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3. Box A is 2 centimeters high, 3 centimeters wide, and 5 centimeters long. It can hold 40 grams of clay. a. Box B has twice the height, three times the width, and the same length as the first box. How many grams of clay can it hold?b. Box C has double the length, width, and height of the original box. How many grams of clay can it hold? Explain your thinking.c. By changing just one dimension of box A, design a box that can hold 80 grams of clay.

User Hhovhann
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1 Answer

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Step-by-step explanation

In this question, the boxes can hold a particular amount of clay. This implies that the volume of the box is what would contain the grams of clay.

To find the volume of any of the boxes, we will use the formula below;


\text{Volume=length x width x height}

Therefore, the volume of box A is;


\text{Volume of box A =2 x 3 x 5}=30\operatorname{cm}

Part A

Since Box B has twice the height, three times the width, and the same length as the first box;

This would give the dimensions of box B as;


\begin{gathered} \text{length}=5\operatorname{cm} \\ \text{width}=3*3=9\operatorname{cm} \\ \text{height}=2*2=4\operatorname{cm} \end{gathered}

Therefore the volume of the second box is


\text{Volume of the second box = 5 x 9 x 4 = 180}

Hence,


\begin{gathered} \text{Volume of box A = 30cm}^3;\text{ Amount of clay = 40g} \\ \text{Therefore, amount of clay in box B is }(180*40)/(30)=240\text{grams} \end{gathered}

Answer A


240\text{ grams}

Part B

Box C has double the length, width, and height of the original box. This gives the dimension as;


\begin{gathered} \text{length }=\text{ 5 x 2=10cm} \\ wid\text{th }=3*2=6\operatorname{cm} \\ Height=2*2=4\operatorname{cm} \\ \end{gathered}

Therefore the volume of box C is


\text{Volume of Box c=10 x 6 x 4=240 cm}^3

Hence, the amount of clay in box c is


\begin{gathered} \text{Volume of box A = 30cm}^3;\text{ Amount of clay = 40g} \\ \text{Therefore amount of clay in box C is }(240*40)/(30)=320\text{grams} \end{gathered}

Answer B


320\text{ grams}

The idea behind the solutions is to use direct proportion to find the missing values. This implies that an increase in the volume of the box would lead to a corresponding increase in the amount of clay the box can hold.

Part C

We can design a box that can hold 80 grams of clay If we double the height of box A.

Therefore the length becomes


New\text{ length = 2 x 3 =6cm}

User Zhrist
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