NIMCET 2010 — Reasoning PYQ

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Question

NIMCET 2010 — Reasoning PYQ

NIMCET | Reasoning | 2010

Directions : Questions : 220 guests are to be trans- ported from A to B. Any number of buses of the following passenger carrying capacities are available.
Type P : 60, Type Q : 50, Type R : 40, Type S : 30
The cost per trip for a bus of each of these types is given as follows :
Type P : Rs 200, Type Q : Rs 140, Type R : Rs 125,
Type S : Rs 95
No buses can be overloaded and prefer no vacant seats in each trips.

How many buses are needed for the above (Minimum cost trip)

Choose the correct answer:

Explanation

1. Cost Efficiency per Seat:

Type P: $200 / 60 \approx 3.33$ Rs/seat
Type Q: $140 / 50 = 2.80$ Rs/seat (Most Economical)
Type R: $125 / 40 = 3.125$ Rs/seat
Type S: $95 / 30 \approx 3.16$ Rs/seat

2. Finding the Minimum Cost Combination (220 Guests):

Since Type Q is the cheapest per seat, we maximize its use.

Try 4 Type Q buses: $4 \times 50 = 200$ . We need 20 more seats. No bus fits exactly 20.
Try 3 Type Q buses: $3 \times 50 = 150$ . We need 70 more seats.
- One Type R (40) + One Type S (30) = $40 + 30 = 70$ .
- Total Capacity: $150 + 70 = 220$ seats.
- Total Cost: $(3 \times 140) + 125 + 95 = 420 + 125 + 95 = Rs\ 640$ .
Try 2 Type Q buses: $2 \times 50 = 100$ . We need 120 more seats.
- Three Type R buses: $3 \times 40 = 120$ .
- Total Capacity: $100 + 120 = 220$ seats.
- Total Cost: $(2 \times 140) + (3 \times 125) = 280 + 375 = Rs\ 655$ .

The minimum cost is Rs 640, achieved by using 3 Type Q buses, 1 Type R bus, and 1 Type S bus.

3. Total Number of Buses:

\text{Number of buses} = 3 (\text{Type Q}) + 1 (\text{Type R}) + 1 (\text{Type S})

\text{Number of buses} = 3 + 1 + 1 = 5

Conclusion:

A total of 5 buses are needed for the minimum cost trip.

Correct Option: (a) 5