Quarter Wit, Quarter Wisdom: Rates Revisited

Quarter Wit, Quarter WisdomPeople often complain about getting stuck in work-rate problems. Hence, I would like to take some 700+ level questions on rate today. I have discussed the basic concepts of work-rate (using ratios) in a previous post:

Cracking the Work Rate Problems

You  might want to go through that post before you set out to work on these problems. Ensure that you are very comfortable with the relation: Work = Rate*Time and its implications: If rate doubles, work done doubles too if the time remains constant; if one work is done, rate = 1/time etc. Thorough understanding of these implications is fundamental to ‘reasoning out’ the answer.

Question 1: Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

(A)   1/2
(B)   2
(C)   3
(D)   5
(E)    6

Solution: Tricky, eh? It is a little cumbersome if you get into variables. If you just try to reason it out, it could be done rather quickly and easily. Let’s see!

Machine A and B together complete 1 work in 3 hrs i.e. together, they do 1/3rd work every hour.

If machine A’s speed were double, they would do 1/2 work in 1 hour together. How come they do (1/2 – 1/3 =) 1/6th work extra in 1 hour now? Because machine A’s speed is double the previous speed. The extra speed that machine A has allows it to do 1/6th work extra. This means, at normal speed, machine A used to do 1/6 work in an hour (its speed had doubled so work had doubled too). Hence, at usual speed, it will take 6 hrs to produce 1 widget.

Answer (E)

Consider the amount of time and effort you would have spent on this question had you tried to use two variables to figure out the answer. You would have made equations like this: 1/a + 1/b = 1/3 and 2/a + 1/b = 1/2 and then you would have solved them simultaneously to get the value of a. Whereas in the solution above, we have done all the work orally!

Question 2: One woman and one man can build a wall together in two hours, but the woman would need the help of two girls in order to complete the same job in the same amount of time. If one man and one girl worked together, it would take them four hours to build the wall. Assuming that rates for men, women and girls remain constant, how many hours would it take one woman, one man, and one girl, working together, to build the wall?

(A)   5/7
(B)   1
(C)   10/7
(D)   12/7
(E)    22/7

Solution: This question is certainly quite tricky but if you understand the relation between work and rate, you can still solve this question easily. Mind you, we are using variables here only because I don’t want to write man, woman and girl again and again. Notice that there are no ‘=’ signs i.e. we are not making equations so we are not doing any algebraic manipulations.

The question is long so take one line at a time and analyze it. We will keep condensing the information we get from each sentence and figuring out the implications of new and previous information as we go along.

“One woman and one man can build a wall together in 2 hrs,”
1w + 1m -> 2 hrs ……(I)

“but the woman would need the help of 2 girls in order to complete the same job in the same amount of time.”
1w + 2g -> 2 hrs …..(II)

From (I) and (II), we can say that 1m is equivalent to 2g (i.e. 1 man does the same work as 2 girls do in the same amount of time; 1m ≡ 2g)

“If 1 man and 1 girl worked together, it would take them four hours to build the wall.”
1m + 1g -> 4hrs (Since 1m ≡ 2g, we can say that 3g will take 4 hrs to build the wall.)
or 2m + 2g -> 2 hrs …..(III) (If number of workers double, time taken to do the work becomes half)

From (II) and (III), 1w ≡ 2m (i.e. 1 woman does the same work as 2 men do in the same amount of time)
Hence, 1w ≡ 2m ≡ 4g

“Assuming that rates for women, men and girls remain constant, how many hours would it take 1 woman, 1 man and 1 girl working together to build the wall?”
1w + 1m + 1g ≡ 4g + 2g + 1g ≡ 7g. Since 3g take 4 hrs to build the wall, 7g will take 3*4/7 = 12/7 hrs to complete the  wall.

Answer (D)

We have done most of the work while reading the question only. Had we tried to solve it algebraically, we would have made 3 equations using 3 variables and then tried to solve them.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

 

7 Responses

  1. shankar says:

    Hi kairshma,

    The last statement says that Since 3g take 4 hrs to build the wall, 7g will take 3*4/7 = 12/7 hrs to complete the wall.

    3g takes 4 hours
    So 1g takes 4/3 hrs right?
    now 7g would take 28/3 hours right?

    I’m not sure how this calcualtion is feasible
    3*4/7 = 12/7 hrs to complete the wall.
    Please help.

    • Karishma says:

      Tell me, if 3 girls take 4 hrs to complete the work, will 1 girl take 4/3 hrs or 12 hrs to complete the work on her own?

      • shankar says:

        true 12 hrs it is. Completely agree. I was just blindly substituting without thinking.

        Karishma,

        Can you please help me to solve this question in an intuitive way please?
        Machines A, B, and C can either load nails into a bin or unload nails from that bin. Each machine works at a constant rate that is the same for loading and for unloading, although the individual machines may have different rates. Working together to load at their respective constant rates, machines A and B can load the bin in 6 minutes. Likewise, working together to load at their respective constant rates, machines B and C can load the bin in 9 minutes. How long will it take machine A to load the bin if machine C is simultaneously unloading the bin?

        • Karishma says:

          Note here that Speed of loading = Speed of unloading = Speed of machine

          Say, machines A and B are loading.
          Work done by A and B in 1 min = 1/6

          Say, machines B and C are unloading simultaneously
          Work done by B and C in 1 min = 1/9 (because Speed of loading = Speed of unloading)

          What is going on? Machine A is loading, machine C is unloading and B is doing nothing (since it is loading as well as unloading)
          How much work is getting done in a min? 1/6 – 1/9 = 1/18

          Time taken to load the bin = 18 mins

  2. Santhosh says:

    Question 1: Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

    The extra speed that machine A has allows it to do 1/6th work extra. This means, at normal speed, machine A used to do 1/6 work in an hour.

    Usual Speed = 1/6
    Revised Speed = 1/3 (Since the rate of A is doubled)

    The question asks to calculate based on current speed so I believe the answer is 3 (Option C) and not 6(option E). Please correct me if I’m wrong.

    • Karishma says:

      Look at the question again:
      ” If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours ”

      This part describes a hypothetical situation – what would happen if the speed were doubled. This doesn’t actually happen. So the current speed is 1/6 only.

  3. Santhosh says:

    Hi Karishma,

    It makes sense.Thanks a lot!!!

    Your posts are really amazing.It has given me a lot of confidence.Again, Thanks you very much for all your posts.

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