Lawrence has (15/32) of the money before he gave any away....and Kenny had (8/32) =(1/4) of the money before getting any from Lawrence....and John has (9/32) of the money, originally...call T the total...and we know that....

(1/4)T + .40(15/32)T =  (9/32)T + 60    simplify

(1/4)T + (2/5)(15/32)T  = (9/32)T + 60

(1/4)T + (3/16)T  = (9/32)T + 60

(7/16)T  = (9/32)T + 60     subtract (9/32)R from both sides

(7/16 - 9/32)T   = 60

(5/32)T = 60     multiply both sides by (32/5)

T = 60(32/5)  = $384   and this is the total  amount of money

And Lawrence gave Kenny  (3/16)(384)  = $72

Let Lucy's money = L  and Mary's money = M....and we're told that

(5/8)L = (2/3)M     multiply both sides by  (3/2)

(3/2)(5/8)L = M  =  (15/16)L

And we have

L + M = 93    and substituting for M, we have

L +(15/16)L = 93

(31/16)L  = 93     multiply both sides by   (16/31)

L  = (93)(16/31)  =  $48  = Lucy's money

And Mary's is $(93 - 48) =  $45

Yes, we agree, this question has no solutions.   

I looked at this last night, Melody.....I don't believe that a "positive" answer results - at least it didn't for "y"....I didn't consider it any further.....maybe the info is incorrect???

Chris, could you try this one please :)

let there be x short sleeve shirts and y long sleeve shirts.

profit will be $3 for each short sleeve and $6 for each long sleeve

so                3x+6y=1325

cost will be   5x+7y=2450

I have tried solving this simultaneously but it does not work.   

Call the amount in the full tank = A    ....so...

A - 600 = (5/8)A    rearrange

(3/8)A = 600     multiply both sides by (8/3)

A = 1600ml  ....so   (1600 - 600)ml = 1000ml = 1L

Let's call the number of red pens sold = S   so...we have

The number left x (7/9)  = 420...or, mathematically.....

[420 + 160 - S] (7/9) = 420     multiply both sides by (9/7)

[420 + 160 - S]  = 540

580 - S  = 540       rearrange

580 - 540 =  S  = 40   red pens sold

Well if it is a isosceles right angled triangle then the legs would be the same.

$$\\x^2+x^2=(15\sqrt6)^2\\\\
2x^2=225*6\\\\
x^2=225*3\\\\
x=15\sqrt3$$

You cannot give yourself 3 points and I can't give you 3 points either.

You are right, the forum is playing up!  Maybe it just doesn't like you tonight?   :))

Aren't you going to give yourself 3 points Chris ?  :)

Yes it is one for me and one for you tonight Chris.

I saw 'sum' instead of 'difference'.  Sorry Maia.

i think you made a slight error here, Melody....we have

A - B = 10 → B =A - 10     and AB = 504   ...so we have

A(A - 10)  = 504

A^2 - 10A - 504  = 0    factor

(A + 18) (A - 28)  = 0

Then, either  A = -18  and B = -28    or    A = 28 and A = 18

Hi Maia,

a+b=10

so

b=10-a

ab=504

a(10-a)=504

-a^2+10a-504=0

a^2-10a+504=0

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{504}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{479}}}}{\mathtt{\,\times\,}}{i}\\
{\mathtt{a}} = {\sqrt{{\mathtt{479}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{21.886\: \!068\: \!628\: \!239\: \!289\: \!2}}{i}\\
{\mathtt{a}} = {\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21.886\: \!068\: \!628\: \!239\: \!289\: \!2}}{i}\\
\end{array} \right\}$$

There are no real solutions to this .

120%   = 6/5

Call the amount save by Charles = A

The amount saved by Bala = (5/4)A

The amount saved by Ali = (6/5)(5/4)A =  (3/2)A

And we're told that

A + 120 = (3/2)A     subtract A from both sides

120 = (1/2)A    multiply both sides by 2

$240 = A  this is what Charles saved

And Baa saved (5/4)(240)  = $300

And Ali saved (3/2)A = $360

Thanks Chris,

Of course the boys wouold not run at exactly the same speed the whole way but if they did then what I have done makes some sense to me.   

Yeah, Melody....I hadn't comsidered that.....it seems logical....!!!

it is either 19 or 21 ( roman numerals )

I will assume the boys have all run at a constans speed right rom the beginning of the race.

When Robert has run 50, Sam has run 40 metres 

so Sam has has run 4/5 as far as Robert.

If this ratio stays the same then when Robert has run 60m Sam will have run  4/5 of 60 = 48 metres.

So I think that Sam will be 12m behind when Robert crosses the finish line.  :)

This IS hard to solve  !!!!!

2z(3-5)  =   2a(-2)   = -4a      we always work inside the parentheses, first - if possible.....

what the heck?

1 and 4/15

no the answer is 24

the answer is 36