Grades 3-4 Video Solutions 2014-Grades 3-4 Video Solutions 2014

Grades 3-4 Video Solutions 2014

Video Transcription

This is the Math Kangaroo Solutions Video Library,  presenting solution suggestions for levels 3 and 4 from the year 2014.  These solutions are presented by Lucas Naleskowski.  The purpose of the Math Kangaroo Solutions Video Library is to help you learn how to solve math problems, such as those presented in the Math Kangaroo Competition.  It is important that you make sure to read the problem as well as listen as I read the problem.  After reading and listening to the question, pause the video and try to solve the question.  Question 1. Which small figure could be the central part of the larger figure with the star?  To solve this problem, we just have to look at the star and count how many points it has. The star has 9 points.  We look at all the options A, B, C, D, and E. There is only one which has 9  segments. With that, we know the answer. You notice that the fourth star has 9 segments. So the answer is D.  Question 2. Jackie wants to place the digit 3 somewhere in the number  2014. Where should she place the digit 3 to make the resulting 5-digit number as small as possible?  So we have our number 2014. We have a number of different spots that we can put the 3 in,  such as 32,014, 23,014,  20,314, 20,134, 20,143. Out of all these, the smallest number is  20,134. So to make the smallest 5-digit number possible, placing the 3, we want to make  20,134. So the 3 will be between the 1 and the 4. Answer D.  Question 3. Which houses are made using exactly the same triangular and rectangular pieces?  So we can start off by crossing off the fifth house, since its roof is different from all the others.  Next, we cross out the third house, since it has four green triangles.  Finally, we cross out the second house, because it only has two green triangles. All of the other houses, 1 and 4, have three green triangles,  one blue rectangle, one orange triangle,  one orange rectangle, and one yellow rectangle. So our answer is houses 1 and 4.  Answer A. Question 4. When Coco the Koala is not sleeping, he eats 50 grams of leaves per hour.  Yesterday, he slept 20 hours.  How many grams of leaves did he eat yesterday?  Now, we know that one day has 24 hours.  If we know that Coco slept for 20 hours, we can take 24 and subtract 20, giving us 4 hours when Coco was awake.  And now when he's not sleeping, he eats 50 grams of leaves per hour. So we can do 4 hours times 50 grams. When we do this, we find out how many grams of leaves he ate yesterday.  4 times 50 is 200. So our answer is D.  Question 5. Maria performs the subtractions shown to the right and gets as results numbers from 0 to 5.  She connects the dots, starting at the dot with the result 0, and ending at the dot with result 5, in an increasing order.  Which figure does she draw as a result?  So if you take a look at the dots and the numbers, you get 2 minus 2, 6 minus 5, 8 minus 6,  11 minus 8, 13 minus 9, and 17 minus 12.  We simplify these. We get 0, 1, 2, 3, 4, and 5. We know that she connected them in increasing order, starting from 0.  So we connect 0 to 1, 1 to 2, 2 to 3, 3 to 4, and 4 to 5.  This gives us the same shape as the one in answer A. So the answer is A.  Question 6.  Adam built fewer sandcastles than Martin, but more than Susan. Lucy built more sandcastles than Adam, and more than Martin.  Dana built more sandcastles than Martin, but fewer than Lucy.  Which of the children built the most sandcastles?  So if we make a chart organizing from least to greatest,  we know that Adam built fewer sandcastles than Martin, but more than Susan.  But more than Susan. And we can set up Susan building the least, then Adam, and then Martin.  Next, we know that Lucy built more sandcastles than Adam, and more than Martin.  So she will be greater than both of them.  We know that Dana built more sandcastles than Martin, but less than Lucy. So she'll be before Lucy, but after Martin.  So which of the children built the most sandcastles? The answer is Lucy.  E. Question 7.  Monica writes numbers in the diagram to the right,  so that each number is the product of the two numbers below it.  Which number does she write in the gray cell?  So if we look at the diagram, we have several numbers filled in. 64, 2, 1, 2, and 1.  Now, we know that any number is the product of the two numbers below it.  Since we have a 2 right above a 1, the number multiplied by 1 to get 2 has to be 2.  So we can put in 2s on the bottom right here.  And when we do 2 times 2, we get 4. So 4 will be put in the next row.  Now, to get to the gray cell, we just have to multiply the two cells below it with 4 and 2.  When we multiply 2 and 4, we get our answer.  8.  So our answer is 8, letter E.  Question 8.  Anne has the four gray pieces shown to the left.  She needs to completely cover the white shape shown to the left. Where does she need to place the T-shaped piece shown below  so that she can use the other three pieces to cover the rest of the shape?  So if we take a look at the white shape, we can start putting in pieces.  On the bottom right, only the L shape can fit.  Next, the best place to put the long 4 shape will be here. Finally, we can put our zigzag shape over here, leaving a space for the T-shape.  This is shown in answer C, which is the correct answer.  Question 9.  Mr. Brown painted some flowers on the store window.  See picture on the right.  What do these flowers look like from the other side of the window?  We know when something is drawn on a window,  it'll be reflected in the opposite version on the other side of the window.  So if we take a look at our flowers, we take a closer look at this flower, the one with the five petals, and the little twig with the three  balls on it, and we can compare it to the different answers.  Now, answer A has the twig pointing downwards.  Which would not be the case on the other side of the glass.  So this cannot be the answer.  The next picture has the flower with the petals in the same spot  as it does on the same other side of the glass. So this cannot be the answer.  The next answer has the flower with the petals on the right side. However, here, the leaf and the twig have switched spots.  So this cannot be the answer.  The same issue is seen in the next one. So this also cannot be the answer. We are only left with one correct solution. So the answer to this question is E.  Question 10. There were some pieces of candy in a bowl.  Sally took half of the pieces of candy.  Then Tom took half of the pieces left in the bowl.  After that, Clara took half of the remaining pieces.  In the end, there were six pieces of candy in the bowl.  How many pieces of candy were in the bowl at the beginning?  So if we start off with the bowl, we know that Sally took half of the pieces. So we'll have the bowl divided by half.  Then Tom takes another half. So we have the bowl divided by half again.  Finally, Clara takes half of the candy. So we'll divide by two again.  And then we know the bowl has six pieces of candy.  So if we have six, and we know this is only after half of the candy has been taken,  we can multiply this number by two to reverse the operation. We have 12 pieces of candy before half are taken. We can do this again.  We take 12, and we know that there was double the candy left since half of it was taken. So we have 24. And finally, when we have 24 pieces of candy,  we know this is only after half of the candy was taken. So before it was taken, there would be twice as much candy.  Two times 24 gives us 48.  So the answer is 48.  E.  Question 11. Which tile must be added to the picture so that the total light gray area  is the same size as the total dark gray area?  Take a look at the diagram.  We can start counting the light and the dark gray areas.  Start off with the light gray. We can count each full square as one and each half of a square as half.  Doing so, we get one plus one plus one.  Then we get half, another half, and another half.  So together, it's one plus one plus one plus a half plus a half plus a half,  which gives us a total area of 4.5 light gray.  Now we want to calculate the dark gray area.  This we have one, one, half, half, and another half.  Do this, we have one plus one plus a half plus a half plus a half,  giving us 3.5 dark gray area.  Now we want the total light gray area to be the same as the total dark gray area.  To do this, both these numbers must be the same.  Now to get 3.5 to equal 4.5, we have to add one.  To do this, we'll add one dark gray area box, giving us answer B.  Question 12.  Paula shoots arrows at the target shown on the left.  When she misses the target, she gets zero points.  Paula shoots two arrows and adds the number of points. Which of the following sums cannot be her score?  So when Paula shoots an arrow, there are four options. She gets the 30, 50, 70, or misses and gets zero points.  I want to see which of the answers cannot be her score.  Answer A, which is 60, can be attained if she hits the 30 twice, since 30 plus 30 equals 60.  70 can be attained if she hits the 70 and then misses.  80 will be achieved if she hits the 50 and the 30,  and 100 is attained if she hits the 50 twice. The only option that is impossible to achieve with two arrows is 90, so the correct answer is B.  Question 13.  Mary had equal numbers of red, yellow, and green tokens.  She used some of these tokens to make a pile.  You can see all the tokens she used in the figure.  After making the pile, she still had five tokens which were not used. How many yellow tokens did she have at the beginning?  We take a look at the pile she made.  We notice that she had four yellow, four green, and five red tokens in the pile.  Now, we know that after this pile, she has five tokens left.  Since we know at the beginning she has an equal number of tokens,  we want to have the same amount of yellow, green, and red tokens.  To do this, we have to say one of the five tokens will be yellow. So, five yellow tokens with four miscellaneous tokens left.  Then, five green tokens with three miscellaneous tokens left.  We have three tokens left. We know that each one will be a different color, so there will be six yellow, six green, and six red.  So, she had six yellow tokens at the beginning. Answer B. Question 14. Peter Rabbit likes cabbage and carrots very much. In one day, he can eat  only 9 carrots, only 2 cabbages, or 1 cabbage and 4 carrots. During one week, Peter ate 30 carrots. How many cabbages did he eat during that week?  So we know that Peter can eat 9 carrots, 2 cabbages, or 1 cabbage and 4 carrots. We know that one week has 7 days. To start off, let's say Peter will eat 9 carrots on the  first day, 9 carrots on the second day, then 4 carrots and 1 cabbage on the third day, 4 carrots and 1 cabbage on the fourth day, and 4 carrots and 1 cabbage on the fifth day.  With this, we get a total of 30 carrots and 3 cabbages. Now, this was only 5 days, so we need to take our 3 cabbages and assume that the last 2 days of the week, Peter eats  only cabbages. So on the sixth day, he'll eat 2 more cabbages, and then on the seventh  final day, he will eat 2 more cabbages, giving us a total of 7 cabbages. So Peter ate 7 cabbages during the week, or answer B.  Question 15. The solid in the picture to the right was made by sticking 8 identical cubes together. What does this solid look like from directly above? If we look at the shape, we  can start noting the top facing sides. We'll start off with red dots, and then green dots. If we looked at this perspective from the top, we would see something like this, each  dot representing one square. Now, the solid looks directly from above, it looks like 3 and 2 squares together, so the answer will be C.  Question 16. How many dots are there in this picture? If we take a look at the picture, we can start counting the dots. Count them like so, we have 5 dots, plus 5, plus 5, plus  5. Total, that is 20 dots. We count these dots, we get 3. Now we can repeat this like so, getting 20 and 3 dots. And we recognize that there is a pattern like so, that continues  with 20 dots, and then 3 dots. And this continues on for the entire picture. Next, we will have  to total up all the 20s and the 3s. So when we do this, we get 20 plus 3, plus 20, plus  3, plus 20. Simplify this by making all of these 23, and then 120. And when we simplify this further, we get 46, plus 46, plus 46, plus 43. Simplify this even further, we get  92, plus 89, which will give us the total number of dots in the picture. When we add these together, we get our answer, which is 181, or answer B.  Question 17. On the planet Kangaroo, each Kang year has 20 Kang months. Each Kang month has 6 Kang weeks. How many Kang weeks are there in one quarter of a Kang year? Now,  we know that one Kang year is 20 Kang months. If we divide both of these by 4, to get one quarter of a Kang year, we get the fact that one quarter of a Kang year is equal to 5 Kang  months. Next, we know that each Kang month has 6 Kang weeks. Now, we want to find out how many Kang weeks there are in a quarter of a Kang year, which is 5 Kang months. So  if we multiply each side by 5, we get 5 Kang months equal to 30 Kang weeks. And we want to know how many Kang weeks there are in one quarter of a Kang year. Since we know a quarter  of a Kang year is 5 Kang months, and that is also equal to 30 Kang weeks, there will be 30 Kang weeks in a quarter of a Kang year. So the answer is B.  Question 18. 7 children are standing in a circle. No 2 boys are standing next to each other. No 3  girls are standing next to each other. Which of the statements below about the number of  girls standing in the circle is true? If we draw out a circle with a space for each person, 7  spaces, we can start putting them down. Let's say we put a boy here, then we can put a boy here.  The only other place we can put is a boy here. So that way, no 2 boys are standing next to each  other. It means every other person has to be a girl. Now, knowing this, it is possible to have  3 boys. That means that 3 is not the only possible number, since in this example we can have 4 girls.  Also, B cannot be true. Now, if we take away the girls, take away one boy, and try this with only  2 boys, and fill out all the other spots with girls, then we notice that having 5 girls does  not work, since we have more than 3 girls standing next to each other. So there cannot be 5 girls.  So E and D cannot be answers. The only other left is 4 is the only possible number. So the answer is C.  Question 19. Eve arranged cards in a line as shown below. In one move, Eve can switch the places of  any two cards. What is the smallest number of moves Eve needs to make to get the word kangaroo?  So we start off with the letters in this order. We want to end up with the word kangaroo. So we want to switch the K and the O, so as to start off with the K and the kangaroo.  Next, we will want to switch N and R, so as to get king. And finally, we only need one step to  reach kangaroo, and that is switching the O and the A. With this, we have kangaroo in three steps.  So the answer is B. Question 20. We are making a sequence of triangles out of diamonds. The first  three steps are shown. In each step, a line of diamonds is added to the bottom. In the bottom  line, the two outside diamonds, one on each side, are white. All the other diamonds in the triangle  are black. How many black diamonds will the figure have in step six? So we know what the first  three steps will look like. If we follow this pattern, we just have to add a row of diamonds  at the bottom each time. So step four will look like so, step five will look like so, and step six  will look like so. Now we also know that only the bottom two left and right diamonds will be white.  All the others will be black. So once we count up all the black diamonds in the sixth step,  we get our answer, which is 26C. Question 21. Kangaroo Hamish bought some toys shown in the  picture and gave the cashier 150 king coins. He received 20 king coins back. Then he changed his  mind and exchanged one of the toys for another. He got back an additional five king coins.  What toys did Hamish leave the store with? So we know that one toy plus another toy, which will  represent my A and B, plus 20 coins equals Hamish's total starting 150 coins. And we also know that  when he switches out one of the toys, he gets back an extra five coins. So that means toy A plus toy C plus 25 will be 150. With this, we can set these two equal to each other.  We get A plus B plus 20 equal to A plus C plus 25. When we simplify, subtracting A from both sides  and then subtracting 20 from both sides, we get toy B is equal to toy C plus five.  This means we know that the two toys have a monetary difference of five coins.  There are only two toys that have a difference of five between them. It is the horse and the duck  at 52 and 57 coins. So that means toy B will be 57 coins and toy C will be 52 coins,  like so. Now if we put this back into our second formula, A plus C plus 25 equals 150.  Put in 52, then we get A plus 52 plus 25 equals 150. When we combine them, it's A plus 77 equals  150. When we subtract, we get toy A is equal to 73 coins. So that means that the toy A is the bear.  And we know that the other toy that Hamish bought was 52 coins. So the two toys he left with were at 73 and 52 coins, which were the bear and the horse. Answer A. Question 22.  Write each of the numbers 0, 1, 2, 3, 4, 5, and 6 in the squares to make the addition on the right correct. Which digit will be in the gray square?  If we take a look at the addition, we can start putting in different values.  Now we'll start off by putting 6 in here and then 4 here, since we know that when we add 6 and 4,  we get 10. So the 0 will go here and the 1 will go here. Now we only have three numbers left,  2, 3, and 5. Now 2 and 3 are smaller than 5, so we'll put them here and here. And with 2 plus 3,  we get 5. Just to make sure, we know that 62 plus 43 gives us 105. So the digit in the gray square  is 5. Answer D. Question 23. What is the largest number of small squares that can be shaded in the  figure on the left so that no square like the one shown on the right, made of four small shaded  squares, appears in the figure? Take a closer look at the figure. We can start shading different  squares. Now the best way to shade as many squares as possible is to shade each of the squares on the outside perimeter of our figure, since these cannot make a small four shaded  square. Now we can do this one more time in the next part of the perimeter, without having  anything create a two by two shaded square. Now there is only one place left we can put a shaded  square without making the undesired square, and that is in the middle. And when we count up all  the shaded squares, we get our total, which is 21. Answer D. Question 24. Nick wrote each of the numbers  from 1 to 9 in the cells of a 3 by 3 table shown to the right. Only four of the numbers can be  seen in the figure. Nick noticed that for the number 5, the sum of the numbers in the neighboring  cells is equal to 13. Now that neighboring cells are cells that share a side. He noticed that the  same is also true for the number 6. Which number did Nick write in the shaded cell? So if we look  at the figure, we have cells with 1, 2, 3, and 4 filled in. Now we can notice that 1 plus 4  and 2 plus 3 are both equal to 5. So we know that the neighboring cells of those cell with 5 and 6  will have a total of 13. If we put 5 and 6 in the spots next to 1 and 4, and then next to 2 and 3,  then they will both have a total of 5, and we'll have to add one more cell to be next to them.  So we know that 13, the total number of sum of cells minus 5, will give us 8. So if we put 5  next to the 2 and the 3, and 6 next to 1 and the 4, then we can put 8 next to them. So that way,  the 6 is next to a 1, a 4, and an 8, which totals 13. And the 5 is next to a 2, a 3, and an 8, which also totals 13. So the number in the shaded cell must be 8. Answer D.

Video Summary

The Math Kangaroo Solutions Video Library provides solutions for math problems from the 2014 Math Kangaroo Competition for levels 3 and 4, presented by Lucas Naleskowski. Viewers are instructed to read and listen to the problem and attempt to solve it before watching the provided solution. Solutions for various questions are showcased, focusing on critical thinking and logical reasoning to solve diverse mathematical problems. Each solution highlights the step-by-step process of identifying patterns, performing arithmetic calculations, understanding geometric shapes, and utilizing logical deductions to find the correct answer, illustrated through various questions involving counting segments, arranging numbers, identifying shapes, and solving logical puzzles. The aim is to enhance the ability to tackle similar math competition problems effectively.