Trucks A, B, and C, working together, can move a load of sand in "t" h

Question

Trucks A, B, and C, working together, can move a load of sand in "t" hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, "t" extra hours. What is the value of t?

A. 1/6
B. 1/4
C. 1/3
D. 2/3
E. 3/4

A. 1/6
B. 1/4
C. 1/3
D. 2/3
E. 3/4

sthahvi · Answer

what is wrong with this method:
1/a= t+1, 1/b=t+6, 1/c=2t 
1/a+1/b+1/c=t 
substitute values as mentioned above then we get:
t+1 + t+6+ 2t = t 
 what is wrong with this ?
why am I not getting the answer ?

DudeBeCool · Answer

Time taken by A = t+1
Time taken by B = t+6
Time taken by C = 2t

Adding in the equation, R*T = W, we get

* t = 1

solve for 't' and you will get your answer.

Answer: 2/3

jrk23 · Answer

One more solution :

as per question:

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

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

= 3t^2-7t-6=0
=t is -3,2/3

t can't be negative so t=2/3

MBAB123 · Answer

sthahvi , the highlighted portion is wrong. You've taken t+1,t+6 and 2t to be the rates of A,b,c. That is the total time. A=t+1, b=t+6 and c=2t.

Hence, 
1/a=1/t+1
1/b=1/t+6
1/c=1/2t.
1/a +1/b +1/c = t.

^solve the above for t and you'll get your answer!

Riccardo2505 · Answer

I arrived no problem at

3t^2 - 7t - 6 = 0 --> t1,2 =  /2a --> t = (7 + 11)/6 = 3

What am I missing? 
Thanks

philipssonicare · Answer

Trucks A, B, and C, working together, can move a load of sand in "t" hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, "t" extra hours. What is the value of t?

A. 1/6
B. 1/4
C. 1/3
D. 2/3
E. 3/4

Work=Rate·Time; Rate=Work/Time
Rates are additive. When things work together their respective rates are added to make one single rate (their working together rate)

A completes 1 work in A time, B in B, and C in C. Together they complete 1 work in t time.

Set up the basic formula
 \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{t}

\frac{1}{A}=\frac{1}{(t+1)} 
 \frac{1}{B}=\frac{1}{(t+6)} 
 \frac{1}{C}=\frac{1}{(2t)}

\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t} 
 \frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{1}{t}-\frac{1}{2t} 
 \frac{(t+6)}{(t+1)(t+6)}+\frac{(t+1)}{(t+1)(t+6)}=\frac{2}{2t}-\frac{1}{2t} 
 \frac{(t+6)+(t+1)}{(t+1)(t+6)}=\frac{2-1}{2t} 
 \frac{(2t+7)}{(t+1)(t+6)}=\frac{1}{2t} 
Cross multiply
(t+1)(t+6)=(2t+7)2t
t^2+7t+6=4t^2+14t
0=3t^2+7t-6
0=3t^2+9t-2t-6
0=3t(t+3)-2(t+3)
0=(3t-2)(t+3)
t=-3 or t=2/3

effatara · Answer

Once you get the equation 3t^2 +7t - 6 = 0, the rest is easy but I think the thought process involved in arriving at this equation should be explained. 
C alone does the work in '2t' days. So, in 't' days he can do 1/2 of the work. When all three work together for 't' days, they can do the full work. So it stands to reason A and B does the remaining half of the work. So:

t/(t+1) + t/(t+6) = 1/2.....> 3t^ + 7t - 6 = 0......> 3t^2 + 9t - 2t - 6 = 0......> 3t(t+3) - 2(t+3) = 0. Since 't' can't be negative, 3t-2=0. Therefore t=2/3

Kinshook · Answer

Given: Trucks A, B, and C, working together, can move a load of sand in "t" hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, "t" extra hours.

Asked: What is the value of t?

Truck A takes (t+1) hours to move the sand
Truck B takes (t+6) hours to move the sand
Truck C takes 2t hours to move the sand

Trucks A, B, and C, working together, can move a load of sand in "t" hours. 
1/(t+1) + 1/(t+6) + 1/2t = 1/t
(2t + 7)/(t+1)(t+6) = 1/2t
2t(2t +7) = (t+1)(t+6)
4tˆ2 + 14t = tˆ2 + 7t + 6
3tˆ2 +7t - 6 = 0
3tˆ2 + 9t - 2t - 6 = 0
3t (t+3) - 2(t+3) = 0
(3t-2)(t+3) = 0
t = 2/3 hours since t = -3 hours is not feasible

IMO D

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Trucks A, B, and C, working together, can move a load of sand in "t" h

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Trucks A, B, and C, working together, can move a load of sand in "t" h

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