CAT 2017 Slot 2 — DILR Question 3

DistributionEasy

Passage / Data

Directions for next 4 questions.

Funky Pizzaria was required to supply pizzas to three different parties. The total number of pizzas it had to deliver was 800, 70% of which were to be delivered to party 3 and the rest equally divided between Party 1 and Party 2.

Pizzas could be of Thin Crust (T) or Deep Dish (D) variety and come in either Normal Cheese (NC) or Extra Cheese (EC) versions. Hence, there are four types of pizzas: T-NC, T-EC, D-NC and D-EC. Partial information about proportions of T and NC pizzas ordered by the three parties is given below:

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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?

Answer & solution

A
18
12
C
30
D
24

Solution

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:

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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).