Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour

Could please anyone help me :
How did you find x in the first place?
Thank you

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We can let their combined rate = 1/x; thus:

Alpha’s rate = 1/(x + 6)

Beta’s rate = 1/(x + 1)

Gamma’s rate = 1/(2x)

Let’s solve for x, using the equation:

1/(x + 6) + 1/(x + 1) + 1/(2x) = 1/x

To rid the equation of fractions, we can multiply the equation by 2x(x + 1)(x + 6):

2x^2 + 2x + 2x^2 + 12x + x^2 + 7x + 6 = 2x^2 + 14x + 12

5x^2 + 21x + 6 = 2x^2 + 14x + 12

3x^2 + 7x - 6 = 0

(3x - 2)(x + 3) = 0

x = 2/3 or x = -3

Since x can’t be negative, x = 2/3. Therefore, Alpha’s rate = 1/(2/3 + 6) = 1/(20/3) = 3/20 and Beta’s rate = 1/(2/3 + 1) = 1/(5/3) = 3/5. Their combined rate is 3/20 + 3/5 = 3/20 + 12/20 = 15/20 = 3/4. Therefore, h = 1/(3/4) = 4/3 hours.

Answer: C

Can anyone help me with this, Where did I go wrong?

(1)ta+tb+tc = ta - 6
(2)ta+tb+tc = tb - 1
(3)ta+tb+tc = tc/2

h= ta + tb

(4)ta+tb+(tc/2) = 0 [from (3)]
Therefore: h + (tc/2) = 0

Adding (1) + (2)
2(ta+tb) + 2tc = (ta+tb) - 7
2h + 2tc = h-7
tc = (-h-7)/2

Substituting tc in (4)

We get h + [(-h-7)]/4 = 0
h = 7/3

Did I do algebra wrong or is there a concept problem?

Thanks.

chetan2u, Gladiator59, Bunuel, how do we solve this question by substituting values?

Let the time taken for all three men working together to complete the job be T hrs.
Then the time taken by Alpha (A), Beta (B) or Gamma (G), working alone, is (T+6), (T+1) and (2T) hrs respectively.
When all three men work together to complete the job in T hrs, G's contribution is only half of the job (since he takes 2T hrs to do the whole job). Therefore, the remaining half of the job is done by A and B. So A and B together can do half of the job in T hrs or the whole job in 2T hrs (h=2T) or (1/2T) of the job in 1 hr. So, A and B's combined rate is (1/2T).
We know that A's and B's individual rates are [1/T+6)] and [1/(T+1)] respectively.
So we have another expression for their combined rate which is [1/(T+6)] + [1/(T+1)]
Therefore, [1/(T+6)] + [1/(T+1)]=(1/2T)---> (2T+7)/(T+6))(T+1)=(1/2T)---> 3T^2 + 7T - 6 = 0.
The value of T can be calculated either by factorizing or employing the quadratic equation formula.
T=2/3
h=2T=4/3 Ans: C

This approach is slightly faster since it saves some time while substituting the value of T to get 'h'.

Given: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone.

Asked: Let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

Let the time taken by Alpha , Beta & Gamma to complete the job alone be a, b & c hours respectively

1/(1/a + 1/b + 1/c) = a - 6 = b - 1 = c/2
(1/a + 1/b + 1/c) = 1/(a-6) = 1/(b-1) = 2/c
1/a + 1/b = 1/c
a-6 = b-1 = c/2
a = 6 + c/2 = (c+12)/2
b = 1 + c/2 = (c+2)/2

1/a + 1/b = 1/h
c = h

2/(c+12) + 2/(c+2) =1/c
2(2c + 14)/(c+2)(c+12) = 1/c
4c^2 + 28c = c^2 + 14c + 24
3c^2 + 14c - 24 = 0
3c^2 + 18c - 4c - 24 = 0

(c+6)(3c - 4) = 0
c = h = 4/3 hours


IMO C