Grades 5-6 Video Solutions 2012-2012Grades56part2

2012Grades56part2

Video Transcription

Problem 11. Which three of the numbered puzzle pieces should you add to the picture to complete the square?  In this problem we are trying to complete this puzzle. Let's start with the piece in the center.  We are looking for a piece that has a bump over here and a bump in and then a bump in. So you'll see that out of all our possibilities only piece 2 fits.  That means that then we can cross this out of our choices. Let's look for the piece next to next here.  You'll see that it's a piece that has to have a bump out here, a bump out here and a straight line here.  So you'll see that the only piece that fits that description is piece number 3. So then we can cross it out over here.  Finally we are looking for a corner piece that has a bump out and then a bump in and then is a corner condition.  So out of all our possibilities you'll see that only piece number 6 fits that description. So the three pieces that you need are 2, 3 and 6.  Problem 12. Lisa has eight dice with the letters A, B, C and D with the same letter on all sides of each die. She builds a block with them.  Two adjacent dice always have different letters. What letter is on the die that cannot be seen in the picture?  So this is a pretty tricky problem but you can solve it. Basically you have to remember that this cube is made out of, that every side of this cube has two die on it.  And you have to remember that two adjacent dice always have different letters.  So we know that the cube that we are looking for, which is somewhere back here, is right next to this dice with the letter C on it. Right next to this dice with the letter A on it.  And right next to this dice with the letter D on it. And since we know that all adjacent dice have to have different letters,  the only letter that can be on this cube that's right back here is the letter B. And so the correct answer is B. Problem 13. There are five cities in Wonderland.  Each pair of cities is connected by one road, either visible or invisible. On the map of Wonderland there are only seven visible roads as shown. Alice has magical glasses.  When she looks at the map through these glasses, she only sees the roads that are otherwise invisible. How many invisible roads can she see?  To start this problem, I think it's best to figure out what is the maximum amount of connections between every city that is in Wonderland. So let's draw the five cities.  And then let's label them. City A, B, C, D, and E. And then we now have to connect them. And we know that there is a connection between A and B, A and C, A and D, and A and E.  Let's just write those down. A, B, A, C, A, D, and A, E. Now let's move on to the next city. There's a connection B, C. There's a connection B, D. And there's a connection B, E.  Connection B, A already exists. So then let's write down those extra connections. There's B, C, B, D, and B, E.  And now let's move on to city C. There's already a connection C, B, and there's already a connection C, A.  So then let's draw the connection C, D, and let's draw the connection C, E. And let's write those down. C, D, and C, E. And now let's move on to city D.  There's already a connection D, C, D, B, and D, A. So let's draw in the connection D, E.  And since we've gotten to city E, you see that it's already fully connected. So let's count up how many different connections there are. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 connections.  Now, if we compare that to this map that's over here, which are the visible roads, you see that there are seven visible roads on this map.  And the ones that are invisible are this one, this one, and this one. There are three invisible roads. So the correct answer is D. Problem 14.  The positive integers have been colored red, blue, or green. 1 is red, 2 is blue, 3 is green, 4 is red, 5 is blue, 6 is green, and so on.  Renata calculates the sum of a red number and a blue number. What color can the resulting number be?  In this problem, the integers have been colored different colors. The number 1 is red, the number 2 is blue, the number 3 is green, and then it goes on.  Number 4 is red again, number 5 is blue, number 6 is green, and then so on and so forth. So to understand this relationship between the colors and the numbers,  you can look that there is a red number once every three numbers. There is a blue number once every three numbers. There is a green number once every three numbers.  You can write this relationship as red is equal to 1 plus a multiple of 3. Blue is equal to 2 plus a multiple of 3. And you can write green is equal to 3 plus a multiple of 3.  So if you want to calculate the sum of a red number and a blue number, you'll see that it's going to be 1 plus 2 plus a multiple of 3.  Which equals green. 3 plus a multiple of 3. So the correct answer is C. Problem 15.  The perimeter of the figure to the right, made up of identical squares, is equal to 42 centimeters. What is the area of the figure?  Now let's look at the figure that is part of this problem. It is made up of 8 identical squares. Now let's look at one of these squares.  And let's say that the sides of this square are all equal to S. Now, we know that the perimeter of this figure is equal to 42 centimeters. What does that mean?  That means that if we add all of the sides of these squares together, we should be able to get 42.  And since all the sides are the same, because they're made up of the same kind of square, that should be a pretty easy thing to do.  So let's count up how many sides are along the perimeter of this figure. That's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14.  So that we know that 42 divided by 14 is equal to 3, which is equal to S. So we know that the area of one of these squares is equal to the square of its sides.  So that means that it's equal to 3 squared, which is equal to 9, which is the area of one square. And we know that there are 1, 2, 3, 4, 5, 6, 7, 8 squares.  So we know that 9 times 8 is equal to 72. So that the answer is that the area of this figure is equal to 72 centimeters squared. Problem 16. Look at the pictures.  Both shapes are formed from the same five pieces. The rectangles measure 5 centimeters by 10 centimeters, and the other parts are quarters of two different circles.  The difference between the lengths of the perimeters of the two shapes is...  To solve this problem, you have to realize that both shapes are formed from the same five pieces,  and that the basis of these five pieces is the same rectangle, marked as you can see. And that this rectangle measures 5 centimeters by 10 centimeters.  Now, the first step of the problem is to understand what the difference in the perimeter is of these two shapes. And you can see that since they are made out of the same pieces,  you can say that this perimeter here is equal to this perimeter here. And you can say that this perimeter here is equal to this part of the perimeter here,  this part is equal to this perimeter here, and this perimeter is equal to this perimeter. And so, what you're left with are the pieces that are different.  You can see that on this shape, there's an edge here, an edge here, an edge here, an edge here, and on this shape, there's only one edge here.  And since you know that this rectangle is 5 centimeters by 10 centimeters, you know that for this shape, this is equal to 10 centimeters.  And since you know that these are quarter circles, you know that since this shape here, right here, is a quarter circle that has 10,  for one side, you know that this is 10 over here. So you know that this is 10 over here. You know that for this shape, since this side is equal to 5,  you know that this side is equal to 5, and so you know that this side is equal to 5. So the difference in perimeters is going to be 10 plus 10 plus 5 plus 5,  which is equal to 30 for this. And as we already know, for this shape, the perimeter is 10. So if you subtract 10 from 30, you get a difference of 20 centimeters,  and so you know the correct answer is D. Problem 17. Place the numbers from 1 to 7 in the circles in such a way  that the sum of the numbers on each of the indicated lines of the three circles is the same. What is the number at the top of the triangle?  Let's begin by what we know in this problem. First, let's say that T is equal to the sum of the numbers from 1 to 7.  So 1 plus 2 plus 3 plus 4 plus 5 plus 6 plus 7, and that's equal to 28. And then we know that we want a sum that would,  we want each of these three circles and each of these three lines to add up to the same number. So let's give each circle a letter value  so that we can understand a little bit better what's going on. Let's label these A, B, C, D, E, and F. And let's say that the letter S is equal to A plus B plus C.  It's also equal to D plus E plus F. It's going to be equal to D plus A plus X and X plus B plus E, right, and X plus C plus F.  So we can write an equation that says 2S is equal to T minus X because we know that A plus B plus C, right, plus D plus E plus F is 2S  and that the only number we're not including is X right here. So we can say 2S is equal to T minus X. Another equation that we can write down is 3S is equal to T plus 2X.  Now, we can write that down, right, because we see that there are three lines here. Right, there is D plus A plus X and E plus B plus X and F plus C plus X.  And we know that those are, right, 3 times S then because those are three sums that will equal S, but we're repeating the integer X  and that's why we have to say it's equal to T plus 2X. Now, to make these two equations the same, let's multiply this equation by 3 and let's multiply this equation by 2.  And so you get 6S is equal to 3T minus 3X and you're going to get 6S is equal to 2T plus 4X. Since 6S is equal to 6S, we can say 3T minus 3X is equal to 2T plus 4X  and then we can say T is equal to 7X and then we can say 28 is equal to 7X and we can say X then is equal to 4. And so to get the correct answer then is C, which is equal to 4.  Problem 18. A rubber ball falls vertically through a height of 10 meters from the roof of a house.  After each impact on the ground, it bounces back up to 4 fifths of the previous height. How many times will the ball appear in front of a rectangular window  whose bottom edge is at a height of 5 meters and whose top edge is at a height of 6 meters?  To begin, let's draw a sketch of what is going on in this problem. So we have a building with a window in it and this window is from 5 meters to 6 meters high  and the building is 10 meters high and we start at the ground on 0. So first of all, there is a ball that drops and then it bounces up to 4 fifths of the height that it started at.  So let's take 10 and multiply that by 4 fifths and we get 8 meters. So we know it goes up past the window and then it falls back down again  and then it bounces up again and it goes to a height that's 4 fifths the height of the 8 meters. So let's multiply 8 times 4 fifths and we'll get 6.4 meters.  So it reaches 6.4 meters and it falls back down again and it bounces up again and it will bounce up again to 4 fifths the height of the previous point it was at.  So let's multiply that 6.4 by 4 fifths and we get 5.12 meters and then the ball will fall down again and it will bounce up again  and it will only reach a height that's 4 fifths of the previous height which is equal to 4.096 meters. So it doesn't reach the window and so we are no longer interested in it.  And so if we count the amount of times that this ball gets in front of the window, we'll see that it passes the window once, it disappears, it comes back again,  it disappears up here, it comes back again, it bounces down, it reappears here, it reappears again and then it reappears a final time  before not being able to bounce as high as the window. And if we count those, we'll count 1, 2, 3, 4, 5, 6 times that the ball will appear in front of the window.  So the answer is D. Problem 19. There are four gear wheels on fixed axles next to each other as shown.  The first one has 30 gears, the second one 15, the third one 60 and the last one 10. How many revolutions does the last gear wheel make  when the first one turns through one revolution? So to begin with, we have to know that in a gear wheel, one movement of one gear means one movement of one gear  in the wheel that is next to it, like this. So that means that if this circle makes one full revolution and if that circle has 30 gears,  that means that one revolution of it means that 30 gears have moved and that means that this circle has also moved 30 gears  and this circle has moved 30 gears and this circle has moved 30 gears and since the last circle has only 10 gears, that means that we have to divide the 30 gears of circle 1  by the 10 gears of circle 4 and so we get three revolutions of the last circle and so the correct answer is A. Problem 20. A regular octagon is folded in half exactly three times  until a triangle is obtained as shown. Then the apex is cut off at a right angle as shown in the picture. If the paper is unfolded, what will it look like?  This problem is best understood graphically. So let's begin with what's going on here. We've been asked to cut off the apex, that is the tip,  of an octagon that has been folded into one-eighth of its original size.  So let's do that. Let's cut off the apex. So what we get is a shape that looks like this. And then if we start unfolding this shape, first we unfold another eighth  and when we unfold the eighth, we unfold the quarter and when we unfold that, we have only a half left and we get a square within the octagon.  This is the shape that you get in the end and that means that the correct answer is C.

Video Summary

The video provides solutions to a series of mathematical problems related to puzzles, dice configuration, city connections, color patterns, shapes, and other logical puzzles. Specific solutions include finding appropriate puzzle pieces to complete a square (2, 3, and 6), determining that the unseen die in a block is 'B', calculating the invisible roads in a city map (3 roads), and identifying color results from color-patterned sums (result is green). Furthermore, the video discusses calculating the area of a figure with a given perimeter (72 cm²), determining the difference in perimeter of transformed shapes (difference is 20 cm), and problems involving sequences, gear movements, and geometry transformations of a regular octagon into a specific patterned shape upon folding and cutting. Contact with each problem is approached through step-by-step logical analysis, focusing on mathematical reasoning and pattern recognition.