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A and B can complete a task in 20 days. B and C can c
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Updated on: 07 Aug 2018, 02:24
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Q. A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order. A. A>B>C B. A>C>B C. B>A>C D. B>C>A E. C>A>B Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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A and B can complete a task in 20 days. B and C can c
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Updated on: 07 Aug 2018, 02:26
Method 2 : The method shown in the previous post is a lengthy method. Especially when we are not asked to find the efficiency. We are just asked to arrange them in descending order. • Let us try to compare two statements at a time and see if we can infer anything from it?
o A and B can complete the task in 20 days. o B and C can complete the task in 30 days. o Let us take the above two statements. Now notice carefully, that B is common in both the statements, right?
A is just replaced by C in the second statement. And what do we notice, when C comes in place of A?
• We see that the time taken to complete the work increases from 20 days to 30 days. • This means that C works slower than A. • Hence, A is more efficient than C. (A>C) • Now let us compare
o B and C can complete the task in 30 days. o C and A can complete the task in 40 days. o Now notice carefully, that C is common in both the statements.
B is just replaced by A in the second statement. And what do we notice, when A comes in place of B?
• We see that the time taken to complete the work increases from 30 days to 40 days. • Thus, that means that A works slower than B. • Hence, B is more efficient than A.(B>A) • Combining both the inferences, we can conclude that B>A>C. • Thus, you can see that even without calculating a single value, we found the answer, by simply observing the data and comparing them. • That’s the power of observation!! :D Thanks, Saquib Quant Expert eGMATAiming to score Q50 or higher in GMAT Quant? Attend this webinar on 2nd April to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Register
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Re: A and B can complete a task in 20 days. B and C can c
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04 Apr 2017, 21:08
Less than 20 second solution:
A + B > 20 C + B > 30 So, A >> C (1)
B + C > 30 A + C > 40 So, B >> A (2)
(1) & (2): B >> A >> C Hence, option C.



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Re: A and B can complete a task in 20 days. B and C can c
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02 Apr 2017, 07:46
EgmatQuantExpert wrote: Q. A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order. A. A>B>C B. A>C>B C. B>A>C D. B>C>A E. C>A>B Thanks, Saquib Quant Expert eGMATSaquib or anyone: I got the correct answer, but I am not sure my method is correct. I wrote equations for combined rates (efficiency, i.e. how many days to complete). A + B = 20 (i) B + C = 30 (ii) A + C = 40 (iii) Then added (ii) B + C = 30 (iii)A + C = 40, which yields (A + B) + 2C = 70 We know (A + B) = 20, so (20) + 2C = 70 C = 25 Solve for B. B + C = 30 B = 5 Solve for A. A + B = 20 A = 15 B takes 5 days. A takes 15 days. C takes 25 days. Therefore, from most efficient to least: B > A > C, where "greater than" means "more efficient." Answer: C I looked at Saquib's approaches and mine seemed flabby by comparison. Did I do it wrong? Help?
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Re: A and B can complete a task in 20 days. B and C can c
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03 Apr 2017, 05:10
genxer123 wrote: Saquib or anyone: I got the correct answer, but I am not sure my method is correct. I wrote equations for combined rates (efficiency, i.e. how many days to complete). A + B = 20 (i) B + C = 30 (ii) A + C = 40 (iii) Then added (ii) B + C = 30 (iii)A + C = 40, which yields (A + B) + 2C = 70 We know (A + B) = 20, so (20) + 2C = 70 C = 25 Solve for B. B + C = 30 B = 5 Solve for A. A + B = 20 A = 15 B takes 5 days. A takes 15 days. C takes 25 days. Therefore, from most efficient to least: B > A > C, where "greater than" means "more efficient." Answer: C I looked at Saquib's approaches and mine seemed flabby by comparison. Did I do it wrong? Help? Hey, The method that you have used is incorrect. Let me explain by asking you a simple question. Say A completes a piece of work in 20 days, so A = 20. And B completes the same work, in 30 days, so B = 30.
Now tell me if A and B work together will they take (20+30) = 50 days? Think about it carefully, when two people will work together, will they complete the work faster or will the take more than the individual time taken by them to complete the work?? Answer the above question and I am sure you will understand why your method is flawed. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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A and B can complete a task in 20 days. B and C can c
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Updated on: 07 Aug 2018, 02:28
There are a number of ways to solve this. Let me show you two. Method 1 : Given: • A and B can complete the task in 20 days. • B and C can complete the task in 30 days. • C and A can complete the task in 40 days. Approach and Working Out: • Let us assume the task to be 120 units. [LCM of 20,30 and 40] • Therefore, we can say
o Efficiency of A + B = Work/ Time = \(\frac{120}{20}\) = 6 units/day……….(i)
o Efficiency of B + C = \(\frac{120}{30}\) = 4 units/day…………(ii)
o Efficiency of C + A = \(\frac{120}{40}\) = 3 units/day…………(iii) • If we add these up we get
o 2 (Efficiency of A + B + C) = (6+4+3) = 13
o Efficiency of A + B + C = 6.5 ………..(iv) • Subtracting (i) from (iv) we get
o Efficiency of C = 0.5 units/day • Subtracting (ii) from (iv), we get
o Efficiency of A = 2.5 units/day • And efficiency of B = 3.5 units/day
• Thus, we can arrange the efficiency of A, B and C as B>A>C.
• And the correct answer is Option C. Thanks, Saquib Quant Expert eGMATAiming to score Q50 or higher in GMAT Quant? Attend this webinar on 2nd April to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Register
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Re: A and B can complete a task in 20 days. B and C can c
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05 Apr 2017, 08:27
I applied the following method. Please let me if the concept is correct.
1. In 20 day; A and B complete the work. Hence in 1 day work done is: 1/20
Rate of A + rate of B = 1/20 (Similarly) Rate of A + Rate of C = 1/ 40 Rate of B + Rate of C = 1/30
Now comparing the equations; 1/20> 1/40, so : Rate of A + rate of B > Rate of A + Rate of C (Minus "rate of A from both sides; we get): Rate of B > Rate of C
1/30> 1/40, so : Rate of B + rate of C >. Rate of A + Rate of C (Minus rate of C from both sides we get): Rate of B> Rate of A
1/20> 1/30, so : Rate A> Rate C
Combining all inequalities: B> A> C



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Re: A and B can complete a task in 20 days. B and C can c
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03 May 2017, 19:57
EgmatQuantExpert wrote: A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order.
A. A>B>C B. A>C>B C. B>A>C D. B>C>A E. C>A>B
Let's call \(A,B,C\) are the time for each to complete that work. Hence \(\frac{1}{A},\frac{1}{B},\frac{1}{C}\) are the efficiencies of each. \(\frac{1}{A}+\frac{1}{B}=\frac{1}{20} \quad (1)\) \(\frac{1}{B}+\frac{1}{C}=\frac{1}{30} \quad (2)\) \(\frac{1}{C}+\frac{1}{A}=\frac{1}{40}\quad (3)\) \((1) > (2) \implies \frac{1}{A} > \frac{1}{C} \) \((2) > (3) \implies \frac{1}{B} > \frac{1}{A} \) \((1) > (3) \implies \frac{1}{B} > \frac{1}{C} \) Hence \(\frac{1}{B}>\frac{1}{A}> \frac{1}{C}\) The answer is C
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Re: A and B can complete a task in 20 days. B and C can c
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15 May 2017, 03:04
I have used below approach. Please let me know if this is correct.
A+B = 20 days  (1) B+C = 30 days  (2) A + C = 40 days  (3)  2 (A+B+C) = 90 days 
A+B+C = 45 days  (4)
I have substituted the value of value of eq (1) , eq (2) and eq (3) in eq (4) one by one.
20 + C = 45 [A+B = 20 days] C= 25 [C alone can complete work in 25 days]
B + 40 = 45 [ A+C = 40 days] B=5 days [B alone can complete work in 5 days]
A+ 30 = 45 [B+C = 30] A=15 days [A alone can complete work in 15 days]
Hence, B>A>C
Ans : C



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Re: A and B can complete a task in 20 days. B and C can c
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The official solution has been posted. Looking forward to a healthy discussion..
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Re: A and B can complete a task in 20 days. B and C can c
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04 Apr 2017, 16:03
EgmatQuantExpert wrote: genxer123 wrote: Saquib or anyone: I got the correct answer, but I am not sure my method is correct. I wrote equations for combined rates (efficiency, i.e. how many days to complete). A + B = 20 (i) B + C = 30 (ii) A + C = 40 (iii) Then added (ii) B + C = 30 (iii)A + C = 40, which yields (A + B) + 2C = 70 We know (A + B) = 20, so (20) + 2C = 70 C = 25 Solve for B. B + C = 30 B = 5 Solve for A. A + B = 20 A = 15 B takes 5 days. A takes 15 days. C takes 25 days. Therefore, from most efficient to least: B > A > C, where "greater than" means "more efficient." Answer: C I looked at Saquib's approaches and mine seemed flabby by comparison. Did I do it wrong? Help? Hey, The method that you have used is incorrect. Let me explain by asking you a simple question. Say A completes a piece of work in 20 days, so A = 20. And B completes the same work, in 30 days, so B = 30.
Now tell me if A and B work together will they take (20+30) = 50 days? Think about it carefully, when two people will work together, will they complete the work faster or will the take more than the individual time taken by them to complete the work?? Answer the above question and I am sure you will understand why your method is flawed. Thanks, Saquib Quant Expert eGMATSaquib, thank you! Every so often I invert rates. Your question about people working faster together will remind me to check my hereAWOL logic on hard problems. Thanks again.
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Re: A and B can complete a task in 20 days. B and C can c
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29 Apr 2017, 22:18
Gargie.D wrote: I applied the following method. Please let me if the concept is correct.
1. In 20 day; A and B complete the work. Hence in 1 day work done is: 1/20
Rate of A + rate of B = 1/20 (Similarly) Rate of A + Rate of C = 1/ 40 Rate of B + Rate of C = 1/30
Now comparing the equations; 1/20> 1/40, so : Rate of A + rate of B > Rate of A + Rate of C (Minus "rate of A from both sides; we get): Rate of B > Rate of C
1/30> 1/40, so : Rate of B + rate of C >. Rate of A + Rate of C (Minus rate of C from both sides we get): Rate of B> Rate of A
1/20> 1/30, so : Rate A> Rate C
Combining all inequalities: B> A> C Hey Gargie, Your method is absolutely correct. People do tend to make the mistake of comparing time instead of efficiencies, i.e. rate of doing work. But you have done it exactly the way I expect a student to do it. :D Just one suggestion, in questions of high difficulty level, solving the question using fractions sometimes gets cumbersome, so instead of assuming the work to be 1 units, try to assume the work to be the LCM of the time taken by each person, that way you will be solving the question mostly using integers and for me it is easier to solve using integers than fraction/decimals. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: A and B can complete a task in 20 days. B and C can c
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03 May 2017, 14:25
lcm of 20,30,40=120 unit A+B=120 units/20hr= 6 unit per hr....(1) B+c=120/30=4......(2) A+C=120/40=3......(3) (1)(2) AC=2....(4) (3)+(4) A=2.5 B=3.5 C=0.5 B>A>C answer:C



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Re: A and B can complete a task in 20 days. B and C can c
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03 May 2017, 18:29
EgmatQuantExpert wrote: Q. A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order. A. A>B>C B. A>C>B C. B>A>C D. B>C>A E. C>A>B adding, rate of A+2B+C=20/240 rate of A+C=6/240 subtracting, 2B=14/240 rate of B= 7/240rate of A=12/2407/240= 5/240rate of C=6/2405/240= 1/240B>A>C C



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Re: A and B can complete a task in 20 days. B and C can c
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15 Feb 2018, 10:38
A + B = 1/20 (1) B + C = 1/30 (2) (1)(2) => A  C = (1/20)  (1/30) = 1/60 A + C = 1/40 solving => A = 1/48 so A alone takes 48 days to complete.
computing C: so two As => A + A >will take 24 days to complete, but given that A + C > 40 days, this means C is slower than A => A > C computing B: two As => A + A > 24 days to complete, but given that A + B > 20 days, this mean B is faster than A => B > A
B > A > C => option (C)



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Re: A and B can complete a task in 20 days. B and C can c
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29 Sep 2018, 17:11
EgmatQuantExpert wrote: Q. A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order. A. A>B>C B. A>C>B C. B>A>C D. B>C>A E. C>A>B Since A and B can complete a task in 20 days but B and C can complete the same task in 30 days, A must be more efficient than C. Since B and C can complete the same task in 30 days but A and C can complete the task in 40 days, B must be more efficient than A. Therefore, their efficiencies in descending order is B, followed by A, followed by C. Answer: C
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Re: A and B can complete a task in 20 days. B and C can c
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30 Sep 2018, 20:15
Hi All, We're told that A and B can complete a task in 20 days. B and C can complete the same task in 30 days and A and C can complete the task in 40 days. We're asked to arrange the 'efficiencies' of A, B, and C in descending order. This is actually more of a 'logic question' than a 'math question'  and you can solve it with some basic logical comparisons. To start, we know: A and B take 20 days to do a task B and C take 30 days to do a task Thus, 'swapping' A for C makes the job take 10 days LONGER, meaning that A is MORE efficient than C (re: A > C). Next, we know: A and B take 20 days to do a task A and C take 40 days to do a task Thus, 'swapping' B for C makes the job take 20 days LONGER, meaning that B is MORE efficient than C (re: B > C). Finally, we know: B and C take 30 days to do a task A and C take 40 days to do a task Thus, 'swapping' B for A makes the job take 10 days LONGER, meaning that B is MORE efficient than A (re: B > A). Combining the three inequalities, we have B > A > C Final Answer: GMAT assassins aren't born, they're made, Rich
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A and B can complete a task in 20 days. B and C can c
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28 Jan 2020, 23:17
Total 3 equations: Consider common factor: A+B = 20 (When B was with A, it took 20 days) C+B = 30 (When B was with C, it took 30 days) Means A is faster than C A>C Again consider 2nd equation for B B+C = 30 (When C was with B, it took 30 days) A+C = 40 (When C was with A, it took 40 days) So, B is a faster companion than A Hence, B>A>C Option C
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Re: A and B can complete a task in 20 days. B and C can c
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30 Jan 2020, 16:00
20(1/a + 1/b) = 1 30(1/b + 1/c) = 1 40(1/a + 1/c) = 1
ab faster than bc bc faster than ac
b>a>c




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