Question

A rectangular box is completely filled with dice. Each dice has a volume of 1 cm³. The length of the box is 3 cm greater than its width and its height is 5 cm. Suppose the box holds at most 140 dice. What are the possible dimensions of the box?​

Answer 1

Answer:

Dimensions, w+3, w, 5

Volume, 140 cubic centimeters

5w%28w%2B3%29=140

w%28w%2B3%29=28

Factors for 28 may be 1&28, 2&14, 4&7.

 

The 4 & 7 factorization looks like the right one.

system%28w=4%2Cw%2B3=7%29

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Width, 4 cm

Length, 7 cm

Height, 5 cm

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Step-by-step explanation:

Answer 2

Given: A rectangular box is completely filled with dice. Each dice has a volume of 1 cm³. The length of the box is 3 cm greater than its width and its height is 5 cm. The box holds at most 140 dice.

To find: Possible dimensions of the box

Solution: Volume of a dice= 1 cm³

Volume of 140 such dices

= 140 × Volume of one dice

= 140 × 1 cm³

= 140 cm³

Since the rectangular box is completely filled with 140 dice, volume of this box= 140 cm³

Let the width of the box be w.

Since the length is 3 cm greater than width,

Length= w+3

Height= 5 cm

Volume of the box is given by the formula

= length × width × height

= w × w+3 × 5

= 5w(w+3)

$= 5 {w}^{2} + 15w$

This must be equal to 140 cm³.

Therefore,

$5 {w}^{2} + 15w = 140$

$= > {w}^{2} + 3w - 28 = 0$

$= > {w}^{2} + 7w - 4w - 28 = 0$

$= > w(w + 7) - 4(w + 7) = 0$

$= > (w + 7)(w - 4) = 0$

=> w+7 = 0 or w-4=0

=> w = -7 or w = 4

But since width cannot be negative, w= 4 cm

Length = w+3

= 4+3

= 7 cm

Therefore, the length of the box is 7 cm, width is 4 cm and height is 5 cm.