CEMC Euclid, 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.