SAT Question db88a933: Distributions, Center & Spread answer and explanation with Desmos solution

Choice B is correct. The mean of a data set is the sum of the values of the data set divided by the total number of values in the data set. When a data set is represented in a frequency table, the sum of the values in the data set is the sum of the products of each value and its frequency. For data set Q, the sum of the products of each value and its frequency is $110a+30b+0c$, or $110a+30b$, and the number of values is $110+30+0$, or $140$. Therefore, the mean of data set Q is $\frac{110a+30b}{140}$. For data set R, the sum of the products of each value and its frequency is $0a+3b+11c$, or $3b+11c$, and the number of values is $0+3+11$, or $14$. Therefore, the mean of data set R is $\frac{3b+11c}{14}$. Since $a$, $b$, and $c$ are positive, consecutive integers and $a<b<c$, it follows that $b=a+1$ and $c=a+2$. Substituting $a+1$ for $b$ in the mean of data set Q yields $\frac{110a+30(a+1)}{140}$, or $\frac{110a+30a+30}{140}$, which is equivalent to $\frac{140a+30}{140}$, or $a+\frac{3}{14}$. Substituting $a+1$ for $b$ and $a+2$ for $c$ in the mean of data set R yields $\frac{3(a+1)+11(a+2)}{14}$, or $\frac{3a+3+11a+2(11)}{14}$, which is equivalent to $\frac{14a+3+2(11)}{14}$, or $a+\frac{3}{14}+\frac{2(11)}{14}$. Subtracting the mean of data set Q from the mean of data set R yields $(a+\frac{3}{14}+\frac{2(11)}{14})-(a+\frac{3}{14})$, or $\frac{2(11)}{14}$. Therefore, the mean of data set R is $\frac{2(11)}{14}$ greater than the mean of data set Q.