Mathematical Thinking I, II, .... (previously Puzzle: Marrie

[ayw]

A little mathematical one...

In a certain town, 5/7 of the men are married to 3/8 of the women. What fraction of the adult population is married? Assume monogamy.


(Edit: topic title changed)

[akai-kun]

51/56 ??

well, this is false, or? but this is the only solution I can think of now, well, I'm a little bit tired, so...
Say, false or true?

[ayw]

@akai-kun, false but judging from your answer, you might be on the right track. It's slightly tricky, but elementary fractions.

I thought I saw a reply by x64_02 that was correct.. where's it gone?

[akai-kun]

Spoiler:

Could it be ~49% ?

[ayw]

Spoiler:

Could it be ~49% ?

I'm pretty sure you got it. Can you give me the exact fraction (in spoilers)?

[akai-kun]

Spoiler:

Could it be ~49% ?

I'm pretty sure you got it. Can you give me the exact fraction (in spoilers)?

Really? Yeah!!

Well, the fraction( not sure if it is mathematical correct, but...):

Spoiler:

According to the question, there can be made this rule:

5/7 of the men = 3/8 of the women

that means:  40/56 Men = 21/56 Women --> so, from 21/56 -> 21 * 40/21 and 56 * 40/21 --> so you get 40 and 40 above and now you can count the number of inhabitants of the town:

56 Men + 107 Women = 163 Inhabitants

Because of the 40s above the fractures, you can calculate (40+40)/163 which has as a result:

---> ~ 49%

That should be it...

Correct calculation? 

[ayw]

Hmm... you got nearly the right result, and rounding down to two digits gives the right answer. Your first step

Spoiler:

5/7 of the men = 3/8 of the women

is correct and is very important but I can't follow the maths beyond that. The final fraction

Spoiler:

(40+40)/163

is not correct. Hint:

Spoiler:

It is not possible to find out the actual number of inhabitants with the given information, since you could e.g. double the number of inhabitants and still keep the same factions in the original question. Nonetheless, you were right about writing down an expression for the number of inhabitants, but use variables e.g. M for "all men" and W for "all women" instead. Then use the information from your first step. (Perhaps that's what you were attempting in your subsequent steps. I couldn't quite tell.) After that there's one more step to get the fraction of married people.

[cancerkani]

Just for fun... Not sure if this is right...

Spoiler:

Lowest common multiple between 3 and 5: 15

5/7 of men versus 3/8 of women translates to:

15/21 of men versus 15/40 of women

Total population: 21 + 40 = 61

Fraction married: 30/61

Percentage married: 49.1803278%

[akai-kun]

Hmm... you got nearly the right result, and rounding down to two digits gives the right answer. Your first step

Spoiler:

5/7 of the men = 3/8 of the women

is correct and is very important but I can't follow the maths beyond that. The final fraction

Spoiler:

(40+40)/163

is not correct. Hint:

Spoiler:

It is not possible to find out the actual number of inhabitants with the given information, since you could e.g. double the number of inhabitants and still keep the same factions in the original question. Nonetheless, you were right about writing down an expression for the number of inhabitants, but use variables e.g. M for "all men" and W for "all women" instead. Then use the information from your first step. (Perhaps that's what you were attempting in your subsequent steps. I couldn't quite tell.) After that there's one more step to get the fraction of married people.

Well, yeah, you're right... Variables would be better and from what I understand, what you wrote was my calculation...
And the last step?
Well, this is the calculation I wanted to show:

Just for fun... Not sure if this is right...

Spoiler:

Lowest common multiple between 3 and 5: 15

5/7 of men versus 3/8 of women translates to:

15/21 of men versus 15/40 of women

Total population: 21 + 40 = 61

Fraction married: 30/61

Percentage married: 49.1803278%

right or not? (...)

[ayw]

@akai-kun, Ah, then you were correct!Sorry about that. Well done!!

@cancerkani, Thats right, although

Spoiler:

21+40=61 is not necessarily the population, but is indeed the smallest population satisfying the question, since you chose the smallest common multiple between 3 and 5.

Well done!!

[akai-kun]

Next quiz!

I think we could make a game like in other threads/forums that the one who anwered the question at first can start a new quiz, how about that idea? (But it should be with maths...)

And if you like, here's the next quiz(easy one):

"The sum of two positive rational numbers is 15.
For which two numbers the product of them is at the highest point?
Does the result change when the numbers also could be negative? Explain your answer!"

Good luck!

[ayw]

Next quiz!

I think we could make a game like in other threads/forums that the one who anwered the question at first can start a new quiz, how about that idea? (But it should be with maths...)

Good Idea! I'll change the name of the topic title accordingly.

[Sayumi]

Okay here is my answer...

Spoiler:

Highest Product: 7.5 x7.5=56.27
I can't really think of a way to prove this right now (except for a way which would involve Pythagoras and derivatives.... and I can't be bothered to formulate that!)
but it can be explained using approximation.
0x15= 0
1x14= 14
2X13= 26
...
6x9= 54
7X8= 56
8X7= 56
9X6= 54
...
from here we could continue this with 7.9x7.1=56.09 ...
And would eventually end up with 7.5x7.5

The result won't change if we can use negative numbers as well. When the sum is +15 and one of numbers is negative the other one has to be positive. When multiplying a positive and a negative number the result will always be negative and therefore lower than 56.25.

Now... another maths puzzle....

John is celebrating his 36th birthday. One of the guests asks him how old his brother Joe is.
John answers: I am now double as old, as Joe was, when I was as old, as Joe is now.
How old is Joe? (assuming that John's statement is correct of course)

[kat1214young]

Spoiler:

6?

here' one

Suppose we are given an ordinary chessboard (8x8) and 32 dominoes (2x1).
Obviously, the dominoes can be arranged on the board to cover it completey.
Now, two opposite corner squares are removed. Determine whether or not 31
dominoes can be arranged on the reduced board to cover it exactly.

enjoy~

[allstar1234]

@akai-kun, Ah, then you were correct!Sorry about that. Well done!!

@cancerkani, Thats right, although

Spoiler:

21+40=61 is not necessarily the population, but is indeed the smallest population satisfying the question, since you chose the smallest common multiple between 3 and 5.

Well done!!

Spoiler:

as long as he set up a system of to equations or anyone for that matter, the population can be set to the answer's choice as long as they still have the fraction of men=fraction of women, and a total of both population = total population.