17 replies to this topic

[Bikerman]

Here's the first in a few maths questions I'll post to test yourself and/or keep your hand in with basic maths/trig.

This one is not too tricky, but you need to remember basic trig.



Simple question - find x

PS - Mods: If this is best in another forum please feel free to either shift it or get me to repost. I tried to find the best fit for it, and I didn't think it would be right in the jokes forum, so this seemed the best....

Edited by Bikerman, 14 August 2010 - 10:00 AM.

[deadmad7]

What do you mean by "find X" ? You never gave any equations or even any other number... what the heck are we supposed to do?


EDIT: I just saw the image, i had to switch to Chrome to see it, LOL sorry... i will try to solve and ill reply

Edited by deadmad7, 14 August 2010 - 10:20 AM.

[Bikerman]

deadmad7, on 14 August 2010 - 10:19 AM, said:

EDIT: I just saw the image, i had to switch to Chrome to see it, LOL sorry... i will try to solve and ill reply

No problem, I figured you must have a problem with the image. Which browser did it not work in, just for interest?

[rvalkass]

I'll make a wild stab in the dark that x is approximately equal to 8.722

I am likely very wrong

[Shahrukh]

I think I am wrong, but I tried using the little math I remebered:

[attachment=2142:questionfour.jpg]

h² = 36 + b²
g² = 9 + b²

h² + g² = 81
(36 + b²) + (9 + b²) = 81
45 + 2b² = 81
2b² = 36
b² = 18
b = √18

tan ß = 3/b
tan ß = 6/x

3/b = 6/x
3x = 6b
x = 2b
x = 2√18
x = 8.458

*TADA*

[Bikerman]

Shahrukh, on 14 August 2010 - 08:13 PM, said:

I think I am wrong, but I tried using the little math I remebered:

questionfour.jpg

h² = 36 + b²
g² = 9 + b²

h² + g² = 81
(36 + b²) + (9 + b²) = 81
45 + 2b² = 81
2b² = 36
b² = 18
b = √18

tan ß = 3/b
tan ß = 6/x

3/b = 6/x
3x = 6b
x = 2b
x = 2√18

x = 8.458

******OUCH... So good until there. But I'll forgive and forget the minor error
8.485 not 8.458 (you transposed 2 digits)

Good effort, but you can do it more simply.......look to do it using only the top big right triangle but using the two right triangles that make it up. Look for a common angle which will allow solution in 2 expressions....

BUT before you can, you need to spot a relationship between some triangles....think back....congruent?....similar?.....

See if my crappy sketch helps

Edited by Bikerman, 14 August 2010 - 08:43 PM.

[Shahrukh]

I did mean to write 8.485. Mistyped by mistake.

I did know that the triangles had a relation but couldn't remember the name nor the relation.
So, I did what little I remembered.
I was good at Maths back then. Scored 98% in my O' Levels and 96% in Additional Mathematics.

[Bikerman]

Shahrukh, on 14 August 2010 - 09:02 PM, said:

I did mean to write 8.485. Mistyped by mistake.

I did know that the triangles had a relation but couldn't remember the name nor the relation.
So, I did what little I remembered.
I was good at Maths back then. Scored 98% in my O' Levels and 96% in Additional Mathematics.

Yes, you did solve it so QUDOS to you.

BEA and CED are smiliar triangles (same angles sides twice as big) that immediately gives you EB - 1/2x
From there it is much easier...

[Bikerman]

Problem 2 - these from my maths cohort at Frih

They asked the baker, Mr. Wis, how many breads he had sold on monday. The baker answered:

My first customer bought half of the stock + half a bread. My Second customer, bought half of what was left + half a bread. The third customer asked for half of the stock that was still available + half a bread to feed the ducks. The last Customer, Mr Salomon, wanted to buy a bread, but there were no more breads.

Mr Bread didn't sold any half breads. Can you tell me how many breads he sold?

Problem 3
In a farm house there are some hens and pigs. There are altogether 60 eyes and 86 feet. So, give me the total number of hens and pigs in the farm.

Edited by Bikerman, 14 August 2010 - 09:34 PM.

[Shahrukh]

Bikerman, on 14 August 2010 - 09:26 PM, said:

Yes, you did solve it so QUDOS to you.

BEA and CED are smiliar triangles (same angles sides twice as big) that immediately gives you EB - 1/2x
From there it is much easier...

Hey, I reached to that too. But I thought I was doing it wrong.
Its been about 4 years since I did any Maths.

Answer 2:

Total : x
1st: x/2+.5
2nd: x/4+.75
3rd: x/8+.875
Since inventory ended then:
x/2 + x/4 + x/8 + 2.125 = x
.875x +2.125 = x
2.125 = .125x

x = 17

Edited by Shahrukh, 14 August 2010 - 09:58 PM.

[Bikerman]

Yes correct answer: well worked out.

[Shahrukh]

Bikerman, on 14 August 2010 - 11:00 PM, said:

Yes correct answer: well worked out.

Thank you.

Answer 3:
h = total hens
p = total pigs

2h + 2p = 60
2h + 4p = 86

60 - 2p = 86 - 4p
2p = 26
p = 13

2h = 60 - 26
2h = 34
h = 17

[Bikerman]

Shahrukh, on 15 August 2010 - 12:27 AM, said:

Thank you.

Answer 3:
h = total hens
p = total pigs

2h + 2p = 60
2h + 4p = 86

60 - 2p = 86 - 4p
2p = 26
p = 13

2h = 60 - 26
2h = 34
h = 17

Yet again correct....applause rings out...
OK as the final one of this series, something a bit more tricky perhaps...

Question 4

In my street, there's only one side with houses. The sum of all the housenumbers smaller than mine is equaly to the sum of all housenumbers bigger than mine. The 'connection/relation' between my housenumber x and all houses in my street is?

[rvalkass]

Bikerman, on 15 August 2010 - 03:27 AM, said:

In my street, there's only one side with houses. The sum of all the housenumbers smaller than mine is equaly to the sum of all housenumbers bigger than mine. The 'connection/relation' between my housenumber x and all houses in my street is?

Is that you're the only house on the street

Sum of all the house numbers smaller than yours = 0
Sum of all the house numbers greater than yours = 0

0 = 0

Q.E.D.

I think this is why my teachers hated me at school

[Bikerman]

rvalkass, on 15 August 2010 - 07:19 AM, said:

Is that you're the only house on the street

Sum of all the house numbers smaller than yours = 0
Sum of all the house numbers greater than yours = 0

0 = 0

Q.E.D.

I think this is why my teachers hated me at school

That could be a solution, but it is a bit trivial :-) There is another less trivial one

[rvalkass]

Fine, if I must actually make use of maths

Let the total number of houses on the street = N
Your house number = x

Sum of all the house numbers lower than yours, S_low = 1/2 * (x-1) * x
Sum of all the house numbers higher than yours, S_high = 1/2 * (N-x) * (N+x+1)

To satisfy the question S_low = S_high
x(x-1) = (N-x)(N+x+1)
x²-x = N² + N - x² - x
2x² = N² - N
x = sqrt(1/2 * (N² + N))

So, if there are 42 houses on the road, you'd be number 30. Of course this should really be limited to the case where whole integer values of x are solutions, as few people number their house 30.05. This also assumes the houses are numbered in the sequence 1, 2, 3... rather than you living on the odd or even side of the street, houses having been demolished or converted since original construction, and general screwups by Royal Mail.

[Bikerman]

Neatly done. It took me two attempts (I couldn't remember the expression for the sum of numbers < n and had to work it out from basics. I worked it out for 0<n<N OK then I stupidly forgot to subtract the 1 when doing the problem - dumbass - I only saw the error when I crunched it - still, got it right the second time ).

Quote

Of course this should really be limited to the case where whole integer values of x are solutions, as few people number their house 30.05. This also assumes the houses are numbered in the sequence 1, 2, 3... rather than you living on the odd or even side of the street, houses having been demolished or converted since original construction, and general screwups by Royal Mail.

Typical physicist - always the specific case and never the general solution. No rigour at all....bleedin atom smashers...

Edited by Bikerman, 15 August 2010 - 09:17 PM.

[Bikerman]

Here are some interesting problems of a more philosophical nature.

* Link 1
* Link 2

I normally rate myself as pretty good but some of the questions caught me out fair and square...