Solving PSLE Math proves a challenge for pupils, but it isn’t as difficult as they seem to find it. Many pupils seem to get fed up with trying to learn and understand the math problem sums because the methods and steps for solving them are simply too many, but there is a way to group these problem sums into their proper solving methods. And they won’t seem so bulky or tedious for the pupils.
In this article, there are identified some methods that pupils will do well to be able to solve math problem sums in the PSLE. Tutoring your children using these methods will help them improve in their PSLE math results, and solving problem sums will not be a problem for them again.
In solving PSLE Math problem sums, there are basically four steps to follow:
1. Read and understand the question/problem
2. Identify the right method for solving the question
3. Use the identified method to solve the problem
4. Make sure your answer is correct
These are the only steps necessary in solving Math problem sums. Following these steps with the proper method will always make you arrive at the right answers.
Methods for Solving PSLE Math Problem Sums
• Model Drawing
Model drawing is the method that is the most commonly used in solving Math problem sums. Model drawing is most useful when solving problem sums that contain fractions, especially those where you have to divide the model into different pieces and hypothetically hand them to another person.
The main problem pupils face in solving problem sums using the model drawing method is that they do not understand how to cut up the model into different parts. You should note that, to solve problems using model drawing, ensure that all the models you have drawn are cut up into the same number of pieces so as to maintain consistency.
• Grouping
The grouping method is used when items need to be grouped into sets. Solving problem sums this way involves dividing the total cost of the sets by the cost of each individual set to get to the total number of sets.
For example:
Jessica bought 2 more oranges than apples and paid a total of $124. Each apple costs $8 and each orange costs $12. So, (a) How many apples did she buy? (b) How many oranges did she buy?
Solution
2 × 12 = 24
124 – 24 = 100
8 + 12 = 20 (1 apple and 1 orange)
Answer for (a): 100 ÷ 20 = 5
Answer for (b): 5 + 2 = 7
• Excess And Shortage
The excess and shortage method is used to solve problem sums that give you an excess of items when you group them into smaller groups; and a shortage of items when you group them into larger groups.
For example:
Paula packed 5 pencils into each cup and discovered that she has 7 pencils left over. If she packed 6 balls into each cup, she would need 5 more pencils. So, (a) How many cups did she have? (b) How many pencils did she have altogether?
Solution
6 – 5 = 1
7 ÷ 1 = 7
Answer for (a): 7 + 5 = 12 cups
Answer for (b): 12 × 5 + 7 = 67 pencils
• Units And Parts
Units and parts is an algebraic method of solving problem sums in which you have been giving a starting ratio. When solving problems using units and parts, both sides of the ratio change by differing amounts, after which you are given a final ratio. A common mistake that pupils make is that they see the starting ratio and the final ratio as the same units. The right thing to do is to treat the starting ratio as units, and the final ratios as parts.
For example:
Martha had some mangoes and bananas in a ratio of 9:3. She bought 30 more mangoes and 6 bananas. As a result, the ratio of mangoes to bananas become 4:1. How many mangoes did she have at first?
Solution
A: 9u + 30 —-> 4 parts
O: 3u + 6 —-> 1 part
12u + 24 —-> 4 parts
9u + 30 = 12u + 24
12u – 9u = 30 – 24
3u = 6
1u = 2
9u = 18 mangoes
• Constant Part, Constant Total, Constant Difference
Also used in solving Math problem sums that involve ratios, this method is used in three parts:
Constant Part: one side of the ratio changes, while the other side of the ratio remains the same.
Constant Total: while the total amount remains constant or the same, but there is an internal transfer from one side to the other side.
Constant Difference: both sides of the ratio change by the same amounts. For example:
In a bus, 30% were girls. After 9 boys and 9 girls joined the bus, there were 3/5 as many girls as boys. How many pupils were on the bus in the end?
Solution
Start
B : G : Difference
= 3 : 7 : 4
End
B : G : Difference
= 3 : 5 : 2
= 6 : 10 : 4
3 units = 9 pupils
Answer: 16 units = 48 pupils
• The X Value
This method is used when you have to group items into different sets which each have a predetermined set. The number in the ratio is to be multiplied by the value that it represents: this will provide the total value in the set. Divide the total value by the answer obtained to get the total number of sets. For example:
The ratio of the number of 50 cents to 1 dollar is 3:1. The total value of the coins is $12.50. What is the total number of coins?
Solution
3u × $0.5 = $1.5u
1u × $1 = $1u
$1.5u + $1u = $2.5u
$2.5u = $12.50
1u = 5
Answer: 4u = 20 coins
• Work Backwards
This method is used to solve problem sums when you are given the final answer, and you are expected to work backwards to get the original numbers. For example:
A train left an interchange, taking some passengers in it. At the train stop, ¼ of the passengers got off and 5 people got on. At the 2nd stop, ½ of the people got off and 20 people got on the train. When the train left the 2nd stop, there were 60 people on it. How many people were on the train when it left the interchange?
Solution
60 – 20 = 40
40 × 2 = 80
80 – 5 = 75
Answer: 75 ÷ 3 × 4 = 100 people
Your child only requires to continuously practice on these methods to achieve stress-free attempts on his Math papers. An alternative is to engage a private tutor to have 1 on 1 time with your child and to drill him on these methods using different sets of problem sums each time. In time to come with tremendous effort, your child’s Maths scores will improve.
Problem sums is a minefield for careless mistakes, find out how your child can avoid them: Careless Mistakes In Math: How They Kill Your PSLE Score
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