Math fun book answers

math fun day activities and math fun dividing fractions and fun math 5th grade worksheets
Nataliebarry Profile Pic
Nataliebarry,New Zealand,Researcher
Published Date:13-07-2017
Your Website URL(Optional)
Y Ya ak ko ov v Perelman can be f mir Pubishers Moscow1 1 B Brain rain t teasers easers l lunch unch It was raining .... We had just sat down for lunch at our holiday home when one of the guests asked us whether we would like to hear what had happened to him in the morning. Everyone assented, and he began. 1 A Squirrel in the Glade "I had quite a bit of fun playing hide-and-seek with a squirrel," he said. "You know that little round glade with a lone birch in the centre? It was on this tree that a squirrel was hiding from me. As I emerged from a thicket, I saw its snout and two bright little eyes peeping from behind the trunk. I wanted to see the little animal, so I started circling round along the edge of the glade, mindful of keeping the distance in order not to scare it. I did four rounds, but the little cheat kept backing away from me, eyeing me suspiciously from behind the tree. Try as I did, I just could not see its back." "But you have just said yourself that you circled round the tree four times," one of the listeners interjected. "Round the tree, yes, but not round the squirrel." "But the squirrel was on the tree, wasn't it?" "So it was." "Well, that means you circled round the squirrel too." "Call that circling round the squirrel when I didn't see its back" 13Part One Figures for Fun. "What has its back to do with the whole thing? The squirrel was on the tree in the centre of the glade and you circled round the tree. In other words, you circled round the squirrel." "Oh no, I didn't. Let us assume that I'm circling round you and you keep turning, showing me just your face. Call that circling round you?" "Of course, what else can you call it?" "You mean I'm circling round you though I'm never behind you and never see your back?" "Forget the back You're circling round me and that's what counts. What has the back to do with it?" "Wait. Tell me, what's circling round anything? The way 1 understand it, it's moving in such a manner so as to see the object I'm moving around from all sides. Am I right, professor?" He turned to an old man at our table. "Your whole argument is essentially one about a word," the professor replied. "What you should do first is agree on the definition of 'circling'. How do you understand the words 'circle round an object'? There are two ways of understanding that. First, it's moving round an object that is in the centre of a circle. Secondly, it's moving round an object in such a way as to see all its sides. If you insist on the first meaning, then you walked round the squirrel four times. If it's the second that you hold to, then you did not walk round it at all. There's really no ground for an argument here, that is, if you two speak the same language and understand words in the same way." "All right, I grant there are two meanings. But which is the correct one?" "That's not the way to put the question. You can agree about anything. The question is, which of the two meanings is the more generally accepted? In my opinion, it's the first, and here's why. The sun, as you know, does a complete revolution in a little more than 25 days.... "Does the sun revolve?" Part One "Of course, it does, like the earth around its axis. Just 14Chapter 1 Brain-Teasers for Lunch 1 imagine, for instance, that it would take not 25 days, but 365 / 4 days, i. e. a whole year, to do so. If this were the case, the earth would see only one side of the sun, that is, only its 'face'. And yet, can anyone claim that the earth does not revolve round the sun?" "Yes, now it's clear that I circled round the squirrel after all." "I've a suggestion, comrades" one of the company 15Part One Figures for Fun. shouted. "It's raining now, no one is going out, so let's play riddles. The squirrel riddle was a good beginning Let each think of some brain- teaser." "I give up if they have anything to do with algebra o: geometry," a young woman said. "Me too," another- joined in. "No, we must all play, but we'll promise to refraii from any algebraical or geometrical formulas, except perhaps, the most elementary ones. Any objections?" "None" the others chorussed. "Let's go." "One more thing. Let the professor be our judge." School groups 2 "We have five extra-curricular groups at school, a schoolboy began. "They're fitters', joiners', photo graphic, chess, and choral groups. The fitters' grouf meets every other day, the joiners' every third day, the photographic every fourth day, the chess every fifth da\ and the choral every sixth day. These five groups firs met on January 1 and thenceforth meetings were helc according to schedule. The question is, how many time; did all five groups meet on the same day in the firs quarter (January 1 excluded)?" "Was it a Leap Year?" "No." "In other words, there were 90 days in the firs' quarter." "Right." "Let me add another question," the professor broke in. "How many days were there when none of the groups met in the first quarter?" "So, there's a catch to it There won't be another da when all the five groups meet, and there won't be one day when none meet. That's clear" "Why?" "Don't know. But I've a feeling there's a catch." "Comrades" said the man who had suggested the game. "We won't reveal the results now. Let's have more 16Chapter 1 Brain-Teasers for Lunch time to think about them. The professor will anounce the answers at supper." A Log Problem 3 "It happened at a summer cottage. A household problem, you might call it. The cottage is shared by three persons-let's call them X, Y, and Z. It's an old house with an old-fashioned cooking stove. X put three logs into the stove, Y added five and Z, who had no firewood, paid them eight kopecks as her share. How were X and Y to divide the money?" "Equally," some hastened to say. "Z used the fire produced by logs supplied by both, didn't she?" "You're wrong," another protested. "X and Y invested, as it were, different amounts of wood. Therefore, X should .receive three kopecks and Y the rest. That would be fair, in my opinion." "Well, you have a lot of time to think about it," the professor said. "Who's next?" Who Counted More? 4 "Two persons, one standing at the door of his house and the other walking up and down the pavement, were counting passers-by for a whole hour. Who counted more?" "Naturally the one walking up and down," said somebody at the end of the table. "We'll know the answer at supper," the professor said. "Next" Grandfather and Grandson 5 In 1932 I was as old as the last two digits of my birth year. When I mentioned this interesting coincidence to my grandfather, he surprised me by saying that the same applied to him too. I thought that impossible...." "Of course that's impossible," a young woman said. "Believe me, it's quite possible and grandfather proved it too. How old was each of us in 1932?" 17Part One Figures for Fun. Railway Tickets 6 "I'm a railway ticket seller," said the next person, a young lady. "People think this job is easy. They probably have no idea how many tickets one has to sell, even at a small station. There are 25 stations on my line and I have to sell different tickets for each section up and down the line. How many different kinds of tickets do you think I have at my station?" "Your turn next," the professor said to a pilot 7 A Helicopter s Flight "A helicopter took off from Leningrad and flew north. After five hundred kilometres it turned and flew 500 kilometres east. After that it turned south and covered another 500 kilometres. Then it flew 500 kilometres west, and landed. The question is, where did it land: west, east, north, or south of Leningrad?" "That's an easy one," someone said. "Five hundred steps forward, 500 to the right, 500 back, and 500 to the 18Chapter 1 Brain-Teasers for Lunch left, and you're naturally back where you'd started from" "Easy? Well then, where did the helicopter land?" "In Leningrad, of course. Where else?" "Wrong" "Then I don't understand." "Yes, there's some catch to this puzzle," another joined in. "Didn't the helicopter land in Leningrad?" "Won't you repeat your problem?" The pilot did. The listeners looked at each other. "All right," the professor said. "We have enough time to think about the answer. Let's have the next one now." 8 Shadow "My puzzle," said the next man, "is also about a helicopter. What's broader, the helicopter or its perfect shadow?" "Is that all?" "It is." "Well, then. The shadow is naturally broader than the helicopter: sun-rays spread fanlike, don't they?" "I wouldn't say so," another interjected. "Sun-rays are parallel to each other and that being so, the helicopter and its shadow are of the same size." "No, they aren't. Have you ever seen rays spreading from behind a cloud? If you have, you've probably noticed how much they spread. The shadow of the helicopter must be considerably bigger than the helicopter itself, just as the shadow of a cloud is bigger than the cloud itself." "Then why is it that people say that sun-rays are parallel to each other? Seamen, astronomers, for instance." The professor put a stop to the argument by asking the next person to go ahead with his conundrum. 9 Matches The man emptied a box of matches on the table and 19Part One Figures for Fun. divided them into three heaps. "You aren't going to start a bonfire, are you?" someone quipped. "No, they're for my brain-teaser. Here you are-three uneven heaps. There are altogether 48 matches. I won't tell how many there are in each heap. Look well. If I take as many matches from the first heap as there are in the second and add them to the second, and then take as many from the second as there are in the third and add them to the third, and finally take as many from the third as there are in the first and add them to the first-well, if I do all this, the heaps will all have the same number of matches. How many were there originally in each heap?" The "Wonderful" Stump 10 "My puzzle is the one I was once asked by a village mathematician to solve," the next person began. "It was really a story, and quite humorous at that. One day, a peasant met an old man in a forest. The two fell into a conversation. The old man looked at the peasant attentively and said: '"There's a wonderful little stump in this forest. It helps people in need.' "'Does it? What does it do, cure people?' "'Not exactly. It doubles one's money. Put your \ pouch among the roots of the stump, count one hundred and-presto-the money's doubled. It's a wonderful stump, that' ' "Can I try it?' the peasant asked excitedly. ' "Why not? Only you must pay.' ' "Pay whom and how much?' " 'The man who shows you the stump. That's me. As to how much, that's another matter.' "The two men began to bargain. When the old man learned that the peasant did not have much money, he agreed to take 1 ruble 20 kopecks every time the money doubled. 20Chapter 1 Brain-Teasers for Lunch "The two went deep into the forest where, after a long search, the old man brought the peasant to a moss-covered fir stump in bushes. He then took the peasant's pouch and shoved it among the roots. After that they counted one hundred. The old man took a long time to find the pouch and returned it to the peasant. "The latter opened the pouch and, lo The money really had doubled He counted off the ruble and 20 kopecks, as agreed upon, and asked the old man to repeat the whole thing. "Once again they counted one hundred, once again the old man began his search for the pouch and once again there was a miracle-the money had doubled again. And just as they had agreed, the old man got his ruble and 20 kopecks. "Then they hid the pouch for the third time and this time too the money doubled. But after the peasant had paid the old man his ruble and 20 kopecks, there was nothing left in the pouch. The poor fellow had lost all his money in this process. There was no more money to be doubled and he walked off crest- fallen. "The secret, of course, is clear to all-it was not for nothing that the old man took so long to find the pouch. But there is another question I would like to ask you: how much did the peasant have originally?" 11 The December Puzzle "Well, comrades," began the next man. "I'm a linguist, not a mathematician, so you needn't expect a mathematical problem. I'll ask you one of another kind, one close to my sphere of activity. It's about the calendar." "Go ahead." "December is the twelfth month of the year. Do you know what the name really means? The word comes from the Greek 'deka'-ten. Hence, decalitre which means ten litres, decade-ten days, etc. December, to all appearances, should be the tenth month and yet it isn't. How d'you explain that?" 21Part One Figures for Fun. An Arithmetical Trick 12 "I'll give you an arithmetical trick and ask you to explain it. One of you-you, professor, if you like-write down a three-digit number, but don't tell me what it is." "Can we have any noughts in it?" "I set no reservations. You can write down any three numerals you want." "All right, I've done it. What next?" "Write the same number alongside. Now you have a six- digit number." "Right." "Pass the slip to your neighbour, the one farther away from me, and let him divide this six-digit number by seven." "It's easy for you to say that, and what if it can't be done?" "Don't worry, it can." "How can you be so sure when you haven't seen the number?" "We'll talk after you've divided it." "You're right. It does divide." "Now pass the result to your neighbour, but don't tell me what it is. Let him divide it by 11." "Think you'll have your own way again?" "Go ahead, divide it. There'll be no remainder." "You're right again. Now what?" "Pass the result on and let the next man divide it, say, by 13." "That's a bad choice. There are very few numbers that are divisible by 13. You're certainly lucky, this one is" "Now give me the slip, but fold it so that I don't see the number." Without unfolding the slip, the man passed it on to the professor. "Here's your number. Correct?" "Absolutely." The professor was surprised. "That is the number I wrote down .... Well, everyone has had his turn, the rain has stopped, so let's go out. We'll know the answers tonight. You may give me all the slips then." 22Chapter 1 Brain-Teasers for Lunch Answers 1 to 12 The squirrel puzzle was explained earlier, so we'll 1 pass on to the next. We can easily answer the first question: how many 2 times did all the five groups meet on the same day in the first quarter (January 1 excluded) by finding the least common multiple of 2, 3, 4, 5, and 6. That isn't difficult. It's 60. Therefore, the five will all meet again on the 61st day-the fitters' group after 30 two-day intervals, the joiners' after 20 three-day intervals, the photographic after 15 four-day intervals, the chess after 12 five-day intervals and the choral after 10 six-day intervals. In other words, they can meet on the same day only once in 60 days. And since there are 90 days in the first quarter, it means there can be only every other day on which they will meet. It is much more difficult to find the answer to the second question: how many days are there when none of the groups meet in the first quarter? To find that, it is necessary to write down all the numbers from 1 to 90 and then cross out all the days when the fitters' group meets: e. g. 1, 3, 5, 7, 9, etc. After that one must cross out the joiners' group's days: e.g. 4, 7, 10, etc. When all the photographic, chess and choral groups' days have also been crossed out, the numbers that remain are the days when there is no group meeting. Do that and you'll see that there are 24 such days-eight in January, i. e. 2, 8,12,14,18, 20, 24, and 30, seven in February and nine in March. It is not right to think, as many do, that eight 3 3 kopecks were paid for eight logs, a kopeck for a log. The money was paid for one-third of eight logs because the fire they produced was used equally by all three. Consequently, the eight logs were estimated to be worth 8 x 3, i. e. 24 kopecks, and the price of a log was therefore three kopecks. 23Part One Figures for Fun. It is now easy to see how much was due to each. Y's five logs were worth 15 kopecks and since she had used 8 kopecks worth of fire, she would have to receive 15 — 8, i. e. 7 kopecks. X would have to receive 9 kopecks, but if you subtracted the 8 kopecks due from her for using the stove, you would see that she had to receive 9— 8, i. e. 1 kopeck. Both of them counted the same number of passers- 4 by. While the one who stood at the door counted all those who passed both ways, the one who was walking counted all the people he met going up and down the pavement. There is another way of putting it. When the man who was walking and counting the passers-by returned for the first time to the man who was standing at the door, they had counted the same number of passers-by - all those passing by the standing man encountered the walk¬ing man either on the way there or back. And each time the one who was walking was returning to the one who was standing he counted the same number of passers-by. It was the same at the end of the hour, when they met for the last time and told each other the final number. At first it may seem that the problem is incorrectly 5 worded, that both grandfather and grandson are of the same age. We shall soon see that there is nothing wrong with the problem. It is obvious that the grandson was born in the 20th century. Therefore, the first two digits of his birth year are 19 (the number of hundreds). The other two digits added to themselves equal 32. The number therefore is 16: the grandson was born in 1916 and in 1932 he was 16- The grandfather, naturally, was born in the 19th century. Therefore, the first two digits of his birth year are 18. The remaining digits multiplied by 2 must equal 132. The number sought is half of 132, i. e. 66. The grand- 24Chapter 1 Brain-Teasers for Lunch father was born in 1866 and in 1932 he was 66. Thus, in 1932 the grandson and the grandfather were each as old as the last two digits of their birth years. At each of the 25 stations passengers can get tickets 6 for any of the other 24. Therefore, the number of different tickets required is: 25 x 24 = 600. There may also be round-trip tickets. If so, the number doubles and there are then 1200 different tickets. There is nothing contradictory in this problem. The 7 helicopter did not fly along the perimeter of a square. It should be borne in mind that the earth is round and that the meridians converge at the poles (Fig. 2). Therefore, when the helicopter flew the 500 kilometres along the parallel 500 kilometres north of Leningrad, it covered more degrees going eastward than it did when it was returning along the Leningrad latitude. As a result, the helicopter completed its flight east of Leningrad How many kilometres away? That can be calculated. Figure 3 shows the route taken by the helicopter: ABCDE. N is the North Pole where meridians AB and DC meet. The helicopter first flew 500 kilometres northward, i. e. along meridian AN. Since the degree of a meridian is 111 kilometres long, the 500-kilometre-long arc of the meridian is equal to 500 1114°5'. Leningrad lies on the 60th parallel. B, therefore, is on 60° + 4°5' - 64°5'. The helicopter then flew eastward, i. e. along the BC parallel, covering 500 kilometres. The length of one degree of this parallel may be calculated (or learned from tables); it is equal to 48 kilometres. Therefore, it is easy to determine how many degrees the helicopter covered in its eastward flight: 500 -f- 48 x 10°4'. Continuing, the aircraft flew south¬ward, i. e. along meridian CD, and, having covered 500 kilometres, returned to the Leningrad parallel. Thence the way lay westward, i. e. along DA; the 500 kilometres 25Part One Figures for Fun. of this way are obviously less than the distance between A and D. There are as many degrees in AD as in BC, i. e. 10°4'. But the length of 1° at the 60th parallel equals 55.5 kilometres. Therefore, the distance between A and D is equal to 55.5 x 10.4 = 577 kilometres. We see thus that the helicopter could not have very well landed in Leningrad: it landed 77 kilometres away, on Lake Ladoga. Fig 2 Fig 3 26Chapter 1 Brain-Teasers for Lunch In discussing this problem, the people in our story made 8 several mistakes. It is wrong to say that the rays of the sun spread fanlike noticeably. The Earth is so small in comparison to its distance from the Sun that sun-rays falling on any part of its surface radiate at an almost absolutely imperceptible angle; in fact, rays may be said to be parallel to each other. We occasionally see them spreading fanlike (for instance, when the sun is behind a cloud). This, however, is nothing but a case of perspective. Parallel lines, as they recede from the station point, always appear to the eye to meet far away at a point, e. g. railway tracks or a long avenue. But the fact that sun-rays fall to the ground in parallel beams does not mean that the perfect shadow of a helicopter is as broad as the helicopter itself. Figure 3 shows that the perfect shadow of the helicopter narrows down in space on the way to the surface of the earth and that, consequently, the shadow the helicopter casts should be narrower than the helicopter, i. e. CD is shorter than AB. It is quite possible to compute the difference, provided, of course, we know at what altitude the helicopter is flying. Let us assume that the altitude is 100 metres. The angle formed by lines AC and BD is equal to the angle from which the sun is seen from the earth. We know that this angle is equal to 1/2°. On the other hand, we know that the distance between the eye and any object seen from an angle of 1/2° is equal to the length of 115 diameters of this object. Hence, section MN (the section seen from the surface of the earth at an angle of 1/2°) should be 1/115 of AC. Line AC is longer than the perpendicular distance between point A and the surface of the earth. If the angle formed by the sun's rays and the surface of the earth is equal to 45°, then AC (given the altitude of the helicopter is 100 metres) is approximately 140 metres long and section MN is consequently equal to 140 / 115 = 1.2 metres. But then the difference between the helicopter and its shadow, i. e. section MB, is bigger than MN (1.4 times, to be exact), because angle MBD is almost equal to 45°. Therefore, MB is equal to 1.2 x 1.4, and that gives us almost 1.7 metres. 27Part One Figures for Fun. But then the difference between the helicopter and its shadow, i. e. section MB, is bigger than MN (1.4 times, to be exact), because angle MBD is almost equal to 45°. Therefore, MB is equal to 1.2 x 1.4, and that gives us almost 1.7 metres. All this applies to the perfect- black and sharp- shadow of the helicopter, and not to the penumbra, which is weak and hazy. Incidentally, our computation shows that if instead of a helicopter we had a small balloon about 1.7 metres in diameter, there would be no perfect shadow. All we would see would be a hazy penumbra. This problem is solved from the end. Let us proceed 9 from the fact that, after all the transpositions, the number of matches in each heap is the same. Since the total number of matches (48) has not changed in the process, it follows that there were 16 iri each heap. And so, what we have in the end is: First Heap Second Heap Third Heap 16 16 16 Immediately before that we had added to the first heap as many matches as there were in it, i.e. we had doubled the number. Thus, before that final trans¬position, there were only 8 matches in the first heap. In the third heap, from which we took these 8 matches, there were: 16 + 8 = 24 Now we have the following numbers: First Heap Second Heap Third Heap 8 16 24 Further, we know that from the second heap we took as many matches as there were in the third heap. That means 24 was double the original number. This shows us how many matches we had in each heap 28Chapter 1 Brain-Teasers for Lunch after the first transposition: First Heap Second Heap Third Heap 8 16+12-28 12 It is clear now that before the first transposition (i. e. before we took as many matches from the first heap as there were in the second and added them to the second) the number of matches in each heap was: First Heap Second Heap Third Heap 22 14 12 This riddle, too, can best be solved in reverse. We know that 10 when the money was doubled for the third time, there was 1 ruble 20 kopecks in the pouch (that is the sum the old man received the last time). How much was there in the pouch before that? Obviously 60 kopecks. That was what remained after the peasant had paid the old man his second ruble and 20 kopecks. Therefore, before the payment there was: 1.20 + 0.60 = 1.80. Further: 1 ruble 80 kopecks was the sum after the money had been doubled for the second time. Before that there were only 90 kopecks, i.e. what remained after the peasant had paid the old man his first ruble and 20 kopecks. Hence, before the first payment there were 0.90+ 1.20 = 2.10 in the pouch. That was after the first operation. Originally, therefore, there was half that amount, or 1 ruble 5 kopecks. That was the sum with which the peasant had started his unsuccessful get-rich-quick operation. Let us verify: Money in the pouch: After the first operation . . . 1.05 x 2 = 2.10 After the first payment . . . 2.10- 1.20 = 0.90 After the second operation . . 0.90 x 2 = 1.80 After the second payment . . . 1.80- 1.20 = 0.60 After the third operation . . . 0.60 x 2 = 1.20 After the third payment . . . 1.20- 1.20 = 0. 29Part One Figures for Fun. Our calendar comes from the early Romans who, before 11 Julius Caesar, began the year in March. December was then the tenth month. When New Year was moved to January 1, the names of the months were not shifted. Hence the disparity between the meaning of the names of certain months and their sequence: Month Meaning Place September 9th (septem- seven) October 10th (octo - eight) November 11th (novem nine) December 12th (deka - ten) Let us see what happened to the original 12 12 number. First, a similar number was written alongside it. That is as if we took a number, multiplied it by 1 000 and then added the original number, e.g.: 872 872 = 872 000 + 872. It is clear that what we have really done was to multiply the original number by 1 001. What did we do after that? We divided it in turn by 7, 11, and 13, or by 7 x II x 13, i.e. by 1001. So, we first multiplied the original by 1 001 and then divided it by 1001. Simple, isn't it? ? ? ? Before we close our chapter about the brain- teasers at the holiday home, I should like to tell you of another three arithmetical tricks that you can try on your friends. In two you have to guess n umbers and in the third, the owners of certain objects. These tricks are very old and you probably know them well, but I am not sure that all people know what they are based on. And if you do not know the theoretic basis of tricks, you cannot expect to unravel 30Chapter 1 Brain-Teasers for Lunch them. The explanation of the first two will require an absolutely elementary knowledge of algebra. The Missing Digit 13 13 Tell your friend to write any multi-digit number, say, 847. Ask him to add up these three digits (8 + 4 + + 7) = 19 and then subtract the total from the original. The result will be: 847 - 19 = 828. Ask him to cross out any one of the three digits and tell you the remaining ones. Then you tell him the digit he has crossed out, although you know neither the original nor what your friend has done with it. How is this explained? Very simply: all you have to do is to find the digit which, added to the two you know, will form the nearest number divisible by 9. For instance, if in the number 828 he crosses out the first digit (8) and tells you the other two (2 and 8), you add them and get 10. The nearest number divisible by 9 is 18. The missing number is consequently 8. How is that? No matter what the number is, if you subtract from it the total number of its digits, the balance will always be divisible by 9. Algebraically, we can take a for the number of hundreds, b for the number of tens and c for the number of units. The total number of units is therefore: 100a + 10 b + c. From this number we subtract the sum total of its digits a + b + c and we obtain: 100a + 10& + c - (a + b + c) = 99a + 9b = 9 (11a + b). But 9 (11a + i) is, of course, divisible by 9. There¬fore, when we subtract from a number the sum total of its digits, the balance' is always divisible by 9. It may happen that the sum of the digits you are told is divisible by 9 (for example, 4 and 5). That shows that the digit your friend has crossed out 31

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.