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mathematics
Maths, the horror of (almost) every student ! (but the Bible says: you are to love also your enemies!) - or expressed with a nice parable (the German word �Parabel� also means parabola*):
Many years back a Chinese emperor intended to conquer the country of his enemies and destroy them all.
They saw him later eating and joking with his enemies.
" Didn�t you intend to destroy your enemies?�, they asked highly surprised. The emperor answered: " I destroyed my enemies. I made them my friends. "
(*a parabola is, by the way, a mathematical, more exactly geometrical item: a curve, which - after a turn - runs back into the direction it came from, thus forming a type of " U ", which continuously gets wider at the open side; the formula of the standard parabola being: y = x�)
1. The football
This one might be a standard story of many maths teachers: 3 boys buy a football. According to the price tag in the shop window it costs 30 �. They share the price, i. e. each boy puts 10 � on the counter. Afterwards the shop owner realizes that the ball cost only 25 � and assigns his apprentice shop assistant, who happens to know the boys, to return to them the difference of 5 �. Since the apprentice has a mathematical problem in how to divide 5 by 3 he returns 1 � to each boy = 3 �), so that the boys paid only 27 � (each 9 �) and keeps 2 � for himself. The maths teacher�s triumphant question: Where is the missing 1 � ? (According to the (German) slogan: The expert remains surprised, the layman just wonders why! )
2. The rope
A further highlight of each maths teacher besides his maths books is the following: If a rope was put around the globe at the equator (circumference: approx. 40,076 km) but was 1 km too long, how far would be away from the earth on its whole length ? (one must know actually only the formula of circle circumference.)
3. The total
The following anecdote is told of the great German mathematician Carl Friedrich Gauss: His teacher had to leave the class urgently for a few minutes. He therefore presented an arithmetic problem to his pupils, so that they would be busy for a while: They were to add all numbers from 1 to 100. The teacher was just about to leave the classroom, when the young Gauss presented him the solution in his exercise book: 5050. The teacher was astonished. Since he had already let another class do the job count, he knew that the result was correct. How had Carl calculated it so fast?
Solution 1: The football
The task contains a mean mistake in reasoning. The question is misleading. Actually the thing is that: the 3 boys paid (30 � minus 3 � back = ) 27 �. If the shop assistant had returned the 2 � to them, too (i.e. each 2/3 = 0.67 �), they would have paid (27 - 2 =) 25 �, as much as the ball actually cost. (Each boy would have paid: 8,33 �) (multiplied by 3 = 24,99 �). (additional question: Where is the missing 1 Cent ?? O.K., you�re smart: of course, it�s a �mistake� in rounding).
Solution 2: The rope
The distance between the rope and the globe is the difference between the radius (= half diameter) of the rope positioned around the earth which is too long by 1 km (R) and the radius of the earth (r). With the formula U = d � π ("pi") or (because of: d = 2 times r) U = 2 � r � π (conversion: r = U : 2 π) you could calculate with a good pocket calculator R - r = (U rope divided by 2 π) minus ( U earth divided by 2 π) = (U rope minus U earth ): 2 π. But here one realizes without calculation that independently of the size of the ball or the length of the rope: radius distance = circumference divided by 2 π (2 π = approx. 6.283). That means with a rope extension of 1 km everywhere along the equator the rope would be 1000 m divided by 6,283 = 159.16 m from the ground ! (height of Cologne cathedral!), and an extension of only 1 meter would still produce a distance everywhere along the equator of 15,9 cm !
Solution 3: The total
Gauss may have thought: adding the first and the last number (1 + 100) results in 101, likewise the second and last but one (2 + 99), and so on. Altogether there are 50 "pairs" (= half of the total of 100 number ), that all result in 101. As a formula: total = (first number + last number) multiplied by (amount of numbers)/2 or : (a + b) � n/2 .
Example 1: Total of the numbers from 1 to 1000 : (1 + 1000) � 1000/2 = 1001 � 500 = 500500.
Example 2: Total of the numbers from 100 to 200 : (100 + 200) � (200 - 99)/2 = 300 � 101/2 (= 300 � 50.5) = 15150.
4. The age
You ask someone to tell you in which columns of the following table his or her age (only up to the age of 63 years!) (or any number between 1 and 63) is located. With a quick glance view at the table (with something exercise also without) you can say immediately, how old that person is (or which number he had chosen).
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c6.
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c5.
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c4.
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c3.
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c2.
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c1.
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32
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16
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8
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4
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2
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1
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33
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17
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9
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5
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3
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3
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34
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18
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10
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6
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6
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5
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35
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19
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11
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7
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7
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7
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36
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20
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12
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12
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10
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9
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37
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21
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13
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13
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11
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11
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38
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22
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14
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14
|
14
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13
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39
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23
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15
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15
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15
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15
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40
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24
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24
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20
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18
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17
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41
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25
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25
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21
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19
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19
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42
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26
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26
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22
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22
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21
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43
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27
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27
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23
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23
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23
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44
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28
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28
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28
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26
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25
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45
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29
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29
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29
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27
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27
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46
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30
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30
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30
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30
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29
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47
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31
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31
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31
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31
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31
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48
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48
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40
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36
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34
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33
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49
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49
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41
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37
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35
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35
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50
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50
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42
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38
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38
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37
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51
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51
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43
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39
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39
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39
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52
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52
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44
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44
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42
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41
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53
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53
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45
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45
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43
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43
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54
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54
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46
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46
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46
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45
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55
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55
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47
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47
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47
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47
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56
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56
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56
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52
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50
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49
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57
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57
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57
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53
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51
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51
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58
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58
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58
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54
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54
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53
|
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59
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59
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59
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55
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55
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55
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60
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60
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60
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60
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58
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57
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61
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61
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61
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61
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59
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59
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62
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62
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62
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62
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62
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61
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63
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63
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63
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63
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63
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63
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Table (download)
5. The chess match
From the orient comes a story that many years ago in an Arab country an old wise man lived in poverty, who was known for being unbeaten in the game of chess. One day the Sultan heard of him, who judged himself being the best chess player in the country. So he invited the old man to a memorable chess match in his palace. In order to prevent the old person from letting the Sultan win due to a sense of modesty and humility towards his ruler, the Sultan determined the following conditions: The old person would in the case of his defeat be beheaded, in the case of a victory, however, he would have the right to any desire. How did the match end? The old man won. The anger of the Sultans was somewhat moderated by the fact that the desire of the old person was apparently extremely modest: He just required 1 grain of rice on the first chessboard field, the double quantity, thus 2 grains on the 2nd field, again the double quantity, i. e. 4 grains on the 3rd field, 8 on the 4th field, 16 on 5th, and so on. But could the Sultan really meet the desire of the old person in the end?
6. The tower to the moon
If one folds a DIN A4 format sheet ( 29.7 cm times 21 cm ) or cuts it into 2 sections and puts the sections one on top of the other, the pile will be twice as high and has now the format of DIN A5. How often would one - theoretically - have to fold or cut 1 sheet of writing paper (80 g/m�) of the DIN A4 format, and put the piles one on top of the other, until a pile develops, which reaches beyond the distance between the earth to the moon? Which DIN A format would that be (theoretically, as the smallest official DIN A format is DIN A10)?
Solution 4: The age
Assertion : We normally write numbers in the decimal system (one-digit numbers, two-digit numbers, three-digit numbers, etc., i. e.: with every �times 10� there is going to be a new digit in front of it). One can write numbers also in "binary code" (i. e. with every �times 2� (doubling) there is going to be a new digit in front of it) (computers operate on this basis). The 1st (smallest) binary number is 1, the 2nd is 2, the 3rd is 4, the 4th is 8, the 5th is 16, the 6th is 32, etc. (these, by the way, being the numbers of the very first line of the table). One can convert a whole decimal number into binary notation, by dividing them into binary numbers (1, 2, 4, 8, 16, 32, etc.). One begins with the largest binary number contained in the decimal number. One writes from the left to the right. All binary number places are considered. For binary numbers which are contained in the decimal number or the rest, the code receives a 1, for such, which are not contained, a 0 (zero). (a binary code always begins with a 1, i.e. a 0 for binary numbers, which are larger than the decimal number and therefore not contained, are not written.)
Example 1 : the number 25 (largest binary number contained: 16) = 1 time 16 + 1 time 8 + 0 times 4 + 0 times 2 + 1 time 1 � binary code: 11001
Example 2 : the number 45 (largest binary number contained: 32) = 1 time 32 + 0 times 16 + 1 time 8 + 1 time 4 + 0 times 2 + 1 time 1 � binary code: 101101
In the table now - unexpectedly - the decimal numbers from 1 to 63 are located in those columns, in which they would have a 1 when written in binary code!
Solution : If the asked one says, his age or number is written in columns 6, 4 and 1, it tells you, he means the number with the binary code 101001. Since the binary numbers are located in the first line of the table, the decimal number in question can be calculated fast by adding the binary numbers of the columns mentioned, also without knowledge of the binary notation, for example
101001: 32 + 8 + 1 = 41 !
Solution 5: The chess match
On the 1st field 1 grain of rice was to lie, 2 of them on the 2nd, 4 (= 2�) on the 3rd, 8 on that 4th (= 2�), etc., 262 on the 63rd, 263 on the 64. (263 = 9,223,372,036,854,775,808! ). The total of all grains would be 1 (= 20) + 2 (= 21) + 22 + 23 + 24..... + 263 = double the number of grains on the last field minus 1 =
--> 18,446,744,073,709,551,615
Solution 6: The tower to the moon
A sheet of paper with 80g/m� has a thickness ("height") of approx. 0.106 mm (e.g. to be calculated by measuring a pile with 500 sheets and dividing the height by 500). After the 1st division the pile is 0.212 mm high (DIN A5), after the 2nd division (DIN A6) 0.424 mm, after the 3rd division (DIN A7) 0.848 mm, after the 4th division (DIN A8) 1.696 mm, etc. The distance between earth and moon amounts to approx. 384,000 km. A calculation results in that after the 41st division ("DIN A45" !) the paper tower made from that 1 DIN-A4 sheet would have reached a height of 233 095.9 km and after the 42nd division ("DIN A46" !) a height of 466 191.8 km, thus exceeding the distance between earth and moon by approx. 80,000 km ! This "tower" would by the way consist of 4,398,046,511,000 paper particles, which would be naturally much higher than wider (height, like said, 0.106 mm; width 0.0001 mm and 0.000141 mm, the proportions of the sides of the DIN-A format always being 1:1,414! ), i.e. approx. 883 times as high as wide. By the way: This gigantic total of 4,398,046,511,000 particles would be in the case of the �grains on a chessboard� only the quantity of rice grains on the 42nd field !!!
7. The roller and the snail
A road roller drives along a road 6 m wide and covers a quarter of a meter in 24 seconds. a snail 50 meters ahead begins to cross the road with a speed of 8 cm per minute. Question: Will the road roller crush the snail?
8. The ancestors
In the year 1 A.D. ( there was no year 0 ! ) Lydia and Alexander had a child. For them it was the 2nd generation. Under the prerequisite that his ancestors always had their first child (the following generation) on the average at the age of 20, Lucas, their descendant of the 102nd generation, would be born in the year 2001 (generations: in the year 20 A.D.: no. 3, in the year 40 A.D.: no. 4, in the year 60 A.D.: no. 5 generation, and so on). Lucas has 1 father and 1 mother. Those also have or had 1 father and 1 mother (Lukas has or had 2 grandfathers and 2 grandmothers), and so on. Question: How many greatgreatgreat... grandfathers and - mothers (99 times great !) of Lucas lived in the year 1 A.D. (including Lydia and Alexander)?
9. The foil
Aluminum foil is producible in a thickness of 0.06 mm (= 60 micrometers). The usual kitchen aluminum foil has a width of 29,7 cm (= the long side of the DIN A4 format! Why that? ). How many meters of kitchen foil could be obtained from 1 kg of aluminum (density of aluminum: 2.7 grams per cubic centimeter)? How much does a kitchen foil role of 30 m weigh (without the inner cardboard role!)?
Solution 7: The roller and the snail
Twice a typical proportional function (rule of three): The necessary time is wanted in both cases. Roller : 25 cm in 24 s ; 5000 cm in ? s calculation: 24 : 25 (= time for 1 cm) � 5000 = 4800 s = (: 60) 80 min = 1 h 20 min. The roller reaches the "snail crossing" after 1 hour and 20 minutes. To the snail applies: 8 cm in 1 min ; 600 cm in ? min calculation: 1 : 8 (= time for 1 cm) � 600 = 75 min = 1 h 15 min . The snail already crossed the road after 1 hour and 15 minutes, thus 5 minutes before the roller reached the place. In order to determine, where the roller is located, at the very moment the snail is in security, you must calculate: The roller needs 24 s for 25 cm ; 75 min = (times 60) 4500 s for ? m calculation: 25 : 24 � 4500 = 4687.5 cm = 46.88 m (i.e. the distance to the snail�s tail�s end amounted to still safe 50 - 46.88 = 3.12 m)
Solution 8: The ancestors
In the year 1 AD lived 2 to the power of 101 greatgreatgreat.... - grandfathers and - mothers. This number cannot be calculated simply. The mathematically less talented can resolve the power according to the power rules (2 to the power of 20 = 2 to the power of 10 � 2 to the power of 10) in: 2 to the power of 101 = (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � (2 to the power of 10) � 2 (= 2 to the power of 1). 2 to the power of 10 = 1048 = 1.048 � 1000, thus the value being 1,048 to the power of 10 � 1,000 to the power of 10 � 2 (= 2 to the power of 1). The amount of 1,048 to the power of 10 can be calculated with a pocket calculator (= approx. 1.598132658, depending upon pocket calculator accuracy). The result is to be multiplied by 1000 to the power of 10 and be multiplied by 2. The calculation reads: 1.598132658 � 1000 000,000,000,000,000,000,000,000,000 � 2 = 3,196,265,316,000,000,000,000,000,000,000 ! Actually, at the most a few millions humans lived in the whole world in the year 1 A.D.! The unbelievably large number is to be understood only in such a way that even over the relatively small period of 2,000 years almost everyone is actually related to everyone else!
Solution 9: The foil
The volume formula for a right parallelepiped is needed: volume = height � width � length or : V = a � b � c . Since the length is wanted, the formula must be changed to: length c = V : a: b or c = V: (a � b). Now V must be calculated with the formula: mass = volume � density or: m = V � ρ (rho), changed to: V = m: ρ . With: V = 1 kg : 2.7 kg/dm� = 0.37 dm� = (1 dm = 100 mm , therefore 1 dm� = 100 mm � 100 mm � 100 mm = 1,000,000 mm� ) 0.37 � 1,000,000 mm� = 370,000 mm�. Thus c = 370,000 mm� : (0.06 mm � 297 mm) = 370,000 mm� : 17.82 mm� = 20,763 mm = 207.63 m .
10. The wine
Out of a barrel of white wine a spoon of white wine is poured inadvertently into a red wine barrel with the same quantity. Thereupon a spoon from the barrel of red wine is given to the barrel of red wine. Is there now
a) more white wine in the white wine barrel than there is red wine in the red wine barrel, or
b) less white wine in the white wine barrel than there is red wine in the red wine barrel, or
c) as much white wine in the white wine barrel as there is red wine in the red wine barrel?
11. The disks
Out of a sheet of metal with the measures of 1 m by 1 m disk-type pieces at a diameter 10 cm are to be punched out . For technical reasons each disk must be arranged beneath each other and next to each other. Thus a relatively high loss of material develops . The disks may touch each other, also there are no margins. What is the loss of material per disk and of the whole metal sheet (in m�, dm� or cm�)? How many per cent is that? (as far as the percentage goes the task can also be performed in inches, by the way: 1 inch = 2.54 cm)
12. The ball
Balls at a diameter of 10 cm are to be packed. Unfortunately it is only possible in cube- shaped boxes. How many cm� of space is there in every box? How many per cent of the box volume is that?
Solution 10: The wine
An example calculation : The ladle seizes 1/20 of the barrel contents of 60 litres, thus 3 litres of white wine are taken out of the white wine barrel and poured into the red wine barrel. So there are 60 litres of red and 3 litres of white wine in the red wine barrel (63 litres as a mixture of 60 : 3 = 20 : 1). 1 ladle of this mixture (= 3 litres) contains (theoretically, if mixed properly) 3 � 60/63 = 20/7 litres of red and 3 � 3/63 = 1/7 litres of white wine). Thus 20/7 litres of red wine and 1/7 litres of white wine (altogether 21/7 = 3 litres) are added to the remaining 57 litres of white wine in the white wine barrel. In the white wine barrel are 57 1/7 litres of white wine (+ 20/7 litres of red wine). In the red wine barrel there are still 60 litres minus 20/7 litres = 57 1/7 litres of red wine left (+ 20/7 litres of white wine from the first ladle). That means, response c) is correct!
For friends of equations: with W = white wine, R = red wine, a= contents of barrel and b = contents of ladle -->:
inside red wine barrel: aR + bW - (aR + bW) b / a+b -->
(aR + bW)a /(a+b)
inside red wine barrel: aW - bW + (aR + Wb) b /(a+b) -->
(aW + bR)a /(a+b)
It is obvious that the relations inside the barrels are the same.
Solution 11: The disks
From the area of a square with the side length of 10 cm the area of a circle with the diameter of 10 cm must be substracted. The area of the square: A = a � a (= a�) = 10 cm � 10 cm = 100 cm�. The circular area: A = d � d � π : 4 (= d� � π : 4) (also: r� � π : 4) = 10 cm � 10 cm � 3,14 : 4 = 78.5 cm�. The difference = the waste per every disk punched out is: 100 - 78.5 = 21.5 cm�. From a sheet of 1 m� 10 � 10 = 100 disks are to be punched out, leading to a waste of 2150 cm� per sheet. The total area of 1 piece of sheet metal amounts to 100 cm � 100 cm = 10,000 cm�. 2150 of 10,000 = 2150/10 000 = 2150 : 10,000 = 0,215; multiplied by 100 % = 21.5 %. (the same percentage results naturally, if one calculates the blend of only one disk � as "per cent is per cent": 21,5 : 100 = 0,215; multiplied by 100 % = 21.5 %.
Solution 12: The ball
In principle it is similar to problem no. 11, only there being 2 dimensions involved (area) and here 3 dimensions (volume). The box has the volume: V = 10 cm � 10 cm � 10 cm = 1000 cm�. The volume of a ball (according to a maths book or book of tables or dictionary ): V = 1/6 � d�� π = 0,1666... � 1,000 cm� � 3.14 = 523.3 cm�. Thus 1,000 - 523.3 = 476.7 cm� of air in the cube- shaped box. That is 476,7 : 1,000 � 100 % = 47.7 % (thus almost half = 50 % of the volume ! )
13. The booty
Eddy wants to break into a coat shop. It needs 3 buddies for the coup. But Joe wants half of the booty, Willy 2 thirds of the remainder and Harry 3/4 of the still remaining booty. Is Harry's burglary worthwhile at all? How many coats does he have to steal at least, in order to keep 1 only for himself?
14. The pumps
After a strong tempest a large cellar has to be pumped out. 4 pumps with different performances are available: Alone, pump A would have to run 30 h , pump B 20 h, pump C 15 h and pump D 10 h to pump out the whole cellar. Therefore all 4 pumps are used at the same time. After how many hours will the cellar be pumped out?
15. The garden
A farmer wants to fence in as large a garden as possible with 1,000 m of fencing material. He heard that a square format would provide the largest enclosable (rectangular!) area (only demonstrable with differential calculus; otherwise to be seen by trying out with different values of the number). Now a mate at the table reserved for regulars has told him, if he created a circular garden, he would have a still larger fenced-in areal. Is that correct? If so, how much more fenced-in area would he have?
13. Solution: The booty
One could set up a difficult equation with x = booty, in order to calculate, how much remains for the poor Eddy: (x - 1/2 x) - (x - 1/2 x) � 2/3 - ( (x - 1/2 x) - (x - 1/2 x) � 2/3) � 3/4 = ? (remains for Eddy). However, it works more simply, according to the idea: Eddy will get: x � 1/2 � 1/3 � 1/4 = 1/24 x. Of each stolen coat Eddy will get 1/24 (a twenty-fifth) as his remaining share. That is, in order to get 1 coat to keep for himself, he must steal 24 coats, a bad deal for him!
14. Solution: The pumps
Pump A: 30 h/ V (V = volume of the water in the cellar), pump B: 20 h/ V, pump C: 15 h/ V, pump D: 10 h/ V. The pumping performances are expressed by reciprocal value formation into V/ h, then the deliveries of the pumps per hour can be added --> V/ 30 h + V/ 20 h + V/ 15 h + V/ 10 h = (after finding the common denominator considering the smallest multiple of the denominators = 60 h) --> 2V/ 60 h + 3V/ 60 h + 4V/ 60 h + 6V/ 60 h = 15V/ 60 h = (after reducing by 15) 1V/ 4 h (= 1/4 V/ h) --> (after reciprocal value formation): 4 h/ V , i. e. the 4 pumps had pumped out the cellar after 4 hours.
15. Solution: The garden
The range of a square is the quadruple side length a (U = 4 � a; changed over, since U is given: a = U : 4). The side length of a square garden would thus be with a fence length of U = 1,000 m : 250 m. The area of a square is the square of the side length (A = a � a = a� = 250 m � 250 m = 62,500 m�). For calculating the circumference of the circle the diameter has to be multiplied by π (= approx. 3.14) (U = d � π; changed over, since U is given: d = U : π). With the fence length of U = 1,000 m --> d = 1,000 m : 3.14 = 318.47 m. the area of a circle is calculated with the formula: A = π �d� / 4 = 3.14 � 318.47 m � 318.47 m : 4 = 79,617 m�. Possibly as having been expected, the round garden is larger, around 17,117 m� or 17,117/ 62,500 = 27.4 % !
16. The duplication
There is said to be a " general rule ", with whose assistance one could estimate fast, in how many years a sum of money will have doubled with a given interest rate . It is said to be a number, which is divided by the interest rate and the result gives the duration in years. How could one find this number (or prove that it might be only a rumor)?
17. The cube
A painted cube is sawed into 4 times 4 times 4 = 64 small cubes. How many of the small cubes have 3, how many have 2 , how many have 1 and how many have no painted surfaces?
18. The pumpkins
Mathematics teacher Eule is a hobby gardner. He tells his friends at the table reserved for regulars that he harvested 4 pumpkins with the total weight of 12 kg. The second weighed 50 % more than the first, the third even as much as the first two together and two and a half times as much as the fourth. How much did each of his 4 pumpkins weigh?
Solution 16: The duplication
With an interest rate of 5 % a duplication is reached not only after 20years (5% times 20 = 100%) but a while before, since the interest left in the account is paid interest on, too. One calls this the compound interest effect. One could project 2- 3 examples with different interest rates using the compound interest formula for as long as approx. a duplication of the output capital occurs and then compare duration periods of all examples. Since the compound interest formula is relatively difficult, one can calculate by hand:
Example 1: After 1 year an initial amount of 1,000 � - at an interest rate of 5% - amounts to 1000 times 1.05 (= 100% + 5% = 100/100 + 5/100 = 105/100 = 1.05) = 1,050 �. After the 2nd year: 1,050 times 1.05 = 1,102.50 �, etc. (with some pocket calculators one can press easily the times key until the result amounts to approx. 2000 = duplication of 1000) --> after 3 years: 1157,63; after 4 years: 1215,51; after 5 years: 1,276.28.; after 6 years: 1,340.10.; after 7 years: 1,407.10; after 8 years: 1,477.46; after 9 years: 1,551.33; after 10 years: 1,628.89; after 11 years: 1,710.34; after 12 years: 1,795.86; after 13 years: 1,885.65; after 14 years: 1,979.93; after 15 years: 2,078.93.
Example 2: 1,000 � at 3% (without compound interest effect: 3% times 33.3 years = 100%!): after 1 year: 1,030 �,... after 23 years: 1,973.59 �; after 24 years: 2,032.79.
Example 3: 1,000 � at 7%: after 1 year: 1,070 �; after 10 years: 1,967.15 �; after 11 years: 2,104.85.
Compilation: ex. 1 : at 5% --> approx. 14.2 years (by "intrapolation": difference between 14 and 15 years: 2,078.93 - 1,979.93 = 99 --> 2,000 - 1,979.93 = 20.07 --> 20.07/99 = 0.2 years --> 14 years + 0.2 years); ex. 2: at 3% --> approx. 23.45 years (23 years + 26.41 / 59.2 years = 23 years + 0.45 years); ex. 3: at 7% --> approx. 10.24 years (10 years + 32.85 / 137.70 years --> 10 years + 0.24 years). Check of the existence of a generally valid factor (since a small interest rate means a longer doubling period, only multiplication could lead to similar factors): ex. 1: 5 times 14.2 = 71 ex. 2: 3 time 23.45 = 70.35 ex. 3: 7 times 10.24 = 72.8 it seems as if one could count at least with one-digit interest rates on a factor of approx. 71 (the error could be situated in the area of linear interpolating), which is to be divided by the interest rate, i.e. at an interest rate of 6% assets double in (71 : 6) approx. 11.8 years.
Solution 17: The cube
A cube has 8 corners. Thus there are 8 small corner cubes with 3 painted surfaces each. Furthermore a cube has 12 edges, which consist of cubes with 2 painted surfaces, in the case of a division into 4 sections per side there would be actually 4 cubes per edge. Since the corners belong in each case to three edges and have 3 painted surfaces, they must be deducted from the edge cubes. Thus with each edge only the inner edge cubes are to be considered, i.e. in the example give there are altogether 12 �2 = 24 small edge cubes with 2 painted surfaces. Furthermore, a cube has (as well-known of cubes) 6 sides. In the example given 1 side would consist of 4 �4 = 16 cubes. But the small corner and edge cubes already considered are to be deducted. Thus there are (inside) only 4 �6 = 24 small side cubes with 1 painted surface each. All remaining small cubes (of the inside of the cube) are unlacquered. By subtracting the lacquered cubes (4 �4 �4 = a total of 64 small cubes; 64 - 8 - 24 - 24 =) there are 8 unlacquered small cubes left. General formula (n = number of small cubes per side length, in the example: n = 4): 3 painted surfaces: always 8 corner cubes; 2 painted surfaces: 12 edges �(n-2 corners) edge cubes; 1 painted surface: 6 surfaces �(n-2 corners) �(n-2 corners) = 6 �(n-2)� side cubes; no painted surface: (n-2) �(n-2) �(n-2) = (n-2)� internal unlacquered cubes.
Solution 18: The pumpkins
Mathematics teacher Owl makes it relatively simple for his friends: The function can be solved "in mind" by trying out, particularly if you assume, that Mr. Owl surely counted only in whole kilograms : If K1 = 2 kg; then K2 = 2 times 1.5 = 3 kg; then K3 = K1 + K2 = 2 + 3 = 5 kg; then K4 = 5 : 2.5 = 2 kg. Check: 2 + 3 + 5 + 2 = 12 kg. By formula (with K1 = x): K2 = 1.5 x; K3 = x + 1.5 x; K4 = (x + 1.5 x): 2.5 --> x + 1.5 x + (x + 1.5 x) + (x + 1.5 x): 2.5 = x + 1.5 x + (2.5 x) + (2.5 x: 2.5) = 6 x = 12 kg --> x (= K1) = 12 kg: 6 = 2 kg.
19. The balls
Among 9 equally large balls there is 1 lighter ball. All the others are the same weight. Through just weighing twice, the lighter ball is to be found out.
20. The lines
3 next to each other standing newly-built houses are to be supplied with in each case 3 lines for gas, electricity and water. The lines of the gas works, power station and water company are not to overlap, however! Draw a wiring scheme.
PS: You can try here: http://www.chilloutzone.de/files/07051804.html
21. The price discount
On a complaint of the customer a painter master grants a deduction of 10 % on the invoice sum. He stresses that the deduction be made before the addition of the value added tax. But - isn't that all the same? 10 % are 10 %, anyway??
Solution 19: The balls
One puts three balls each on the 2 scale pans of a beam balance (remainder = 3). 1st weighing: The balance shows an equilibrium --> = searched for = lighter ball is in the remainder! 2nd weighing: One puts a ball each from the remainder on the 2 scale pans (new remainder = 1). In case of an equilibrium --> the remaining ball is the one searched for! In case there is no equilibrium --> it is the ball in the higher scale pan!
The balance was not in the equilibrium at the 1st weighing --> the ball searched for is among the 3 balls in the higher scale pan. 2nd weighing: One of these three balls each is put into the 2 scale pans (remainder = 1). If the balance shows an equilibrium, the remaining ball is the one searched for! Otherwise it is the ball in the higher scale pan! (one could execute the weighing with used up tennis balls, whereby one makes one heavier by an injection with adhesive or so - and searches for the heavier one accordingly.)
Solution 20: The lines
Still trying? As far as I know there has not been found any solution so far without at least 1 overlap. Perhaps that could be proven even mathematically. Sorry!
Solution 21: The price discount
It doesn't matter whether the price discount is deducted before or after the addition of the value added tax. Because a deduction of e.g. 10% of a sum can be represented as a sum times 0.9 (S � 0.9). The addition of the VAT can be represented as a sum times 1.16 (S � 1.16). After the permutation law: S � 0.9 � 1.16 = S � 1.16 � 0,9.
22. The return trip
A businessman drives from Frankfurt to Hanover. He had counted on an average speed of 110 km/h. Due to traffic jams he only reaches an average of 100 km/h on his way there. He decides to reach an average of 120 km/h on his way back in order to fully compensate for the time delay on his way there. Is this calculation correct??
23. The 3 daughters
A travelling salesman rings at a door and wants to sell a magazine subscription. A woman opens and says, "I'll subscribe to one of your magazines, if you can figure out how old my 3 daughters are." The representative consents and gets the following hints from the mother: the product of the respective ages of my 3 daughters is 36. - the sum of the respective ages of my 3 daughters is our house number. Thereupon the representative looks at the house number, returns and says: "That's not enough !". Thereupon the mother adds, "True. Well, my eldest daughter's name is Rose."
Question: How old are the 3 daughters ?
24. The 3 Nomads
3 Nomaden count - like their ancestors did - the animals of a large herd, which trod by them, with their fingers. How many animals can they count and how does their thousands-of-years old counting method work?
Solution 22: The return trip
The time delay cannot be caught up. Reason: The formula for a homogeneous speed reads: v = s : t. The changed over formula reads: t = s : v. If he had caught up the lost time on the return journey, the sum of the time used on his way there (at 100 km/h) and the return journey (at 120 km/h) would have to equal the travel duration at the planned average speed of 110 km/h.
--> t1 + t2 = td.
--> s : v1 + s : v2 = 2 � s : vd
with a travel distance of approx. 400 km one way the calculation results in:
400 : 100 + 400 : 120 = 800 : 110 for the addition of the fractions on the left side the common denominator must be found. It is 600 (both 100 and 120 are divisors of 600). The fractions must be extended accordingly:
400 � 6 : 600 + 400 � 5 : 600 = 800 : 110
2400 : 600 + 2000 : 600 = 800: 110 (= 2400/600 + 2000/600 = 800/110)
--> 4400/600 = 800/110??
The calculation results in: 7.33 =/= 7.27 (hours) result: The businessman did not catch up the lost time. He needs 7.33 hours instead of the planned 7.27 hours.
Addition: Conversion of the decimal numbers in h/min/sec:
7.33 h = 7 h + 20 min (0.33 h = 0.33 � 60 min = 20 min). 7.27 h = 7 h + 16.4 min (0.27 h = 0.27 � 60 min = 16.4 min). I.e. he loses approx.. 3 1/2 minutes.
Solution 23: The 3 daughters
A very tricky task, as I find. I found it in a chat forum and thought at first it was a joke (like the one: there was a train accident. So and so many people died. What was the engine driver's name ?). When several forum participants made proposals for a solution, I also did it, first assuming there was more than one solution. I was however convinced that there is only ONE SOLUTION...
The data:
A times B times C = 36;
A + B + C = (unknown) house number;
The name Rosi is mathematically uninteresting, allows, however, to exclude solutions.
1. The divisors of 36: 1; 2; 3; 4; 6; 9; 12; 18; 36
2. the possible combinations:
1 time 1 time 36 (house number 38)
1 time 2 times 18 (house number 21)
1 time 3 time 12 (house number 16)
1 time 4 times 9 (house number 14)
1 time 6 times 6 (house number 13)
2 times 2 times 9 (house number 13)
2 times 3 time 6 (house number 11)
3 times 3 times 4 (house number 10)
3. The logic: If the house number was 38, 21, 16, 14, 11 or 10, the solution would be CLEAR and the salesman would not have have to inquire. Thus the house number must have been 13. Due to the information that there is only ONE (eldest) daughter, the solution 1 + 6 + 6 can be dropped, since there would be 2 eldest daughters in this case (twins) - and the woman would have said "elder" instead of "eldest".
Thus the correct solution is : 2 + 2 + 9. (so it doesn't matter, what the eldest daughter's name is or what hobby she has, main thing is, it is mentioned.)
Solution 24: The 3 Nomads
They can count up to 1000 animals with their fingers. The first nomad counts from 1 - 10 . The second counts, how often the first counted to 10, i.e. each finger of the second represents 10. The third nomad finally counts, how often the second used all his fingers (10 fingers of the second = 10 times 10 = 100).
When the third stretched all of his fingers --> 1000.
(here might be the cradle of the decimal system: 10 = 10 times 1 ; 100 = 10 times 10; 1,000 = 10 times 100, etc....)
IF you know similar interesting, impressing maths tasks, let me know via woko50@hotmail.com Thanks !
(NOTE: SINCE I AM NOT A NATIVE SPEAKER OF ENGLISH THERE MIGHT BE A COUPLE OF MISTAKES IN THE TEXT: PLEASE FEEL FREE TO TELL ME/ CORRECT ME via woko50@hotmail.com Thanks !)
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