Q&A - PSLE Math

[TT2]

Hi, how to do this qn?


Given that p:(q+r)=1:3, r:(q+s)=1:2 and p:s=2:5.
(a) find p:q:r:s
(b) find (r+q):(p+s)

Thank you

[TT2]

Hi, need help on another speed qn…


Ahmad started traveling from town P towards town Q at 7am. After traveling for some time, he passed Steve who was traveling at an average speed of 80 km/h on the opposite direction. After traveling for another 2 hrs, Ahmad reached town Q while Steve was still 40km away from town P.

(a) if Ahmad reached town Q at 1pm, what was his average speed?
(b) if Steve started his journey from town Q, at what time did he leave town Q?

Thank you.

[tianzhu]

TT2:

Hi, how to do this qn?


Given that p:(q+r)=1:3, r:(q+s)=1:2 and p:s=2:5.
(a) find p:q:r:s
(b) find (r+q):(p+s)

Thank you

Hi

Please refer to Vanilla Cake’s post @ page 432.

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=4310

Best wishes

[TT2]

tianzhu:

TT2:

Hi, how to do this qn?


Given that p:(q+r)=1:3, r:(q+s)=1:2 and p:s=2:5.
(a) find p:q:r:s
(b) find (r+q):(p+s)

Thank you

Hi

Please refer to Vanilla Cake’s post @ page 432.

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=4310

Best wishes

Got it. Thank you, tianzhu

[MathIzzzFun]

TT2:

Hi, need help on another speed qn...


Ahmad started traveling from town P towards town Q at 7am. After traveling for some time, he passed Steve who was traveling at an average speed of 80 km/h on the opposite direction. After traveling for another 2 hrs, Ahmad reached town Q while Steve was still 40km away from town P.

(a) if Ahmad reached town Q at 1pm, what was his average speed?
(b) if Steve started his journey from town Q, at what time did he leave town Q?

Thank you.

Hi

Draw a diagram to show the journey - name the meeting point X.
From X, after Steve travelled for 2h, he was 40 km from P.
So, distance PX = 40 km + 2 x 80 km/h = 200 km

Ahmad reached Town Q at 1pm, so total journey time = 6h, time taken to travel from Town P to meeting point X = 4h.

So, Ahmad's speed = 200km/4 h = 50 km/h

Steve met Ahmad at 7am + 4h = 11am
Time taken for Steve to travel from Town Q to meeting point X
= 50/80 x 2h = 1h 15min

He left Town Q at (11am - 1h 15min) = 9.45 am

cheers.

cheers.

[pensiveowl]

MathIzzzFun:

pensiveowl:

Could someone pls help with this one as well - thanks


The ratio of the number of green marbles to the red marbles in a box is 6 : 5. After 60 green marbles and 60 red marbles were removed from the box. the ratio of the number of green marbles to red marbles became 9 : 7. How many red marbles were there in the box at first?

Hi

the two questions are similar, but for q2, the concept of \"same gap\" can be applied to solve.

q1. Mr Ho and Mrs Lee took part in a chess competition. Mr Ho played against 1/3 as many men as women. Mrs Lee played against 2/5 as many men as women. How many women took part in the chess competition?

Mr Ho played 1/3 as many men as women --> men : women = 1u : 3u
..total number of men = 1u + 1

Mrs Ho played 2/5 as many men as women
--> men : women = 1u + 1 : 3u -1 = 2 : 5

cross-multiply: 5u+5 = 6u -2, 1u = 7

Number of men = 1u + 1 = 8
Number of women = 3u = 21

q2. The ratio of the number of green marbles to the red marbles in a box is 6 : 5. After 60 green marbles and 60 red marbles were removed from the box. the ratio of the number of green marbles to red marbles became 9 : 7. How many red marbles were there in the box at first?

at first, green : red = 6u : 5u
in the end, green : red = 6u - 60 : 5u - 60 = 9 : 7
so, 42u - 420 = 45u - 540, 3u = 120, 1u = 40

At first, red marbles = 5u = 200

using \"same gap/difference\":

in the end, green : red = 9u : 7u (gap 2u)
at first, green : red = 6 : 5 = 12u : 10 u (gap 2u)
so 3u = 60, 1u = 20

Number of red marbles at first = 10u = 200

cheers.

Thanks MathIzzzFun!

[TT2]

MathIzzzFun:

TT2:

Hi, need help on another speed qn...


Ahmad started traveling from town P towards town Q at 7am. After traveling for some time, he passed Steve who was traveling at an average speed of 80 km/h on the opposite direction. After traveling for another 2 hrs, Ahmad reached town Q while Steve was still 40km away from town P.

(a) if Ahmad reached town Q at 1pm, what was his average speed?
(b) if Steve started his journey from town Q, at what time did he leave town Q?

Thank you.

Hi

Draw a diagram to show the journey - name the meeting point X.
From X, after Steve travelled for 2h, he was 40 km from P.
So, distance PX = 40 km + 2 x 80 km/h = 200 km

Ahmad reached Town Q at 1pm, so total journey time = 6h, time taken to travel from Town P to meeting point X = 4h.

So, Ahmad's speed = 200km/4 h = 50 km/h

Steve met Ahmad at 7am + 4h = 11am
Time taken for Steve to travel from Town Q to meeting point X
= 50/80 x 2h = 1h 15min

He left Town Q at (11am - 1h 15min) = 9.45 am

cheers.

cheers.

:thankyou: MathIzzzfun

[kavisuresh]

Pls help to solve.

1. There are ___ hundredths in 186.42

2. Gerald had $20 more than Tommy at first. After Gerald gave Tommy $60, Gerald had 1/3 as much money as Tommy. Find amt of money Gerald had at first ?

3. Krishnan spent $84 on some plates and 1/3 of remaining money on some cups. If he had 1/5 of original sum of money left, how much did she have at first ?

4. http://i56.tinypic.com/104hxrp.jpg\">

Thanks a lot.

[zsqchx]

Please help to solve this question. Thank you


At Carpark P, the number of lorries to that of vans in the ration 3:7. At Carpark Q, the number of lorries to that of vans was in the ratio 8: 9. When 40% more lorries from an industrial park entered Carpark P. and 20% of the vans at Carpark Q moved to Carpark P. there were 76 fewer lorries at Carpark p than at Carpark Q. How many vehicles were there altogether at the carparks finally?

[smileplease]

This is the 2009 PSLE Math Q, the Jim and Ken’s Q.


My daughter’s school is clearing the PSLE 2006-2010 book NOW… but still do not know how to do this Q. Please HELP.

Jim bought some chocolates and gave half of them to Ken. Ken bought some sweets and gave half of them to Jim.

Jim ate 12 sweets and Ken ate 18 chocolates. After that, the number of sweet and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.

How many sweets did Ken buy?


thanks.