=1000\text{ m}\)
   \(12\text{ km }18\text{ m}-5\text{ km }30\text{ m}\)
= \(11\text{ km }1018\text{ m}-5\text{ km }30\text{ m}\)
= \(6\text{ km }988\text{ m}\)

In conclusion, distance between Arvin's house and Chong's house is \(6\text{ km }988\text{ m}\).

A practice book is \(2\text{ cm }5\text{ mm}\) thick. A textbook is \(5\) times thicker than a practice book. Find the thickness of the textbook.

SOLUTION:
Practice book thickness: \(2\text{ cm }5\text{ mm}\)
Text book thickness: \(5\times\text{thicker than practice book}\)

\(5\times2\text{ cm }5\text{ mm}=10\text{ cm }25\text{ mm}\)

\(\color{orange}1\text{ cm}=10\text{ mm}\)
\(10\text{ cm }25\text{ mm}\rightarrow\color{blue}{12}\text{ cm }\color{red}5\text{ mm}\)

In conclusion, text book thickness is \({12}\text{ cm }5\text{ mm}\).

Independence Run with a track distance of \(10\text{ km }500\text{ m}\) will be held following Independence Day celebration. \(4\) runners need to run while carrying Jalur Gemilang with the same distance divided for each runner. How far, in m, will be run by each of the runners?

SOLUTION:
Track distance: \(10\text{ km }500\text{ m}\)
Number of runner: \(4\) people
Run distance for each runner: \(10\text{ km }500\text{ m}\div4=\fbox{\color{white}1\,\,\,\,\,\,\,\,\,\,}\text{ m}\)

\(10\text{ km }500\text{ m}\rightarrow(10\times1\,000)+500\text{ m}=10\,500\text{ km}\)

\(10\,500\text{ m}\div4=2\,625\text{ m}\)

In conclusion, the run distance for each runner is \(2\,625\text{ m}\).

ADDITION
&
SUBTRACTION

A pack of peanuts has the mass of \(3\text{ kg }620\text{ g}\). A dozen eggs has a mass of \(640\text{ g}\). A pack of snacks a the mass of \(980\text{ g}\) less than the total mass of a pack of peanuts and a dozen of eggs. What is the mass of a pack of snacks?

SOLUTION:
Mass of peanuts: \(3\text{ kg }620\text{ g}\rightarrow3\,620\text{ g}\)
Mass of eggs: \(640\text{ g}\)
Mass of snacks: (Mass of peanuts + Mass of eggs) - \(980\text{ g}\)

\(3\,620\text{ g}+640\text{ g}-980\text{ g}=3\,280\text{ g}\)

Convert to kg and g.
\(3\,280\text{ g}\rightarrow3\text{ kg }280\text{ g}\)

In conclusion, mass of snacks are \(3\text{ kg }280\text{ g}\).

\(5\) boxes of biscuits has a total mass of  \(5\text{ kg }20\text{ g}\). Find the mass for \(3\) boxes of biscuits.

SOLUTION:
Convert to kg and g.
Total mass of \(5\) boxes of biscuits: \(5\text{ kg }20\text{ g}\rightarrow5\,020\text{ g}\)

Mass of \(1\) box of biscuits: \(5\,020\text{ g}\div5=1\,004\text{ g}\)
Mass of \(3\) boxes of biscuits: Mass of \(1\) box of biscuits \(\times3\)

Convert to kg and g.
\(1\,004\text{ g}\times3=3\,012\text{ g}\rightarrow3\text{ kg }12\text{ g}\)

In conclusion, the total mass of \(3\) boxes of biscuits is \(3\text{ kg }12\text{ g}\).

Azmi has \(18\,\ell\,50\,\text{m}\ell\) of soy drinks. He gives \(16\,\ell\,180\,\text{m}\ell\) to his neighbours. Then, he restocks his soy drinks with \(12\,\ell\) of soy drinks. What is the final volume of soy drinks, in \(\ell\) and \(\text{m}\ell\), that Azmi has?

SOLUTION:
Initial soy drinks volume: \(18\,\ell\,50\,\text{m}\ell\)
Volume of soy drinks given to neighbours: \(16\,\ell\,180\,\text{m}\ell\)
Volume of soy drinks added: \(12\,\ell\)

\(18\,\ell\,50\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell=\fbox{\color{white}1}\,\ell\,\fbox{\color{white}1}\,\text{m}\ell\)

     \(18\,\ell\,50\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell\)
\(=17\,\ell\,1050\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell\)
\(=1\,\ell\,870\,\text{m}\ell+12\,\ell\)
\(=13\,\ell\,870\,\text{m}\ell\)

In conclusion, the final volume of doy drinks that Azmi has is \(13\,\ell\,870\,\text{m}\ell\). 

A bucket of paint was poured into \(9\) small cans. The volume of paint in \(2\) small cans is \(2\,\ell\,700\,\text{m}\ell\). Find the volume of paint that the bucket initially holds.

SOLUTION:
Volume of paInt in \(2\) small cans: \(2\,\ell\,700\,\text{m}\ell\)
Original volume of paint: \(9\times\text{Volume of 1 small can}\)

Volume of \(1\) small can: \(2\,\ell\,700\,\text{m}\ell\div2=1\,\ell\,350\,\text{m}\ell\)
Original volume of paint: \(9\times1\,\ell\,350\,\text{m}\ell=9\,\ell\,3150\,\text{m}\ell\rightarrow\color{blue}12\,\ell\,\color{red}150\,\text{m}\ell\)

In conclusion, the initial volume of the paint is \(12\,\ell\,150\,\text{m}\ell\).

Tarmizi drew \(3\) straight lines with lengths of \(12\text{ cm }4\text{ mm}\) , \(8\text{ cm }6\text{ mm}\) and \(15\text{ cm }9\text{ mm}\). What is the sum of length for the \(3\) straight lines?

SOLUTION:

\(\begin{aligned} 12\,\text{cm}\;4\,\text{mm}\\8\,\text{cm}\;6\,\text{mm}\\ \underline{+\;15\,\text{cm}\;9\,\text{mm}}\\ 35\,\text{cm}\;19\,\text{mm} \\\hline\end{aligned}\)

\(\color{orange}1\text{ cm}=10\text{ mm}\)
\(35\text{ cm }19\text{ mm}\rightarrow\color{blue}{36}\text{ cm }\color{red}9\text{ mm}\)

In conclusion, sum of the length for the \(3\) lines are \({36}\text{ cm }9\text{ mm}\).

Distance between Arvin's house and Bunga's house is \(12\text{ km }18\text{ m}\). Distance between Arvin's house and Chong's house is \(5\text{ km }30\text{ m}\) shorter than the distance between Arvin's house and Bunga's house. Find the distance between Arvin's house and Chong's house.

SOLUTION:

Distance between Arvin's house and Bunga's house: \(12\text{ km }18\text{ m}\)
Distance between Arvin's house and Chong's house: \(12\text{ km }18\text{ m}-5\text{ km }30\text{ m}\)

\(\color{orange}1\text{ km}=1000\text{ m}\)
   \(12\text{ km }18\text{ m}-5\text{ km }30\text{ m}\)
= \(11\text{ km }1018\text{ m}-5\text{ km }30\text{ m}\)
= \(6\text{ km }988\text{ m}\)

In conclusion, distance between Arvin's house and Chong's house is \(6\text{ km }988\text{ m}\).

A practice book is \(2\text{ cm }5\text{ mm}\) thick. A textbook is \(5\) times thicker than a practice book. Find the thickness of the textbook.

SOLUTION:
Practice book thickness: \(2\text{ cm }5\text{ mm}\)
Text book thickness: \(5\times\text{thicker than practice book}\)

\(5\times2\text{ cm }5\text{ mm}=10\text{ cm }25\text{ mm}\)

\(\color{orange}1\text{ cm}=10\text{ mm}\)
\(10\text{ cm }25\text{ mm}\rightarrow\color{blue}{12}\text{ cm }\color{red}5\text{ mm}\)

In conclusion, text book thickness is \({12}\text{ cm }5\text{ mm}\).

Independence Run with a track distance of \(10\text{ km }500\text{ m}\) will be held following Independence Day celebration. \(4\) runners need to run while carrying Jalur Gemilang with the same distance divided for each runner. How far, in m, will be run by each of the runners?

SOLUTION:
Track distance: \(10\text{ km }500\text{ m}\)
Number of runner: \(4\) people
Run distance for each runner: \(10\text{ km }500\text{ m}\div4=\fbox{\color{white}1\,\,\,\,\,\,\,\,\,\,}\text{ m}\)

\(10\text{ km }500\text{ m}\rightarrow(10\times1\,000)+500\text{ m}=10\,500\text{ km}\)

\(10\,500\text{ m}\div4=2\,625\text{ m}\)

In conclusion, the run distance for each runner is \(2\,625\text{ m}\).

ADDITION
&
SUBTRACTION

A pack of peanuts has the mass of \(3\text{ kg }620\text{ g}\). A dozen eggs has a mass of \(640\text{ g}\). A pack of snacks a the mass of \(980\text{ g}\) less than the total mass of a pack of peanuts and a dozen of eggs. What is the mass of a pack of snacks?

SOLUTION:
Mass of peanuts: \(3\text{ kg }620\text{ g}\rightarrow3\,620\text{ g}\)
Mass of eggs: \(640\text{ g}\)
Mass of snacks: (Mass of peanuts + Mass of eggs) - \(980\text{ g}\)

\(3\,620\text{ g}+640\text{ g}-980\text{ g}=3\,280\text{ g}\)

Convert to kg and g.
\(3\,280\text{ g}\rightarrow3\text{ kg }280\text{ g}\)

In conclusion, mass of snacks are \(3\text{ kg }280\text{ g}\).

\(5\) boxes of biscuits has a total mass of  \(5\text{ kg }20\text{ g}\). Find the mass for \(3\) boxes of biscuits.

SOLUTION:
Convert to kg and g.
Total mass of \(5\) boxes of biscuits: \(5\text{ kg }20\text{ g}\rightarrow5\,020\text{ g}\)

Mass of \(1\) box of biscuits: \(5\,020\text{ g}\div5=1\,004\text{ g}\)
Mass of \(3\) boxes of biscuits: Mass of \(1\) box of biscuits \(\times3\)

Convert to kg and g.
\(1\,004\text{ g}\times3=3\,012\text{ g}\rightarrow3\text{ kg }12\text{ g}\)

In conclusion, the total mass of \(3\) boxes of biscuits is \(3\text{ kg }12\text{ g}\).

Azmi has \(18\,\ell\,50\,\text{m}\ell\) of soy drinks. He gives \(16\,\ell\,180\,\text{m}\ell\) to his neighbours. Then, he restocks his soy drinks with \(12\,\ell\) of soy drinks. What is the final volume of soy drinks, in \(\ell\) and \(\text{m}\ell\), that Azmi has?

SOLUTION:
Initial soy drinks volume: \(18\,\ell\,50\,\text{m}\ell\)
Volume of soy drinks given to neighbours: \(16\,\ell\,180\,\text{m}\ell\)
Volume of soy drinks added: \(12\,\ell\)

\(18\,\ell\,50\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell=\fbox{\color{white}1}\,\ell\,\fbox{\color{white}1}\,\text{m}\ell\)

     \(18\,\ell\,50\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell\)
\(=17\,\ell\,1050\,\text{m}\ell-16\,\ell\,180\,\text{m}\ell+12\,\ell\)
\(=1\,\ell\,870\,\text{m}\ell+12\,\ell\)
\(=13\,\ell\,870\,\text{m}\ell\)

In conclusion, the final volume of doy drinks that Azmi has is \(13\,\ell\,870\,\text{m}\ell\). 

A bucket of paint was poured into \(9\) small cans. The volume of paint in \(2\) small cans is \(2\,\ell\,700\,\text{m}\ell\). Find the volume of paint that the bucket initially holds.

SOLUTION:
Volume of paInt in \(2\) small cans: \(2\,\ell\,700\,\text{m}\ell\)
Original volume of paint: \(9\times\text{Volume of 1 small can}\)

Volume of \(1\) small can: \(2\,\ell\,700\,\text{m}\ell\div2=1\,\ell\,350\,\text{m}\ell\)
Original volume of paint: \(9\times1\,\ell\,350\,\text{m}\ell=9\,\ell\,3150\,\text{m}\ell\rightarrow\color{blue}12\,\ell\,\color{red}150\,\text{m}\ell\)

In conclusion, the initial volume of the paint is \(12\,\ell\,150\,\text{m}\ell\).