There are 4 machines namely P, Q, R and S in a factory. P and Q running together can finish an order in 10 days. If R works twice as P and S works 1/3 as much as Q then the same order of work can be finished in 6 days. Find the time taken by P alone to complete the same order.
a) 11.5 days b) 12.5 days c) 13.5 days d) 14.5 days

• 1/p+1/q=1/10
  1/p= 1/10-1/q ------->(1)
  
  given "R" works twice as "P"
  
  so r's 1 day work= 2(1/p) ====> 2(1/10-1/q) ==> 1/5-2/q
  
  "S" works 1/3 as much as "Q"
  
  so s's 1 day work= 1/3(1/q)
  
  r+s's 1 day work
  
  1/5-2/q+1/3q=1/6
  
  Solving this v ll get
  
  q=50
  
  substitute q=50 in eq(1)
  
  v ll get
  
  1/p=2/25
  
  Therefore "P" ll finish the work alone in 25/2= 12.5 days
• 10 years agoHelpfull: Yes(39) No(0)
• ans:12.5
  p+q 1 day work = 1/10 .......(1)
  R+s 1 day work = 1/6 ........(2)
  given R=2p , S=Q/3 substitute in eqn (2) then 2P+Q/3=1/6 .....(3)
  solving 1 and 3 we get
  P's 1 day work = 2/25
  p can complete work in 25/2 days =12.5 days
• 10 years agoHelpfull: Yes(8) No(0)
• Let P's 1 day work be X and Q's 1 day work be Y ...(1)
  Given that, the time taken to complete the order by (P+Q) = 10 days.
  Then (P+Q)'s 1 day work = 1/10 …(2)
  Therefore, from (1) and (2), X+Y = 1/10 …(3)
  
  Suppose, R works twice as P then R's 1 day work 2X.
  And, S works 1/3 as much as Q then S's 1 day work be Y/3.
  
  Now, the time taken to complete the order by (R+S) = 6 days
  2X+Y/3 = 1/6
  12X+2Y = 1 ….(4)
  
  Solving the above two equations(3) and (4), we get X = 4/50.
  Thus, P's 1 day work = 4/50.
  Hence P alone can complete the entire order of work in 50/4 days = 12.5 days.
• 10 years agoHelpfull: Yes(6) No(0)
• 12.5 days
  
• 10 years agoHelpfull: Yes(2) No(2)
• Let P's 1 day work be X and Q's 1 day work be Y ...(1)
  Given that, the time taken to complete the order by (P+Q) = 10 days.
  Then (P+Q)'s 1 day work = 1/10 …(2)
  Therefore, from (1) and (2), X+Y = 1/10 …(3)
  
  Suppose, R works twice as P then R's 1 day work 2X.
  And, S works 1/3 as much as Q then S's 1 day work be Y/3.
  
  Now, the time taken to complete the order by (R+S) = 6 days
  2X+Y/3 = 1/6
  12X+2Y = 1 ….(4)
  
  Solving the above two equations(3) and (4), we get X = 4/50.
  Thus, P's 1 day work = 4/50.
  Hence P alone can complete the entire order of work in 50/4 days = 12.5 days.
• 10 years agoHelpfull: Yes(0) No(0)
• p+q=1/10
  r+s=1/6(2p+q/3=1/6)according to ques
  so solving both equ. finds
  p=8/100(p's one day work)
• 10 years agoHelpfull: Yes(0) No(0)
• let P's efficiency= x% and that of Q=y%
  10(x+y)=100 => x+y=10 -----(1)
  R's eff= 2x% and S's eff= y/3%
  6(2x + y/3)=100 => 6x+y=50 ----(2)
  Eq(2)- Eq(1), we get,
  5x=40 => x=8%
  so, P alone do the work in (100/8)= 12.5 days.
• 10 years agoHelpfull: Yes(0) No(0)