CPhill

OK

When we cut all the washers out of the sheet   (  and before we cut any holes in the washers )  the area occupied by each washer  is just  pi  ( outside radius)^2  = pi (1 cm)^2  = pi cm^2

[Note that if the diameter of each washer is 2 cm....then the radius of each is just 1 cm  ]

So....the area occupied by each washer is that of a circle with a radius of 1 cm  =

pi (radius)^2  = pi  (1 cm)^2   =  pi  cm^2

So.....the total are occupied by these washers is  just  7500* pi  cm^2

And the area of the sheet  is  just  (300 cm * 100 cm)  = 30000 cm^2

So....the amount of excess metal surrounding the washers   is just the total area  of the sheet  less the total area occupied by all the washers   = [ 30000  - 7500 pi ] cm^2    (1)

When we take all the washers off the original sheet  [ 7500 of them ] we will cut the centers out  of each of them.....the amount of metal in the center  (the area of metal that will form the hole in the washer )  is just  pi * ( inside radius)^2  = pi  (1/2 cm)^2  = pi/4 cm^2

Since we have 7500 of these....the total area of excess metal from this part is just

7500 *[ pi /4 cm^2]  =   1875 pi  cm^2     (2)

So  the total amout of metal  that we have  to cut washers from is now just  the sum of (1)  and (2)  =

[30000 cm^2  - 7500 pi cm^2]  + 1875 pi cm^2  =  30000cm^2  - 5625 pi cm ^2  =

[ 30000  - 5625 pi ] cm^2   = area of the new sheet 

And each washer  cut from this sheet will  occupy the same area as before  = pi cm^2

So....the total number   of washers we can cut from the new sheet  is

[ Area  of new sheet  ] / [ Area of each washer ] =  [ 30000 - 5625pi] cm^2  /  pi  cm^2   =

[ 30000  - 5625 pi ]  / pi   =

about   3924  washers

Does that make sense  ??

Yeah...I see what you mean, Gavin...your answer is probably better....mine might just be coincindence....I don't know....!!!

Here is my attempt, Ghosty.. however..I don't come up with  "3050"

Your answer to "a"   is correct  !!!

Assuming  a uniform thickness of the sheets  throughout...

We have 7500 washers .......

Note that the amount of leftover metal that is cut from the center of each washer  is just

pi (inside radius)^2   =  pi (1/2 cm)^2  =  pi/ 4  cm^2

So.....the amount of this excess metal for 7500 washers is just  7500 (pi/ 4)cm^2  =

1875 pi cm^2    (1)

However....we must also figure the additional amouint of excess metal on the sheet that surrounds all of the washers

Note that the area of the sheet is just  (300cm  * 100cm)  = 30000cm^2

And the are occupied by each washer is just  pi (outside radius)^2  = pi (1 cm)^2  = pi  cm^2

So......7500 washers occupy  7500pi  cm^2

So....the excess metal  surrounding all the washers is  just

Area of the sheet  - area  occupied by the washers =

(30000  - 7500pi) cm^2   (2)

So....the total amount of excess metal available  for forming new washers  is just  the sum of   (1)  and (2)  =  [  1875 pi  +  (30000 - 7500pi) ] =  [30000 - 5625 pi ]  cm^2

So....the total number of   possible washers   that can be made with a new sheet   is just

[  Area  of metal available  ]  / [ Area  occupied by each washer ]   =

[ 30000 - 5625pi ] cm^2  / [ pi] cm^2  ≈  3924 additional washers 

Thanks, Gavin....!!!!

Here's another way to see this....

Note that there are 9  faces on each side of the large cube that are painted....so....the total number of painted faces is  just  9 * 6  =  54  painted faces

And the total number  of faces on the smaller cubes  is just  27 cubes * 6  faces on each  = 162

If we put all the cubes in a bag and selected only one, the probability that it would have a painted face  is just  54  /162  =  1 / 3

Thaks for that recap, X^2......you certainly have been a valuable member, here  !!!!

Here's my best effort

We have 4 ways to color A, 3 ways to  color  B, 2 ways to color C  and one way to color D

So.....the the total ways to color the rectangles  = 4 * 3 * 2 * 1  = 4!  =  24 ways

(ax + b) ( bx + a)  =  ab x^2  +  ( a^2 + b^2) x  +  ab

So..equating coefficients, then  ab  = 26   and  a^2 + b^2  = v

If  ab   = 26   and a,b are distinct intergers....then we have the following possibilities for a,b

a         b             a^2  +  b^2

26       1             26^2 + 1^2  =  677

13       2             13^2  + 2^2  = 173

So...the minimum value for v  =  173

I'm assuming that we have the following :

Let A  = (0,0)

Since BE   = 2, then AB  =  3BE  = 6...so....AE  = 4

And since angle BAC  = 60°, the point B  is  (6 cos 60°, 6 sin 60°)  = (3, 3√3)

And point  E  is   ( 4cos 60°, 4 sin 60°)   = (2, 2√3)

And  since  AD  = 3, then AC  = 3AD = 9

So  D  is  (3, 0)

So.....the  area  of BCDE  =  area  of triangle ABC  -  area of triangle  AED

Area of triangle ABC  =  (1/2)DB * AC  = (1/2) (3√3) (9)  = (13.5)√3  units^2

And area  of triangle AED  = (1/2) (AD) (2√3)  = (1/2)(3)(2√3)  = 3√3 units^2

So...the area  of  quadrilateral BCDE  = [ (13.5)√3  - 3√3] units^2  = (10.5√3) units^2

Here's the second one

2x^2 - 8x  +  7  =   0

Let's write this in the form  mx^2 + nx  + q  = 0    to avoid confusion with the roots

And we can write 

1 / (2a)   +  1 / (2b)  =    [a + b ] / [ 2ab ]

Using a similar idea as in the first one

The sum of the roots ( a + b)  = -n/m =  - (-8) / 2  = 8/2  = 4

The product of the roots  ab  = q/m    = 7/2

So...

1/(2a) + 1/ (2b)   =[ a + b ] / [2ab]  =  4 / [ 2 (7/2) ] =  4 / 7 

This isn't as difficult as  it seems...we just need to use some "creative" Algebra

4x^2 +9x - 6 =  0

In the form  ax^2  + bc  + c  = 0

The sum of the roots,  ( s + t )  =   -b/ a =    -9 / 4

The product of the roots,   (st )  = c /a = -6/4  = -3/2

Note that  we can write

s/t  +  t /s     =   [ s^2  + t^2 ] / st

And note that  ( s + t)^2  = s^2  + 2st + t^2

So

(-/9/4)^2  = s^2  + 2(-3/2) + t^2

81/16  =  -3  +  [s^2  + t^2]  .....    add 3 to both sides

81/16 + 3  = s^2 + t^2

81/16 + 48/16 = s^2 + t^2

129/16  = s^2 + t^2

So......

s/t + t/s  =   [ s^2 + t^2 ]/ st   =  [ 129/16] / [-3/2)]  =  [129/16] * [ -2/3]  = -43 / 8