Fractions and ratio

Let the gift be $x$ , the amount Avery paid be $a$ , the amount Ben paid be $b$ , the amount Carly paid to be $c$ .

$a$ = $\frac{1}{5}$ $x$

$b$ = $\frac{5}{7}$ $a$ = $\frac{5}{7}$ $\times$ $\frac{1}{5}$ $x$ = $\frac{1}{7}$ $x$

$c$ =1- $\frac{1}{5}$ $x$ - $\frac{1}{7}$ $x$ = $\frac{23}{35}$ $x$

$c$ - $a$ = $\frac{23}{35}$ $x$ - $\frac{1}{5}$ $x$ = $\frac{16}{35}$ $x$

$\frac{16}{35}$ $x$ =192

$\frac{1}{35}$ $x$ = $\frac{192}{16}$ =12

$x$ = $12\times 35$ = $\boxed{420}$

Let the cost of the gift be $x$ , the amount paid by Avery be $a$ , the amount paid by Ben be $b$ and the amount paid by Carly be $c$ .

$a=\dfrac{1}{5}x \implies$ Eq. (1)

$b=\dfrac{5}{7}a = \dfrac{5}{7} \times \dfrac{1}{5}x = \dfrac{1}{7}x \implies$ Eq. (2)

$c:(a+b) = 23:12$

$\dfrac{c}{a+b} = \dfrac{23}{12} \implies$ Eq. (3)

$c-a = 192 \implies$ Eq. (4)

We now have 4 linear equations with 4 unknowns.

From Eq. (4):

$c=a+192$

Substitute this into Eq. (3):

$\dfrac{a+192}{a+b} = \dfrac{23}{12}\\ 12(a+192) = 23(a+b)\\ 12a+2304 = 23a +23b\\ 11a+23b = 2304$

Now, substitute Eq. (1) and Eq. (2) in here:

$11\left( \dfrac{1}{5}x \right) + 23 \left( \dfrac{1}{7}x \right) = 2304\\ \dfrac{11}{5} x + \dfrac{23}{7}x = 2304\\ \dfrac{77 + 115}{35}x = 2304\\ \dfrac{192}{35}x = 192(12)\\ x=\dfrac{192(12)(35)}{192} = 12 \times 35 = \boxed{420}$