Q&A - PSLE Math

[Andrew Lee]

P6ians:

Hi

Q.1:
1h 30min--1.5h
1.5hx90=135km
135+25=160km
160km/(4-1.5)=64km/h
64x4h=256km(answer)
Q.2:(in this question,total remains the same)
Before. After
D:B. D:B
4:5.|9u 10:17|27u
12:15|27u
12u-10u=2u
2u----(100/2)
15u---$375(answer)

Thank you so much :rahrah:

[Yvonne Tan]

Nanyang prelim 2011

Ali and Ben started jogging from Park X to Park Y at the same time. Ali jogged at an average speed of 7.5km/h while Ben jogged at an average speed of 4.5km/h. After jogging for 1.2h, Ali decided to turn back and jog towards Ben. Ali met Ben at point Z which was halfway btw Park X and Park Y. Ali and Ben then continued their jog together towards Park Y at an average speed of 4.5km/ h. What was the distance between park X and park Y.

Age problem

When Kok’s age was twice as Bala’s age, Mike’s age was 30. When Mike’s age was twice Kok’s age, Bala’s age was 21. Given that Mike was the oldest among the three persons, find Bala’s age when Mike was 62 years old. Thanks.

Thanks.

[Andrew Lee]

P6ians:

Hi

Q.1:
1h 30min--1.5h
1.5hx90=135km
135+25=160km
160km/(4-1.5)=64km/h
64x4h=256km(answer)
Q.2:(in this question,total remains the same)
Before. After
D:B. D:B
4:5.|9u 10:17|27u
12:15|27u
12u-10u=2u
2u----(100/2)
15u---$375(answer)

Sorry, for Q2, the question asking how many notes? not how much... :oops:

[MathIzzzFun]

Muku:

MathIzzzFun:

[quote=\"Andrew Lee\"]

Hello, can help me solve this headpain sch question ?
http://i53.tinypic.com/2s6sjmp.jpg\">

Hi

You can use MD as done for a similar question - http://www.flickr.com/photos/62167097@N02/5679183503/in/photostream

or we can also use \"units\" & \"parts\" ie simultaneous equations/algebra...

X = 6 units
Y = 9 parts
Shaded = 1/2 X + 1/3 Y = 3 units + 3 parts

X+Y = 3/4 C, C = 4/3 (X+Y) = 8 units + 12 parts ...(eq 1)

3/10 C is shaded, so 3 units + 3 parts = 3/10 C
C = 10 units + 10 parts = 8 units + 12 parts (from eq 1)
so 1 unit = 1 part

X : Y = 6 units : 9 parts = 6 units : 9 units = 2 : 3

cheers.

Hi hi,

For your second equation, \"3/10 C is shaded, so 3 units + 3 parts = 3/10 C\", it doesn't look correct to me. Looks more like 3 parts = 3/10C. The 3 units is not part of circle Z.

I can't seem to get a logical answer. Can help reconfirm if there's anything wrong with this sum?

Thanks![/quote]Hi

Oops... I was in a rush and I forgot to add the \"modified\" diagram - I have swapped Y and Z in the original diagram.

With Y and Z as is, we will end up with an illogical relation between 1 unit and 1 part.

cheers.

[Andrew Lee]

Andrew Lee:

P6ians:

Hi

Q.1:
1h 30min--1.5h
1.5hx90=135km
135+25=160km
160km/(4-1.5)=64km/h
64x4h=256km(answer)
Q.2:(in this question,total remains the same)
Before. After
D:B. D:B
4:5.|9u 10:17|27u
12:15|27u
12u-10u=2u
2u----(100/2)
15u---$375(answer)

Sorry, for Q2, the question asking how many notes? not how much... :oops:

Hi MathizzFun, can you show me your \"modifed digram\"? tks

[MathIzzzFun]

Andrew Lee:

MathIzzzFun:

[quote=\"Andrew Lee\"]

Hello, can help me solve this headpain sch question ?
http://i53.tinypic.com/2s6sjmp.jpg\">

Hi

You can use MD as done for a similar question - http://www.flickr.com/photos/62167097@N02/5679183503/in/photostream

or we can also use \"units\" & \"parts\" ie simultaneous equations/algebra...

X = 6 units
Y = 9 parts
Shaded = 1/2 X + 1/3 Y = 3 units + 3 parts

X+Y = 3/4 C, C = 4/3 (X+Y) = 8 units + 12 parts ...(eq 1)

3/10 C is shaded, so 3 units + 3 parts = 3/10 C
C = 10 units + 10 parts = 8 units + 12 parts (from eq 1)
so 1 unit = 1 part

X : Y = 6 units : 9 parts = 6 units : 9 units = 2 : 3

cheers.

Thank you so much :imcool:[/quote]u r welcome.. pls note that the solution was worked with Y & Z swapped in the diagram, otherwise it is not possible to get a sensible answer - can you the diagram to confirm that Y & Z is as indicated or it should be swapped?

cheers.

[MathIzzzFun]

Andrew Lee:

Andrew Lee:

[quote=\"P6ians\"]Hi

Q.1:
1h 30min--1.5h
1.5hx90=135km
135+25=160km
160km/(4-1.5)=64km/h
64x4h=256km(answer)
Q.2:(in this question,total remains the same)
Before. After
D:B. D:B
4:5.|9u 10:17|27u
12:15|27u
12u-10u=2u
2u----(100/2)
15u---$375(answer)

Sorry, for Q2, the question asking how many notes? not how much... :oops:

Hi MathizzFun, can you show me your \"modifed digram\"? tks[/quote]http://i52.tinypic.com/2mmcu4i.jpg\">

cheers.

[MathIzzzFun]

ozora:

Need help.Jim has 3 cubicle tanks of different sizes. The length of Tank A is twice the length of Tank B and the length of Tank B is twice the length of Tank C. Jim filled 20% of Tank A,35%of Tank B and 40% of Tank C with water at first. He then poured all the water from Tank B and Tank C into Tank A. What percentage of Tank A is filled with water now? Can you please solve this question?

Hi

To remove any ambiguity, \"length\" should be changed to \"volume\". Otherwise, it has to be assumed that all the tanks have same base area.

Tank A : Tank B : Tank C = 4u : 2u : 1u
water in Tank A = 0.2 x 4u = 0.8u
water in Tank B = 0.35 x 2u = 0.7u
water in Tank C = 0.4 x 1u = 0.4u
In the end, water in Tank A = 0.8u + 0.7u + 0.4u = 1.9u

1.9u /4u x 100% = 47.5% of Tank A will be filled with water.

cheers.

[MathIzzzFun]

Yvonne Tan:

Nanyang prelim 2011

Ali and Ben started jogging from Park X to Park Y at the same time. Ali jogged at an average speed of 7.5km/h while Ben jogged at an average speed of 4.5km/h. After jogging for 1.2h, Ali decided to turn back and jog towards Ben. Ali met Ben at point Z which was halfway btw Park X and Park Y. Ali and Ben then continued their jog together towards Park Y at an average speed of 4.5km/ h. What was the distance between park X and park Y.

Age problem

When Kok's age was twice as Bala's age, Mike's age was 30. When Mike's age was twice Kok's age, Bala's age was 21. Given that Mike was the oldest among the three persons, find Bala's age when Mike was 62 years old. Thanks.

Thanks.

Hi

q1. After 1.2h,
Ali's distance from X = 1.2h x 7.5km/h = 9 km
Ben's distance from X =1.2h x 4.5 km/h = 5.4 km

Ali was ahead of Ben by (9-5.4) = 3.6km

Distance ran by Ben (after Ali turned back) when they meet at Z = 4.5/(4.5+7.5) x 3.6km = 1.35 km

distance XZ = 5.4km + 1.35km = 6.75 km
So, distance between Park X and Park Y = 6.75 km x 2 = 13.5 km

q2 is same as this question which was discussed earlier - http://www.flickr.com/photos/62167097@N02/5679985631/in/photostream

cheers.

[P6ians]

Sorry for the error.I accidentally added the $ sign. :imsorry: .The answer is :

2u---($100/2)
=50 notes
1u----50/2
=25
15u---375