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$A$, $B$, $C$ do a job $6$ hours faster than $A$ alone, $1$ hour faster than $B$ alone, and in half the time needed by $C$ alone. … [closed]
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Three men $A$, $B$, $C$ working together can do a job in $6$ hours less time than $A$ alone, in $1$ hour less time than $B$ alone, and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in how many hours?
Could we solve this without a quadratic equation?
arithmetic
edited Jul 27 at 18:40
Blue
43.6k868141
asked Jul 27 at 18:27
user160370
1297
closed as off-topic by Martin R, Delta-u, Taroccoesbrocco, John B, John Ma Jul 28 at 18:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Delta-u, Taroccoesbrocco, John B, John Ma
add a comment |Â
up vote
3
down vote
favorite
Three men $A$, $B$, $C$ working together can do a job in $6$ hours less time than $A$ alone, in $1$ hour less time than $B$ alone, and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in how many hours?
Could we solve this without a quadratic equation?
arithmetic
edited Jul 27 at 18:40
Blue
43.6k868141
asked Jul 27 at 18:27
user160370
1297
closed as off-topic by Martin R, Delta-u, Taroccoesbrocco, John B, John Ma Jul 28 at 18:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Delta-u, Taroccoesbrocco, John B, John Ma
6
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Three men $A$, $B$, $C$ working together can do a job in $6$ hours less time than $A$ alone, in $1$ hour less time than $B$ alone, and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in how many hours?
Could we solve this without a quadratic equation?
arithmetic
edited Jul 27 at 18:40
Blue
43.6k868141
asked Jul 27 at 18:27
user160370
1297
Three men $A$, $B$, $C$ working together can do a job in $6$ hours less time than $A$ alone, in $1$ hour less time than $B$ alone, and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in how many hours?
Could we solve this without a quadratic equation?
arithmetic
edited Jul 27 at 18:40
Blue
43.6k868141
asked Jul 27 at 18:27
user160370
1297
edited Jul 27 at 18:40
Blue
43.6k868141
edited Jul 27 at 18:40
Blue
43.6k868141
edited Jul 27 at 18:40
Blue
43.6k868141
43.6k868141
asked Jul 27 at 18:27
user160370
1297
asked Jul 27 at 18:27
user160370
1297
asked Jul 27 at 18:27
user160370
1297
1297
closed as off-topic by Martin R, Delta-u, Taroccoesbrocco, John B, John Ma Jul 28 at 18:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Delta-u, Taroccoesbrocco, John B, John Ma
closed as off-topic by Martin R, Delta-u, Taroccoesbrocco, John B, John Ma Jul 28 at 18:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Delta-u, Taroccoesbrocco, John B, John Ma
6
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10
add a comment |Â
6
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10
6
6
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it. 
answered Jul 28 at 5:57
user160370
1297
add a comment |Â
up vote
0
down vote
I think it is not possible without a qudratic equation, here is my solution.
Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.
We know that
$$frac100x+y+z=frac100x-6 implies100x=100(x+y+z)-6x(x+y+z)$$
$$frac100x+y+z=frac100y-1implies100y=100(x+y+z)-y(x+y+z)$$
$$frac100x+y+z=frac12frac100zimplies z=x+y$$
then summing up the two first equations we obtain
$$100z=400z-(5x+z)(2z)implies5x+z=150 implies z=150-5x$$
and from the first one
$$100x=200z-12xz implies100x=30000-1000x-1800x+60x^2 $$
$$implies 3x^2-145x+1500=0$$
$$implies x=15 implies z=75 implies y=60$$
Thus $A$ and $B$ working togheter need $frac10075=frac43$h to complete the work.
answered Jul 27 at 18:57
gimusi
64.9k73483
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it.
answered Jul 28 at 5:57
user160370
1297
add a comment |Â
up vote
1
down vote
I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it.
answered Jul 28 at 5:57
user160370
1297
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it.
answered Jul 28 at 5:57
user160370
1297
I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it.
answered Jul 28 at 5:57
user160370
1297
answered Jul 28 at 5:57
user160370
1297
answered Jul 28 at 5:57
user160370
1297
answered Jul 28 at 5:57
user160370
1297
1297
add a comment |Â
add a comment |Â
up vote
0
down vote
I think it is not possible without a qudratic equation, here is my solution.
Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.
We know that
$$frac100x+y+z=frac100x-6 implies100x=100(x+y+z)-6x(x+y+z)$$
$$frac100x+y+z=frac100y-1implies100y=100(x+y+z)-y(x+y+z)$$
$$frac100x+y+z=frac12frac100zimplies z=x+y$$
then summing up the two first equations we obtain
$$100z=400z-(5x+z)(2z)implies5x+z=150 implies z=150-5x$$
and from the first one
$$100x=200z-12xz implies100x=30000-1000x-1800x+60x^2 $$
$$implies 3x^2-145x+1500=0$$
$$implies x=15 implies z=75 implies y=60$$
Thus $A$ and $B$ working togheter need $frac10075=frac43$h to complete the work.
answered Jul 27 at 18:57
gimusi
64.9k73483
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
add a comment |Â
up vote
0
down vote
I think it is not possible without a qudratic equation, here is my solution.
Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.
We know that
$$frac100x+y+z=frac100x-6 implies100x=100(x+y+z)-6x(x+y+z)$$
$$frac100x+y+z=frac100y-1implies100y=100(x+y+z)-y(x+y+z)$$
$$frac100x+y+z=frac12frac100zimplies z=x+y$$
then summing up the two first equations we obtain
$$100z=400z-(5x+z)(2z)implies5x+z=150 implies z=150-5x$$
and from the first one
$$100x=200z-12xz implies100x=30000-1000x-1800x+60x^2 $$
$$implies 3x^2-145x+1500=0$$
$$implies x=15 implies z=75 implies y=60$$
Thus $A$ and $B$ working togheter need $frac10075=frac43$h to complete the work.
answered Jul 27 at 18:57
gimusi
64.9k73483
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think it is not possible without a qudratic equation, here is my solution.
Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.
We know that
$$frac100x+y+z=frac100x-6 implies100x=100(x+y+z)-6x(x+y+z)$$
$$frac100x+y+z=frac100y-1implies100y=100(x+y+z)-y(x+y+z)$$
$$frac100x+y+z=frac12frac100zimplies z=x+y$$
then summing up the two first equations we obtain
$$100z=400z-(5x+z)(2z)implies5x+z=150 implies z=150-5x$$
and from the first one
$$100x=200z-12xz implies100x=30000-1000x-1800x+60x^2 $$
$$implies 3x^2-145x+1500=0$$
$$implies x=15 implies z=75 implies y=60$$
Thus $A$ and $B$ working togheter need $frac10075=frac43$h to complete the work.
answered Jul 27 at 18:57
gimusi
64.9k73483
I think it is not possible without a qudratic equation, here is my solution.
Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.
We know that
$$frac100x+y+z=frac100x-6 implies100x=100(x+y+z)-6x(x+y+z)$$
$$frac100x+y+z=frac100y-1implies100y=100(x+y+z)-y(x+y+z)$$
$$frac100x+y+z=frac12frac100zimplies z=x+y$$
then summing up the two first equations we obtain
$$100z=400z-(5x+z)(2z)implies5x+z=150 implies z=150-5x$$
and from the first one
$$100x=200z-12xz implies100x=30000-1000x-1800x+60x^2 $$
$$implies 3x^2-145x+1500=0$$
$$implies x=15 implies z=75 implies y=60$$
Thus $A$ and $B$ working togheter need $frac10075=frac43$h to complete the work.
answered Jul 27 at 18:57
gimusi
64.9k73483
edited Jul 27 at 20:30
answered Jul 27 at 18:57
gimusi
64.9k73483
answered Jul 27 at 18:57
gimusi
64.9k73483
answered Jul 27 at 18:57
gimusi
64.9k73483
64.9k73483
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
add a comment |Â
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@gmusi: the quadratic equation has two roots, why did you pick $15$?
– Vasya
Jul 27 at 19:01
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
@Vasya The other one (100/3) gives a negative value for $z$.
– gimusi
Jul 27 at 19:02
1
1
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
A and B working toghether $= frac 100x+y = frac 10075 = frac 43$ hours.
– Doug M
Jul 27 at 20:29
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
@DougM Ops...yes I summed up A and C...thanks I fix
– gimusi
Jul 27 at 20:30
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
Oh, how many times have I done great work right up to the last step and then lost my focus.
– Doug M
Jul 27 at 20:32
add a comment |Â
6
If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach
– Blue
Jul 27 at 18:42
Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic.
– fleablood
Jul 27 at 19:10