Q&A - P5 Math

[MathIzzzFun]

Winx5015:

Thanks MathIzzzFun. You really save my day. Your model drawing and solution is very clear and my son can understand once he looks at it. Once again, thank-you!


Cheers

u r welcome

cheers.

[pixiedust]

Appreciate help with these 2 questions :

(1) What is the missing number and why :
2, 4, 8, 10, 20, 22, _____, 46

(2)

http://i1197.photobucket.com/albums/aa424/pixie_dust8/mq2.jpg\">

[MathIzzzFun]

pixiedust:

Appreciate help with these 2 questions :

(1) What is the missing number and why :
2, 4, 8, 10, 20, 22, _____, 46

(2)

http://i1197.photobucket.com/albums/aa424/pixie_dust8/mq2.jpg\">

Hi

q1's an interesting one.. is the question from school or assessment books ?

1) .. series goes +2,x2,+2,x2
2,4,8,10,20,22,44,46,92,94...

2)
a) AG=BC, so AB=AG, angle ABG = angle AGB = (180-24)/2 = 78
b) angle ACB = (180 - angle ABG)/2 = 51
c) angle EFG = Angle FCG + angle CGF = 34 + angle AGB = 34 + 78 = 112

cheers.

[RRMummy]

Hi sunset_dae,


You should be able to get help in this thread:
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&start=110

Hi mods, please merge.

TIA

[Herbie]

B:W

140: 120
-p:-p
7u:2u

hence 140-p =7u – equation 1
120- p= 2u ---- equation 2

280-2p = 14u
840-7p = 14u

280-2p= 840-7p
p= 112

she gave away 112 x 2 = 224.

[pixiedust]

MathIzzzFun:

Hi, q1's an interesting one.. is the question from school or assessment books ?

Thanks for your help. I looked at it for so long any tips for cracking this kind of questions ?

I should have stated earlier : Both questions are from NYPS 2010 P5 SA2 exam.

[pixiedust]

MathIzzzFun,


Sorry I have to go back to question (2) :
AG=BC, so AB=AG <------- how to concluded AB=AG with the given fact that AG=BC ?

Also, appreciate any help on this question, from the same SA2 paper :
(3) A coin box contained some twenty-cent and fifty-cent coins in the ration of 4:5. When 16 fifty-cents coins were taken out and replaced by some twenty-cent coins, the ratio then became 8:7. The total value of the twenty-cent coins added was the same as the total value of the fifty-cent coins taken out. Find the sum of mone in the coin box.

I understands from the answer key this far :
16 * 50cents = $8
$8 / 20cents = 40 coins

20cents : 50cents
4:5

4U+50 : 5U-16 <— this is the after situation
40U-128->28U+280 <— I don’t understand how to get this, I suppose it is to link back to ratio 8:7

or maybe you have your own solution …thanks in advance !

[PiggyLalala]

pixiedust:

MathIzzzFun,


Sorry I have to go back to question (2) :
AG=BC, so AB=AG <------- how to concluded AB=AG with the given fact that AG=BC ?

Also, appreciate any help on this question, from the same SA2 paper :
(3) A coin box contained some twenty-cent and fifty-cent coins in the ration of 4:5. When 16 fifty-cents coins were taken out and replaced by some twenty-cent coins, the ratio then became 8:7. The total value of the twenty-cent coins added was the same as the total value of the fifty-cent coins taken out. Find the sum of mone in the coin box.

I understands from the answer key this far :
16 * 50cents = $8
$8 / 20cents = 40 coins

20cents : 50cents
4:5

4U+50 : 5U-16 <--- this is the after situation
40U-128->28U+280 <--- I don't understand how to get this, I suppose it is to link back to ratio 8:7

or maybe you have your own solution ...thanks in advance !

Hi pixiedust,
ABCD is a rhombus,
Therefore all the sides are equal AB = BC.
Hence given AG = BC => AB = AG

For questions on angles, it is always good to mark down all the properties of the quadrilateral on the diagram. In this case, it is a rhombus, therefore mark on the diagram all sides are equal and opposite sides are parallel.

[pixiedust]

Thank you PiggyLalala ! Alamak…I must be blind !

[PiggyLalala]

pixiedust:

Thank you PiggyLalala ! Alamak......I must be blind !

I always tell my ds to highlight the key word in the question and next must mark all the properties onto the diagram. If not, CAN'T see the hidden information.