Q&A - P3 Math

[concerned1970]

ksi:

concerned1970:

Gd evening! A ques from Taonan paper.


Apple x Pear=60
Apple x Apple x Pear=360
Find the value of Pear x Pear x Pear? Ty in adv!

Apple x (Apple x pear) = 360
Apple x 60 = 360
Apple = 360/60 = 6

Pear = 60/6 = 10

Pear x Pear x Pear = 10 x 10 x 10 = 1000

Thanks Ksi and Teh_oh! Sry now i understand!

[teh_oh]

concerned1970:

ksi:

[quote=\"concerned1970\"]Gd evening! A ques from Taonan paper.


Apple x Pear=60
Apple x Apple x Pear=360
Find the value of Pear x Pear x Pear? Ty in adv!

Apple x (Apple x pear) = 360
Apple x 60 = 360
Apple = 360/60 = 6

Pear = 60/6 = 10

Pear x Pear x Pear = 10 x 10 x 10 = 1000

Thanks Ksi and Teh_oh! but why is that pear (60) still has to divide by 6?[/quote]Because Apple x Pear = 60.
Apple is 6, so to find pear, have to 60/6.

[concerned1970]

Ty ty!

[cimman]

tianzhu:

hypergatak:

Thanks for prompt response. My son is so dissappointed as he got a different answer. He is perstering me to ask another 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?

Thanks.

Hi

This is a question on Gap and Difference or Excess and Shortage.

15 – 3 ------ 12
6 – 4 ------ 2
12/2 ------6

There are 6 packets
Number of sweets ------- (4*6) + 15 ------- 39

Best wishes

just some musings on the Gap and Difference method.

it seems your answer is different from ksi's.

I was more familiar with ksi's method, since it is widely used in assessment books, and they normally categorise it under LCM (lower common multiple) method.
The Gap and Difference type of problem is not often seen, but it still lurks around and normally found in more difficult problems, a higher order problem of the LCM category.

When I used it, I found that I got a different answer, as you've done in the above example, from the LCM method of listing multiples.

Upon further investigation, I found that unless the 2 groups shares the same multiple, one can't really use the Gap and Difference method.
In the above example, the 15 excess sweets belongs to a packet of 12 (3rd multiple of 4), while the 3 excess sweets belongs to a packet of 24 (4th multiple of 6), so they have different multiples. Thus the Gap and Difference method cannot be used here.

In most cases, Gap and Difference problems requires you to answer 2 parts, (a) and (b). As in the example below:
Mrs Lee had to give some beads to some girls. If each girl had 9 beads, Mrs Lee would need another 48 beads. If each girl had 12 beads, Mrs Lee would need another 72 beads.
a) how many girls are there ?
b) how many beads does each girl get ?

The (a) and (b) gives you a clue that this is a Gap and Difference problem, and most children should be able to pick this up. For the more advanced child with greater analytical ability, they can analyse and surmise that the above problem can't really have 2 different multiples in each group, ie. the number of girls that Mrs Lee gives beads to is the same, whether it's 9 beads or 48 beads. Or in a case where the unit cost of an item is given, and that item is seperated into 2 groups, the unit cost of the item is the same in both groups.

In cases where it is possible to have different multiples in the 2 groups, as in the above Mr. Lim example, then the LCM method of listing multiples is the only way to solve it.

The LCM method can also be used to solve Gap and Difference problems. Just that it requires slightly more time and calculations in listing out the various multiples. The Gap and Difference method is so much more efficient when it can be used.

However, if one just wants to use a single method to solve both types of problems, the LCM method would be the one.

[cimman]

ksi:

concerned1970:

Gd evening! A ques from Taonan paper.


Apple x Pear=60
Apple x Apple x Pear=360
Find the value of Pear x Pear x Pear? Ty in adv!

Apple x (Apple x pear) = 360
Apple x 60 = 360
Apple = 360/60 = 6

Pear = 60/6 = 10

Pear x Pear x Pear = 10 x 10 x 10 = 1000

I find that problems like this is so unfair. This is a thinly diguised algebra problem. If they want students to solve algebra problems, they should just teach algebra in the lower block, instead of all this beating around the bush.

[ttyh]

meiling wants to do 100 skips on a skipping rope a day. She starts with 20 skips on day1 and increases 8 each day.

Find the number of days it will take Meiling to reach 100skips

[corneyAmber]

ttyh:

meiling wants to do 100 skips on a skipping rope a day. She starts with 20 skips on day1 and increases 8 each day.

Find the number of days it will take Meiling to reach 100skips

8x10=80
80+20 = 100

So she takes 10+1=11 days.

[MathIzzzFun]

cimman:

tianzhu:

[quote=\"hypergatak\"]Thanks for prompt response. My son is so dissappointed as he got a different answer. He is perstering me to ask another 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?

Thanks.

Hi

This is a question on Gap and Difference or Excess and Shortage.

15 – 3 ------ 12
6 – 4 ------ 2
12/2 ------6

There are 6 packets
Number of sweets ------- (4*6) + 15 ------- 39

Best wishes

just some musings on the Gap and Difference method.

it seems your answer is different from ksi's.

I was more familiar with ksi's method, since it is widely used in assessment books, and they normally categorise it under LCM (lower common multiple) method.
The Gap and Difference type of problem is not often seen, but it still lurks around and normally found in more difficult problems, a higher order problem of the LCM category.

When I used it, I found that I got a different answer, as you've done in the above example, from the LCM method of listing multiples.

Upon further investigation, I found that unless the 2 groups shares the same multiple, one can't really use the Gap and Difference method.
In the above example, the 15 excess sweets belongs to a packet of 12 (3rd multiple of 4), while the 3 excess sweets belongs to a packet of 24 (4th multiple of 6), so they have different multiples. Thus the Gap and Difference method cannot be used here.

In most cases, Gap and Difference problems requires you to answer 2 parts, (a) and (b). As in the example below:
Mrs Lee had to give some beads to some girls. If each girl had 9 beads, Mrs Lee would need another 48 beads. If each girl had 12 beads, Mrs Lee would need another 72 beads.
a) how many girls are there ?
b) how many beads does each girl get ?

The (a) and (b) gives you a clue that this is a Gap and Difference problem, and most children should be able to pick this up. For the more advanced child with greater analytical ability, they can analyse and surmise that the above problem can't really have 2 different multiples in each group, ie. the number of girls that Mrs Lee gives beads to is the same, whether it's 9 beads or 48 beads. Or in a case where the unit cost of an item is given, and that item is seperated into 2 groups, the unit cost of the item is the same in both groups.

In cases where it is possible to have different multiples in the 2 groups, as in the above Mr. Lim example, then the LCM method of listing multiples is the only way to solve it.

The LCM method can also be used to solve Gap and Difference problems. Just that it requires slightly more time and calculations in listing out the various multiples. The Gap and Difference method is so much more efficient when it can be used.

However, if one just wants to use a single method to solve both types of problems, the LCM method would be the one.[/quote]Hi cimman,

I would like to share my view on this question. Firstly, I agree with you that the LCM method or the Listing method is a more generic method to use for solving this type of questions.

For this particular case, the Gap & Difference method can be applied because one can infer from the question that same number of packets (ie same multiple, in this case 6 packets) were used to pack the sweets. This inference can be made from the fact that when packing 4 sweets to a packet, there would be 15 sweets remaining. This means that all packets were used to pack the sweets ie same number of packets used.

If it is intended that number of packets used need not be the same, then there are multiple answers - 27, 39, 51, 63 etc and in this case, the last sentence of the question should be worded as \"What is the least number of sweets that Mr Lim has?\" in order to have a unique answer - 27.

Hope this will not create more confusion.

cheers.

[Just4Fun]

I hv come across this Q in previous exam question… I thought it was easy but yet got conflicting answers !

Here’s the question :
Child A has 1000 sweets. He shares them among his 4 friends. How many sweets does each friend get ?
Ans : ???

Simple question ? Right answer pls? Anyone can try ?

[CoffeeCat]

Just4Fun:

I hv come across this Q in previous exam question.. I thought it was easy but yet got conflicting answers !

Here's the question :
Child A has 1000 sweets. He shares them among his 4 friends. How many sweets does each friend get ?
Ans : ???

Simple question ? Right answer pls? Anyone can try ?

\"He shares them among his 4 friends.\"
This probably means that only the 4 friends (not including A) shares the 1000 sweets. So the ans is 250.
If the author wants to include A, then he can say \"He shares them with his 4 friends\".