Q&A - PSLE Math

[MathIzzzFun]

atutor2001:

MathIzzzFun:

[quote=\"atutor2001\"]
Mathematically, there is no faster way because the number of unknowns (i.e. The total no. of cadets; The no. of groups of 4 in case 1; The no. of groups of 6 in case 2 - 3 unknowns) is more than the number of relationship (equations) given by the question.

Hi

newbie here. Just to share that there is a quick method for this type of problem.

Look at the numbers carefully and you will see that by adding 3 to the total will make the total divisible by both 7 and 9. This means that the new total is a multiple of both 7 and 9. So the number is question is 7 x 9 - 3 = 63 - 3 = 60.

cheers.

Hi MathIzzzFun

Thank you for sharing this technique. The addition of '3' to make the total divisible by both 7 and 9 because their remainders are 4 (which needs 3 more to form 7) and 6 (also needs 3 more to form 9) is a short cut that students who a trained in math olympiad may know.

Thanks for highlighting, I have forgotten about this technique - a very powerful tool for solving questions whereby the remainders need the same amount to form the exact group. If the above condition does not exist, we still need to list to get the answer.
Cheers[/quote]Hi atutor2001

Avec plaisir

cheers.

[atutor2001]

Just to share a question which needs to use the method that MathIzzzFun shown earlier :


What is the smallest number when divided by :
2 will give a remainder 1
3 will give a remainder 2
4 will give a remainder 3
…
8 will give a remainder 7
9 will give a remainder 8

By using the same method with slight modification, the answer should be 2519.

[pixiedust]

MathIzzzFun & atutor2001, thanks for bringing a new dimension to the solution.


I don't have enough maths sense to see :
\" Look at the numbers carefully and you will see that by adding 3 to the total will make the total divisible by both 7 and 9. This means that the new total is a multiple of both 7 and 9. So the number is question is 7 x 9 - 3 = 63 - 3 = 60 \"

I'll have to stick with the listing with intelligence.
Confirm I am no math olympiadien

[ozora]

tianzhu:

ozora:

hi Tianzhu
is the answer: 6:5:3?

Hi ozora

Good Morning

The answer listed in the worksheet is 20:15:1

Best wishes

then i am stuck.
using your method.
I had the following:
Circle A: shaded= 1u ; unshaded=5 u; total : 6units
Circle B: shaded= 1u ; unshaded=4 u; total : 5units
Circle C:: shaded= 1u ; unshaded=2 u; total : 3units

area of A: Area of B: Area of C
6:5:3
unless i missed out the steps?

[tianzhu]

ozora:

then i am stuck.
using your method.
I had the following:
Circle A: shaded= 1u ; unshaded=5 u; total : 6units
Circle B: shaded= 1u ; unshaded=4 u; total : 5units
Circle C:: shaded= 1u ; unshaded=2 u; total : 3units

area of A: Area of B: Area of C
6:5:3
unless i missed out the steps?

Hi

You’ve started correctly, but how about this condition?

The total area of B and C is 4/5 that of A.

Best wishes

[tisha]

tianzhu:

WXYZ is counted twice when we are calculating the areas of the two triangles.

So we need to subtract 40 from the total areas of the two triangles.
There are some unnecessary steps in the WS’s solution.
You may work it this way.

1/2 X 12 X 18 = 108 ----- this gives the areas of the two triangles.
108-40=68cm^2

Thanks tianzhu. Agree there are additional unnecessary steps, which causes confusion. Now I understand it.

[tianzhu]

tisha:

Thanks tianzhu. Agree there are additional unnecessary steps, which causes confusion. Now I understand it.

Hi tisha

Good Afternoon.

You're welcome.

Best wishes

[anneshirleygilbert]

tianzhu:

ozora:

then i am stuck.
using your method.
I had the following:
Circle A: shaded= 1u ; unshaded=5 u; total : 6units
Circle B: shaded= 1u ; unshaded=4 u; total : 5units
Circle C:: shaded= 1u ; unshaded=2 u; total : 3units

area of A: Area of B: Area of C
6:5:3
unless i missed out the steps?

Hi

You’ve started correctly, but how about this condition?

The total area of B and C is 4/5 that of A.

Best wishes

Hello tianzhu,
Like ozora, \"I'm stuck.\" I can see how area of A would be 20 units, but how did you get the breakdown for B to be 15 units and C to be 1 unit? Been trying to crack it all weekend. It's like a bone in my throat. Please help.

[Dharma]

anneshirleygilbert:

tianzhu:

[quote=\"ozora\"]
then i am stuck.
using your method.
I had the following:
Circle A: shaded= 1u ; unshaded=5 u; total : 6units
Circle B: shaded= 1u ; unshaded=4 u; total : 5units
Circle C:: shaded= 1u ; unshaded=2 u; total : 3units

area of A: Area of B: Area of C
6:5:3
unless i missed out the steps?

Hi

You’ve started correctly, but how about this condition?

The total area of B and C is 4/5 that of A.

Best wishes

Hello tianzhu,
Like ozora, \"I'm stuck.\" I can see how area of A would be 20 units, but how did you get the breakdown for B to be 15 units and C to be 1 unit? Been trying to crack it all weekend. It's like a bone in my throat. Please help.[/quote]Area B + C = 4/5 of Area of A
Area of A = 5/4 of Area of (B + C) = 5/4 of B + 5/4 of C ----------- ( 1 )

Shaded area of A = Shaded areas of (B + C)
1/6 of A = 1/5 of B + 1/3 of C
6/6 of A = 6/5 of B + 6/3 of C ------------- ( 2 )

Equating ( 1 ) and ( 2 )
5/4 of B + 5/4 of C = 6/5 of B + 6/3 of C
5/4 of B – 6/5 of B = 6/3 of C – 5/4 of C
25/20 of B – 24/20 of B = 24/12 of C – 15/12 of C
1/20 of B = 9/12 of C
B = 15C

A = 5/4 (15C + C) = 20C

Ratio of areas of A : B : C = 20 : 15 : 1


OR


B ( shaded ) : B ( unshaded ) = 1 : 4 => 1u : 4u
C ( shaded ) : C ( unshaded ) = 1 : 2 => 1p : 2p
A (shaded ) : A ( unshaded ) = 1 : 5 = (1u + 1p) : (5u + 5p)

Total area of (B + C ) = 5u + 3p
Total area of A = 6u + 6p

Total area of (B + C) = 4/5 of total area of A
5u + 3p = 4/5(6u + 6p)
25/5u – 24/5u = 24/5p – 15/5p
1u = 9p

Ratio of A : B : C = (6u + 6p) : 5u : 3p = 60p : 45p : 3p = 20 : 15 : 1

[anneshirleygilbert]

Thank you, Dharma. You’re so kind to offer two options in the solutions. Understand both of them. Was just wondering, is this level of math expected of Spore P5/P6 students preparing for PSLE? Seems quite a tall order to me.


Appreciate your help.