Question

For Party 2, if 50% of the Normal Cheese pizzas were of Thin Crust variety, what was the difference between the numbers of T-EC and D-EC pizzas to be delivered to Party 2?

• A. 18
• B. 12
• C. 30
• D. 24

Answer 1

The correct option is B 12

Given, Number of pizzas to be delivered to Party $3=0.7\times 800=560$
Number of pizzas to be delivered to Party 2 = Number of pizzas to be delivered to Party $1=\frac{(800-560)}{2}=120$
Refer to the table below:
$\begin{array}{ccc}Party& ThinCrust\left(T\right)& NormalCheese\left(NC\right)\\ 1& 0.6\times 120=72& 416-36-364=16\\ 2& 0.55\times 120=66& 0.3\times 120=36\\ 3& 300-72-66=162& 0.65\times 560=364\\ Total& 0.375\times 800=300& 0.52\times 800=416\end{array}$
For Party 2,
Given, $T-NC=\frac{36}{2}=18$
From the table, T-EC + T-NC = 66 & D-NC + T-NC = 36
$\Rightarrow T-EC=66-18=48$
T-EC + T-NC + D-NC + D-EC = 120
48 + 36 + D-EC = 120
D-EC = 36
T-EC - D-EC = 48 - 36 = 12.