Thanks all for the answers.

Ps for my typo...

Thks all.

Teachers, I still don't understand...why isn't 12 apples at first? What have I missed out? Please enlighten. Stacey Tan, you illustrated total no of fruit is 56. Not total 56 Apples at the end? A + P=> 40 total 56 / 2 => 28 pears 40-28 = 12 apples (at first)

Yes ... All the 56 fruits are apples (since we have exchange all extra pears to apples) We cannot divide 56 / 2 because not all the 56 were pears exchanged for apples. There were already some apples there initially before the exchange.

Total number of fruits is 40. According to your answer, you have 12 apples, which means you have (40-12=28 pears). If we change all the pears to apples (28x2=56), then you will have (56+12=68 apples) in the end. But the question says that you only have 56 after the exchange.

Ahh...I missed the first step, I'll be misled! Thank you, Stacey Tan.

Can't be so simple, right... :D It's good to work backward to counter check answer. Got it!

If he gives each friend 6 marbles, he will have 4 marbles left. If he gives each friend 7 marbles, he will be short of 5 marbles. The umber friends increased by 1, yet the demand for marbles increased by 4(left over) + 5(short) =9.

So new number of friends = 9/1=9 Marcus will have 9x7-5=58 marbles.

It's a typical multiples qn which they learnt in p4. They are suppose to present ans in a table form for each multiple. Otherwise qn can be solve with assumption method mentioned above.

There is a name to this type of question, commonly known as "Shortage-excess" question.

Thks all

Edlynn Rose, for the first method, how do we know when to stop? Why not stop at 16?

well actually, we wouldn't know when to stop unless we work on both concurrently and stop as soon as we see a match.

16 is a match too, right?

as in there are 16 marbles?

Yes. There's a match there. I saw the same method in the assessment book. Similar question asking for number of sweets. They took the first match as the answer.

Number of friends must be the same in both multiples. For 16 marbles, you have 2 friends for multiple of 6 and 3 friends for multiple of 7....

Like your sweets question, you stop at 6 friends for both multiples of 7 and 8.

Stacey, how about in this case?

I'm sorry. A little confused here.

Judy Low, in your question on paper clips, number of bags is not a requirement, which means there is no need to put in the equal number of bags.

So she can put in 13 bags of 6 clips and have 3 left or 10 bags of 8 clips and have 1 left. this means that she must have 81 clips in all.

In your sweet question, the number of friends is the same whether you give 7 sweets or 8 sweets. It will not be logical if the number of friends changes with the number of sweets you give out.

Ok. So gotta if it's abt giving to friends, must be the same in both multiples since the number of friends is a fixed number?

Yes

Ok. Thanks, Stacey Tan!

The total no of marbles do not change. So we only need to find how many marbles there must in A at the end, and then subtract 120 from it.

total no of marbles ＝120+120+260＝500 500+238＝438 438÷2＝219 219-120＝99.

The coreect ans will be 189

will calculate again when i reach hm 😊

Hi Stacey, I like your working steps is a very clear steps to understand.