Question 1172934: Case study based-1
Mr. Jacson hired a warehouse for rent to hold unloaded goods before
further distribution. All the goods were packed into boxes of
dimensions 1.5m, 1.25m and 0.5m. The warehouse measured 40m
x 25m x 15m. Before storing the goods, he planned to whitewash and
clean the warehouse.
a) Mr. Jacson painted the four walls and the ceiling of the warehouse.
Calculate the area painted?
i) 3950𝑚2
ii) 2950𝑚2
iii) 1290𝑚2
iv) 1950𝑚2
b) How many boxes of paint would be required if 1 box of paint
contains 2 litres and 1 litres of paint can be used to paint an area of
5𝑚2?
i)590 boxes
ii)200 boxes
iii)390 boxes
iv)295 boxes
c) Find the maximum number of boxes that can be kept in the room.
i)160
ii) 2000
iii) 16000
iv)4000
d) How high must the warehouse be to hold 500 cubical boxes of side
4m?
i)1.6m
ii)32m
iii)8m
iv)33m
e) Find the diagonal of the warehouse.
i)50√2
ii)70
iii)35√2
iv) 2√35
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It helps to draw this.
Two walls are 25 x 15 for a total area of 750 m^2.
Two other walls are 40 x 15 for a total area of 1200 m^2.
The ceiling is 40 x 25 or 1000 m^2
that is 2950 m^2.
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1 box paints 10 liters, so 295 boxes are needed.
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The volume of each box is 0.9375 m^3, and the volume of the warehouse is 15000 m^3, so that would contain 16,000 boxes.
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Side 4 m will be 10 along the long wall and 6 more along the width for 60 per row. He needs 8 1/3 rows of 4 m and that comes closest to 33 m, BUT, fractional rows don't work here. He needs a ninth row for the last 20 and that means 36 m high.
Another way is to say the boxes are 64 m^3 and 500 of them are 32,000 m^3.. The volume of the warehouse is 1000 m^3*h, and this would mean the height has to be 32 m to contain the equivalent material. Note, the actual boxes would not fit in that, only the area they contain could.
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There are two diagonals, one of the floor, which is the hypotenuse with legs 40 and 25 whose summed squares are 2225. The other diagonal is the three dimensional one, which has legs squared of 2225 and 225, the diagonal across open space. That sums to 2450 and the square root of that is 35 sqrt (2) (break it up into 1225 and 2).
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