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    Hi guys,
    I'm not great with maths and I'm really stuck on a few of these "logic" problems. Could you please help? If you could figure out how to do any one of the three questions below then I would love to hear how you did it.
    Xxx

    1. A, B and C can do a piece of work in 20, 30, and 60 days respectively. In how many days can A do work if he is assisted by B and C on every third day?
    a)12 days
    b)15 days
    c)16 days
    d)18 days
    e)20 days

    2. The average weight of 8 people increases by 2.5kg when a new person changes places with one of the original 8, who weighs 65kg. What is the weight of the new person?
    a)76 kg
    b)85 kg
    c)76.5
    d) Data inadequate
    e) 80kg

    3. From a group of 7 men and 6 women, five people are to be selected to form a committee. The committee must have at least 3 men. In how many different ways can this be done?
    a)564
    b)645
    c)735
    d) 535
    e) 756

    I would really appreciate your help on this
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    These types of problems are quite nice. Just looking at them two seems to be the easiest.

    You know that the average weight is the total weight/8. Lets say this is denoted by Wt (total weight)/8.

    So Wt/8 increases by 2.5kg. This means that the total weight (Wt) increases by 8*2.5=20.

    You know that the previous person weighed 65kg, so the new person must weight (65+20)kg, which is 85kg. Do you have the answers to confirm this is correct?

    If you dont like the algebraic method, an easy way to do this is to just come up with some numbers. Lets take eight people of 45, 50, 60, 65, 70, 80, 90 and 100. The average weight is (sum of weights)/8= (560/8)=70kg. Now replace the 65kg person with an 85kg person. The average weight is now (580/8)=72.5kg. This is 2.5kg heavier, as per the question.

    Hope that helps :P Ill try and work on the others
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    1. In 60 days, A can do 3 pieces of work, B can do 2 and C 1. So B and C working together can do the same as A. In 2 days, A can do 1/10 pieces of work. On the third day, B and C join A which doubles the productivity. So A, B and C do 1/10 pieces of work on the third day. Meaning, in the first three days, they do 1/5 pieces of work (2 x 1/10). If they do a fifth on three days, they will do it all in 15 days.

    2. Let A be the average weight of all the people. Then the total weight of all is equal to 8A. If the average is increased by 2.5kg, the total becomes 8(A+2.5) = 8A+20. So the new person must be 20kg heavier than the old person or 65+20=85kg.


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    For number 1 I got an answer of 15 days.

    A does work at a rate of 1/20 per day, with B at 1/30 and C at 1/60.

    So every three days, the amount of completion of work is as follows:

    Day 1+day 2+ day 3

    = (1/20) + (1/20) + (1/20 + 1/30 + 1/60)

    = 1/5

    So every three days, they complete 1/5 of the work. Hence, the full work is completed in five lots of three days, which is fifteen days.
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    Thanks everyone! You're the best.

    A friend of mine suggested that number 3 has something to do with permutations. I'll google now and try to post the solution if i figure it out.
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    (Original post by Pirateprincess)
    Thanks everyone! You're the best.

    A friend of mine suggested that number 3 has something to do with permutations. I'll google now and try to post the solution if i figure it out.
    Number 3.
    3 men and 2 women can be selected in ^7C_3 \times ^6C_2 ways
    Repeat for 4 men and 1 woman and all 5 men and add.
 
 
 
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