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GATE CS 2011 · General Aptitude

Question 64

MCQPrinted as GA Q.642 marksWrong answer: −2/3Medium (estimate)

Question

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?

Choose one option
Official answer and solution
Official answer
C
From
Official GATE 2011 answer key

Solution

Set up a system of linear equations involving initial backlog, daily incoming order rate, and truck daily clearance capacity.

  1. Let:
  • BB be the initial pending orders (backlog).
  • rr be the number of incoming orders received per day.
  • cc be the clearance capacity of one truck per day.
  1. Formulate equations for the two given operational conditions:
  • Using 7 trucks for 4 days:
    Total capacity=4×(7c)=28c=B+4r— (1)\text{Total capacity} = 4 \times (7c) = 28c = B + 4r \quad \text{--- (1)}
  • Using 3 trucks for 10 days:
    Total capacity=10×(3c)=30c=B+10r— (2)\text{Total capacity} = 10 \times (3c) = 30c = B + 10r \quad \text{--- (2)}
  1. Subtract equation (1) from equation (2):
30c−28c=(B+10r)−(B+4r)30c - 28c = (B + 10r) - (B + 4r)
2c=6r  ⟹  c=3r2c = 6r \implies c = 3r
  1. Substitute c=3rc = 3r back into equation (1) to express the backlog BB in terms of rr:
B=28(3r)−4r=84r−4r=80rB = 28(3r) - 4r = 84r - 4r = 80r
  1. Let NN be the number of trucks required to clear all orders by the end of day 5.
    The total orders to be shipped over 5 days equal the initial backlog plus 5 days of incoming orders:
Total orders=B+5r=80r+5r=85r\text{Total orders} = B + 5r = 80r + 5r = 85r

In 5 days, NN trucks can ship:

Capacity=5×N×c=5×N×3r=15N⋅r\text{Capacity} = 5 \times N \times c = 5 \times N \times 3r = 15N \cdot r

To ensure there are no pending orders remaining:

15N⋅r≥85r  ⟹  15N≥85  ⟹  N≥8515=173≈5.6715N \cdot r \ge 85r \implies 15N \ge 85 \implies N \ge \frac{85}{15} = \frac{17}{3} \approx 5.67
  1. Since the number of trucks must be an integer, the minimum number of trucks required is ⌈5.67⌉=6\lceil 5.67 \rceil = 6.
Final answer

C

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2011 Computer Science and Information Technology (CS)
Question
64 of 65, printed as GA Q.64
Section
General Aptitude

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