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Math for fun thread


tk102

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Posted

Okay, here's me in action:

 

700/6 = 116.6(recurring), rounded off to 117

700 - 117 = 583, this the amount Mike pays in dollars.

583/5 = 116.6 :eyeraise:

Therefore, 583 - 116.6 = 466.4

 

Final Answer: $466.4. If that is right, this was easier than pie. :p

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Posted
  Sabretooth said:
Okay, here's me in action:

 

700/6 = 116.6(recurring), rounded off to 117

700 - 117 = 583, this the amount Mike pays in dollars.

583/5 = 116.6 :eyeraise:

Therefore, 583 - 116.6 = 466.4

 

Final Answer: $466.4. If that is right, this was easier than pie. :p

 

Close enough, if you don't round off until the end, your answer should be $466.66. But, you did it correctly, and probably still deserve full marks.

 

Your turn. ;)

  • 2 weeks later...
Posted

Using only lower case letters of the English alphabet, how many 4-letter "words" can be created using unique letters with at least one vowel(a,e,i,o,u,y) in each word? No other rules apply.

Posted
  tk102 said:
Using only lower case letters of the English alphabet, how many 4-letter "words" can be created using unique letters with at least one vowel(a,e,i,o,u,y) in each word? No other rules apply.

When I saw this thread I could only think of one four letter word with at least one vowel and no repetitions: "nerd" :xp: (j/k)

 

Sorry for the unorthodox method but I don't remember my math :p

 

Total words with 4 letters (26x25x24x23) - Total words with no vowels (20x19x18x17)

 

So the answer should be 242520 if I am correct.

Posted

Darth333 takes the short cut to the right answer! :D

 

Here's the long method:

 

Words with 1 vowel (V) in them can have the vowel appear in one of the 4 positions with the others being taken by consonants ©.

VCCC

CVCC

CCVC

CCCV

 

and there are 6 possible vowels. The first consonant has 20 possibilities, the 2nd has 19, and the 3rd has 18 so:

 

1 vowel words = (6*20*19*18) + (20*6*19*18) + (20*19*6*18) + (20*19*18*6) = 164160

 

2 vowel words:

VVCC

VCVC

VCCV

CVVC

CVCV

CCVV

 

six variations of (6*5*20*19) = 68400

 

3 vowel words:

VVVC

VVCV

VCVV

CVVV

 

4 variations of (6*5*4*20) = 9600

 

and 4 vowel words:

 

1 variation of 6*5*4*3 = 360

 

add them all up:

 

164160+68400+9600+360=242520 :xp:

 

I like D3's way better.

Posted

:eyeraise: "Why make simple when we can make complicated?" If it was on paper that answer would have taken at least one more tree :xp:

 

 

I didn't had the courage to go through that!

  • 3 weeks later...
Posted

You have 50 coins from the USA totaling $10.00.

 

You have more nickels ($0.05) than pennies ($0.01), you have more dimes ($0.10) than nickels, and you have more quarters ($0.25) than dimes.

 

How many coins of each type do you have? (two answers)

Posted

That's one! But you didn't show your work! :D

 

25q + 10d + 5n + p = 1000

q + d + n + p = 50

q > d > n > p

 

p must be divisible by 5 in order to get the nice round number of $10.00

 

Test: can p = 5?

 

 

pennies nickels dimes quarters COUNT VALUE

5 6 7 32 50 $9.05

 

No! Even if we maximize the number of quarters, we can't reach $10 with 5 pennies in play.

 

Therefore pennies = 0.

 

Now we can easily reach $10...

 

pennies nickels dimes quarters COUNT VALUE

0 1 2 37 50 $12.00

 

 

We'll try shifting the coins from quarters to dimes until we fall below $10.00

 

pennies nickels dimes quarters COUNT VALUE

0 1 16 33 50 $9.90

 

So we know we need at least 34 quarters. Try that

 

q=34 (VALUE = $8.50)

 

n+d = 16

5n+10d = 150

5(16-d) + 10d = 150

80 + 5d = 150

d = 14

n = 2

 

 

pennies nickels dimes quarters COUNT VALUE

0 2 14 34 50 $10.00

 

 

 

What about 35 quarters?

 

q=35 (VALUE = $8.75)

 

n+d = 15

5n+10d = 125

5(15-d) + 10d = 125

75 + 5d = 125

d = 10

n = 5

 

 

pennies nickels dimes quarters COUNT VALUE

0 5 10 35 50 $10.00

 

 

 

What about 36 quarters?

 

q=36 (VALUE = $9.00)

 

n+d = 14

5n+10d = 100

5(14-d) + 10d = 100

70 + 5d = 100

d = 6

n = 8 ! no d must be greater than n.

 

 

 

That is all.

Posted

I'll start out with an easy one - mainly because I'm rusty with the longhand of this too. :D

 

T6B-1.gif

 

Calculate the contribution (In Amps) of the 480 V source to Io (the current through the 140 Ohm Resistor).

 

Hint: Super-positioning is the shortcut through this.

Posted

To determine the contribution of the 480V, we isolate it by opening the current source and shorting the other voltage source...

 

 

100 140

+-/\/\/\---/\/\/\---+------+

| | |

-+- | \ 150

480V | /

-+- | \

| | |

+-------------------+------+

 

 

 

240

+----------/\/\/\---+

| |

-+- |

480V |

-+- |

| |

+-------------------+

 

480V through 240ohms = 2A

Posted

Looking at the 2nd voltage source

 

<----

720V through 240ohm = -3A

100 140

+-/\/\/\---/\/\/\---+------+

| | |

| --- \ 150

| 720V /

| --- \

| | |

+-------------------+------+

 

 

The current source is a little trickier. Need to find the equivalent voltage.

 

1/R = 1/R1 + 1/R2

1/R = 1/100 + 1/140

R = 58.33

12A * 58.33 Ohm = 700V

100 140

+-/\/\/\-+-/\/\/\---+------+

| | | |

| ^ | \ 150

| 12A | /

| ^ | \

| | | |

+--------+----------+------+

 

 

58.333 58.333

+-/\/\/\---+ +-/\/\/\---+

| | | |

^ | ---- |

12A | ----> 700V |

^ | ---- |

| | | |

+----------+ +----------+

 

 

 

100 140 700V through 140 ohms = 5A

+-/\/\/\-+-/\/\/\---+------+

| | | |

| ---- | \ 150

| 700V | /

| ---- | \

| | | |

+--------+----------+------+

 

Io: 2A + -3A + 5A = 4A

 

 

 

P.S. Thank you Rhett. I completely forgot all of this stuff the minute I left my EE210 class. Good to know part of my brain could pick it back up again. I'll see if I can come up with another problem unless someone else posts one first.

Posted

On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups?

Posted

Via trigonometry:

 

tan(theta) = 54.663/36.7

theta = arctan(54.663/36.7)

theta = 56.123°

 

sin(theta) = 54.663/x

x = 54.663 / sin(theta)

x = 65.84

 

or by Pythagoras:

x = sqrt(36.7^2 + 54.663^2) = 65.84

 

but what about the snow-cone guy? :(

Posted

(Pi)(R^2)(H) = area of cylinder

 

(1/3)(Pi)(R^2)(H) = area of cone

 

So a comparable cylinder is holding three times as much.

 

(.75)(3) = $2.25

 

So the answer is that he darn well better still be charging 75 cents because there's no way in heck I'm paying that much for a snow cone.

Posted
  tk102 said:
On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups?

 

Assuming that the paper cups are cylindric and having on mind that a cylinder have a volume that's 3 times greater than that of a cone, I would say:

 

Cylinder radius and height = x

Cones radius and height = x

 

Therefore, they're equal. But the cylinder got a volume 3x greater.

 

If logic applies, then he should sell the cups 3 times more expensive.

 

Answer: $2.25

 

Edit: Oh, I didn't saw Rhett's reply...

Posted

Ah you're both right of course. I always thought it was strange that a cone was 1/3 the volume of the equivalent cylinder. After all, if you draw a rectangle and then draw and isosceles triangle inside it, that triangle has 1/2 the area of the rectangle. But if you spin that configuration on the axis, the volume encompassed by the triangle (now a cone) is 1/3 the volume encompassed by the rectangle (now a cylinder). Kind of spooky.

Posted
  tk102 said:
On a hot summer's day a guy selling snow-cones runs out of cones and has to use paper cups instead. If the cups are the same height and same radius as the cones, and the snow-cones cost $0.75 each, how much should he charge for the snow-cups?
The invisible hand of the free market will, like the hand of God Himself, set the price of a paper cup at its optimal price and ensure the that our way of life will not be threatened by the sanguine tentacles of Communism :carms:
Posted

To 1 decimal place? Might as well use trial and error to get that rather than use the cumbersome cubic equation.

 

4^3 = 64

3^3 = 27

 

so we know it's between 3 and 4.

 

3.4 -> 42.7

3.3 -> 39.23

 

so 3.3 close enough.

Posted
  tk102 said:
To 1 decimal place? Might as well use trial and error to get that rather than use the cumbersome cubic equation.

 

4^3 = 64

3^3 = 27

 

so we know it's between 3 and 4.

 

3.4 -> 42.7

3.3 -> 39.23

 

so 3.3 close enough.

Actually, I did find that out (with the help of Microsoft Windows Calculator), but I thought trial and error was evil in real maths. :(

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