With so many apples in rectangular pyramids, your manager is concerned that the

Question

With so many apples in rectangular pyramids, your manager is concerned that the apple display will be so tall that it will interfere with the low-hanging signage in the store; the signage is only 2 feet above the surface of the display table. You think you can get a good estimate of the height of the apple display if you assume that the apples are spherical. You also simplify the height calculation by assuming each layer is 1 apple-diameter thick. Given the approximate apple diameter, the dimensions of the base, and the number of apples: How many layers of apples will fit on the display table without touching the sign? How many apples are left over? How tall is the pyramid (in inches and in number of layers)? Show test cases:

3.25 inch diameter, 10x16 base, max 10 layers, sign height 2 feet, 1000 apples are available,

2.75 inch diameter, 14x32 base, max 13 layers, sign height 2 feet, 1200 apples are available.

Background: Your manager at the local supermarket has asked you to create a produce display; he’s asked you to build a pyramid of apples from the upcoming shipment. The pyramid should be erected such that each square of four apples in one layer supports one apple on the next layer.

Example shown to right:

Top layer (1 apple)

Next layer (2 rows, 2 columns, i.e., 4 apples)

Base layer (3 rows, 3 columns, i.e., 9 apples)

Explanation / Answer

ANSWER

TEST CASE 1 : 3.25 inch diameter, 10x16 base, max 10 layers, sign height 2 feet, 1000 apples are available

Given :

   2 feet = 24 inch

so we have , length = 16 inch , breadth = 10 inch

now calculate apples cover length and breadth

16 inch length = 4 apples ( 3.25 x 4 = 13 inch )

10 inch breadth = 3 apples ( 3.25 x 3= 9.75 inch )

1.Total apple in base = 12 apples ( layer 1= base layer)

base layer = 4 columns and 3 rows i.e 12 apples

2. Now it is said that 4 apples will hold 1 apple for next layer so :

layer 2 ( next layer ) = 6 apples

= 2 rows and 3 columns

3. In layer 3 there will be 2 apples

layer 3 ( top layer )= 2 apples

= 1 row and 2 columns

we have 3 layers and 1 apple diameter = 3.25

height = 9.75 inch

==> So that result of test case :

Layers of apples ( without touching sign )= 3 layers ( total 20 apples )

Apples left over = 980 apples

height of pyramid = 9.75 inch , number of layers = 3

TEST CASE 2 :2.75 inch diameter, 14x32 base, max 13 layers, sign height 2 feet, 1200 apples are available.

Given :

     2 feet = 24 inch

so we have , length = 32 inch , breadth = 14 inch

now calculate apples cover length and breadth

32 inch length = 4 apples ( 2.75 x 11 = 30 inch )

14 inch breadth = 3 apples ( 2.75 x 5= 13.75 inch )

1.Total apple in base = 55 apples ( layer 1= base layer)

base layer = 11 columns and 5 rows i.e 55 apples

2. Now it is said that 4 apples will hold 1 apple for next layer so :

layer 2 ( next layer )  = 40 apples

= 4 rows and 10 columns

3. In layer 3 there will be 27 apples

layer 3 ( next layer )= 27 apples

= 3 row and 9 columns

4. In layer 4 there will be 16 apples

layer 4 ( next layer )= 16 apples

= 2 row and 8 columns

5. In layer 5 ( top layer )there will be 7 apples

layer 5 ( top layer )= 7 apples

= 1 row and 7 columns

we have 5layers and 1 apple diameter = 2.75

height = 5 x 2.75 = 13.75 inch

==> So that result of test case :

Layers of apples ( without touching sign )= 5 layers ( total 145 apples )

Apples left over = 1055 apples

height of pyramid = 13.75 inch , number of layers = 5

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