Some Quick Questions to ask yourself when doing this type of problem solving:
- What is being asked? Underline the specifics in the question.
- Analyse the data, is there anything we can eliminate immediately?
- What information do I need to solve the question? Where can I find it?
- How can I draw or use the graph to help me?
Here, the best approach is to try and disprove the statements.
- Let’s take 3 apples at a cost of €0.30 each = €0.90. We have €1.10 left over from his total €2 spent.
Banana + banana = €0.80, which is not enough. If we add another banana = €1.20 which is too much.
Orange + Orange = €1.00, which is not enough, but adding another orange would be too much.
Orange + Banana = €0.90, which is not enough, if we add another banana or orange it would be too much.
Therefore the original statement for 1. is true.
- This is not true and we can easily disprove this by taking:
Orange + orange + orange + orange = €2.00
Therefore this is false.
- This can be tested by taking the cheapest fruit and the expensive fruit. Then we can have a good range. At the bottom of the range: Apple + Apple + Apple + Apple + Apple = €1.50. Add one final orange to get to the target of €2.00 and we have 6 fruits.
Orange + orange + orange + orange = €2.00. This has four fruits total.
This is probably enough to be conclusive but just to double check we can find a five fruit combination. Apple + Apple + Banana + Orange + Orange = $€2.00.
Therefore this statement is correct.
- Start by taking one of each type: €0.30 + €0.40 + €0.50 = €1.20
We can then add an apple since we are trying to see if we can have more apples than bananas and oranges combined.
€1.20 + €0.30 = €1.50. We have only €0.50 left to spend, therefore we can only get an orange. Any apple or banana would not leave us with a total of €2.00 left. We will have 2 apples, 1 banana, and 2 oranges, which means less apples than oranges and bananas.
Therefore this is true.
So We know that 1, 3, and 4 are correct, which is C.