, 2018

When $x=11$, $$x+(x+1)+(x+2)+(x+3)=4x+6 = 4(11)+6 = 50$$ Alternatively, $$x+(x+1)+(x+2)+(x+3)=11+12+13+14=50$$
We multiply the equation $\dfrac{a}{6}+\dfrac{6}{18} = 1$ by $18$ to obtain $3a+6=18$.

Solving, we get $3a=12$ and so $a=4$.
Solution 1

Since the cost of one chocolate bar is 1.00 more than that of a pack of gum, then if we replace a pack of gum with a chocolate bar, then the price increases by 1.00.

Starting with one chocolate bar and two packs of gum, we replace the two packs of gum with two chocolate bars.

This increases the price by $2.00 from$4.15 to $6.15.

In other words, three chocolate bars cost 6.15, and so one chocolate bar costs $\frac{1}{3}(\$6.15)$$or$\$2.05$.

Solution 2

Let the cost of one chocolate bar be $$$$x$.

Let the cost of one pack of gum be $$$$y$.

Since the cost of one chocolate bar and two packs of gum is $4.15, then$x+2y=4.15$.

Since one chocolate bar costs 1.00 more than one pack of gum, then x=y+1$.

Since $x=y+1$, then $y = x-1$.

Since $x+2y=4.15$, then $x+2(x-1) = 4.15$.

Solving, we obtain $x + 2x - 2 = 4.15$ or $3x = 6.15$ and so $x=2.05$.

In other words, the cost of one chocolate bar is $2.05.