More than one way to solve Math problem sums

Anonymous posted on 14 Apr 2008 PM 10:12

isn't it faster to do this:
Ratio 20 cents: 50 cents : $1
= 2 : 4 : 7
2 x 20 cents + 4 x 50 cents + 7X 100 cents = 940 cents
4700 / 940 = 5 groups
5 x (2 + 4 + 7) = 65 coins

The question and solution that I have worked out, that is referred to, can be found in this link.
http://road-to-psle.blogspot.com/2007/12/anglo-chinese-school-primary-sa2-2006.html

Question - The ratio of the number of 20-cent coins to the number of 50-cent coins to the number of $1 coins in a bag is 2 : 4 : 7. Given that the total amount of money in the bag is $47, how many coins are there altogether?

My reply -
Dear Anonymous,

There is more than one way to solve Math problem sums. My bottom line, as a teacher when I was teaching in schools, and as a tutor now, is that the method used/taught must contain these important elements –

Presentating the solution in a way such that the student is able to understand its mechanics–
This is important. While above average students are able to analyse deeper, average and below average students usually are not able to. As such, the more complex the problem sum, the more there is a need to break down the solution with a step-by-step method.

Transfer of the skill/knowledge used to solve the sum to students –
This is the acid test of your teaching method. If a teacher/tutor is unable to transfer his skill and/or knowledge to the student, such that the student is unable to solve the problem sum, should he come across it again in the exam, all effort is wasted.

The complexity of the math question above, requires the ability on the part of the student to differentiate 2 distinct differences. They are:

1. The number of coins.
2. The value of coins.

Your “short” method assumes the student is able to differentiate the above 2 mentally. This can be seen when you wrote:

2 x 20 cents + 4 x 50 cents + 7X 100 cents = 940 cents
4700 / 940 = 5 groups
5 x (2 + 4 + 7) = 65 coins

However, not all students are able to see that there are 5 groups of 940 cents in 4700 cents, let alone the need to multiply the number of coins by its value before you can come to that.

Your last line also requires the student to understand that within one group of the 3 coins of different values in the (2 + 4 + 7), there are 940 cents – which essentially means 5 groups of the same 3 coins of different values (2 + 4 + 7), will contain 65 coins, because the 5 groups are valued at 4700 cents.

Your method is good for above average students.

However, not all students are able to see what you see.

The biggest challenge some students face with a problem sum like the above is, the failure to realize that the number of coins and the value of coins are 2 different things - or – if they do realize they are 2 different things, they do not know how to proceed to solve the sum.

Hence, I would recommend the longer method as I have worked out for the average and below average student. The longer method I have worked out uses the step-by-step approach I mentioned earlier. The steps are –

Step 1 - Converting the number of coins of different denominations into a single common base, which is to consider the values of all the 3 different coins. This is the key to solve the problem because we are given the total value of the coins. This can be seen as shown below.



Hence, from a ratio of 2 : 4 : 7 of different values, we convert them into a ratio of 2 : 10 : 35, which has a common base (which is the value). Since this ratio concerns value of the coins of each denomination, it can then be compared to the total value of the coins, which is $47. A common base between the ratio and total value has been established.

Step 2 – The next step easy. By comparing the converted ratio (which is based on value) to the total value of the coins, we can establish that critical “1 unit” as shown below.

47 units ----- $47 (Total amount of money in bag is $47)
1 unit ----- $47 divided by 47 = $1

Step 3 – The rest of the solution is even simpler to understand and it is self-explanatory.
Value of 20-cent coins ----- 2 units ----- $2
Value of 50-cent coins ----- 10 units ----- $10
Value of $1 coins ----- 35 units ----- $35

No. of 20-cent coins ----- $2 divided by 20 cents ----- 10 coins
No. of 50-cent coins ----- $10 divided by 50 cents ----- 20 coins
No. of $1 coins ----- $35 divided by $1 ----- 35 coins

Total number of coins in the bag, 10 + 20 + 35 = 65
Answer: There were 65 coins in the bag.

Conclusion -
There is nothing wrong with your “short method”. My only concern is that weaker students may be unable to see as deeply as you are able to. I have noticed that many tuition centres teach “fast methods”.

However, it must be remembered that these “fast methods” are effective if the students are bright. The 2 essential points I mentioned earlier, which are, presenting the solution in a manner such the student understands its mechanics, and transferring that skill to solve the sum to the student such that he will be able to solve the same problem sum should he come across it in exams, may be compromised, if the student is weak in maths.

Average students and below average students need a step-by-step approach.

The purpose of this blog is to help average and below average students. As such, you may notice that most of the solutions follow a step-by-step approach.