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SET 1 — 🔍 CAT Level
Emily's Pizza (EP) and Logan's Pizza (LP) are located next door to each other. Each day, 100 customers buy one whole pizza each from one of the two restaurants. The price of a pizza at each restaurant is set daily and is always a multiple of Rs. 10. If both restaurants charge the same price, the 100 customers are split equally between them. For every Rs. 10 by which one restaurant's price is higher than the other's, it loses one customer to the other restaurant. The cost of making a pizza for each restaurant is Rs. 50. The table below shows the number of customers served by each restaurant from Day 1 to Day 10 (D1 to D10).
The further information is known to us:
(i) The price of a pizza at EP on Day 1 is the same as the price at LP on Day 7. The price at LP on Day 3 is the same as the price at EP on Day 8.
(ii) The price of a pizza at EP on Day 10 is the same as the price at LP on Day 4. The price at EP on Day 7 is the same as the price at LP on Day 9.
(iii) The prices of pizzas at EP on Day 6, Day 9, Day 5, and Day 7 are in an arithmetic progression, in that order. Similarly, the prices of pizzas at LP on Day 5, Day 10, Day 2, Day 8, and Day 3 are in an arithmetic progression, in that order.
(iv) The profit earned by EP on Day 2 is Rs. 4,490 higher than the profit earned by EP on Day 6.
Question 1
On which of the following days did Logan's Pizza charge the second-highest price?
Day 8
Day 3
Day 4
Day 9
Question 2
By how much does the profit (in Rs.) earned by Logan's Pizza on Day 4 exceed the profit (in Rs.) earned by Emily's Pizza on Day 9?
Question 3
On how many days was the price of Emily's Pizza higher than the average price of Emily's pizza over the 10 days?
Question 4
On which of the following days was the ratio of EP's profit to LP's profit the lowest?
Day 1
Day 3
Day 5
Day 9
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✅ Answers & Explanations
Step 1: Find the price difference on each day
If both restaurants charge the same price, each gets 50 customers.
For every Rs. 10 increase in one restaurant's price, it loses 1 customer.
Hence,
LP Price−EP Price=(EP Customers−50)×10
Using the customer table:
Day | EP Customers | LP Customers | LP − EP Price (Rs.) |
|---|---|---|---|
D1 | 48 | 52 | -20 |
D2 | 45 | 55 | -50 |
D3 | 49 | 51 | -10 |
D4 | 54 | 46 | +40 |
D5 | 47 | 53 | -30 |
D6 | 52 | 48 | +20 |
D7 | 48 | 52 | -20 |
D8 | 47 | 53 | -30 |
D9 | 54 | 46 | +40 |
D10 | 47 | 53 | -30 |
Step 2: Use the given conditions
Using
Equal-price conditions,
Arithmetic progression conditions,
Profit condition:
Profit on D2−Profit on D6=4490\text{Profit on D2}-\text{Profit on D6}=4490Profit on D2−Profit on D6=4490
we obtain the unique prices.
Final Price Table
Day | EP Price | LP Price |
|---|---|---|
D1 | 220 | 200 |
D2 | 300 | 250 |
D3 | 320 | 310 |
D4 | 210 | 250 |
D5 | 220 | 190 |
D6 | 180 | 200 |
D7 | 240 | 220 |
D8 | 310 | 280 |
D9 | 200 | 240 |
D10 | 250 | 220 |
Question 1
On which day did Logan's Pizza charge the second-highest price?
LP prices:
200, 250, 310, 250, 190, 200, 220, 280, 240, 220
Highest = 310 (Day 3)
Second-highest = 280 (Day 8)
Answer: Option 1 (Day 8) ✅
Question 2
By how much does LP's profit on Day 4 exceed EP's profit on Day 9?
Profit = (Selling Price − 50) × Customers
LP Day 4
Price = 250
Customers = 46
(250−50)×46=200×46=9200
EP Day 9
Price = 200
Customers = 54
(200−50)×54=150×54=8100
Difference
9200−8100=1100
Answer: Rs. 1100 ✅
Question 3
On how many days was EP's price greater than its average price?
Average EP price
(220+300+320+210+220+180+240+310+200+250)10=2450/10= 245
Prices greater than 245:
D2 = 300
D3 = 320
D8 = 310
D10 = 250
Total = 4 days
Answer: 4 ✅
Question 4
On which day was the ratio of EP's profit to LP's profit the lowest?
Compute profits:
Day | EP Profit | LP Profit | Ratio |
|---|---|---|---|
D1 | 8160 | 7800 | 1.046 |
D3 | 13230 | 13260 | 0.998 |
D5 | 7990 | 7420 | 1.077 |
D9 | 8100 | 8740 | 0.927 |
The smallest ratio is on Day 9.
Answer: Option 4 (Day 9) ✅
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