Q&A - PSLE Math

[Dharma]

[quote="Drdj"]Dear all,


3. A, B, C, D share in a gift equally. D forgot to bring his share, so the rest paid for him first. Cindy paid 3/5 of (A+B)'s share. B paid $10 more than A. The next day, Danny returned $24 to C. How much did D return to A and B?

A paid - 5units
B paid - 5 units + $10
C paid - 3/5x(10units + $10) = 6units + $6

Total paid by A,B and C = 16units + $16

If shared equally , each should bear = 4units +$4

(6units + $6) - (4 units +$4) = $24
2 units +$2 = $24
2 units = $22
1 unit = $11

D must return to A = (5 units) - (4 units +$4) = $7
D must return to B = (5 units +$10) - (4 units +$4) = $17

[Dharma]

http://www.postimage.org/[/quote]


Alternatively, if area of CED = 30% of area of ABCD

Then, area of AEB must be 20% of area of ABCD
# Area of AEB + Area of CED = Area of AEC + Area of BED

20% - 80cm2
100% - 400cm2

Area of ABCD - 400cm2

[Drdj]

Thank you so much, Dharma and Vanilla Cake Your wonderful response is very much appreciated !!


A question for your solution

\"Alternatively, if area of CED = 30% of area of ABCD
Then, area of AEB must be 20% of area of ABCD\"
How do you know Triangle AEC = Triangle BEC ? (by proportion) ?

Similarly for your 1st solution, when you draw a line through, we assume the 2 triangles constitute the area of Triangle AEB.

Apologies for the trouble

Dharma:

http://www.postimage.org/

Alternatively, if area of CED = 30% of area of ABCD

Then, area of AEB must be 20% of area of ABCD
# Area of AEB + Area of CED = Area of AEC + Area of BED

20% - 80cm2
100% - 400cm2

Area of ABCD - 400cm2[/quote]

[Dharma]

Dear Drdj,


There isn’t any triangle BEC … it’s just that areas triangle AEC and BED add up 50% of the area of ABCD …cause they share a common base and the perpendicular dist is the breath of the rect.

Similarly triang. AEB and CED, their areas also add up to the remaining 50% of area of rect. ABCD

[small]

Dharma:

AnnT2009:

Please help,



Q2 Mr Tan bought three times as many badges as toy cars and spent $144 in total. He spent $84 more on toy cars than on badges. Given that a toy car cost $10.40 more than a badge. What is the cost of a badge?
Ans:$1

TIA.

Qty
Badges : 3 units
Toy Cars : 1 unit

Unit Price
Badges : 1 part
Toy car : 1 part + $10.40

Total Value
Badges : (3 units x 1 part)
Toy car : 1 unit x (1 part +$10.40)

(4 units x 1 part) + (1 unit x $10.40) = $144 ------- Eqn 1

(1 unit x $10.40) - (2 units x 1 part) = $84
(2 units x $10.40) - (4 units x 1 part) = $ 168 -------- Eqn 2

Eqn 1 + Eqn 2
(3 units x $10.40) = $ 312
1 unit = 10

40 x 1 part = $144 - $104
1 part = 1

1 part - unit price of badge = $1

Hi Dharma,

Much appreciated if you can elaborate on how to get this:

(1 unit x $10.40) - (2 units x 1 part) = $84

Thanks in advance.

[Dharma]

Hi small,


It's the diff of amount spent on toy cars and badges, ...qn says it is $84

(1 unit) x (1 part +$10.40) - (3 units) x (1 part) = $84
(1 unit)x (1 part) + (1 unit) x ($10.40) - (3 units) x (1 part) = $84
(1 unit) x ($10.40) - (2 units) x (1 part) = $84

Hope it helps ...

[small]

Dharma:

Hi small,


It's the diff of amount spent on toy cars and badges, ...qn says it is $84

(1 unit) x (1 part +$10.40) - (3 units) x (1 part) = $84
(1 unit)x (1 part) + (1 unit) x ($10.40) - (3 units) x (1 part) = $84
(1 unit) x ($10.40) - (2 units) x (1 part) = $84

Hope it helps ...

Hi Dharma,

Thanks...You are great..

Please forgive my ignorance.... :oops:

[tianzhu]

Thank you for your help.


Quite an interesting question, the answer hinges on how we would interpret together took

Mr Yoma and Mr Ula together took 5 hours to paint a room.
Mr Wara and Mr Hito together took 4 hours to paint the same room.
Mr Ula and Mr Wara together took 3 hours to paint the same room.
How long would both Mr Yoma and Mr Hito together take to paint the same room?

Take the case of Mr Yoma and Mr Ula, do you interpret as both men were painting for a total duration of 5 hours. (T1+T2 =5)
or they were painting for 5 hours each (T1=5 and T2 =5)?

[small]

tianzhu:

Thank you for your help.


Quite an interesting question, the answer hinges on how we would interpret together took

Mr Yoma and Mr Ula together took 5 hours to paint a room.
Mr Wara and Mr Hito together took 4 hours to paint the same room.
Mr Ula and Mr Wara together took 3 hours to paint the same room.
How long would both Mr Yoma and Mr Hito together take to paint the same room?

Take the case of Mr Yoma and Mr Ula, do you interpret as both men were painting for a total duration of 5 hours. (T1+T2 =5)
or they were painting for 5 hours each (T1=5 and T2 =5)?

Hi tianzhu,

I will interpret as the both men were painting for a total duration of 5 hours, so the answer is 6 hours???

[Dharma]

small:

tianzhu:

Thank you for your help.


Quite an interesting question, the answer hinges on how we would interpret together took

Mr Yoma and Mr Ula together took 5 hours to paint a room.
Mr Wara and Mr Hito together took 4 hours to paint the same room.
Mr Ula and Mr Wara together took 3 hours to paint the same room.
How long would both Mr Yoma and Mr Hito together take to paint the same room?

Take the case of Mr Yoma and Mr Ula, do you interpret as both men were painting for a total duration of 5 hours. (T1+T2 =5)
or they were painting for 5 hours each (T1=5 and T2 =5)?

Hi Tianzhu and small,

Yoma and Ula took 5 hrs to complete painting the room. T1 + T2 = 5 is not right. You need not bother abt time taken by either Yoma or Ula to paint because their rate of painting are different.

Just take the combinations of times taken by the different groups of men to complete the room.

In 1 hr:
Y+U - 1/5 of room
W+H - 1/4 of room
U+W - 1/3 of room

Therefore, if 4 of them paint together,

In 1 hr:
Y+U+W+H = 1/5 + 1/4 = 9/20 of room

So, Y + H will complete (9/20 - 1/3) = 7/60 of room in 1 hr

So, Y & H will complete painting the room in 60/7hrs