Question

Three rectangular-shaped holes have been drilled passing all the way through a solid 3 × 4 x 5 cuboid. The diagrams show the front, side and top views of the resulting block. What fraction of the original cuboid remains?
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Answer 1

The volume of a cuboid implies the product of its length, width, and height. So that the fraction of the original cuboid that would remain is $\frac{7}{15}$.

A cuboid is a solid derived from a rectangle, thus it has rectangular faces. Its volume can be determined by;

volume of a cuboid = length x width x height

In the given question, the volume of the original cuboid can be determined as;

  volume = 3 x 4 x 5

                = 60 Cubic units

Since holes can not be drilled at the intersection of the holes, then the volume of the hole has to be determined.

To determine the volume of the hole drilled, we have:

(6 x 3) + (3 x 2) + (2 x 2) = 28 Cubic units

So that the fraction of the original cuboid that would remain = $\frac{28 cubic units}{60 cubic units}$

                                                                   = $\frac{7}{15}$

Therefore,  $\frac{7}{15}$ of the original cuboid would remain.

Fro further clarifications on volume of a cuboid, visit: brainly.com/question/46030

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