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

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Math Expert
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Posts: 54372
Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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14 Mar 2019, 23:29
2
7
00:00

Difficulty:

85% (hard)

Question Stats:

38% (02:36) correct 62% (03:22) wrong based on 47 sessions

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

(A) $$\frac{5}{2}$$

(B) $$\frac{3}{2}$$

(C) $$\frac{4}{3}$$

(D) $$\frac{5}{4}$$

(E) $$\frac{3}{4}$$

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Joined: 27 Sep 2018
Posts: 37
Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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15 Mar 2019, 01:15
1
Let time taken by alpha, beta and gamma be = x hrs
Time taken by alpha = x+6 hrs
Time taken by beta = x+1 hrs
Time taken by gamma = 2x hrs
Also,
1/(x+6) + 1/(x+1) + 1/2x = 1/x
Solving for x we get x= 2/3. (i)

Now atq h is no of hours needed by alpha and gamma for job
So, 1/h = 1/(x+6) + 1/(x+1)
Solving for h by substituting x =2/3 (from I)
we get h 4/3 hrs

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Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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16 Mar 2019, 01:19
2
Bunuel wrote:
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. Let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

(A) $$\frac{5}{2}$$

(B) $$\frac{3}{2}$$

(C) $$\frac{4}{3}$$

(D) $$\frac{5}{4}$$

(E) $$\frac{3}{4}$$

let total time of A+B+G= x hrs
so time for A= x+6
B=x+1
and G= 2x

combing all
1/x+6 + 1/x+1 + 1/2x = x
x = 2/3
given
A+B = h
1/x+6 + 1/x+1 = 1/h
x = 2/3
or say
h = 20/15 ; 4/3
IMO C
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Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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18 Mar 2019, 07:28
Could please anyone help me :
How did you find x in the first place?
Thank you

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Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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19 Mar 2019, 19:07
1
Bunuel wrote:
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. Let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

(A) $$\frac{5}{2}$$

(B) $$\frac{3}{2}$$

(C) $$\frac{4}{3}$$

(D) $$\frac{5}{4}$$

(E) $$\frac{3}{4}$$

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.

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Posts: 46
Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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22 Mar 2019, 12:44
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

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.
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Joined: 11 Jun 2018
Posts: 72
GMAT 1: 500 Q39 V21
Re: Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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05 Apr 2019, 06:47
chetan2u, Gladiator59, Bunuel, how do we solve this question by substituting values?
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Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour  [#permalink]

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11 Apr 2019, 04:44
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'.
Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hour   [#permalink] 11 Apr 2019, 04:44
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