Latest TCS Aptitude Question SOLUTION: 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 or

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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

Asked In TCS ADIL (10 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Saraswathy  (10 years ago)

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

Is this solution Helpfull? Yes (39) | No   

sushma  (10 years ago)

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

Is this solution Helpfull? Yes (8) | No   

krishna  (10 years ago)

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.

Is this solution Helpfull? Yes (6) | No   

Jasmeet  (10 years ago)

12.5 days

Is this solution Helpfull? Yes (2) | No (2)  

subramanian.s  (10 years ago)

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.

Is this solution Helpfull? Yes | No   

manoj sahu  (10 years ago)

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)

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Ranadip Saha  (9 years ago)

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.

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