Brief Summary
Alright folks, this video is all about cracking a CAT LRD set from 2017. Vijay sir breaks down a problem involving repair costs for hostel rooms, categorizing them as light, moderate, or extensive. The key is to organize the given info, use the bar graph to figure out cost distributions, and then apply the additional conditions to nail down the exact repair costs for each room. Plus, there's a bit of number crunching and logic to solve the questions at the end.
- 10 hostel rooms need repairs with costs ranging from ₹1 to ₹6 crore.
- Repairs are categorized as Light (₹1-2 crore), Moderate (₹3-4 crore), and Extensive (₹5-6 crore).
- Additional conditions restrict certain repair types for specific room numbers.
- Questions require deducing repair costs and identifying women's dorms based on given conditions.
Intro
Vijay sir welcomes everyone to Day 5 of the Best of LRD series for CAT. He urges viewers to try solving the problem themselves before watching the solution. He also mentions the Lakshya Batch for CAT 24 aspirants, providing details and contact information for those interested.
Problem Setup: Understanding the Data
There are 10 hostel rooms (doms) numbered 1 to 10 that need urgent repairs. The repair costs are integral values ranging from ₹1 to ₹6 crore. Costs of ₹1-2 crore are light repairs (L), ₹3-4 crore are moderate repairs (M), and ₹5-6 crore are extensive repairs (E). Since there are 10 rooms and only 6 cost values, some costs will definitely repeat. The bar graph shows the distribution of repair costs across the 10 rooms. Two rooms need ₹1 crore, one needs ₹2 crore, three need ₹3 crore, one needs ₹4 crore, one needs ₹5 crore, and two need ₹6 crore.
Applying the Conditions
Odd-numbered rooms (1, 3, 5, 7, 9) cannot have light repairs. Even-numbered rooms (2, 4, 6, 8, 10) cannot have moderate repairs. Rooms divisible by 3 (3, 6, 9) cannot have extensive repairs. Based on these restrictions, room 3 must have a moderate repair, room 6 must have a light repair, and room 9 must have a moderate repair. Rooms 4 to 9 must all have different repair costs, meaning they will have costs of ₹1, ₹2, ₹3, ₹4, ₹5, and ₹6 crore exactly once. This means the remaining rooms (1, 2, 3, 10) will have the remaining costs: one ₹1 crore, two ₹3 crore, and one ₹6 crore.
Deductions and Filling the Grid
Since room 3 needs a moderate repair and the remaining moderate repair cost is ₹3 crore, room 3 costs ₹3 crore. Rooms 2 and 10 cannot have moderate repairs, so room 1 must have a moderate repair, costing ₹3 crore. Rooms 2 and 10 can have either light or extensive repairs, meaning they can cost either ₹1 or ₹6 crore. Rooms 4 to 9 must have costs from ₹1 to ₹6 crore, with room 7 needing the maximum cost (₹6 crore) and room 8 needing the minimum cost (₹1 crore). This means room 7 has an extensive repair and room 8 has a light repair.
Final Touches and Solving for 4-9
In rooms 4 to 9, two light repairs, two moderate repairs, and two extensive repairs are needed. One light repair (₹1 crore) and one extensive repair (₹6 crore) are already assigned. The remaining light repair must be ₹2 crore. One extensive and one moderate repair are left. Room 4 cannot have a moderate repair, so it must have an extensive repair (₹5 crore). Room 5 must have a moderate repair. The remaining moderate repairs can be ₹3 or ₹4 crore in any order.
Solving Question 1 and 2
Question 1 asks which statement is NOT necessarily true.
- Option A: Room 1 needs a moderate repair (TRUE).
- Option B: Room 5 repair will cost no more than ₹4 lakh (TRUE).
- Option C: Room 7 needs an extensive repair (TRUE).
- Option D: Room 10 will cost no more than ₹4 (NOT necessarily true, it could be ₹6). So, the answer is option D.
Question 2 asks for the total cost of repairing the odd-numbered rooms (1, 3, 5, 7, 9). The costs are ₹3 crore (room 1), ₹3 crore (room 3), ₹3 or ₹4 crore (room 5), ₹6 crore (room 7), and ₹3 or ₹4 crore (room 9). The sum of rooms 5 and 9 is ₹7 crore. So, the total cost is 3 + 3 + 6 + 7 = ₹19 crore.
Additional Conditions for Questions 3 and 4
Four out of the 10 rooms are "women's dorms" and need a total of ₹20 crore for repair. Only one of the rooms 1, 2, and 5 is a women's dorm. The combination of costs for the women's dorms must be 6 + 6 + 5 + 3 = 20. Since only one of rooms 1, 2, and 5 is a women's dorm, the other three women's dorms must be among rooms 6, 7, 8, 9, and 10.
Deductions with New Conditions
Since rooms 6, 7, 8, 9, and 10 must include three women's dorms with costs of ₹6, ₹6, and ₹3 crore, and room 7 already costs ₹6 crore, one of the ₹6 crore costs is assigned to room 7. Room 10 must also be a women's dorm with a cost of ₹6 crore. Room 9 must be a women's dorm with a cost of ₹3 crore. This means room 8 is NOT a women's dorm. The remaining women's dorm must be among rooms 1, 2, and 5, and it must cost ₹5 crore. Therefore, room 5 is a women's dorm.
Solving Questions 3 and 4 with New Info
Question 3 asks for the cost of repairing room 9. Room 9 costs ₹3 crore.
Question 4 asks which of the following is a women's dorm.
- Room 2: No
- Room 5: Yes
- Room 8: No
- Room 10: Yes So, the answer is option D, room 10.
Conclusion
Vijay sir concludes that this set from CAT 2017 was doable even under exam pressure. He encourages viewers to leave any doubts in the comments and promises to meet in the next video.
