Word Problems. :-)

[GlueStick1]

What is the difference between the sum of the first 2003 even numbers and the sum of the first 2003 odd numbers?

Let d and e denote the solutions to the equation:
2x^2+3x-5=0, what is the value of (d-1)(e-1)?

A solid box, 15cmx10cmx8cm, has a solid cube, 3cm on each side, removed from each corner of the box. What percentage of the original box was removed?

It takes Devon 30 min. to walk the 1 km to school. It then takes him 10 minutes to return home. What is his average speed for the trip?

 

[Unco]

"What is the difference between the sum of the first 2003 [positive] even numbers and the sum of the first 2003 [positive] odd numbers?"

The sum of the first two positive
.....even numbers: 0 + 2 = 2
.....odd numbers: 1 + 3 = 4
Difference: 4 -2 = 2

The sum of the first three positive
.....even numbers: 0 + 2 + 4 = 6
.....odd numbers: 1 + 3 + 5 = 9
Difference: 9 - 6 = 3

The sum of the first five positive
.....even numbers: 0 + 2 + 4 + 6 + 8 = 20
.....odd numbers: 1 + 3 + 5 + 7 + 9 = 25
Difference: 25 - 20 = 5

Can you see the pattern?

 

[Unco]

Let d and e denote the solutions to the equation:
2x^2+3x-5=0, what is the value of (d-1)(e-1)?

What are the solutions to 2x^2+3x-5=0? That'll give us 'd' and 'e'.

Can you factorise it?

If you have trouble...

Multiply 'a' and 'c' terms: 2*-5 = -10, call this the 'd' term.
Find two numbers that sum to the 'b' term, 3, and multiply to give the 'd' term, -10: 5 * -2 = -10; 5 -2 = 3 so 5 and -2 will suffice.

Rewrite the equation with 5 and -2 as the new 'b' terms:
2x^2 + 5x - 2x - 5 = 0

Rewrite to make factorisation more obvious:
2x^2 - 2x + 5x - 5 = 0

Factorise the first two terms, and factorise the last two terms:
2x(x - 1) + 5(x - 1) = 0

There is a common factor of (x-1):
(2x + 5)(x - 1) = 0

Thus
x = -5/2 or x=1

Thus d = -5/2 and e = 1 (or e=-5/2 and d=1, it doesn't matter)

Following that it's a matter of plugging 'd' and 'e' into (d-1)(e-1).