As you'll note, there are frequently many solutions to any problem. As an example, note the creativity employed by first/second graders on the Meanie Genie.
Meanie Genie Solution (textbook)
There is great power to be found in logical and creative thinking.
Alley arranges the stones into three groups of three and places one group on one side of the first balance scale and another group on the other side. What can she conclude? If both sides weigh the same, then she knows that the (heavier) jewel must be in the third group of three. If, however, one side is heavier than the other, then she knows that the jewel is one of the three that weighed more. In either case, after only one weighing, Alley is able to identify a group of only three stones among which is the Rama Nujan.
She then takes two of these three stones and places one on each side of the second scale. If one weighs more than the other, then she knows that this stone is the one containing the jewel. If they both weigh the same, then she knows that the third stone must contain the jewel. Thus, by weighing the stones only twice, Alley is able to find the jewel.
Take partial steps whenever possible. Notice that, instead of trying to identify the jewel immediately, Alley first reduces the pool of choices from nine to three. Thus she first makes the problem easier. “Divide and conquer” is an important and useful technique in both mathematics and life.
Meanie Genie (1st/2nd Graders)
We solved the math problem by saying Allie left one stone out & had 2 balance scales & Allie could put 4 stones on each side of the balance scale and if one end is heavier and Allie could weigh the 4 stones that were heavier 2 on each side of the balance scale and weigh the 2 that are left as the heaviest in your hands 1 on 1. In the beginning we said that Allie left one out and if the stones all were the same you know it was the stone you left out.
Notice how they applied creativity and decided to use their own hands as a set of scales. Cool....
Damsel in Distress (textbook)
Notice that the corner of the moat forms a 20-foot-by-20-foot square. By the Pythagorean Theorem, the distance from the outer corner of the shore to the inner corner of the castle island is equal to the square root of 20^2 x 20^2. Using a calculator, we see that the distance is 28.2842 feet.
Placing the 19-foot beam diagonally across the corner of the moat as far out as it can go creates a triangle that cuts off the corner. If we draw a line from the center of the beam to the outer corner of the moat, we create two identical 45-degree right triangles, as shown. Since the length of half the beam is 9.5 feet, we learn that the center of the beam is also 9.5 feet from the outer corner of the moat.
Since the total diagonal distance from the outer corner of the moat to the corner of the castle island is 28.2842 feet, the distance to the center of the beam is (28.2842 feet - 9.5 feet) = 18.7842 feet. Since that distance is just less than 19 feet, the other beam will just barely span the distance between the beam and the island. In gratitude for her rescue, the damsel provided the good knight with a romantic lesson in geometry.
Damsel in Distress (textbook)
Focusing attention on the corner of the moat suggests using one of the beams to span the corner. Of course, we need to check that the two 19-foot beams are long enough to make the configuration in the picture.
There are at least two ways to verify that this picture is correct. One way is to construct a physical model. The picture shown at left is a physical model scaled down so that 1 foot in the story corresponds to 1 millimeter in the picture. You can now measure and ensure that this configuration is possible.
There are at least two ways to verify that this picture is correct. One way is to construct a physical model. The picture shown at left is a physical model scaled down so that 1 foot in the story corresponds to 1 millimeter in the picture. You can now measure and ensure that this configuration is possible.
An alternative method would be to observe that the picture has some right triangles. This observation foreshadows our look at the Pythagorean Theorem. After we examine good old Pythagoras’s theorem (Chapter 4), the following paragraphs will seem soothing and comforting. If for now you find them less so, feel free to glance through them and just move on.
Placing the 19-foot beam diagonally across the corner of the moat as far out as it can go creates a triangle that cuts off the corner. If we draw a line from the center of the beam to the outer corner of the moat, we create two identical 45-degree right triangles, as shown. Since the length of half the beam is 9.5 feet, we learn that the center of the beam is also 9.5 feet from the outer corner of the moat.
Since the total diagonal distance from the outer corner of the moat to the corner of the castle island is 28.2842 feet, the distance to the center of the beam is (28.2842 feet - 9.5 feet) = 18.7842 feet. Since that distance is just less than 19 feet, the other beam will just barely span the distance between the beam and the island. In gratitude for her rescue, the damsel provided the good knight with a romantic lesson in geometry.
A Shaky Story (my solution)
Draw a picture - works for me a lot of times.
The hostess shook hands eight (8) times.
Vertices, Faces, Edges (Euler Characteristic - my solution)
Easy Doodle
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Complicated Doodle
Watsamattawith U (textbook)
Break a hard problem into easier ones.
Suppose there are 300 students in each of the graduating classes. In last year’s class, 150 were male and 150 were female. Suppose that the average GPA of last year’s men was 2.0 and the average GPA of last year’s women was 3.5. Then the average GPA for last year’s graduating class was 2.75. One way to arrive at that answer is to replace all the men’s GPAs with the male average and all the women’s GPAs with the female average and then average all 300 GPAs as follows:
(2.0 x 150 + 3.5 x 150)/300 = 825/300 = 2.75
Now assume that this year’s graduating class is comprised of 200 men and only 100 women. If the average GPA of the male students rose to 2.1 (higher than last year’s average) and the average GPA of the female students became 3.6 (again higher than last year’s average), then the GPA for this year’s class can be found by computing:
(2.l x 200 + 3.6 x 100)/300 = (420 + 360)/300 = 780/300 = 2.6!
Amazing . . . until we realized we had the ability to change the proportion of men to women. By having a higher proportion of the poorer male students, even though the GPA of the males increased from 2.0 to 2.1, since there were more males this year, they dragged down the GPA of the student body. The number of male students compared to female students was a quantity that could be changed, but we might not have thought about that possibility. Hidden features of a statistical situation like this are sometimes called “lurking variables.”
(2.0 x 150 + 3.5 x 150)/300 = 825/300 = 2.75
Now assume that this year’s graduating class is comprised of 200 men and only 100 women. If the average GPA of the male students rose to 2.1 (higher than last year’s average) and the average GPA of the female students became 3.6 (again higher than last year’s average), then the GPA for this year’s class can be found by computing:
(2.l x 200 + 3.6 x 100)/300 = (420 + 360)/300 = 780/300 = 2.6!
Amazing . . . until we realized we had the ability to change the proportion of men to women. By having a higher proportion of the poorer male students, even though the GPA of the males increased from 2.0 to 2.1, since there were more males this year, they dragged down the GPA of the student body. The number of male students compared to female students was a quantity that could be changed, but we might not have thought about that possibility. Hidden features of a statistical situation like this are sometimes called “lurking variables.”
Be on the lookout for opportunities in which you have the ability to vary a parameter or circumstance.
The Fountain of Knowledge (textbook)
Suppose we fill up the 10-ounce glass with mango juice and slowly pour it into the 6-ounce glass, stopping at the moment the 6-ounce glass is full. Notice that what’s left in the 10-ounce glass is precisely 4 ounces of mango juice. We now empty the 6-ounce glass back into the pool and refill it with the 4 ounces from the other glass. If we now refill the 10-ounce glass from the pool, we can again slowly pour its contents into the 6-ounce glass until the 6-ounce glass is full. Filling it takes exactly 2 more ounces, and now the larger glass contains exactly 8 ounces. Hap- pily, those 8 ounces of mango juice can now be served to Trey (on a tray). If Trey had found a solution, he would have made his first discovery in an area of mathematics known as number theory.
There is more than one solution to this puzzle. For example, we could have begun by filling the 6-ounce glass and pouring its entire contents into the 10-ounce glass. See if you can use this starting point to find an alternative solution.
Let's Make a Deal (textbook)
Keep an open mind and be willing to understand new ideas that first appear counterintuitive.
Warren Piece first picked Door Number 3, and the likelihood of his finding the car there was 1 out of 3, that is, one-third, which is not too likely. Next, Monty opened a door showing a mule. Let’s see what happens if Warren were to switch in each of the three possible scenarios. In Case 1, Monty opens Door Number 2. If Warren switches in this case, he wins the Cadillac. In Case 2, Monty opens Door Number 1. If Warren switches in this case, he would win. In Case 3, Monty could open either Door Number 1 or 2. If Warren switches in this case, sadly, he would be the owner of a mule. Therefore, overall, the likelihood of his winning the car by switching is 2 out of 3, or two-thirds, which is twice as likely as the one- third chance of selecting the car if he sticks to his original guess. This brief encounter with probability illustrates that counterintuitive outcomes can occur during attempts to measure the unknown.
The Fountain of Knowledge (textbook)
Suppose we fill up the 10-ounce glass with mango juice and slowly pour it into the 6-ounce glass, stopping at the moment the 6-ounce glass is full. Notice that what’s left in the 10-ounce glass is precisely 4 ounces of mango juice. We now empty the 6-ounce glass back into the pool and refill it with the 4 ounces from the other glass. If we now refill the 10-ounce glass from the pool, we can again slowly pour its contents into the 6-ounce glass until the 6-ounce glass is full. Filling it takes exactly 2 more ounces, and now the larger glass contains exactly 8 ounces. Hap- pily, those 8 ounces of mango juice can now be served to Trey (on a tray). If Trey had found a solution, he would have made his first discovery in an area of mathematics known as number theory.
There is more than one solution to this puzzle. For example, we could have begun by filling the 6-ounce glass and pouring its entire contents into the 10-ounce glass. See if you can use this starting point to find an alternative solution.
Let's Make a Deal (textbook)
Keep an open mind and be willing to understand new ideas that first appear counterintuitive.
Fortunately, Warren Piece enjoys mathematics as a hobby, so he believes he can solve this conundrum. He thinks carefully, assesses the chances each way, and confidently proclaims (while still jumping up and down, of course), “I switch my guess to Door Number 1, Monty.”
Monty Hall turns and says, “Okay. Let’s see what deal you’ve made. What is behind Door Number 1?” The door swings slowly open, and the crowd gasps as they see behind Door Number 1 the most beautiful finned chassis that General Motors ever painted pink. Bedlam reigns. “How did you know?” asks Monty Hall over the din. Warren Piece explains.
Monty Hall turns and says, “Okay. Let’s see what deal you’ve made. What is behind Door Number 1?” The door swings slowly open, and the crowd gasps as they see behind Door Number 1 the most beautiful finned chassis that General Motors ever painted pink. Bedlam reigns. “How did you know?” asks Monty Hall over the din. Warren Piece explains.
“When I originally guessed Door Number 3, I had a one-third chance of being right and a two-thirds chance of being wrong. Thus it’s more likely I was wrong than right. When you opened Door Number 2 and revealed no car, I hoped I was wrong originally—which, recall, was more likely than not. If I was wrong originally, then the car must be behind the remaining door, Door Number 1. So I switched, knowing that the prob- ability of my winning after switching was 2 out of 3, whereas the chance of my having picked it correctly the first time was only 1 out of 3.” Monty Hall was compelled to ask, “How did you figure that out?”—to which our hero raisin sagely replied, “By studying The Heart of Mathematics: An invitation to effective thinking.” (Not a bad coast-to-coast TV plug for our book, even coming from a guy named Warren Piece in a raisin suit.)
Some people might think that it doesn’t matter if he switches or not. However, the chances of finding the car are indeed greater by switching. One way to demonstrate this is to list all the possible ways the Cadillac and the mules can be placed behind the doors:
Some people might think that it doesn’t matter if he switches or not. However, the chances of finding the car are indeed greater by switching. One way to demonstrate this is to list all the possible ways the Cadillac and the mules can be placed behind the doors:
Warren Piece first picked Door Number 3, and the likelihood of his finding the car there was 1 out of 3, that is, one-third, which is not too likely. Next, Monty opened a door showing a mule. Let’s see what happens if Warren were to switch in each of the three possible scenarios. In Case 1, Monty opens Door Number 2. If Warren switches in this case, he wins the Cadillac. In Case 2, Monty opens Door Number 1. If Warren switches in this case, he would win. In Case 3, Monty could open either Door Number 1 or 2. If Warren switches in this case, sadly, he would be the owner of a mule. Therefore, overall, the likelihood of his winning the car by switching is 2 out of 3, or two-thirds, which is twice as likely as the one- third chance of selecting the car if he sticks to his original guess. This brief encounter with probability illustrates that counterintuitive outcomes can occur during attempts to measure the unknown.