3.5
The previous chapter dealt with problems for which the method of solution was relatively easy to find. In this chapter we are looking at problems where the primary skill in solving them is to develop a method of solution. The way of proceeding to an answer in some problems may not be clear:
a either because it is necessary to find an intermediate solution first,
b or because we need to work
simultaneously forward from the data (to identify what can be calculated) and backwards from the required answer (to identify what needs to be calculated).
Having a strategy for approaching such problems is important. In particular, it can be very useful if you have seen a problem of a similar sort before, which you know how to approach – this is where experience in tackling problem-solving questions can be invaluable.
Sometimes it may be necessary to try different ways of approaching the problem; it is
important to realise quickly if your line of attack is being unsuccessful.
One strategy that can help to solve problems when you are not clear how to proceed is to analyse the problem:
• organise the information you are given
• write down or underline those pieces of information which you feel are important
• simplify (reject unimportant information)
• look at the question and decide what pieces of information could lead to the answer
• make a sketch, list or table.
Sometimes, intermediate answers are
necessary in order to proceed to the complete solution. This may be regarded as similar to the identifying of intermediate conclusions in Chapter 2.6. The solution of a problem can be like an argument that first leads to one conclusion, which then, possibly using further information, proceeds to the final conclusion.
This may be illustrated as in the diagram below. Here, the calculation steps are represented by the arrows. Not all these processes are used in all problem solutions.
A problem that may be solved using an intermediate result is given in the example on the following page. This is similar to the question in the previous chapter in that it involves distances, speeds and times but, because of the nature of the question, the method of proceeding is less obvious.
Answer required Information required
Data
Intermediate
result Solution
Commentary
This is another problem where an
intermediate calculation is necessary. In order to calculate Petra’s bill, we need to know the monthly charge and the rate per unit. We know the difference between the two quarters’
bills - this difference is only due to the reduced consumption, so 1400 fewer units saved $112. This means units cost 8¢ each.
The three-monthly fixed charge, therefore, is:
$250 − 2000 × 8¢ = $250 − $160
= $90 (or $30 per month)
If Petra reduced her January to March
consumption by 25%, this would then be 1500 units, so her bill would be:
$90 quarterly charge plus 1500 × 8¢
= $90 + $120 = $210
In fact, the quarterly charge does not have to be calculated, only the unit rate. The entire process of solving this problem could be speeded up by simply recognising that the relevant three-month bill would be reduced by 500 × 8¢ or $40.
Commentary
From the data given, it is easy to find out when they both arrive at their destinations, but finding when they cross is not so straightforward. The problem can be made much simpler by using an intermediate step.
First calculate where Amy is when David leaves. She has been travelling for 2 hours, so she has covered 240 km – that is, she is 160 km from David’s house. The problem is now quite easy. At 10 a.m. they are 160 km apart and rushing towards each other at a joint speed of 240 km/h. Therefore, they will meet 40 minutes later (160 km/240 km per hour is 23 hour or 40 minutes). The time they pass each other is 10.40 a.m.
If we had been asked to find the place where they pass, the passing time could have been used as a second intermediate value. David travels 120 km in an hour. 40 minutes represents 23 of this, or 80 km, so they cross 80 km from David’s house.
In this case the numbers were very easy but the same method of solution could have been used whatever the distances, times and speeds.
The method of solution, which was not immediately obvious, became easier by using the intermediate step.
Amy and her brother David live 400 km apart.
They are going to have a week’s holiday by exchanging houses. On the day they are starting their holiday, Amy leaves home at 8 a.m. and David at 10 a.m. They both drive at 120 km/h on a motorway that travels directly between their homes.
At what time do they pass each other on the road?
Petra’s electricity supply company charges her a fixed monthly sum plus a rate per unit for electricity used. In the most expensive quarter last year (January to March), she used 2000 units and her bill was $250.
In the least expensive quarter (July to September), she used 600 units and her bill was $138.
She is now adding extra insulation to her home which is expected to reduce her overall electricity consumption by 25%. What can she expect her January to March bill to be next year (if there are no increases in overall tariffs)?
Activity
Activity
In this diagram, the outer box represents all the students (the universal set); in this case they all take physics. The left-hand circle represents those taking chemistry (and physics) and the right-hand circle represents those taking biology (and physics).
We know that the number taking only physics is 12; this is represented by the area outside both circles. Those taking all three sciences are represented by the intersection of the two circles and shown as 9. The number taking both physics and biology is 24; of these 9 take all three, so 15 take only physics and biology. This is shown by the outer section of the right-hand circle. We can now calculate the number in the area marked by the question mark, as this must be all the students in the class minus the numbers in the other three areas, i.e. 45 − 12 − 9 − 15 = 9.
This is the required answer: the number studying chemistry and physics but not biology is 9.
Interestingly, the number studying all three was not used in the original calculation. It is, in fact, not needed to solve the problem. We used it in the Venn diagram solution so we could calculate the numbers in all the areas on the diagram.
Graphs, pictures and diagrams can often be useful in solving problems as they help to clarify the situation and represent the numbers used in a more digestible manner.
This is covered in more depth in Chapter 6.2.
The activity above shows that problems can often be solved in more than one way. It is important to keep the mind open to alternatives and not always to pursue a method which is not apparently leading to a solution.
Another way of approaching problems is to lay out the information in a different way. This is especially so when the information is given verbally – and therefore the connection between the different pieces may not be immediately obvious. Consider, for example, the following problem:
In a group of 45 students at a school, all students must study at least one science.
Physics is compulsory, but students may also opt to study chemistry or biology or both. 9 students take all three sciences. 24 take both physics and biology (with or without chemistry). 12 students take only physics.
How many are studying chemistry and physics but not biology?
Activity
Commentary
With a bit of clear thinking, this may be solved in a direct fashion by making an intermediate calculation (those not studying biology). Since all students take physics, the situation is simplified. 24 study biology and there are 45 in total, so 21 do not study biology. These 21 comprise those studying physics alone and those studying both physics and chemistry.
However, we know that 12 take only physics, so 9 must take physics and chemistry but not biology.
Although this appeared to be an easy calculation, the method of approach was not obvious. The situation can be made a lot easier by using a Venn diagram:
Physics
12 Chemistry Biology
9 15
?
questions – that is, checking that we have the right answer.
If ER takes 24 minutes (including a 3-minute stop) and MR takes 18 minutes (no stop), then EM (no stop) must take 3 minutes (24 − 18 − 3). Similarly, RW must take 36 − 18
− 3 = 15 minutes. We now have the times for all the sections:
EM = 3 minutes Stop at M = 3 minutes MR = 18 minutes Stop at R = 3 minutes RW = 15 minutes
The total of all these is 3 + 3 + 18 + 3 + 15
= 42 minutes as expected.
Check that the times from E to R and M to W agree with the answer of 18 minutes from M to R as given.
This example once again shows that representing the data in a different way can lead to a simple method of solving a problem that at first appears unclear.
Summary
• We have learned the importance of finding methods of solution for problems for which the way of proceeding to an answer is not necessarily obvious.
• We have found that looking for
intermediate results may help to lead to the final answer.
• We also looked at the value of alternative ways of presenting data and considering more than one way of solving problems.
Commentary
A diagram makes this problem much easier to solve.
42 min
24 min
36 min
E M R W
It can now be seen that if we add the time from E to R to the time from M to W, we get the time from E to W plus the time from M to R. The times from E to R and M to W each include one 3-minute stop, whilst the time from E to W includes two 3-minute stops, so when we subtract the EW time from the sum of the ER and MW times, the stops cancel out. Thus the time from M to R is 24 + 36 − 42 = 18 minutes.
We can now look further at an extra useful element in the solution of problem-solving
There is one railway on the island of Mornia, which runs from Enderby to Widmouth. There are two intermediate stops at Maintown and Riverford. The trains run continuously from one end to the other at a constant speed, stopping for three minutes at each station.
From departing Enderby to arriving at
Widmouth takes 42 minutes. From Enderby to Riverford takes 24 minutes. From Maintown to Widmouth takes 36 minutes.
How long does it take from Maintown to Riverford?
Activity
3 From my holiday cottage by the sea I can see two lighthouses. The southern flashes regularly every 11 seconds. The northern lighthouse, after its first flash, flashes again after 3 and 7 seconds. The whole cycle repeats every 17 seconds.
They have just flashed at exactly the same time, the northern one having just started the cycle described above.
When will they both flash again at exactly the same time?
4 The 23 members of a reception class in a school have done a survey of which cuddly toys they own. Pandas and dogs are the most popular, but 5 children have neither a panda nor a dog. 12 have a panda and 13 have a dog. How many have both a dog and a panda?
Answers and comments are on pages 317–18.
1 Aruna’s neck chain has broken into two parts. She has lost the broken link and is having it repaired by a jeweller who will open one of the remaining links and use it to rejoin the chain. The chain is made from metal 2 mm thick and each of the broken pieces has a fitting at the end used for closing the chain which each adds 1 cm to the total length.
One of the broken pieces is 34.2 cm long and has 10 more links than the other, which is 26.2 cm long. Excluding the fittings at the ends, how many links will there be in the complete chain?
2 The distance from Los Angeles to Mumbai is 14,000 km. Flights take 22 hours, whilst the return flight from Mumbai to Los Angeles takes only 17 hours because of the direction of the prevailing wind.
Assuming the aeroplane would fly the same speed in both directions in still air, what is the average wind velocity?