It takes 5 days, 10 days, 20 days, 40 days, and 50 days respectively f

Question

It takes 5 days, 10 days, 20 days, 40 days, and 50 days respectively for workers A, B, C, D, and E to complete the same work independently. The employer wants the work to be inspected within one day of completion. Further, the employer can employ exactly two of these five workers, and due to certain administrative restrictions, the employer mandates that the work takes at least 10 days but not more than 14 days. X is the slower worker of the two workers selected, and Y is the other worker. Select an option for X and an option for Y that are jointly consistent with the information provided. Make only two selections, one in each column.

Yxy · Answer

The slower worker should be the one with greater days to complete the work, so the given answer should be reversed

crimson_king · Answer

The answer should be reversed as D is the slower worker out of the two

poojaarora1818 · Answer

If x is the slower worker of the two workers, how can X be 20 and Y be 40? The slower one should take more days to complete a task.

Can you please walk us through the solution?

Bismuth83 · Answer

Correct! I reversed the order of the answers

Bismuth83 · Answer

Sure thing! I fixed the answers and reversed them. Let me walk you through the solution.

1. There are 5 workers with different speeds. We need to  find a combination of 2 workers that will do the work in 10 to 14 days .

2. Let the amount of work be equal to 1. Then the worker speeds for A, B, C, D, and E are  \frac{1}{5} ,  \frac{1}{10} ,  \frac{1}{20} ,  \frac{1}{40} , and  \frac{1}{50} .

3. That means our question can be rewritten. We need to find X and Y such that the value of t in this equation is between 10 and 14:  t(X_{speed} + Y_{speed}) = 1 . Now, we can look at the extremes.

4. Workers A or B won’t work, since they can do all the work by themselves in 10 days or less. C, D and E are left.

5. For the combination D and E we have:  t(\frac{1}{40} + \frac{1}{50}) = 1 \rightarrow t = \frac{1}{\frac{1}{40} + \frac{1}{50}} > \frac{1}{\frac{1}{40} + \frac{1}{40}} = \frac{1}{\frac{1}{20}} = 20 . This doesn’t work since t must be between 10 and 14.

6. We are left with 2 combinations - C+D and C+E. Since we know that C must be there, we can say that both combinations take more than 10 days and  Y must be C . This is because  10 * (\frac{1}{20} + \frac{1}{20}) = 1  and replacing one 1/20 with something less would make t larger.

7. Because there is only one answer, the combination that takes less time should work and that’s C+D. However, let’s double-check just in case: 
 t(\frac{1}{20} + \frac{1}{40}) = 1 \rightarrow t = \frac{1}{\frac{1}{20} + \frac{1}{40}} = \frac{1}{\frac{3}{40}} = \frac{40}{3} < \frac{42}{3} = 14  and  t(\frac{1}{20} + \frac{1}{50}) = 1 \rightarrow t = \frac{1}{\frac{1}{20} + \frac{1}{50}} = \frac{1}{\frac{7}{100}} = \frac{100}{7} > \frac{98}{7} = 14 .

8. So, our answer is   X = D and Y = C  .

poojaarora1818 · Answer

Thank you so much for the detailed solution. Awesome!

Yangyixi · Answer

why is it not possible to combine A with somebody who is not B?

Bismuth83 · Answer

If A worked with anyone it would take less time than if A worked alone.

A has a speed of  \frac{1}{5} . In 5 days, A would complete the work alone, which is less 10. Thus, working with anyone else would take even less and that doesn't work.

I hope that answered your question!

Yangyixi · Answer

You did, thank you !

Ashok2530 · Answer

A=5 Days 
B=10 Days
C=20Days
D=40 Days
E=50 Days

Total Units of Work = 200 Units

C= 200/20 = 10 Units/Day
D= 200/40 = 5 Units/Day

C+ D = 15 Units/Day

200/15 = 40/3 days which is greater than 10 days and less than 14 days.

ag821 · Answer

Not sure if this is right, but I did it this way -
x < y

In 10 days, each does how much work -
A : 1/5*10 = 2 units
B : 1/10*10 = 1 unit
C : 1/20* 10 = 0.5 unit
D : 1/40*10 = 0.4 unit
E : 1/ 50 * 10 = 0.2 unit

Since we have to make 1 unit of work jointly in min 10 days and max 14 days, C & D is the best combination because they finish 0.9 unit of work in 10 days, and in less than 14 days, they will finish the other 0.1 unit of remaining work.

D & E combination won't make sense because in 10 days they finish only 0.6 unit of work, and the remaining work of 0.4 is still left, and by the 15th day they would have jointly completed only 0.9 unit of work (0.2 of D and 0.1 of E), which is more than 14 days & the job is still not complete.

Hence x = 0.4 i.e. D and y = 0.5 i.e. C

Used this method to solve within 2 mins

Hrley · Answer

I did this question correctly in a bit over 3min, and I used somewhat of a cheeky way for this...

If X is slower than Y, and the time that both need to work is from 10 to 14 days, then:
1/14 <= (1/X + 1/Y) <= 1/10
Or,
0.0714 <= (1/X + 1/Y) <= 0.1

Then I converted all the workers' rates into decimals.
I can use a calculator in DI so this can be done quickly.
1/A = 0.2, 1/B = 0.1, 1/C = 0.05, 1/D = 0.025, 1/E = 0.02

It's D and C, bc their rate total falls between 0.0714 and 0.1. And in that order since D is obviously slower than C, according to their rates.
If I did not round 1/14 correctly, I would have seriously been confused between D/C or D/E.

quamquia · Answer

For easier calculations and approximation i took lcm of days 5, 10 and rest which was 200. I now equated 200 as the work , from rate time concept . For individual rates i divided work by time of all for eg A took 5 days to complete work so 200/ 5 i.e 40 units of work a day. Aim is to complete 200 units of work in more than 10 but less than 14 days , so directly add units of work and divide by 200 . Here C ,D have 15 units of combined rate . 200 /15 gives 13 something days which is what we want hence is the answer.

It takes 5 days, 10 days, 20 days, 40 days, and 50 days respectively f

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