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Alignments to Content Standards: G-GMD.A.3


Janine is planning on creating a water-based centerpiece for each of the 30 tables at her wedding reception. She has already purchased a cylindrical vase for each table. The radius of the vases is 6 cm and the height is 28 cm. She intends to fill them half way with water and then add a variety of colored marbles until the waterline is approximately three-quarters of the way up the cylinder. She can buy bags of 100 marbles in 2 different sizes, with radii of 9 mm or 12 mm. A bag of 9 mm marbles costs \$3 and a bag of 12 mm marbles costs \$4.

  1. If Janine only bought 9 mm marbles how much would she spend on marbles for the whole reception? What if Janine only bought 12 mm marbles? (Note: $1\text{ cm}^3 = 1\text{ ml}$)

  2. Janine's parents, who are paying for the wedding, have told her she can spend at most $d$ dollars on marbles. Write a system of equalities and/or inequalities that she can use to determine how many marbles of each type she can buy.

  3. Based on your answer to part (b), how many bags of each size marble should Janine buy if she has \$180 and wants to mix in as many small marbles as possible.

IM Commentary

The purpose of this task is to use geometric and algebraic reasoning to model a real-life scenario. In particular, students are in several places (implicitly or explicitly) to reason as to when making approximations is reasonable and when to round, when to use equalities vs. inequalities, and the choice of units to work with (e.g., mm vs. cm).

Submitted by Patrick Barringer to the Third illustrative Mathematics Task writing contest.


  1. We are looking to fill one fourth of the cylinder's volume with marbles, a volume given by $$\tfrac{1}{4}\pi r^2h=\tfrac{1}{4} \pi \cdot 6^2 \cdot 28=252 \pi \text{ cm}^3.$$ Using the formula for the volume of the sphere, each of the 9mm ($=.9$cm) marbles has a volume of $\frac43 \pi (.9^3)$ or $.972 \pi \text{ cm}^3$. So to fill the desired volume requires $\tfrac{252 \pi \text{ cm}^3}{.972 \pi \text{ cm}^3}\approx 260$ marbles. To obtain 260 marbles per table for each of 30 tables, Janine needs to purchase $260\cdot 30=7800$ marbles. At 100 marbles per \$3-bag, this requres 78 bags of marbles, for a total cost of $3\cdot 78=234$ dollars. We repeat for the 12mm marbles similarly: Each of the 12mm marbles has a volume of $\frac43 \pi(1.2)^3$ or $2.304 \pi \text{ cm}^3$. To fill that volume requires $\tfrac{252 \pi \text{ cm}^3}{2.304 \pi \text{ cm}^3}$ or approximately 110 marbles. Thus 30 tables requires 3300 marbles, which in turn requires 33 bags of 12mm-marbles. At a cost of \$4 per bag, we arrive at a total cost of \$132.

  2. We have two constraints: 1) that we don't spend more than $d$ dollars, and 2) that we acquire enough volume of marbles to fill the cylinders to their desired level (approximately a quarter of the cylinder, as in part a). Let $s$ be the number of bags of smaller (9mm) marbles and $b$ be the number bags of bigger (12mm) marbles.

    The first constraint corresponds to the inequality $$3 \cdot s + 4 \cdot b \leq d,$$ and the second (using calculations from part (a)) is the approximate equality $$100 \cdot s \cdot .972 \pi + 100 \cdot b \cdot 2.304 \pi \approx 252 \pi \cdot 30,$$ which we can re-arrange more simply as $$97.2 \cdot s + 230.4 \cdot b \approx 7560.$$

  3. Since we learned in part (a) that it would require \$234 to do the reception entirely with small marbles, she will certainly have to spend all \$180 when maximizing the number of small marbles. Thus we can precede by solving the system of linear equations below (we have replaced approximations with equalities for the sake of solving the system, and then consider rounding below.) \begin{aligned} 3 \cdot s + 4 \cdot b &= 180\\ 97.2 \cdot s + 230.4 \cdot b &= 7560. \end{aligned}

    Multiplying the top row by $\frac{97.2}{3}=32.4$, we get \begin{aligned} -97.2 \cdot s - 129.6 \cdot b &= -5832\\ 97.2 \cdot s + 230.4 \cdot b &= 7560 \end{aligned}

    Adding the two equations gives $100.8 \cdot b = 1728$, giving $b=17\frac17$. So Janine can buy 18 bags of the larger marbles, and by substituting $b=18$ into the first equation and solving $$ 3 \cdot s + 4 \cdot 18 \leq 180, $$ we find $s=36$. So Janine can cover the requisite approximate volume by buying 18 bags of large marbles and 36 bags of small marbles for precisely \$180.