Q&A - P3 Math
-
Sakina has 15 coins. There are 20c coins and 50c coins. The total amount is $4.80. How many 20c coins are there?
The above sum can be done using the guess method like
10 20c coins = $2.00
5 50c coins = $2.50
then by increasing 20c coins number & decreasing the 50c coins number. But this method is very time consuming & a P3 student may not able to guess correctly which coin to increase & which one to decrease.
So can any one let me know some method which can be used for these kind of sums.
Thanks in advance. -
mamona:
Perhaps you could explain the following method for P3 :Sakina has 15 coins. There are 20c coins and 50c coins. The total amount is $4.80. How many 20c coins are there?
The first step is to assume that all the coins are either 50c coins or 20c coins.
Assuming all 15 coins are 20c coins, total value is only $3
Find the difference compared to $4.80; i.e. $1.80.
(easier to do show this in a table - see below)
2nd step
Since the total is less than $4.80, there should be more 50c coins and fewer 20c coins ( i.e. need more of those with higher value)
Conversely if the total is more than $4.80, there should be fewer 50c coins and more 20c coins. (This scenario is not possible for we had assumed that all are 20c coins).
3rd step
The 3rd step cuts down the number of iterations required for the 'guess and check' method.
For every 50c coin you increase, you have to decrease the 20c coin as well.
As such, the value increased per coin is only 30c (50c-20c)
As above, assuming all are 20c coins, total value is only $3, we are short of $1.80.
$1.80 / 30c = 6
i.e. increase 6 50c coins and decrease 6 20c coins
As such, the number of 20c coins is 15-6=9
4th step
Complete in table to check
http://www.postimage.org/ -
Many Many Thanks for your reply. Yes I understood the method. :lol:
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mamona:
Many Many Thanks for your reply. Yes I understood the method. :lol:
You are welcome! -
hello comrades,
need your help to solve the following P4 Maths without using ALGEBRA. :?:
detailed explanation much needed. thanks.
Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.
How many apples did they both have at first ? -
Ogima:
It will be easier to see if you draw model. (I am not good at creating nice pictures, so let me try to explain the \"model\" I drew.)hello comrades,
need your help to solve the following P4 Maths without using ALGEBRA. :?:
detailed explanation much needed. thanks.
Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.
How many apples did they both have at first ?
May [----][----][----][----]
{indicate the last unit}<60>
June [----][----][////]
(the last unit for June is the given away part)
Each unit should be drawn same size.
From model, 1 unit = 60
May has 4 x 60 = 240
June has 3 x 60 = 180 initially -
Ogima:
hello comrades,
need your help to solve the following P4 Maths without using ALGEBRA. :?:
detailed explanation much needed. thanks.
Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.
How many apples did they both have at first ?
Let me try, please note it is not drawn in scale
a)assume June has 3 units, therefore May has 3 units +60
May has 3 unit + 60 more [ ][ ][ ][ 60 more ]
June has 3 unit [ ][ ][ ]
b) May gave away 1/3, that mean is 1 unit, hence the model became
May [ ][ ][ ][ 60 more ]
June [ ][ ]
May has twice as many apples as June, hence
[1 st Half ] [2nd half ]
------------- -----------------
May [ ][ ] [ ][ 60 more]
June [ ][ ]
c) Let look at the half of it
May [ ][ 60 more ]
June [ ][ ]
Canceal 1 unit, the other unit is equal to 60, therefore
June has 60+60+60+60 = 240
May has 60+60+60 = 180
Total = 240+180 = 420 apples
They have total of 420 apples
I hope the above is able to help you.
Cheer! -
//Moderator's note: Topics merged.
Can anyone help to solve this qn? tks.
Matthew and Paul have a sum of money at first. If Matthew gives Paul $80, he will have twice as much as Paul. If Matthew gives Paul $20, he will have thrice as much as Paul. How much money did Matthew have at first? -
Anxiousmom:
//Moderator's note: Topics merged.
Can anyone help to solve this qn? tks.
Matthew and Paul have a sum of money at first. If Matthew gives Paul $80, he will have twice as much as Paul. If Matthew gives Paul $20, he will have thrice as much as Paul. How much money did Matthew have at first?
Amended
[....][.....80.....][....][.....80......] = 2 big units
[....][.....80.....] = 1 big unit
[....][20][.60.][....][20][.60.][.60.] = 3 units
[....][20] = 1 unit
1u = 3 x 60 = 180
3u = 3 x 180 = 540
Matthew = 540 + 20 = $560 -
Anxiousmom:
Hi Anxiousmom,//Moderator's note: Topics merged.
Can anyone help to solve this qn? tks.
Matthew and Paul have a sum of money at first. If Matthew gives Paul $80, he will have twice as much as Paul. If Matthew gives Paul $20, he will have thrice as much as Paul. How much money did Matthew have at first?
I drew the models for both parts :
M [--------][--------][-80-]
P [-][-80-]
(note : [-][-80-] is same length as [-------])
M [------][------][------][20]
P [-][20]
(note : [-][20] same length as [-----], and the length for M for both models are the same)
Based on the models,
3(p+20) + 20 = 2 (p+80) + 80
p = 160
m = 2x(160+80) + 80 = 560
Matthew has $560 initially.
(but sorry, my soln seems a bit \"algebraic\", maybe someone can suggest a more \"modelic\" soln :lol: )
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