# Math Help - sequences problem

1. ## sequences problem

A hardware supplier has designed a series of six plastic containers with lids where each container (after the first) can be placed into the next larger one for storage purposes. The containers are rectangular boxes, and the dimensions of the largest one are 120 cm by 60 cm by 50 cm. The dimension of each container is decreased by 10 percent with respect to the next larger one.

Determine the volume of each container, and the volume of all the containers. Write your answers in litres (one litre = 1000 cm3).

Is it 10% from each dimension or from the volume? Hard to figure out from the problem.

If from the each dimension then:
V 1= 50x 60x 120 = 360000 cm3 = 360 litres
V2=45 x 54 x 108 = 262440 cm3 = 262.4 liltres
V 3 = 40.5 x 48.6 x 97.2 = 191.3litres
V4 = 36.45 x 43.74 x 87.48 =139.4 litres
V5 = 32.81 x 39.37 x 78.74 = 101.7 litres
V6 = 29.61 x 35.47 x 70.94 =74.5 litres

I need some feedback.

2. I expect since it says "the dimension of each container is decreased by 10% with respect to the larger one" I would say it's the dimensions and not the volume that is decreased by 10%.

3. Originally Posted by terminator
A hardware supplier has designed a series of six plastic containers with lids where each container (after the first) can be placed into the next larger one for storage purposes. The containers are rectangular boxes, and the dimensions of the largest one are 120 cm by 60 cm by 50 cm. The dimension of each container is decreased by 10 percent with respect to the next larger one.

Determine the volume of each container, and the volume of all the containers. Write your answers in litres (one litre = 1000 cm3).

Is it 10% from each dimension or from the volume? Hard to figure out from the problem.
Why? The problems says "The dimension of each container is decreased by 10 percent". That tells you precisely which is intended

If from the each dimension then:
V 1= 50x 60x 120 = 360000 cm3 = 360 litres
V2=45 x 54 x 108 = 262440 cm3 = 262.4 liltres
V 3 = 40.5 x 48.6 x 97.2 = 191.3litres
V4 = 36.45 x 43.74 x 87.48 =139.4 litres
V5 = 32.81 x 39.37 x 78.74 = 101.7 litres
V6 = 29.61 x 35.47 x 70.94 =74.5 litres

I need some feedback.

4. Hello, terminator!

A hardware supplier has designed a series of six plastic containers with lids
. . where each container (after the first) can be placed into the next larger one.

The containers are rectangular boxes.
. . The dimensions of the largest one are 120 cm by 60 cm by 50 cm.

The dimensions of each container are decreased by 10 percent
. . with respect to the next larger one.

(a) Determine the volume of each container.
(b) Determine the volume of all the containers.
Write your answers in litres (one litre = 1000 cubic cm).

The original box has dimensions $L,\,W,\,H$
. . Its volume is: . $L\!\cdot\!W\!\cdot\!H\text{ cm}^3$

The next box has dimensions: $0.9L,\,0.9W,\,0.9H$
. . Its volume is: . $(0.9L)(0.9W)(0.9H) \:=\:0.729(LWH)$

That is, each box is 0.729 of the volime of the next larger box.

The first box has volume: . $120\cdot60\cdot50 \:=\:360,\!000\text{ cm}^3$

. . That is: . $V_1 \:=\:360\text{ liters.}$

(a) Therefore, the $\,n^{th}$ box has volume: . $V_n \;=\;360(0.729)^{n-1}$

The sum of the volumes of all the boxes is:

. . $S \;=\;360 + 360(0.729) + 360(0.729^2) + 369(0.729^3) + \hdots$

. . $S \;=\;360\underbrace{\bigg[1 + 0.729 + 0.729^2 + 0.729^3 + \hdots\bigg]}_{\text{a geometric series}}$

The sum of the geometric series is: . $\dfrac{1}{1 - 0.729} \:=\:\dfrac{1}{0.271}$

(b) Therefore: . $S \;=\;360\left(\frac{1}{0.271}\right) \;\approx\;1328.4\text{ liters}$