Q.63 A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day?
Transporter Trucks Pending Orders: 6 Trucks for 5 Days
Minimum trucks to clear backlog + daily orders by Day 5
Option (C) – Exactly clears all orders on Day 5
7 trucks × 4 days = 28 truck-days
3 trucks × 10 days = 30 truck-days
6 trucks × 5 days = 30 truck-days ✓
Step-by-Step Algebraic Solution
x = daily orders per truck, y = backlog
4x + y = 28 ← 7 trucks × 4 days
10x + y = 30 ← 3 trucks × 10 days
——————
6x = 2 → x = 1/3 orders/truck/day
y = 28 - 4(1/3) = 80/3 backlog
5 days need: 5x + y = 5(1/3) + 80/3 = 90/3 = 30 truck-days
Trucks required = 30 ÷ 5 = 6 ✓
Option Analysis
| Option | Trucks × 5 Days | Truck-Days | Status |
|---|---|---|---|
| (A) 4 trucks | 4 × 5 | 20 | ❌ Short by 10 |
| (B) 5 trucks | 5 × 5 | 25 | ❌ Short by 5 |
| (C) 6 trucks | 6 × 5 | 30 | ✅ PERFECT |
| (D) 7 trucks | 7 × 5 | 35 | ✅ Works (not minimum) |
Truck-Days: 20
FAILS
Truck-Days: 25
FAILS
Truck-Days: 30
PERFECT
Truck-Days: 35
(Excess)
Lightning Exam Shortcuts
- Pattern: Days₁×Trucks₁ ≈ Days₂×Trucks₂ = Total Work
- Quick Solve: Solve 2 equations → scale to target days
- Verify: 6×5 = 30 matches 3×10 = total work needed
- Traps: Don’t average trucks, solve system properly
- Pro Tip: Truck-days = constant work unit!
Why This Problem Tests GATE Skills
Simultaneous Equations + Work Rates: Two scenarios give exact system to solve x (daily rate) and y (backlog). Target scenario scales linearly. Minimum keyword demands smallest integer satisfying inequality.
