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655-705 Level|   Work and Rate Problems|            
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Workers A, B, and C, working at their individual constant rates, can 30 May 2024, 12:24
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­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
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Workers A, B, and C, working at their individual constant rates, can 31 May 2024, 23:39
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Bunuel
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Show more
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

Let the time taken be a, b and c respectively...... 
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}\)

(1) A is twice as efficient as B. B is three times as efficient as C.
So, a is related to b and b is related to c. Thus, all the variables can be put in terms of one variable, and the value of each of them can be found.
Sufficient

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
We have a in relation to b+c and again the answer can be found as we are looking at b+c.
Sufficient

D

Let us work on each statement, if it were PS.
(1) A is twice as efficient as B. B is three times as efficient as C.
So, b=2a and c=3b=6a....
\(\frac{1}{a}+\frac{1}{2a}+\frac{1}{6a}=\frac{1}{6}......\frac{6+3+1}{6a}=\frac{1}{6}.....\frac{10}{6a}=\frac{1}{6}........a=10\)
We know b=2a = 20, and c=3b = 3*20 = 60.
One day work of B and C = \(\frac{1}{20}+\frac{1}{60}=\frac{4}{60}=\frac{1}{15}\).
Thus, B and C can do the work in 15 days.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Let the time taken be x in both cases, so in x days, combined work = 9+6 or 15 work.
We know 1 work is completed by all three in 6 days, so 15 unit of works will be completed in 6*15 or 90 days.
In 90 days, B and C combined do 6 works, so they will take 90/6 or 15 days for 1 work.

 
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Workers A, B, and C, working at their individual constant rates, can 21 Aug 2024, 03:02
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Bunuel

What does this statement A is twice as efficient as B. B is three times as efficient as C really mean?

Does this mean that the rate of A is twice that of B and the rate of B is thrice that of C?

If the rate of C is let say X, does the below interpretation according to my logic correct?

Rate A=6X
Rate B=3X
Rate C=X


Or

Does A is twice as efficient as B. B is three times as efficient as C gives the relation of the time taken by these 3 workers to eachother?

Iam really confused.

Thanks in advance!
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Rebaz
Bunuel

What does this statement A is twice as efficient as B. B is three times as efficient as C really mean?

Does this mean that the rate of A is twice that of B and the rate of B is thrice that of C?

If the rate of C is let say X, does the below interpretation according to my logic correct?

Rate A=6X
Rate B=3X
Rate C=X


Or

Does A is twice as efficient as B. B is three times as efficient as C gives the relation of the time taken by these 3 workers to eachother?

Iam really confused.

Thanks in advance!
Show more
­yes it means that A can do the same work in half the time as B and B can do the same work in one-third time as C

Lets say C takes 60 mins to pickup 60 bricks thus 1b/m, B is 3 times efficient thus 3b/m which means it takes him one third time or 20 mins to pick 60 bricks and A works twice efficiently as B thus takes 6 b/m taking 10 mins or half the time as B and one-sixth time as C
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Workers A, B, and C, working at their individual constant rates, can 25 Aug 2024, 11:17
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B and C ALONE = B and C together, without A?
shouldn't it be B and C individually?
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Workers A, B, and C, working at their individual constant rates, can 07 Sep 2024, 16:25
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chetan2u

Bunuel
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

Let the time taken be a, b and c respectively......
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}\)

(1) A is twice as efficient as B. B is three times as efficient as C.
So, a is related to b and b is related to c. Thus, all the variables can be put in terms of one variable, and the value of each of them can be found.
Sufficient

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
We have a in relation to b+c and again the answer can be found as we are looking at b+c.
Sufficient

D

Let us work on each statement, if it were PS.
(1) A is twice as efficient as B. B is three times as efficient as C.
So, b=2a and c=3b=6a....
\(\frac{1}{a}+\frac{1}{2a}+\frac{1}{6a}=\frac{1}{6}......\frac{6+3+1}{6a}=\frac{1}{6}.....\frac{10}{6a}=\frac{1}{6}........a=10\)
We know b=2a = 20, and c=3b = 3*20 = 60.
One day work of B and C = \(\frac{1}{20}+\frac{1}{60}=\frac{4}{60}=\frac{1}{15}\).
Thus, B and C can do the work in 15 days.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Let the time taken be x in both cases, so in x days, combined work = 9+6 or 15 work.
We know 1 work is completed by all three in 6 days, so 15 unit of works will be completed in 6*15 or 90 days.
In 90 days, B and C combined do 6 works, so they will take 90/6 or 15 days for 1 work.

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Does a piece of work means 1 unit of a work ?

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Workers A, B, and C, working at their individual constant rates, can 08 Sep 2024, 02:46
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Sazimordecai

chetan2u

Bunuel
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

Let the time taken be a, b and c respectively......
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}\)

(1) A is twice as efficient as B. B is three times as efficient as C.
So, a is related to b and b is related to c. Thus, all the variables can be put in terms of one variable, and the value of each of them can be found.
Sufficient

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
We have a in relation to b+c and again the answer can be found as we are looking at b+c.
Sufficient

D

Let us work on each statement, if it were PS.
(1) A is twice as efficient as B. B is three times as efficient as C.
So, b=2a and c=3b=6a....
\(\frac{1}{a}+\frac{1}{2a}+\frac{1}{6a}=\frac{1}{6}......\frac{6+3+1}{6a}=\frac{1}{6}.....\frac{10}{6a}=\frac{1}{6}........a=10\)
We know b=2a = 20, and c=3b = 3*20 = 60.
One day work of B and C = \(\frac{1}{20}+\frac{1}{60}=\frac{4}{60}=\frac{1}{15}\).
Thus, B and C can do the work in 15 days.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Let the time taken be x in both cases, so in x days, combined work = 9+6 or 15 work.
We know 1 work is completed by all three in 6 days, so 15 unit of works will be completed in 6*15 or 90 days.
In 90 days, B and C combined do 6 works, so they will take 90/6 or 15 days for 1 work.



Does a piece of work means 1 unit of a work ?

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­Yes, in this context, "a piece of work" refers to completing the entire task, which can be considered as 1 unit of work. When the question talks about units of work (such as 9 units or 6 units), it means multiple such pieces of work.
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Sazimordecai
chetan2u

Bunuel
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

Let the time taken be a, b and c respectively......
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}\)

(1) A is twice as efficient as B. B is three times as efficient as C.
So, a is related to b and b is related to c. Thus, all the variables can be put in terms of one variable, and the value of each of them can be found.
Sufficient

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
We have a in relation to b+c and again the answer can be found as we are looking at b+c.
Sufficient

D

Let us work on each statement, if it were PS.
(1) A is twice as efficient as B. B is three times as efficient as C.
So, b=2a and c=3b=6a....
\(\frac{1}{a}+\frac{1}{2a}+\frac{1}{6a}=\frac{1}{6}......\frac{6+3+1}{6a}=\frac{1}{6}.....\frac{10}{6a}=\frac{1}{6}........a=10\)
We know b=2a = 20, and c=3b = 3*20 = 60.
One day work of B and C = \(\frac{1}{20}+\frac{1}{60}=\frac{4}{60}=\frac{1}{15}\).
Thus, B and C can do the work in 15 days.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Let the time taken be x in both cases, so in x days, combined work = 9+6 or 15 work.
We know 1 work is completed by all three in 6 days, so 15 unit of works will be completed in 6*15 or 90 days.
In 90 days, B and C combined do 6 works, so they will take 90/6 or 15 days for 1 work.




Does a piece of work means 1 unit of a work ?

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Show more



Thank you soo much Bunuel
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Workers A, B, and C, working at their individual constant rates, can 03 Oct 2024, 03:01
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shriyashah
B and C ALONE = B and C together, without A?
shouldn't it be B and C individually?
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I had the same confusion but I checked the dic. It says that alone can mean seperated from others and an example sentence is The children alone would eat that much.
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Bunuel,

How can the text "B and C alone" be implying C + B, when clearly A+B+C is stated as "A, B, and C, working at their individual constant rates, can work together"

I assumed its asking for B , C alone. so statement 2 is insufficient as we are given data for C+B, not each individually.

Thanks in advance.
guddo
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Show more
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Workers A, B, and C, working at their individual constant rates, can 19 Apr 2025, 00:48
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Hello
A,B,C working together complete work in six days
so Ra+Rb+Rc=1/6

option 1 gives relation between the all three so then we can get the Rb+Rc ,hence it is sufficient

option B gives the time taken of A and B+C
if we equate time
9\Ra=6\(Rb+Rc)

we can get relation can find the rate , Hence it also sufficient

Hence option D is correct
Thanks
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Workers A, B, and C, working at their individual constant rates, can 28 Apr 2025, 23:07
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Isn’t this a bit confusing? Bunuel chetan2u
In question stem, B and C alone is used for B and C combined.
In statement 2, B and C together is used for B and C combined.
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siddharth_
­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

(1) A is twice as efficient as B. B is three times as efficient as C.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.

Isn’t this a bit confusing? Bunuel chetan2u
In question stem, B and C alone is used for B and C combined.
In statement 2, B and C together is used for B and C combined.
Show more

In the stem, from the context, we can understand that "B and C alone" means B and C working together without A, because earlier it talked about "A, B, and C can work together and complete a piece of work."

In (2), since it is a separate statement, it is framed differently to convey the same meaning independently.

That said, I agree it is a bit confusing but should still be understandable from the context.
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Workers A, B, and C, working at their individual constant rates, can 16 Aug 2025, 21:03
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lets say Rates are Ra, Rb, Rc :
Given : Ra+Rb+Rc = 1/6
To find : Rb+Rc=1/t How much is t then ?

A & B both options give relations in Rates so each option is sufficient individually.
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2) is wrong.
That method only gives you (B + C) as a combined rate, not their individual rates. the stem is asking for b and c alone ? chetan2u am i missing something

chetan2u


­Workers A, B, and C, working at their individual constant rates, can work together and complete a piece of work in 6 days. How long will it take for B and C alone to complete the work?

Let the time taken be a, b and c respectively......
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}\)

(1) A is twice as efficient as B. B is three times as efficient as C.
So, a is related to b and b is related to c. Thus, all the variables can be put in terms of one variable, and the value of each of them can be found.
Sufficient

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
We have a in relation to b+c and again the answer can be found as we are looking at b+c.
Sufficient

D

Let us work on each statement, if it were PS.
(1) A is twice as efficient as B. B is three times as efficient as C.
So, b=2a and c=3b=6a....
\(\frac{1}{a}+\frac{1}{2a}+\frac{1}{6a}=\frac{1}{6}......\frac{6+3+1}{6a}=\frac{1}{6}.....\frac{10}{6a}=\frac{1}{6}........a=10\)
We know b=2a = 20, and c=3b = 3*20 = 60.
One day work of B and C = \(\frac{1}{20}+\frac{1}{60}=\frac{4}{60}=\frac{1}{15}\).
Thus, B and C can do the work in 15 days.

(2) The time taken by A to do 9 units of work is the same as the time taken by B and C together to do 6 units of work.
Let the time taken be x in both cases, so in x days, combined work = 9+6 or 15 work.
We know 1 work is completed by all three in 6 days, so 15 unit of works will be completed in 6*15 or 90 days.
In 90 days, B and C combined do 6 works, so they will take 90/6 or 15 days for 1 work.

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NEYR0N
2) is wrong.
That method only gives you (B + C) as a combined rate, not their individual rates. the stem is asking for b and c alone ? chetan2u am i missing something


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Your doubt is addressed HERE.
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Thank you, Bunuel
Bunuel


Your doubt is addressed HERE.
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