Q&A - P3 Math

[mama_10]

need help urgently.

3 similar oven toasters and 4 similar kettles cost $120. 2 such oven toasters and 1 such kettle cost $95. Find the cost of an oven toater.

thanks

[Niedino]

mama_10:

need help urgently.

3 similar oven toasters and 4 similar kettles cost $120. 2 such oven toasters and 1 such kettle cost $95. Find the cost of an oven toater.

thanks

Hi mama_10,

When doing such questions, teach your child to make one item the same number. If you notice, the second set of items only has 1 kettle. So multiply the set by 4.

(2 oven toasters and 1 kettle) X 4 = 8 oven toasters and 4 kettles
This costs $95 X 4 = $380

Subtract the 2 sets of items together.
8 oven toasters and 4 kettles --> $380
- 3 oven toasters and 4 kettles --> $120
5 oven toasters and 0 kettles --> $260

Now you should only have 1 type of item left.
5 oven toasters --> $260
1 oven toasters --> $260 divided by 5 = $52

Answer is $52.

When your child reaches a higher level, they will ask more difficult questions, such as the cost of the kettle.

Hope it helps!

[mama_10]

Niedino:

mama_10:

need help urgently.

3 similar oven toasters and 4 similar kettles cost $120. 2 such oven toasters and 1 such kettle cost $95. Find the cost of an oven toater.

thanks

Hi mama_10,

When doing such questions, teach your child to make one item the same number. If you notice, the second set of items only has 1 kettle. So multiply the set by 4.

(2 oven toasters and 1 kettle) X 4 = 8 oven toasters and 4 kettles
This costs $95 X 4 = $380

Subtract the 2 sets of items together.
8 oven toasters and 4 kettles --> $380
- 3 oven toasters and 4 kettles --> $120
5 oven toasters and 0 kettles --> $260

Now you should only have 1 type of item left.
5 oven toasters --> $260
1 oven toasters --> $260 divided by 5 = $52

Answer is $52.

When your child reaches a higher level, they will ask more difficult questions, such as the cost of the kettle.

Hope it helps!

:thankyou:

[MathIzzzFun]

ksi:

MathIzzzFun:

Hi

I would think 39 is the only intended answer in this case.

The key here is the remainder of 15 sweets when 4 sweets are packed in a packet, whereas there are only 3 sweets left when 6 are packed in a packet. The only reason that this is case is that he has a limited number of packets and each time, all the packets are used - either filled with 4 or 6 sweets. Otherwise, if there are more than enough packets to pack the sweets, we would expect a remainder of less than 4 when packing 4 sweets into a packet.

Perhaps the other users would like to chip in with their views :lol:

cheers.

This is a P3/P4 level question. We should bear in mind that there should not be so much complex constraint the child has to think about while solving this question. I don't think it is stated there is a limited number of packets, so contraint is not stated. In any case if a question is subjected to interpretaton, I will argue that it is a badly set question. Teachers should vet such type of questions where there are more than one answer. If it is designed as such, then both answers should be acceptable. JMHO.

Supposing the question is:

\"Mr Lim is distributing some sweets to his students. When he gives 4 sweets to each student, he would have 15 sweets left. When he gives 6 sweets to each student, there would be 3 sweets left. How many sweets does he have? \" What would be the answer ?

cheers.

[cimman]

MathIzzzFun:

ksi:

[quote=\"MathIzzzFun\"]
Hi

I would think 39 is the only intended answer in this case.

The key here is the remainder of 15 sweets when 4 sweets are packed in a packet, whereas there are only 3 sweets left when 6 are packed in a packet. The only reason that this is case is that he has a limited number of packets and each time, all the packets are used - either filled with 4 or 6 sweets. Otherwise, if there are more than enough packets to pack the sweets, we would expect a remainder of less than 4 when packing 4 sweets into a packet.

Perhaps the other users would like to chip in with their views :lol:

cheers.

This is a P3/P4 level question. We should bear in mind that there should not be so much complex constraint the child has to think about while solving this question. I don't think it is stated there is a limited number of packets, so contraint is not stated. In any case if a question is subjected to interpretaton, I will argue that it is a badly set question. Teachers should vet such type of questions where there are more than one answer. If it is designed as such, then both answers should be acceptable. JMHO.

Supposing the question is:

\"Mr Lim is distributing some sweets to his students. When he gives 4 sweets to each student, he would have 15 sweets left. When he gives 6 sweets to each student, there would be 3 sweets left. How many sweets does he have? \" What would be the answer ?

cheers.[/quote]if framed in this way, the answer would be 39, because the number of students in a class cannot change.

[corneyAmber]

Yes,this one I also agree is 39. The other one has no constraint stated the same packets are to be used. I think teachers have to be clear when they set such questions.

[tianzhu]

MathIzzzFun:

Supposing the question is:
\"Mr Lim is distributing some sweets to his students. When he gives 4 sweets to each student, he would have 15 sweets left. When he gives 6 sweets to each student, there would be 3 sweets left. How many sweets does he have? \" What would be the answer ?

Hi MathIzzzFun

Good Morning.

Let's take a look at the original question.

Mr Lim has some sweets and packets. When he packs the 4 sweets per packet, there would be 15 sweets left. When he packs 6 sweets per packet, there would be 3 sweets left. How many sweets does he have?

Perhaps, the question should be rephrased in another way so as to have a apple to apple comparison.

Mr Lim has some sweets and students. When he gives 4 sweets per student, there would be 15 sweets left. When he gives 6 sweets per student, there would be 3 sweets left. How many sweets does he have?

Best wishes

[cimman]

Hi MathIzzzFun

I suppose another way to look at problems like this is to determine if there is a physical entity involved ie. packets, students, etc.. vs a virtual entity ie. groups. Most of these questions uses groups, ie. groups of 4, groups of 6, etc...

If a physical item is involved, then the implied assumption should be that all of the physical items should be used in the grouping, ie. in the case of the packets, all of the packets should be used to group 4 sweets or 6 sweets. In this case, then it will constrain the solution to use the same multiple for both groupings.

In the case of students in a class, it is much clearer why the implied assumption holds. The student size in a class does not change, no matter what the grouping is. It doesn't make sense to have some students with no sweets in hand after a distribution, ie. ALL students must have sweets in hand after each distribution.
Thus we can postulate that for all physical items involved in the grouping, the implied assumption holds, ie. All of the physical items must be used after each distribution.

Seen in this light, then we have a very clear guideline on when to use the Gap and Difference method and the answer provided by the Gap and Difference method is the only possible answer.

[MathIzzzFun]

tianzhu:

MathIzzzFun:

Supposing the question is:
\"Mr Lim is distributing some sweets to his students. When he gives 4 sweets to each student, he would have 15 sweets left. When he gives 6 sweets to each student, there would be 3 sweets left. How many sweets does he have? \" What would be the answer ?

Hi MathIzzzFun

Good Morning.

Let's take a look at the original question.

Mr Lim has some sweets and packets. When he packs the 4 sweets per packet, there would be 15 sweets left. When he packs 6 sweets per packet, there would be 3 sweets left. How many sweets does he have?

Perhaps, the question should be rephrased in another way so as to have a apple to apple comparison.

Mr Lim has some sweets and students. When he gives 4 sweets per student, there would be 15 sweets left. When he gives 6 sweets per student, there would be 3 sweets left. How many sweets does he have?

Best wishes

Hi tianzhu,

excellent, that's a better way to rephrase the question.

Basically, the contention here is why would one conceive that number of students will be determinate in this case whereas the number of packets is not as in the original question.

cheers.

[MathIzzzFun]

cimman:

Hi MathIzzzFun

I suppose another way to look at problems like this is to determine if there is a physical entity involved ie. packets, students, etc.. vs a virtual entity ie. groups. Most of these questions uses groups, ie. groups of 4, groups of 6, etc...

If a physical item is involved, then the implied assumption should be that all of the physical items should be used in the grouping, ie. in the case of the packets, all of the packets should be used to group 4 sweets or 6 sweets. In this case, then it will constrain the solution to use the same multiple for both groupings.

In the case of students in a class, it is much clearer why the implied assumption holds. The student size in a class does not change, no matter what the grouping is. It doesn't make sense to have some students with no sweets in hand after a distribution, ie. ALL students must have sweets in hand after each distribution.
Thus we can postulate that for all physical items involved in the grouping, the implied assumption holds, ie. All of the physical items must be used after each distribution.

Seen in this light, then we have a very clear guideline on when to use the Gap and Difference method and the answer provided by the Gap and Difference method is the only possible answer.

Hi cimman,

Let me explain the thought process in me when I read the question.

The first question that I have was \" why is there a remainder of 15 and not a number less than 4 when 4 sweets were packed? why not 11, or 7 for eg \" and then it became clear to me that there was a fixed number of packets and for each scenario, all the packets were used. If there were more than enough packets, then the question is why not fill up 5 packets with 4 sweets each with a remainder of 7 sweets - this would yield a minimum number of 27 as well.

So, for question like this, I would suggest that if the student is unsure, it is best to use the Listing method and then reason and state that all packets were used.

cheers.