Question

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; 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

Answer 1

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

Answer 2

Answer:

b) 12.5 days

Step-by-step explanation:

Let the total work be W, and the rates at which P and Q work be p and q work/day respectively.

Now, given:

$10p + 10q = W$

Next, the rates at which R and S work are twice of P's and a third of Q's respectively, so,

Rate at which R works = 2p

Rate at which S works = q/3

Again, to complete the same work, they take 6 days, hence:

$6(2p) + 6(q/3) = W$

$=> 12p + 2q = W$

Equating:

$10p + 10q = 12p + 2q$

$=> 8q = 2p$, or, $p = 4q$

Substituting in original equation, we get:

$10p + 10q = W => 40q + 10q = W$

$50q = W$

$=> 12.5p = W$

Using our analogy, P's rate of work per day is p, and let's say, it takes x days to complete the same work, so:

$xp = W$

Comparing, we get, x = 12.5 days.