Grades 5-6 Video Solutions 2022-2022_5-6

2022_5-6_29

Video Transcription

Question 29. The three pictures show a structure made from cubes as seen from the top, the front,  and the right. What is the maximum number of cubes that could have been used to build the structure?  Okay, so the structure is part of this box with width 4, depth 4, and height 3. What does that  mean? Well, the structure is part of this box that just, if you can follow the mouse, that goes  like this. This is the base right here, so 1, 2, 3, 4, 1, 2, 3, 4, and then it has a height of  at maximum 3, which is justified by both the top view, sorry, by both the right view and the front view. So we have this rectangular prism and the structure is a part of that box.  So what we can do is we can create a 4x4 table, and the 4x4 table, each one corresponds to one  kind of cell in the base, and each value in each cell indicates the height of the column at that  respective space. So the top view tells us that a few of these cells have zeros, they don't have  any height. So these two, this one, these two, and then these two. That's what I've written out over here.  And then the empty cell in this bottom over here must have a height of 1 as per the right view.  So if you look on the previous one, this is empty. We fill that with this height of 1 as per the  right view, because it tells us that in the front row, the maximum height is 1. Then the  empty cells in the first and fourth column over here, so this empty cell over here, and then this  empty cell over here, also have a height of 1, which we can tell from this front view, because the front view tells us that the first column over here has a maximum  height of 1, and the fourth column also has a maximum height of 1.  Okay, now to maximize the number of cubes. Well, we can say that the empty squares in the third row,  so these two squares over here, can have a height of 2. And that is because in the right view, we see that this is the third row over here, these empty squares have a height of 2.  And then we can also say that the ones in the first and second have a height of 3, and that  kind of corresponds with this front view, which tells us that in this second and third column,  we have a max height of 3. And of course, to maximize it, we can keep all of these values as 3, and then both of these values as 2. Then we can add up the numbers in this grid,  and we get the total sum of 19, so our answer is B.

Video Summary

The problem involves determining the maximum number of cubes in a structure within a 4x4 base and a maximum height of 3. By analyzing the top, front, and right views, we establish that certain cells have specific heights, and others are determined from these views' constraints. Adjusting the empty spaces to maximize cube usage while adhering to height limitations, it’s concluded that the maximum number of cubes that could have been used is 19. Thus, the solution to the problem sums up the calculated heights within the grid, yielding the answer as 19 cubes.