## Problem Solving

 5.4 Problem Solving

LENGTH
 ADDITION 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 $${36}\text{ cm }9\text{ mm}$$. SUBTRACTION 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 $$6\text{ km }988\text{ m}$$. MULTIPLICATION 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 $${12}\text{ cm }5\text{ mm}$$. DIVISION 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 $$2\,625\text{ m}$$.

MASS
 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 $$3\text{ kg }280\text{ g}$$. MULTIPLICATION & DIVISION $$5$$ 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 $$3\text{ kg }12\text{ g}$$.

VOLUME OF LIQUID
 ADDITION & SUBTRACTION 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 $$13\,\ell\,870\,\text{m}\ell$$. MULTIPLICATION & DIVISION 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 $$12\,\ell\,150\,\text{m}\ell$$.

## Problem Solving

 5.4 Problem Solving

LENGTH
 ADDITION 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 $${36}\text{ cm }9\text{ mm}$$. SUBTRACTION 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 $$6\text{ km }988\text{ m}$$. MULTIPLICATION 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 $${12}\text{ cm }5\text{ mm}$$. DIVISION 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 $$2\,625\text{ m}$$.

MASS
 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 $$3\text{ kg }280\text{ g}$$. MULTIPLICATION & DIVISION $$5$$ 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 $$3\text{ kg }12\text{ g}$$.

VOLUME OF LIQUID
 ADDITION & SUBTRACTION 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 $$13\,\ell\,870\,\text{m}\ell$$. MULTIPLICATION & DIVISION 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 $$12\,\ell\,150\,\text{m}\ell$$.