P, Q & R are assigned a work. Working together, they can complete a job in 10 days. However, R worked for only first 3 days and 37/100th of the total work was completed in these 3 days. Remaining work is completed by P & Q in 7 days. The work done by P in 5 days is equal to work done by Q in 4 days. Then the time taken by fastest worker is how much percent less than the second fastest worker?

• 20%
  
  1/P + 1/Q +1/R = 1/10 ---(i)
  3 days work of P, Q & R =37/100 => 3(1/P + 1/Q + 1/R)= 37/100 or 1/P + 1/Q + 1/R = 37/300 ---(ii)
  Remaining work 63/100 is (P+Q)'s 7 days work, so (P+Q)'s 1 day work=(63/100)/7=9/100
  => 1/P + 1/Q = 9/100 ---(iii)
  Subtracting (iii) from (i), 1/R = 1/10 - 9/100 = 1/100
  Also given 5(1/P) = 4(1/Q) => 1/P=4/5Q and 1/Q=5/4P
  Substituting value of 1/P in Eqn. (iii), 4/5Q + 1/Q= 9/100 => 1/Q=1/20
  Substituting value of 1/Q in Eqn.(iii), 1/P + 5/4P =9/100 => 1/P=1/25
  So individually P, Q & R can complete the job in 25, 20 and 100 days respectively.
  Thus fastest worker is Q takes 20 days, whereas second fastest worker P takes 25 days.
  Hence % less time taken by Q than P =100(25-20)/25= 20%
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