A multiple-choice quiz
by treefinger.
Estimated time: 26 mins.

Most Recent Scores

Sep 27 2024
:
Coachpete1: Sep 25 2024 : snadnerb: 3/10

Sep 17 2024 : dee1304:

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Quiz Answer Key and Fun Facts

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.

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.

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.

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.

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.

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.

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.

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.

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.)

**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.)

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

This quiz was reviewed by FunTrivia editor gtho4 before going online.

Any errors found in FunTrivia content are routinely corrected through our feedback system.

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