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THinking skills-- real problem, need help for my exam tomorrow!

please explain the reasoning behind this question... I could not understand it. It's for my THinking Skills exam tomorrow!!

Note by S Zara
3 years, 8 months ago

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Well, assuming the room is cuboid in shape, the answer is option D. My reasoning is as follows:

1. The room has a total volume of $$18 parcels \times (1\times1\times 2) m^{3}/parcel = 36 m^{3}$$
2. You know the floor area is $$9m^{2}$$ and so the height of the room is $$36m^{3} \div 9m^{2} = 4m$$ This invalidates option B, since the height is already known.
3. Now, since you know that the internal dimensions are natural numbers, they can be $$(3m \times 3m)$$ or $$(1m \times 9m)$$.
4. Because you want to fit in cubes of side $$1m$$, these dimensions do not matter, so long as they are natural numbers! The same number of cubes will fit in no matter what the dimensions are - hence, option A in invalidated.
5. Obviously, option C is useless - I don't care what you fit into the parcels, so long as they don't change its dimensions. And so, by elimination, option D is the right answer.

If you have less than or equal to 36 small parcels, you're good. If you've got more, they won't fit.

- 3 years, 8 months ago

He packed 18 parcels of 2 cubic metres so now he know both height and floor area of the store now all that matters is how many small parcels are there so he can calculate whether they will fit or not

- 3 years, 8 months ago

C, the highlighted option is the correct answer btw!

- 3 years, 8 months ago

Now basically we have the info as follows :- base area 9 sq. Mts height 4m hence total volume of container is 36 cubic mts Now we have cubes of volume 1 cubic metre .... we know that no. Of small parcels = volumr of container ÷ vol. Of small parcel (the cube) hence in order to have all the parcels to be exactly fit in the container we must know the number of parcels

- 3 years, 8 months ago