Work problems
Work problems usually involve two different characters who do the same job in different amounts of time. The task is to find out how long they will take to do the job if they work together.
A sample problem
Imagine you are playing a game in which you fight a dragon. A Paladin hacks the dragon to pieces with his sword in 30 seconds, while a Barbarian batters it to death with his club in as little as 20 seconds. The question is how long it would take them to kill the dragon if the Barbarian and the Paladin fought it together. How much “joint work” do they have to put in?
We solve the problem by first finding out how much damage each fearless warrior does to the dragon in one second. Because the Paladin kills the dragon in 30 seconds, in one second he takes off 1/30 of its hit points, one thirtieth. If the dragon had 180 hit points, then for the Paladin to kill it in thirty seconds he must take off 180 / 30 = 6 hit points every second (that is one thirtieth of its hit points). And 180 / 30 is the same as 180 · (1/30).
The Barbarian kills it in 20 seconds, so every second he takes off 1/20 of its hit points. For the same dragon with 180 hit points, he would take off 180 / 20 = 9 hit points every second.
So the Paladin takes off 1/30 of the dragon’s hit points per second and the Barbarian 1/20. Now we ask how many seconds these gentlemen need to kill it by joining forces. We can look at it step by step: how many hit points would they take off in two seconds?
It would be one sixth of its hit points. We add 1/30 + 1/20 because the warriors fight together, and we add it twice in total because they fight for two seconds. We can see that this generalises: if they fight for x seconds, they take off
of its hit points. Now there is just one thing left: the dragon dies when all its hit points have been taken off. All its hit points are represented by the number 1, because the fraction 1/30 stands for one thirtieth of the dragon’s hit points, so the number 1 is the dragon’s total. So we just set the previous expression equal to one:
and look for the solution. We multiply the whole equation by 60, which gives:
We get x = 12. This means they kill the dragon in 12 seconds. Not bad at all.
We can try to check this by substituting actual numbers. We had a dragon with 180 hit points and the result says that together they kill it in 12 seconds. The Paladin takes off 6 hit points per second from this dragon and the Barbarian 9. So in 12 seconds the Paladin takes off 12 · 6 = 72 hit points in total and the Barbarian 12 · 9 = 108 hit points. Together they took off 72 + 108 = 180 hit points, which agrees with the fact that they took off all of them.
Work through some more examples of work problems:
Examples
- Tommy would love to buy a new mobile phone. He agreed with his mum that she would give him pocket money every week and that after 60 weeks he would have exactly enough money for the phone he wants. But Tommy is no fool, so he also went to his dad. Dad also promised him money and said that he would have enough for the new phone after only 30 weeks. How long will it really take Tommy to get the new phone if he gets weekly pocket money from both his mum and his dad?
From his mum, Tommy gets 1/60 of the price of the phone each week, and from his dad 1/30 of the price of the phone. We set up the equation in exactly the same way as in the previous case:
The expression x · 1/60 + x · 1/30 tells us how much of the price of the phone Tommy will have saved after x weeks. Because we are asking when he will have exactly enough for that phone, we set this expression equal to one. If he wanted to buy three phones, there would be a three on the right-hand side.
We solve the equation. First we multiply it by 60:
The result is that Tommy will have the new phone in 20 weeks. Again we can check this by substituting real values: if the phone cost £300, he would get 1/60 of this price from his mum each week, that is £5. From his dad he would get 1/30 of this price, which is £10. Well, he has quite generous parents. In 20 weeks he would receive 20 · 5 = 100 pounds from his mum, while from his dad he would get 20 · 10 = 200 pounds. In total he would get 100 + 200 = 300 pounds.
- Three pipes lead into a swimming pool. One fills the pool in 100 minutes, another in 75 minutes and the last in 50 minutes. How long will it take to fill the pool if all three pipes are used?
This example is again the same as the previous ones, except that instead of two workers we have three. So the equation will look like this:
We multiply the equation by 150 and carry on simplifying:
We cannot get a nicer result than this: the pool fills in 300/13 minutes, which is about 23 minutes.
- The firm Mow and Go can mow a meadow in ten hours. The firm Dynamite Daisies can mow the same meadow in six hours. How many hours would these firms take to mow two such meadows if they worked together?
How much do the two firms mow in x hours? That is the usual expression:
But because we want them to mow two meadows, the right-hand side of the equation will not be 1 but 2:
From here the calculation goes on in the same way. We multiply the whole equation by 30 and simplify:
Together they would mow both meadows in seven and a half hours.
- One woman has a baby in nine months. How many months would it take nine women to have a baby?
