fractions

Let the number of madeleines with Jeff be 'j' and the number of madeleines with Bart be 'b'.

Then, if Bart had 57 fewer madeleines than Jeff: \(b = j - 57 …1\)

 

 

 

Given, Bart saved 3/5 of his madeleines.

He must have spent the rest:                        
    \(1\)\(-\)\({3 \over 5}\frac{}{}\)\( = {2\over 5}\frac{}{}\)

Bart's spending =\( {2 \over 5}b\frac{}{}\)

If Jeff spent 2/3 of his madeleines.

He must have saved the rest:

     \(1 - \)\( {2 \over 3}={}{}\)\({1 \over 3}\frac{}{}\)
 

 

Jeff savings=\( {1 \over 3}\frac{}{}\)

 

Given: the amount Bart spent was 1/4 of Jeff’s savings.

Thus, 

 \( {2 \over 5}{}{}\)\(b\)\(=\)\({1 \over 4}{}{} \)\((\)\( {1 \over 3}j{}{}\)\()\)
 

Cross multiply: 

           2 * 4 * 3 * b = 1 * 5 * j

 

                 24b = 5j ...2

 

Solve using equation 1 and equation 2:

     
        \(b = j - 57 …1\)
             \(24b=5j …2\)

              \(24(j-57) = 5j\)
 

               \(24j-1368=5j\)
 

                   \(19j=1368\)

                 \(j = {1368 \over 19}{}^{}=72\)

 

Hence, Jeff had 72 madeleines originally.

The question is how much did Jeff spend:

Given, Jeff spent 2/3 of the original value: \( {2 \over 3}{}{} * 72 = 48\)

 

Jeff spent 48 madeleines.

Let $a_1$ equal Bart's madeleines, and let $a_2$ equal Jeff's madeleines.

$a_1 + 57 = a_2$

$\left(1-\frac{3}{5}\right) a_1 = \frac{1}{4} \cdot \left(1-\frac{2}{3}\right) a_2$

$\frac{2}{5} a_1 = \frac{1}{12} a_2$

$24 a_1 = 5 a_2$

$24 a_1 = 5(a_1 + 57)$

$24 a_1 = 5 a_1 + 5 \cdot 57$

$19 a_1 = 5 \cdot 57$

$a_1 = \frac{5 \cdot 19 \cdot 3}{19}$

$a_1 = 15$

$a_2 = 15 + 57 = 72$

$\frac{2}{3} a_2 = \frac{2}{3} \cdot 72 =  24 \cdot 2 = \boxed{48}$