Math Is Fun Forum

evene

Member

Registered: 2015-10-18

Posts: 272

Some Problems!

Missed me? (Don't answer that)

(1) If x and y are two different numbers selected from the integers from 500 to 1000, inclusive, what is the greatest value for

(2) Train A and Train B depart at the same time from the same station for a destination 4,000 miles away. If Train A travels at a constant rate of 170 mph and Train B travels at a constant rate of 200 mph, how far will the two trains be from each other, in miles, when Train B reaches its destination?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Some Problems!

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Some Problems!

Odd.

The answer choices for (2) are (a) 30. (b) 90. (c) 450. (d) 600 and (e) 900. I think that the question for (2) also meant that Train A and B's destination are the same station that is 4 000 miles away.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Some Problems!

Hi;

Do they agree with 1) 1999?

Of course now, the answer for 2) is d) 600.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Relentless

Member

Registered: 2015-12-15

Posts: 631

Re: Some Problems!

Hey;

It is 1999, when x is 1000 and y is 999.

Intuitively, you want the denominator extremely low (difference of 1) and the numerator (sum) as high as possible.

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Some Problems!

Thanks!

Yes, I agree with 1999 for (1)

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Some Problems!

Another problem:

Assume that

are real numbers such that

Find the value of

.

My fingers are dead!!

Last edited by evene (2016-01-18 03:40:13)

Relentless

Member

Registered: 2015-12-15

Posts: 631

Re: Some Problems!

Hi, being too lazy at the moment to work this out in detail, I just made a rule for the three values 1,12,123. It predicts that the next number is 334

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Some Problems!

How did you make the rule?

I was hoping it was going to be 1234. But, I guess not!

Relentless

Member

Registered: 2015-12-15

Posts: 631

Re: Some Problems!

Haha, yes that would have been neat, but the numbers do not increase rapidly enough to continue in that manner.

The method is to define a polynomial with a degree one less than the amount of numbers you have (in this case, degree 2). So you get ax^2 + bx + c
You then substitute x=1 for the first number, x=2 for the second, and so on.
a+b+c = 1
4a+2b+c=12
9a+3b+c=123

Solving that system of equations gives the polynomial 50x^2 - 139x + 90, which exactly fits the numbers.

But you asked how I made the rule. I put the numbers in a table in a graphing calculator. As I said: lazy

Last edited by Relentless (2016-01-18 05:40:25)

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Some Problems!

One morning, each member of Angela's family drank an 8 ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Show your work please!