Easy math puzzles
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43 members have voted
Quote: ThatDonGuySince nobody else has published one today, here's a (relatively) easy one for Monday:
A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?
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Thanks for Monday’s question, although I’m answering on Tuesday!
The cross section of a shed is a 10 foot by 10 foot square. The shed is located in the center of an open, level field.
A goat is tethered to one corner of the shed by a forty foot rope. The goat cannot enter the shed.
What is the area over which the goat can graze?
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Picture not to scale.
Quote: GialmereIt's toughie Tuesday. Back to goats...
The cross section of a shed is a 10 foot by 10 foot square. The shed is located in the center of an open, level field.
A goat is tethered to one corner of the shed by a forty foot rope. The goat cannot enter the shed.
What is the area over which the goat can graze?
Picture not to scale.
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I get about 4,847.07 square feet.
This is the result of the following expression:
Quote: GialmereWhat is the area over which the goat can graze?
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I'm getting the answer below, but my confidence is not very high. If I'm right, I'll provide a solution.
Hoping to get a reply to my reply to my reply on the probability of a sister problem.
Quote: camaplQuote: ThatDonGuySince nobody else has published one today, here's a (relatively) easy one for Monday:
A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?
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Thanks for Monday’s question, although I’m answering on Tuesday!Considering your example, where N = 4, as you’ve already done the math, boys have a sister 1/4 + 3/8 + 1/4 = 7/8 or 14/16 of the time, and girls have a sister 3/8 + 1/4 + 1/16 = 11/16 of the time. If you prefer a simpler example, consider N = 2. You only have 2 girls (girl with a sister) 1/2 * 1/2 = 1/4 of the time, while a boy has a sister (1/2 * 1/2) + (1/2 * 1/2) = 1/2 of the time. Answer: more boys than girls have a sister. Consequently, more girls than boys have a brother.
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You are counting how many families with boys versus how families with girls have sisters. I am looking for the actual numbers of boys and girls with sisters.
For example, your 1/16 value for the girls forgets that this is four girls that have one or more sisters.
Quote: ChesterDog
I get about 4,847.07 square feet.
This is the result of the following expression:
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I see we're a little off. For now, let me throw out this diagram for purposes of discussion.
Here is some breakdown of my answer.
| Area | Size | qty | Total |
|---|---|---|---|
| Big quarter slices | 1256.637061 | 3 | 3769.911184 |
| Small slices | 457.5939873 | 2 | 915.187975 |
| Triangle | 103.0776406 | 2 | 206.155281 |
| Total | 4891.254440 |
Quote: WizardQuote: ChesterDog
I get about 4,847.07 square feet.
This is the result of the following expression:
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I see we're a little off. For now, let me throw out this diagram for purposes of discussion.
Here is some breakdown of my answer.
Area Size qty Total Big quarter slices 1256.637061 3 3769.911184 Small slices 457.5939873 2 915.187975 Triangle 103.0776406 2 206.155281 Total 4891.254440
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That's a good diagram!
I found the area of the upper yellow triangle by doing half the product of its base and height. I used 10 feet as the base.
To calculate its height, I used point A as the origin and found the intersection of the line y = x - 10 with the circle x2 + y2 = 900.
For the intersection point at point D I got x = -15.615528 and y = -25.615528. I used 15.615528 as the yellow triangle's height. So, I found one yellow triangle's area is (1/2)(10)(15.615528) = 78.077641
Quote: ChesterDog
That's a good diagram!
I found the area of the upper yellow triangle by doing half the product of its base and height. I used 10 feet as the base.
To calculate its height, I used point A as the origin and found the intersection of the line y = x - 10 with the circle x2 + y2 = 900.
For the intersection point at point D I got x = -15.615528 and y = -25.615528. I used 15.615528 as the yellow triangle's height. So, I found one yellow triangle's area is (1/2)(10)(15.615528) = 78.077641
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Here's what I get:
AC = 10, AD = 30, and angle ACD = 135 degrees
Law of Sines: (sin ADC) / 10 = (sin ACD) / 30, so sin ADC = (sin ACD) / 3 = sqrt(2) / 6; also, cos^2 ACD = 1 - sin^2 ACD = 17/18, so cos ACD = sqrt(34) / 6
Angle CAD = 180 - (ACD + 135) = 45 - ACD, so sin CAD = sin 45 cos ACD - cos 45 sin ACD = (sqrt(2) / 2) (cos ACD - sin ACD)
= (sqrt(2) / 2) (sqrt(34) - sqrt(2)) / 6
= (2 sqrt(17) - 2) / 12 = (sqrt(17) - 1) / 6
Area of CAD = 1/2 AC AD sin CAD = 150 (sqrt(17) - 1) / 6 = 25 (sqrt(17) - 1), which matches ChesterDog's number
Quote: ThatDonGuyQuote: camaplQuote: ThatDonGuySince nobody else has published one today, here's a (relatively) easy one for Monday:
A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?
link to original post
Thanks for Monday’s question, although I’m answering on Tuesday!Considering your example, where N = 4, as you’ve already done the math, boys have a sister 1/4 + 3/8 + 1/4 = 7/8 or 14/16 of the time, and girls have a sister 3/8 + 1/4 + 1/16 = 11/16 of the time. If you prefer a simpler example, consider N = 2. You only have 2 girls (girl with a sister) 1/2 * 1/2 = 1/4 of the time, while a boy has a sister (1/2 * 1/2) + (1/2 * 1/2) = 1/2 of the time. Answer: more boys than girls have a sister. Consequently, more girls than boys have a brother.
link to original post
You are counting how many families with boys versus how families with girls have sisters. I am looking for the actual numbers of boys and girls with sisters.
For example, your 1/16 value for the girls forgets that this is four girls that have one or more sisters.
link to original post
For N=2, boys have a sister 1/2 the time (no change), while girls have a sister 2 * 1/2 * 1/2 = 1/2 the time as well.
| Area | Size | qty | Total |
|---|---|---|---|
| Big quarter slices | 1256.637061 | 3 | 3769.911184 |
| Small slices | 460.5026797 | 2 | 921.005359 |
| Triangle | 103.0776406 | 2 | 206.155281 |
| Less quarter square | -25 | 2 | -50.000000 |
| Total | 4847.071825 |
I also computed angle ADC incorrectly. It is atan(sqrt(2)/sqrt(34)).
