Roundabout Math Quiz | Logic Puzzles

Quiz Answer Key and Fun Facts

1. In my family tree, my great-great grandmother gave birth to a set of twin sisters who each had a set of twins. I then found out that on two separate occasions, both of their two sets of twins each had two sets of twins twice. If this accounts for all of the children in my mother's family, how many children are in her generation? (how many share my mother's great grandmother, including her)

Answer: 64

If my great-great grandmother gave birth to a set of twin sisters, she had 2 daughters (of whom, my great grandmother was one). They each had a set of twins, making 4 children in my grandmother's generation. I then found out that on two separate occasions (which means we have to multiply the final result by 2), both of their two sets of twins (the four in my grandmother's generation) each had two sets of twins twice. Two sets of twins is 4 children, and twice makes it so that each of the four had 8 children, for 32 children total. Since this happened on two separate occasions, the total in my mother's generation is 64.

2. A hexagon is inscribed within a circle, within a square, within a circle, within a triangle, within a circle, within a square, within a circle, within a triangle, within a circle. If hexagon is inscribed within the largest circle, and all of the above polygons are regular, what is the ratio of the area of the larger hexagon to the area of the smaller hexagon?

Answer: 64 to 1

The side of a polygon circumscribed about a circle of radius 'r' is: 2r for a square and 2r*sqrt(3) for a triangle. The radius of a circle circumscribed about a polygon of side 'r' is: r*sqrt(2)/2 for a square, r*sqrt(3)/3 for a triangle, and r for a hexagon.
Since it's all multiplication, we can cancel the terms directly from these ratios if we make the original hexagon's side = 1.
There are 5 circles - we will call their radii R1, R2, R3, R4 and R5 (which will be the largest circle). R1 will equal 1, since it is circumscribed around a hexagon. R2 will equal sqrt(2), since the side of the square inscribed in the circle 2 will equal the diameter of circle 1. The diameter of circle 1 = 2, making R2 = 2*sqrt(2)/2, or sqrt(2). R3 will equal 2*sqrt(2). This is because the side of the triangle circumscribed around circle 2 has a side equal to 2*R2*sqrt(3). Circle 3 will have a radius equal to the side of the triangle: (2*R2*sqrt(3)) times sqrt(3), then divided by 3. Since sqrt(3)*sqrt(3) = 3, the square roots of 3 will cancel when divided by 3, making R3 = 2*R2. Note here that a circle circumscribed about a triangle that is circumscribed about a smaller circle will have a radius twice the length of the smaller circle. Since R2 = sqrt(2), R3 = 2*sqrt(2). This makes the side of the square circumscribed about circle 3 = 2*2*sqrt(2), or 4*sqrt(2). Therefore, R4 will equal (4*sqrt(2))*sqrt(2)/2. The same thing as with the triangle will happen here, as the sqrt(2)'s will cancel each other once divided by 2, leaving us with R4 = 4. As we saw before, since circle 5 is a circle circumscribed about a triangle that is circumscribed about a smaller circle, the ratio of the radius of the larger circle (circle 5) to the radius of the smaller circle (circle 4) is 2:1, so since R4 = 4, R5 = 8. Therefore, since the side of a hexagon inscribed within a circle is equal to the radius of the circle, the side of the smaller hexagon will equal R1, and the side of the larger hexagon will equal R5. Therefore, the ratio of the SIDES of the hexagons is 8:1.
The area of a hexagon with side 'S' is equal to 3*(S^2)*sqrt(3)/2. This means that the area of a hexagon is in a direct proportion to the SQUARE of its side. This means that the ratio of the AREAS of the hexagons will be equal to the ratio of the SQUARES of the sides of the hexagons, or 64:1.

3. I was 2/3 of the way to work when I realized that I had forgotten my lunch, so I turned around and headed home. When I was 2/3 of the way home from where I had turned around, I realized that I didn't need it because I had a lunch meeting, so I turned around and headed back toward work. When I was 3/4 of the way to work from where I had just turned around, I realized that I had forgot my briefcase. I turned around and went all the way back home. I averaged 60mph for the entire trip and had now wasted 90 minutes since I had originally left. If I drove the same route the entire time (switching directions three times), how many miles apart are the two places where I turned around to come back home?

Answer: 5

Let my work be 'x' miles away from home.
I traveled 2/3x on my way there, and then turned around (at point A - 2/3x away from home). I then turned around and traveled 2/3 of 2/3x (or 4/9x) back. When I turned around to go back to work, I had the remaining 1/3x that I had not originally traveled plus the 4/9x that I had just traveled, or 7/9x for my trip back to work. I then traveled 3/4 of that, which is 21/36x.
When I turned around the second time (to head back to work), I had NOT traveled the remaining 2/9x back home (the remainder of the 7/9x that I had to travel back to work from home after I turned around the second time), so I turned around a third time (at point B, 2/9x + 21/36x, or 29/36x away from home) and went home.
I had traveled 2/3x + 4/9x + 21/36x + 29/36x, which equals 90/36x.
If I traveled 60mph for 90 minutes, I traveled 90 miles. Therefore, since 90/36x=90, then x=36 miles. Since I turned around to go home at 2/3x (24 miles) and 29/36x (29 miles), the distance between the two is 5 miles.

4. My friend and I were gambling and he had lost all of his money. I had only lost 1/3 of my money, so I gave him half of mine. He then doubled his money, while I lost half of mine, so he then gave me half of what he had. He then lost half of his money, while I tripled mine. At this point I was ahead, but my friend wasn't doing well, so I gave him half of my money. He turned right around and said that he didn't need all of it, so he gave me back what I had originally loaned him plus $10 for loaning him money in the first place. If we now have $100 between us, who has more money and how much?

Answer: I have $50 more than my friend

I began with $x. When this began, I had 2/3x, and my friend had 0x. After I gave him half of my money, I had 2/6x and he had 2/6x. After he doubled his money and I halved mine, I had 1/6x and he had 4/6x. After he gave me half of his money, I had 3/6x and he had 2/6x. Then, after I tripled my money and he halved his, I had 9/6x and he had 1/6x. After I then gave him half of my money, I had 9/12x and he had 11/12x. He didn't need all of that money, so hegave me back what I originally loaned him +$10, but we don't know how much each of us has except that after this the total doesn't change (and equals $100).
If our total is $100, then 9/12x + 11/12x = $100. Therefore, 20/12x = $100, and x=$60.
That means that right before the end, I had $45 and he had $55. I had loaned him 2/6 of $60 (or $20) originally, so he gave me that +$10 back, making my final total $75, and his $25. I ended up with $50 more than my friend.

5. I have a farm where an accident occured (I won't go into it - it's horrible) and now, most of my pigs have lost their legs. I have chickens and pigs on my farm, and after the accident, the ratio of pigs to chicken legs is the same as the ratio of chicken legs to pig legs. Luckily, all of my chickens still have their legs. If there are three pig legs missing for each leg my chickens have, how many pigs do I have for each chicken leg?

Answer: 1

**NOTE: This problem does not ask how many chickens I have.
The amount of chicken legs I have is 'x'. The amount of pigs I have is 'y'. The amount of pigs' legs missing is '3x' (three missing for each leg the chickens have). Therefore, since pigs normally have 4 legs, the amount of pig legs I have is '4y-3x'.
The ratio of pigs to chicken legs is y/x. the ratio of chicken legs to pig legs is x/(4y-3x). If you cross multiply, you will get x^2 = 4y^2 - 3xy, or 4y^2 - 3xy - x^2 = 0.
If you use the quadratic equation to solve this, where a=4, b=-3x and c=-x^2, you will arrive at the result y=x.
Therefore, the ratio of pigs to chicken legs, and chicken legs to pig legs (and therefore pigs to pig legs) is 1:1. (I told you the accident was horrible.)

6. You can generate a cypher by moving each letter 'x' number of places further in the alphabet 'x' number of times, where 'x' is it's numerical position in the alphabet. In that cypher, you would move A (the 1st letter) 1 letter further in the alphabet once, making it 'B' in the encoded cypher. You would move B 2 letters away twice, making it 'F' in the encoded cypher. You would move C three letters away three times, making it 'L' and so on... What would be the result of the word "MONOPOLY" if you encrypted it, then encrypted the result, then encrypted the result, and so on until you had encrypted the encrypted result once for each letter in the word "MONOPOLY"?

Answer: ZZZZZZZZ

If you apply the cypher to the alphabet, interestingly enough, you get a cyclically repeating sequence of letters: A=B, B=F, C=L, D=T, E=E, F=P, G=D, H=T, L=F, K=B, L=Z, M=Z, N=B, O=F, P=L, Q=T, R=D, S=P, T=D, U=T, V=L, W=F, X=B, Y=Z, Z=Z.
There are only 7 letters in the encrypted alphabet: B, D, F, L, P, T and Z.
For two of these letters, T and D, T=D and D=T. Any letters that become T or D in the cypher will endlessly toggle between T and D, depending on their parity after the first encryption in which they become T or D.
For the rest of the letters in the encrypted alphabet, B=F, F=P, P=L, L=Z, and Z=Z. Therefore, any letters that DO NOT get encrypted into a T or D will become a Z after the fifth encryption, and anything that becomes encrypted into a Z will remain a Z ever after.
For MONOPOLY, none of the letters become a T or a D, so the encryption goes as follows: MONOPOLY, ZFBFLFZZ, ZPFPZPZZ, ZLPLZLZZ, ZZLZZZZZ, ZZZZZZZZ... after that, all further encryptions will result in ZZZZZZZZ.

7. The sum of the two digits in my age is the same as it was when my age was half of what it will be the next time that the digits of my age total the same as they do now. What is the total of the digits in my age?

Answer: 9

In order for my age to be able to be cut in half, the ones digit must be even. Since any age in which the tens digit is also even can be written as 10*(2x) + (2y), cutting it in half will automatically result in a value less than the total of the digits in the original number, since (unless x=y=0) 2x+2y will always be greater than x+y. Therefore, the tens digit is an odd number and the ones digit is an even number.
As you increase in value, the only two digit numbers in which the digits total to a number that is equal to the digits of its half are even multiples of 9 (ie- 18, 36, 54, 72, 90). Since these can all be written as 9*2x, half of each of those numbers is also a multiple of 9, since it will equal 9*x.
My age could be any odd multiple of 9 with two digits and this problem will work, but the total of the digits in my age will always be 9.
In every two digit multiple of 9, the digits total 9.

8. I have three beakers, marked 'A', 'B' and 'C'. Beaker 'A' is 2/3 full of red liquid, beaker 'B' is 2/3 full of water, and beaker 'C' is empty. When the red liquid mixes with water, it immediately mixes completely, and the redness of the liquid is directly proportional to the ratio of red liquid to water in the mixture. I take beaker A and pour half of it into each of the other two beakers. I then pour half of the contents of beaker B into each of the other two beakers, and then I pour half of beaker C into each of the other two beakers. Then I pour off liquid from beaker A evenly into the other two beakers (half into each of beakers B and C) until beaker A has 1/3 of the total liquid left in it. I then pour from beaker B into beaker C until they, too contain 1/3 of the total liquid. Now each of the beakers has the same amount of liquid in it. Which beaker contains the reddest liquid?

Answer: Beaker B

After Beaker A has had its contents poured into the other two, beaker A is empty, beaker B is full and 33% red, and beaker C is 1/3 full and 100% red. When Beaker B has had its contents poured into the other two, Beaker A is 1/2 full and 33% red, beaker B is empty, and beaker C is 5/6 full and 60% red. When beaker C has had its contents poured into the other two, beaker A is 11/12 full and 45% red, beaker B is 5/12 full and 60% red, and beaker C is empty.
Now, each beaker is then to be 32/72 full when they have been equalized. Beaker A is 66/72 full, beaker B is 30/72 full and beaker C is empty. If you divide the 34/72 of the liquid that needs to be poured off of beaker A in half and pour that amount into each of the other two, beaker A will be 32/72 full and be just over 45% red, beaker B will be 47/72 full and be just under 55% red and beaker C will be 17/72 full and also be just over 45% red.
Once you pour 15/72 from beaker B into beaker C, all three beakers will have the same amount in them, and beaker A will be just over 45% red (45.454545%), beaker B will be just under 55% red (54.574468%), and beaker C will be just under 50% red (49.729509%).
Therefore, beaker B is the reddest.

9. After an interview in an office building in New York, I left the office and looked east out at the view of the Atlantic Ocean while waiting for the elevator. When the elevator opened, I turned around and walked into it. I took it down three floors, exited and turned left. I followed the hallway around two corners, turning to the left each time. I then headed down the hall a little way and stopped at the next bank of elevators on my right. I entered one of the elevators, took it down fifteen floors, exited the elevator and went straight across into another elevator that I took to the ground level. I then exited the elevator, turned left and went straight out the front of the building, where I caught a bus. I got on the bus, and the seats facing the door were taken, so I went up and sat facing out the front of the bus. The bus turned left and stopped, then turned right and then right again. It continued fourteen blocks and then turned left, where it stopped and let three people off, one of whom was sitting across from the door. I got up and took his seat. Then, because there is a median in the road, the bus then did a U-turn, went down one block and turned right. It continued on, turned left and stopped. What direction am I facing?

Answer: South

When I turn around and get in the elevator, I am facing west (since I turned around from facing east). I exit the elevator facing east, and turn left, to face north. I then turn left twice more, now facing south. I enter the new bank of elevators to my right, facing west. I exit facing east, go into another elevator, and then exit that elevator facing west. I turn left and exit the building facing south. The doors on public buses face the curb side of the street, and in New York they drive on the right side of the street.

When I get on the bus, I am facing south. I sit down facing west, since the bus is heading west. It turns left, so I am now facing south. It then turns right, so I am facing west again. It then turns right again so I am facing north.

It then turns left so that I am facing west again. The doors are on the north side of the bus, so when I get up and take the seat of the man who gets off, I am then facing north. When the bus does a U-turn, I am then facing south.

It turns right, and I am then facing west. When it turns left, I am facing south.

10. One dinner costs $7. Al, Ben and Carol decide to pool their money and get something to eat. With all of Al's money, 1/2 of Ben's money and 1/3 of Carol's money, they can each get a dinner. With 1/2 of Al's money, all of Ben's money and 5/6 of Carol's money, only one of them can get a dinner. With 2/3 of Al's money, 3/4 of Ben's money and all of Carol's money, they can only get two dinners. With 3/4 of Al's money, 1/3 of Ben's money and 1/2 of Carol's money, they still can only get two dinners. They decide on the first option, and decide to split the remaining money evenly. If none of them has any change (all of the money is in whole dollar bills), and each of the proposed fractions each of them would pay is also in whole dollar amounts, how many dollars do each of them wind up with once they split the remaining money evenly?

Answer: $1

If all of the proposed fractions must be whole dollar amounts, then Al's and Ben's amounts of money must be divisible by 12 and Carol's amount of money must be divisible by 6. Also, the amount left over must be divisible by 3. Let us suppose that each of them has the minimum amount of money possible to fulfill the criteria. Al and Ben would each have $12 and Carol would have $6.

The three dinner agreement would mean Al would give $12, Ben would give $6, and Carol would give $2, for a total of $20, which is $1 short of 3 dinners (3 * $7 = $21).

Therefore, at least one of them has more than the minimum. If Ben has twice his minimum amount of money, then Al would contribute $12, Ben would contribute $12, and Carol would contribute $2, for a total of $26, but that would leave $5 left over which is not evenly divisible between three people. If Ben has more than twice his minimum amount, he would contribute enough that they would be able to afford more than 3 dinners. If Carol has twice her minimum amount of money, they can afford 3 dinners, but would leave only $1 at the end.

She would have to have 3 times her minimum amount to leave $3. 4 times her minimum amount would leave $5 at the end, and 5 times her minimum amount would afford them more than 3 dinners. But, if Al and Ben have the minimum and Carol has 3 times her minimum, then they can afford 3 dinners in two of the other three scenarios. Since Carol and Ben can not have more than their minimum amounts, they must have less, and the only multiple of $12 that would fulfill the criteria and work out in each scenario is $24. Therefore, Al has $24, and Ben and Carol each have $0. After paying $21 for the meals, they have $3 left, which they split among them, at $1 each.

Source: Author treefinger

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Roundabout Math Trivia Quiz

This is a quiz where the answers are meant to be unravelled from the word problem before they are solved. Good ;luck, some of them can be pretty tricky!

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