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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?

Explanation:

 

Now 70% of the pizzas were delivered to party 3 and the balance 30% of the pizzas were delivered to the remaining 2 parties i.e., Party 1 and Party 2. So each of Party 1 and Party 2 will receive

$\Rightarrow \frac{100-70}{100}\times \frac{1}{2}$ × 800 = 120 pizzas.

Party 3 will receive ⇒ $\frac{70}{100}$ × 800

= 560 pizzas.

Now number of Thin crust pizzas received by Party 1 = 0.6 × 120 = 72

So number of Deep dish pizzas received by Party 1 = 120 – 72 = 48

Similarly, number of Thin crust pizzas received by Party 2 = 0.55 × 120 = 66 and number of deep dish pizzas received by party 2 = 120 – 60 = 54

Number of normal cheese pizzas received by party 2 = 0.3 × 120 = 36

Number of extra cheese pizzas received by party 2 = 120 – 36 = 84

Now number of normal cheeze pizzas received by party 3 = 0.65 × 560 = 364

So number of extra cheese pizzas received by party 3 = 560 – 364 = 196

Now total number of Thin crust pizzas delivered to the 3 parties = 0.375 × 800

= 300

So, number of Deep Dish Pizzas delivered to the 3 parties = 800 – 416 = 384

So number of thin crust pizzas received by party B = 300 – (72 + 66) = 162 and number of deep dish pizzas received by party 3 = 500 – (48 + 54) = 398

Let number of thin crust normal cheese pizzas received by party 1, 2 and 3 be a, b and c.

Now number of deep dish normal cheese pizzas received by party 2 = 36 – b

So, the number of deep dish extra cheese pizzas received by party 2 = 54 – (36 – b) = 18 + b

Now total number of extra cheese deep dish pizzas received by party 3 = 196 – (162 – c) = 34 + c

Let number of deep pan normal cheese pizzas received by party 1 be ‘d’.

∴ Number of deep dish extra cheeze pizzas will be ‘48-d’

Now let us represent all this information in a table given below:

​​​​​​​

If 50% of the normal cheese pizzas delivered to party 2 are of thin crust variety, then b : 36 – b

= 1 : 1

$\frac{b}{36-b}$ = 1 ⇒ 2b = 36 or b = 18

So number if T-EC pizzas = 66 – b = 66 – 18 = 48

Also, number of D-EC pizzas = 18 + b = 18 + 18 = 36

So difference between number of T-EC and D-EC pizzas = 48 – 36 = 12

Hence, option (b).