Dating made easier
If a sum invested gains 10% each year how long before it has doubled its value?
Problem
Dating Made Easier printable sheet
If a sum invested gains $10\%$ each year how long will it be before it has doubled its value?
If an object depreciates in value by $10\%$ each year how long will it take until only half of the original value remains?
Why aren't these two answers the same?
Is there a rate, used for both gain and depreciation, for which those two answers would actually be the same? How do you know?
If this problem caught your interest and you know some Stage 5 mathematics, this Plus article on Carbon Dating could be a good next step for you.
Getting Started
What would you need to multiply by to get the same answer as calculating $10\%$ and then adding it on?
Try some different values for the sum invested. It might help to use a spreadsheet.
Student Solutions
We have yet to receive a solution that uses only Level 4 mathematics, (so a solution that doesn't use logs) and so this problem has become Toughnut! See if you can crack it and send us your solution!
We have now received some excellent solutions for this Toughnut question. Patrick from Woodbridge School solved the first question by making a list of each years values:
Let x be the original value.
To add 10%, we multiply by 1.1, and we are trying to reach 2x. Thus, we get
x
1.1x
1.21x
1.331x
1.4641x
1.61051x
1.771561x
1.9487171x
2.14358881x
So, after 8 years, the sum has doubled.
To reduce a value by 10%, we multiply by 0.9, and we must reach 0.5x. Thus, we get
x
0.9x
0.81x 11
0.729x
0.6561x
0.59049x
0.531441x
0.4782969x
so after 7 years the sum has halved in value.
Alexandra, Shannon, Gemma, Katie, Ruby and Caitlin from Herts and Essex High School for Girls also worked out the problem in this way correctly. Well done!
Iain from St. Gregory's Catholic High and Daniel from King's College, Madrid used logarithms to solve the problem:
If a sum increases by 10% each year this means that the value of the previous year is multiplied by 1.1 (it is 110% of what it was before). Therefore for an initial value "a" the progression will be:
Year: 0 1 2 3 ...
$a$ $a\times1.1$ $(a\times1.1)\times1.1$ $((a\times1.1)\times1.1)\times1.1)$
= $a$ $a\times(1.1)$ $a\times(1.1^2)$ $a\times(1.1^3)$ ...
A sum invested gains 10% each year:
$1.1^n = 2$ (1)
$n log 1.1 = log2$
$n = \frac{log2}{log1.1} = 7.273$ (4.s.f.)
It will therefore be 8 years before the sum is effectively doubled.
An object depreciates in value by 10% each year:
$0.9^m = 0.5$ (2)
$mlog0.9 = log0.5$
$m = \frac{log0.5}{log0.9} = 6.579$ (4.s.f.)
It will therefore be 7 years before the sum is effectively halved.
Patrick from Woodbridge School went on to explain why it takes different time for the value to double and half:
Why aren't these two answers the same?
The answers are not the same as, when 10% is added to a value, less than 10% must be taken off as the 10% of the new number is larger. This can be summed up as:
If $10$%$\times x + x = y$
then $y - 10$%$\times y < x$
Is there a rate, used for both gain and depreciation, for which those two answers would actually be the same?
For the above reason, there is no such rate as this is true for all numbers and all percentages, except 0% and £0.
For a non-zero initial value, we cannot find a non-zero rate for which the time taken for the value to double and half would be the same. This is because for any rate, the ratio of values between two successive years will be different for gain and depreciation. To get the same time we really need these ratios to be the same.
Well done everyone!
Teachers' Resources
Why do this problem?
This problem offers students to explore ideas involving percentage change, in particular using the link between multiplication and percentage change. In particular, students explore equivalence of change (ie. if a $10\%$ decrease isn't the opposite of a $10\%$ increase, what is?). The problem is designed to move students towards the Stage 5 concepts of geometric sequences and exponential growth and decay.
Possible approach
You could warm up by finding some repeated 10% increases or decreases, for example "The value of a car decreases by 10% each year. If the original value was £5000, how much is it worth after 3 years?"
"The annual interest on a savings account is 10%. By what percentage does the balance increase over four years?"
Discuss the methods that students have been using. The "multiplier method" (representing a 10% increase as 110% of the original quantity, so multiplication by $\frac{110}{100} = \frac{11}{10}$ or by 1.1) will hopefully emerge.
Ask the students, If a sum invested gains 10% each year how long will it be before it has doubled its value? They could use trials, or a table, to come to the answer which should be 8.
Then ask the second question, If an object depreciates in value by 10% each year how long will it take until only half of the original value remains? The answer to this should be 7.
This can now be a discussion point. Are they surprised the answers are different? Why don’t they get the same answer? If you changed the 10% to a different number, would the answers be the same? Could they ever be the same?
Allow students to explore, using different numbers if they want to. You could encourage students to use spreadsheet and graphical software. You could encourage students to represent the problem algebraically. Graphically, they should see that the curves are not the same. Algebraically, the sticking point is that the expressions are not one another’s reciprocal, which ultimately provides a proof (that is, $\frac12$ and $2$ are reciprocals but $1.1^n$ and $0.9^n$ are not).
Invite students to share their work and encourage them to make connections between the different representations that have been used. Can they explain why the two answers are different? Has anybody found a number such that the increase and decrease do give the same answer? Or, better, has anybody proved that that would be impossible?
Key questions
Is a 10% increase the opposite of a 10% decrease? What happens if you do one and then the other?
Is mutliplying by 1.1 the opposite of mltiplying by 0.9? Why?
Possible support
If students haven’t seen it before, you might want to give them a chance to practice finding some percentages using the multiplier method.
Students may find it difficult to calculate percentage changes when a starting value is not given. If this is the case, tell them to choose a starting value and use that one. Then choose another starting value and check they get the same answer. Then they could use a letter of their choice (which could be any number) as a starting value and check they still get the same answer. For the rest of the problem, you could let students use a letter as their starting value, or a number of their choice (you might recommend using 1).
Possible extension
The problem The Legacy explores the idea of exponentials in the context of long term investments.