Given:
1. At their usual rates, machines A and B together finish a task in 12 days.
2. If A's rate is halved, and B's rate is tripled, the task is done in 9 days.

Asked: How many days would A take to finish the task at its usual rate?

Let machine A and machine B finish the task at a normal rate in a & b days respectively

1/a + 1/b = 1/12 (1)
1/2a + 3/b = 1/9 (2)
1/a + 6/b = 2/9 (2a)

(2a) - (1)
5/b = 2/9 - 1/12 = (8-3)/36 = 5/36
b = 36
1/a = 1/12 - 1/36 = 2/36 = 1/18
a = 18

IMO D
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given
a+b/ab= 1/12
ab= 12*(a+b)--(1)

now a rate is halved and b rate is tripled
1/2a+3/b = 1/9
we get
9b+54a= 24a+24b
or say
2a=b
now plugin in (1)
we get
a= 18
IMO D

r = w/t

Since the amount of work isn't specified, you can set it equal to 1 and set up some equations.

Let A = the rate of machine A and B = the rate of machine B

A + B = (1/12)
(1/2)A + 3B = (1/9)
A = (1/t)

Rearrange either of the first two equations to be B = ... and you can use substitution to solve for A. A = (1/18), and since A = (1/t), t must be equal to 18. (D) is the correct answer.

Assume work = 12
so
A+B = 1 - (1)

(A/2) + 3B = 4/3 - (2)

Solving (1) & (2)
A = 2/3

Therefore T(A) = 12/(2/3) => 18

IMO OA D

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