Q&A - P5 Math

[happyheart]

tianzhu:

happyheart:

Tianzhu, could you help me with one more question please?


3 children traded cards with one another.
At first, Eliza had more cards than Farhan and Michelle had 2 times as many cards as Farhan. After Farhan bought another 33 cards. Michelle lost half what she ha and Eliza lost 14 cards, Farhan now had 6 more cards than Eliza. In the end, the children had a total of 150 cards. How many cards did each child have at first?

Hi

You may present the solution using MD or using units.

Consider the end scenario first.

Michelle ------ 1 unit

Farhan ------ 1 unit + 33

Eliza ------- 1 unit + 27 (Farhan now had 6 more cards than Eliza.)

3 units + 33 + 27 ------- 150

3 units -------- 90

1 unit ------- 30

In the beginning

Michelle ------ 2 units -------- 2*30 --------- 60

Farhan ------ 1 unit ------ 30

Eliza ------- 1 unit + 27 + 14 -------- 30 + 27 + 41 -------71

Best wishes

Thank you!

[happyheart]

tianzhu:

happyheart:

Thanks Tianzhu! I understand now

one question, Would you think this is a typical p5 question? I checked through DS worksheet and cannot find similar question to work on.

Hi

Where did you get this question from?

From my experience during my kid's PSLE, questions involving \"Working Backwards\" are popular.

Best wishes

Tianzhu, The question is from DS school, June holiday worksheet.

[MathIzzzFun]

beanbear:

I have such a struggle with the following type of problem sums. Is there a particular approach to such sums? What is the critical thing to understand about such problems?


(1) Mr Lim bought 4 pears & 5 lemons. Mr Tan bought 5 pears and 4 lemons. Mr Tan paid 20cents more than Mr Lim. If Mr Lim paid $4.40, how much did 4 pears cost?

(2) 5 similar files cost $3 more than 3 similar pens. Mrs Lee purchased 3 files and 6 pens for $17.40. What was the cost of each file?

Hi

For both questions, one needs to find out the difference in price in the two items.

You can use MD, or picture models, units to find the difference first.

Q1.
Lim --> 4P + 5L = (4P + 4L) + 1L
Tan --> 5P + 4L = (4P + 4L) + 1P
Tan paid 20 cents more --> 1P is 20 cents more than 1L
1L --> 1 unit
1P --> 1 unit + 0.20
Lim --> 4 units + 0.80 + 5 units = 9 units + 0.80 --> $ 4.40
1 unit --> $ 0.40
1P --> $0.40 + $ 0.20 = $ 0.60
4P --> $ 2.40
4 pears cost $ 2.40

cheers.

[MathIzzzFun]

beanbear:

I have such a struggle with the following type of problem sums. Is there a particular approach to such sums? What is the critical thing to understand about such problems?


(1) Mr Lim bought 4 pears & 5 lemons. Mr Tan bought 5 pears and 4 lemons. Mr Tan paid 20cents more than Mr Lim. If Mr Lim paid $4.40, how much did 4 pears cost?

(2) 5 similar files cost $3 more than 3 similar pens. Mrs Lee purchased 3 files and 6 pens for $17.40. What was the cost of each file?

Hi

Q2.
lowest common multiple of 5 & 3 --> 15

3 pens --> 15 units, 1 pen --> 5 units
5 files --> 15 units + $3.00, 1 file --> 3 units + $ 0.60
3 files + 6 pens --> (9 units + $1.80) + 30 units = 39 units + $ 1.80 --> $ 17.40
1 unit --> $0.40

1 pen --> $ 2.00
1 file --> $ 1.20 + $ 0.60 = $1.80

cheers.

[the shadowed snake]

thanks tianzhu

[beanbear]

MathIzzzFun:

beanbear:

I have such a struggle with the following type of problem sums. Is there a particular approach to such sums? What is the critical thing to understand about such problems?


(1) Mr Lim bought 4 pears & 5 lemons. Mr Tan bought 5 pears and 4 lemons. Mr Tan paid 20cents more than Mr Lim. If Mr Lim paid $4.40, how much did 4 pears cost?

(2) 5 similar files cost $3 more than 3 similar pens. Mrs Lee purchased 3 files and 6 pens for $17.40. What was the cost of each file?

Hi

Q2.
lowest common multiple of 5 & 3 --> 15

3 pens --> 15 units, 1 pen --> 5 units
5 files --> 15 units + $3.00, 1 file --> 3 units + $ 0.60
3 files + 6 pens --> (9 units + $1.80) + 30 units = 39 units + $ 1.80 --> $ 17.40
1 unit --> $0.40

1 pen --> $ 2.00
1 file --> $ 1.20 + $ 0.60 = $1.80

cheers.

Thanks so much for your clear explanation!! I see the light!

[cimman]

beanbear:

I have such a struggle with the following type of problem sums. Is there a particular approach to such sums? What is the critical thing to understand about such problems?


(1) Mr Lim bought 4 pears & 5 lemons. Mr Tan bought 5 pears and 4 lemons. Mr Tan paid 20cents more than Mr Lim. If Mr Lim paid $4.40, how much did 4 pears cost?

(2) 5 similar files cost $3 more than 3 similar pens. Mrs Lee purchased 3 files and 6 pens for $17.40. What was the cost of each file?

Hi Beanbear
if you are open to using algebra and simultaneous equations, there is a consistent way to solve such problems. It does not rely on heursitcs, just equation generation and solving the equations.

I categorise such problems as Unit Value problems, because they all require the same basic formula: Quantity x Unit Cost = Total Cost

Mr Lim bought 4 pears & 5 lemons. Mr Tan bought 5 pears and 4 lemons. Mr Tan paid 20cents more than Mr Lim. If Mr Lim paid $4.40, how much did 4 pears cost?

the first step is to fill up the table with the values from the problem sum. Here, the fonts in black represents the values from the problem sum. Next, put in the unknown variables, in this case, u and p for the respective unit values. The fonts in red represents the calculated or derived values.

http://i50.tinypic.com/2hwk77q.png\">

5 similar files cost $3 more than 3 similar pens. Mrs Lee purchased 3 files and 6 pens for $17.40. What was the cost of each file?

in the next question, the same Unit Value table is used. The process is the same, fill up the table with values from the problem sum, then put in the unknown variables, then circle the relevant equations.
http://i49.tinypic.com/rgwg44.png\">

the filling up of the table with the values from the problem sum is very straight forward, so the child will not be confused. The table itself guides you on the relationships between the different values.

if you do not use simultaneous equations concepts, then you'll have to use heuristics to remove one unknown from a 2 unknown problem.
In the above example, 5u - 3p = 3. You'll find that if you use Lowest Common Multiple of 5 and 3, you'll remove one unknown from the equation, the u and p will have the same units. This is a techniqe in solving simultaneous equations.

The advantage of using simultaneous equations is that the child do not have to contend with learning multiple heuristics and uses just one technique to solve a wide range of problems. The downside of simultaneous equations is that schools don't teach it, so parents or tutors will have to teach this concept.

Each category of problem type have a different table, because the table enforces the formula required to solve that particular problem type. If you're interested in such technique, I conduct a workshop for parents, you can get the infomation from here: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=32278&start=110
the next workshop would be around end June to early July.

[Udon]

Please help…THANK U…

Appreciate with illustration, thankz for your time.

Q1. Container A has a mass of 1 000g when it is 3/5 filled with oil. Its mass is 40g more when it is 5/8 filled with oil. What is the mass of the empty container?


Q2. Roland had $160 less than Jim. After Jim spent $200 on a shirt and tie, Roland had twice as much money as Jim. Find the amount of money Jim had at first.

[tianzhu]

Udon:

Q2. Roland had $160 less than Jim. After Jim spent $200 on a shirt and tie, Roland had twice as much money as Jim. Find the amount of money Jim had at first.

Hi

Hope this helps.

Best wishes

http://farm8.staticflickr.com/7077/7179026399_d7b10983b0_z.jpg\">

[happyheart]

Hi, I need help on this…please help with illustration. I am confused with multiple variables and the 3 equations. Thank you!


The cost of 6 identical chairs , 5 identical lamps and 1 table is $380.
The cost of 2 such chairs and 1 table is $188, while 1 chair and 1 lamp cost $44.
a) What is the cost of 1 chair and 1 table?
b) What is the total cost of 1 table, 1 lamp and 1 chair?