GATE CS Mastery
Cheat sheets

GATE CS 2011 · General Aptitude

Question 62

MCQPrinted as GA Q.622 marksWrong answer: −2/3Medium (estimate)

Question

P, Q, R and S are four types of dangerous microbes recently found in a human habitat. The area of each circle with its diameter printed in brackets represents the growth of a single microbe surviving human immunity system within 24 hours of entering the body. The danger to human beings varies proportionately with the toxicity, potency and growth attributed to a microbe shown in the figure below:

Scatter plot of microbe Toxicity versus Potency with circular bubbles of diameter 50mm for P, 40mm for Q, 30mm for R, and 20mm for S

A pharmaceutical company is contemplating the development of a vaccine against the most dangerous microbe. Which microbe should the company target in its first attempt?

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

Solution

Calculate the composite danger index for each microbe using the proportionality relation Danger ~ (Potency * Growth) / Toxicity.

  1. From the problem statement:
  • Potency is given along the horizontal axis.
  • Toxicity is given along the vertical axis, defined as 'milligrams of microbe required to destroy half of the body mass in kilograms'. A smaller required mass means the substance is more lethal; thus danger is inversely proportional to the required dose (mg).
  • Growth is represented by the area of the circle, where area is proportional to d2d^2 (square of diameter).
  1. The danger index is given by:
Danger∝Potency×d2Toxicity (mg)\text{Danger} \propto \frac{\text{Potency} \times d^2}{\text{Toxicity (mg)}}
  1. Calculate the relative danger index for each microbe:

  • Microbe P:

    • Diameter d=50 mm  ⟹  d2=2500d = 50\text{ mm} \implies d^2 = 2500
    • Potency=0.4\text{Potency} = 0.4
    • Toxicity=800 mg\text{Toxicity} = 800\text{ mg}
    • Index=0.4×2500800=1000800=1.25\text{Index} = \frac{0.4 \times 2500}{800} = \frac{1000}{800} = 1.25

  • Microbe Q:

    • Diameter d=40 mm  ⟹  d2=1600d = 40\text{ mm} \implies d^2 = 1600
    • Potency=0.5\text{Potency} = 0.5
    • Toxicity=600 mg\text{Toxicity} = 600\text{ mg}
    • Index=0.5×1600600=800600≈1.33\text{Index} = \frac{0.5 \times 1600}{600} = \frac{800}{600} \approx 1.33

  • Microbe R:

    • Diameter d=30 mm  ⟹  d2=900d = 30\text{ mm} \implies d^2 = 900
    • Potency=0.4\text{Potency} = 0.4
    • Toxicity=300 mg\text{Toxicity} = 300\text{ mg}
    • Index=0.4×900300=360300=1.20\text{Index} = \frac{0.4 \times 900}{300} = \frac{360}{300} = 1.20

  • Microbe S:

    • Diameter d=20 mm  ⟹  d2=400d = 20\text{ mm} \implies d^2 = 400
    • Potency=0.8\text{Potency} = 0.8
    • Toxicity=200 mg\text{Toxicity} = 200\text{ mg}
    • Index=0.8×400200=320200=1.60\text{Index} = \frac{0.8 \times 400}{200} = \frac{320}{200} = 1.60
  1. Microbe S has the highest danger index (1.601.60). Therefore, the company should target Microbe S in its first attempt.
Final answer

D

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
62 of 65, printed as GA Q.62
Section
General Aptitude

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.