Algebra Cafe: Solve Math Problems Here

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Author Topic: Algebra Cafe: Solve Math Problems Here  (Read 2304 views)
kobe (m)
Algebra Cafe: Solve Math Problems Here
« on: March 29, 2008, 07:38 PM »

Forget the meaning of life--the real search is for the value of x!


Basic Algbebra:

A number has two digits.  When the two digits that make up the number are added, their sum is 15.
When the number is reversed ( for example 12 becomes 21), the new number is 9 more than the initial number.

What is the number?  Provide mathematical details of your answer.

cheers Grin
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #1 on: March 29, 2008, 07:48 PM »

A boy is three years younger than his sister. If his age three years ago was two-thirds her age at that time, what are their present ages?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #2 on: March 29, 2008, 08:13 PM »

Solve this equation:

Solve: sqrt(x-4) + 10 = sqrt(x+4)
drrionelli (m)
Re: Algebra Cafe: Solve Math Problems Here
« #3 on: March 30, 2008, 10:06 PM »

The answer to the first question is 78 and 87.

87-78=9

8+7 or 7+8 equals 15
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #4 on: March 30, 2008, 11:22 PM »

Let x be the first digit and y the second digit of the first number.  Thus the first number could be written xy.

Since this is a two digit number it is actually composed thus;

xy => 10x + y

When the digits are reversed, we get the number yx, which really translates to;

yx => 10y +x

The sum of the digits of the first number is 15, thus

x + y = 15           ------------------------------------------- (1)

The second number is 9 more than the first, thus

10x + y  + 9  = 10y + x, which reduces to

x - y = -1  --------------------------------------------------- (2)

So we have two equations, (1) and (2)

From whence x = 7 and y = 8.

So the first number is 78.

QED


4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #5 on: March 30, 2008, 11:29 PM »

Excellent answer Huxley . . . i just couldnt figure out how to work with the "two digit number" information.
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #6 on: March 30, 2008, 11:30 PM »

sqrt(x-4) + 10 = sqrt(x+4)

Square both sides;

(x-4) +20sqrt(x-4) +100 = (x+4)

20sqrt(x-4) = -92

x = (-92/20)^2 + 4
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #7 on: March 30, 2008, 11:32 PM »

Quote from: kobe on March 29, 2008, 07:48 PM
A boy is three years younger than his sister. If his age three years ago was two-thirds her age at that time, what are their present ages?

Let sisters age now = a
boy's age now = a - 3

3 years ago

Sisters age = a - 3
boy's age = (a - 3) - 3 = a - 6

therefore (a - 6) = 2/3(a - 3)

Solving equations for a . . . sister's age now = 12, brother's age now = 9.
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #8 on: March 30, 2008, 11:39 PM »

Quote from: kobe on March 29, 2008, 08:13 PM
Solve this equation:

Solve: sqrt(x-4) + 10 = sqrt(x+4)

I expected the answer would be a round figure. Arrived at the same answer as Huxley . . .
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #9 on: March 30, 2008, 11:45 PM »

Quote from: 4Him on March 30, 2008, 11:29 PM
Excellent answer Huxley . . . i just couldnt figure out how to work with the "two digit number" information.

Thanks for the compliments. I remembers setting a similar problems to a class of students learning mathematical modelling some years ago.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #10 on: March 31, 2008, 05:46 AM »

Wow good job guys.  you've certainly more than met expectations, way to go.  I'll post more below.

Enjoy  Wink
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #11 on: March 31, 2008, 06:29 AM »

A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 100 seniors, 150 juniors, and 200 sophomores who applied. Each senior's name is placed in the lottery 3 times; each junior's name, 2 times; and each sophomore's name, 1 time. What is the probability that a senior's name will be chosen?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #12 on: March 31, 2008, 06:47 AM »

A number P is divisible by 2, 3, and 15.  Find a number P, such that P + 30 is also divisible by 2, 3, and 15.
Show your work.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #13 on: March 31, 2008, 06:52 AM »

Of the 6 courses offered by the music department at her college, Kay must choose exactly 2 of them. How many different combinations of 2 courses are possible for Kay if there are no restrictions on which 2 courses she can choose?

kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #14 on: March 31, 2008, 06:58 AM »

Let the function f be defined by f(x) = x2 - 7x + 10 and f(t + 1) = 0, what is one possible value of t?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #15 on: March 31, 2008, 10:35 AM »

If the positive integer y leaves a remainder of 2 when divided by 8, what will the remainder be when y + 9 is divided by 8?
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #16 on: March 31, 2008, 05:45 PM »

Quote from: kobe on March 31, 2008, 10:35 AM
If the positive integer y leaves a remainder of 2 when divided by 8, what will the remainder be when y + 9 is divided by 8?

y = 8n + 2

so, if we add 6 to y, it will make it a multiple of 8. So we are left with a 3.

So the remainder is 3.
his_grace (m)
Re: Algebra Cafe: Solve Math Problems Here
« #17 on: March 31, 2008, 05:49 PM »

before i sleep with my lover - the number of hair on my di ck is 53, after withdrawing my di ck from the pu ssy, the number of hair on my di ck has reduced to 45, but on careful look, 10 is inside my lovers' pu ssy, and five is on the bed - what is the total hair altogether?
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #18 on: March 31, 2008, 06:00 PM »

Quote from: kobe on March 31, 2008, 06:58 AM
Let the function f be defined by f(x) = x2 - 7x + 10 and f(t + 1) = 0, what is one possible value of t?

t = 4 or 1

(t+1)2 -7(t+1) +10 = 0

t2 - 5t + 4 = 0

(t-1)(t-4) = 0

whence t = 1
      or    t = 4

QED

huxley
Re: Algebra Cafe: Solve Math Problems Here
« #19 on: March 31, 2008, 06:05 PM »

Quote from: kobe on March 31, 2008, 06:47 AM
A number P is divisible by 2, 3, and 15.  Find a number P, such that P + 30 is also divisible by 2, 3, and 15.
Show your work.

I don't think this one is very well defined.  In fact, P can be any multiple of 30, thus

30, 60, 90, 120, ,
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #20 on: March 31, 2008, 07:49 PM »

Quote from: kobe on March 31, 2008, 06:29 AM
A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 100 seniors, 150 juniors, and 200 sophomores who applied. Each senior's name is placed in the lottery 3 times; each junior's name, 2 times; and each sophomore's name, 1 time. What is the probability that a senior's name will be chosen?

this is tough . . . don't have a good background in statistics but i think the answer is 5/18

Anyone solved it yet?
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #21 on: March 31, 2008, 07:53 PM »

Quote from: kobe on March 31, 2008, 06:52 AM
Of the 6 courses offered by the music department at her college, Kay must choose exactly 2 of them. How many different combinations of 2 courses are possible for Kay if there are no restrictions on which 2 courses she can choose?

i think this is 6 P 2 = 30
donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #22 on: March 31, 2008, 08:05 PM »

Quote from: kobe on March 31, 2008, 07:49 PM
A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 100 seniors, 150 juniors, and 200 sophomores who applied. Each senior's name is placed in the lottery 3 times; each junior's name, 2 times; and each sophomore's name, 1 time. What is the probability that a senior's name will be chosen?


let me try. There are 100 seniors each with three opportunities. that makes a total of 300 opportunities. same goes for the sophomores and juniors bring it to a total of 800 opportunities. So, probablity that a senior will be chosen is number of senior opportunities diviided by total number of opportunities. which is 300/800 or 3/8


i hope we r not doing your assignment for u Grin
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #23 on: March 31, 2008, 08:09 PM »

Quote from: donchichi on March 31, 2008, 08:05 PM
let me try. There are 100 seniors each with three opportunities. that makes a total of 300 opportunities. same goes for the sophomores and juniors bring it to a total of 800 opportunities. So, probablity that a senior will be chosen is number of senior opportunities diviided by total number of opportunities. which is 300/800 or 3/8

Your answer is the correct one.
dinozzo
Re: Algebra Cafe: Solve Math Problems Here
« #24 on: March 31, 2008, 08:22 PM »

Quote from: kobe on March 31, 2008, 06:29 AM
A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 100 seniors, 150 juniors, and 200 sophomores who applied. Each senior's name is placed in the lottery 3 times; each junior's name, 2 times; and each sophomore's name, 1 time. What is the probability that a senior's name will be chosen?
I think its 3/8, been a while though so I'm not sure. Since there are 100 seniors and each seniors name entered 3 times so thats 300 entries, 150 juniors entered twice thats another 300 entries then sopohomore's once that makes 800 entries in all, 300 of which are seniors. So probability will be 300/800 which will be 3/8.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #25 on: April 01, 2008, 05:12 AM »

Quote from: 4Him on March 31, 2008, 07:53 PM
i think this is 6 P 2 = 30

should be 15.

woohoo, you guys are getting them all right.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #26 on: April 01, 2008, 07:22 AM »

How many liters of a 10% solution of acid should be added to 20 liters of a 60% solution of acid to obtain a 50% solution?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #27 on: April 01, 2008, 07:27 AM »

I have $11.60, all dimes and quarters, in my pocket. I have 32 more dimes than quarters. How many dimes, and how many quarters, do I have?

For non-Americans, a dime is $0.10 and a quarter is $0.25.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #28 on: April 01, 2008, 07:30 AM »

I was able to balance a weightless beam, ten meters long, on a fulcrum by putting a 6 kg mass on one end, and an unknown mass on the other end. To balance this, I had to place the fulcrum 1.5 meters from the 6kg mass. What is the unknown mass?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #29 on: April 01, 2008, 07:31 AM »

Rasheed can mow his mother's lawn in 6 minutes. His brother Peter can mow it in 9 minutes. How long will it take them to do it together, if each has his own lawnmower.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #30 on: April 01, 2008, 07:52 AM »

The following is invalid, explain why.

a = x         
a+a = a+x       
2a = a+x     
2a-2x = a+x-2x     
2(a-x) = a+x-2x     
2(a-x) = a-x       
2 = 1       
donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #31 on: April 01, 2008, 08:09 AM »

Quote from: kobe on April 01, 2008, 07:52 AM
The following is invalid, explain why.

a = x
a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
 2 = 1


if we reverse the eqn from the start;

x = a
x+x = x+a
2x = x+a

Now referring to your equation at exactly
 
2a-2x = a+x-2x

if we substititute the value fro 2x on the RHS, we get

a+x-(x+a) which gives 0

the 2x in the equation is a not a single entity but a sum of two entities
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