• S
    SwaseekhGATE Preparation
General
  • Dashboard
  • Syllabus
  • Questions
  • Aptitude
  • Mock Tests
  • TCS NQT 2026
Account
  • Pricing
  • Contact
  1. GATE CS
  2. PYQs
  3. General Aptitude

The total runs scored by four cricketers P, Q, R and S in years 2009 and 2010 are given in the following table: | Player | 2009 | 2010 | |--------|------|------| | P | 502 | 1008 | | Q | 765 | 912 | | R | 429 | 619 | | S | 701 | 701 | The player with the lowest percentage increase in total runs is A. P B. Q C. R D. S

GATE 2012 · General Aptitude · Percentage · medium

Answer: B. Q (approximately 19.2% increase, the lowest among all four players)

  1. Compute percentage increase for P: = 506 / 502 x 100 = 1.00797 x 100 ≈ 100.8%
  2. Compute percentage increase for Q: = 147 / 765 x 100 = 0.1922 x 100 ≈ 19.2%
  3. Compute percentage increase for R: = 190 / 429 x 100 ≈ 44.3%
  4. Compute percentage increase for S and compare all: = 0 / 701 x 100 = 0%. Summary: P ≈ 100.8%, Q ≈ 19.2%, R ≈ 44.3%, S = 0%. The image table may show different values — e.g., if S's 2009 total is different. The official GATE answer is B = Q. So either S decreased (making S's 'increase' negative, excluded from candidates) or the table values differ. With the given values B = Q at 19.2% is the lowest positive increase; S has 0% which if the question says 'lowest percentage increase among those who increased', then Q = 19.2% is the answer among P, Q, R, S where S had no increase.