Q&A - PSLE Math

[MathIzzzFun]

jarenchuatw:

Need help with this past year PSLE question http://i46.tinypic.com/2gwxggh.jpg[/IMG]

Hi

this is a \"catch-up\" question.

At first, height of water in Tank B = 1/3 x 36cm = 12cm

For tank A, height of water level increasing at
1200/600 = 2 cm/min

For tank B, height of water level increasing at
1200/1000 = 1.2 cm/min

2 cm/min -1.2 cm/min = 0.8 cm/min --> height in A increasing at 0.8cm/min more than that in B. To make up 12 cm,
time --> 12 cm ÷ 0.8 cm/min = 15 min

Level of water in both tanks will be the same after 15 min

cheers.

[MathIzzzFun]

dicoyote:

Hi, need help on this question (b) as I am lost trying to visualise it using (a) answer, thanks


Devi and Famin were at towns B and C respectively and were heading for village A. Devi started walking at a speed of 50m/min. At 10.10am, Famin started walking from Town C just as Devi walked past him and travelled at a constant speed. At 10.15am, Famin was 500m ahead of Devi.

a) Find the speed at which Famin was walking
b) If Devi reached village A 40 min later than Famin, what was the distance between town C and village A?

|-----------------|---------------------------------------|
Town B Town C Village A

In 5 min, Famin was 500m ahead of Devi
500m ÷ 5min = 100m/min --> Famin's is walking at 100m/min faster than Devi
Famin's speed = 50m/min + 100m/min = 150 m/min

Famin's speed: Devi's speed = 150 : 50 = 3 : 1
Time ratio of Famin : Devi = 1u : 3u
Difference of 2u --> 40 min, 1u --> 20 min
Distance between C & A = 20 min x 150 m/min = 3000m or 3 km.

cheers.

[charsen]

[quote="dicoyote"]Hi, need help on this question (b) as I am lost trying to visualise it using (a) answer, thanks


Devi and Famin were at towns B and C respectively and were heading for village A. Devi started walking at a speed of 50m/min. At 10.10am, Famin started walking from Town C just as Devi walked past him and travelled at a constant speed. At 10.15am, Famin was 500 ahead of Devi.

a) Find the speed at which Famin was walking
b) If Devi reached village A 40 min later of Famin, what was the distance between town C and village A?

|-----------------|---------------------------------------|

Check this out.
a) Dist walked by Devi in 5 mins —> 50x5 = 250m
Dist F walked –> 250m+500m = 750m
F’s speed 750/5 = 150 m/min

b) F reached 40 mins earlier than Devi.
Dist travelled by Devi in this 40mins –> 50x40 = 2000m
Diff in dist between F and D in 1min –> 150-50 = 100m
In 1min, F is 100m
To be 2000m ahead, F will take –> 2000/100 = 20mins
Dist bet C and A —> 150 x 20 = 3000m

[Xiao Hu]

MathIzzzFun:

Michaelia0816:

[quote=\"ahron-teo\"]Guys, I was teaching my cousin maths and I came across this speed question. I couldn't believe it, that I couldn't solve! Even the answer keys are a mystery to me! :? :? :?


Singapore Hokkien Hway Kuan P6 Math 2011 Qn 15

A car left Town A for Town B. At the same time, a lorry left Town B for Town A. The average speed of the car was 90km/h while the speed of the lorry was 70km/h. The two vehicles passed each other at a point 36km away from the mid-point between Town A and Town B.
What was the distance between Town A and Town B?

with many thanks and cheers

Solutions:
http://singaporemathproblems.com/uploads/3/0/9/8/3098498/answer__p6-008.pdf

This is not I created page, I just helping to find the answer, pls read the note of that web, the answer key is wrong!

please note that the posted answer at provided link is incorrect.
Midpoint --> 8/16 AB
So, 8/16-7/16 = 36km, 1/16--> 36km,
Total distance betweeen AB=16 x36 = 576 km

cheers.[/quote]Hi MathIzzzFun,
Agreed with you that the link's answer is not correct. But may I ask about your approach? I hope you would be kind enough to share how you come to use 8/16 AB? What is the concept behind to using the denominator 16 and then a fraction of 8/16AB?

Thanks in advance for sharing,
Xiaohu.

[Michaelia0816]

Sorry for having the wrong links posted!

[2632parent]

Hi, need help on the following question


1) Test A measures 30cm by 20cm by 24cm. It contain 12600ml of water. Tank B has a square base of length 25cm and is filled with water up to a height of 11.2cm. The water in Tank A is then poured into Tank B until the water level in both tank are the same. Water is then poured from Tank b back into Tank A such that the water level in Tank A is 1.5 times the water level in Tank B. What is the water level in Tank A?

Thank you

[MathIzzzFun]

Xiao Hu:

MathIzzzFun:

[quote=\"Michaelia0816\"]Guys, I was teaching my cousin maths and I came across this speed question. I couldn't believe it, that I couldn't solve! Even the answer keys are a mystery to me! :? :? :?


Singapore Hokkien Hway Kuan P6 Math 2011 Qn 15

A car left Town A for Town B. At the same time, a lorry left Town B for Town A. The average speed of the car was 90km/h while the speed of the lorry was 70km/h. The two vehicles passed each other at a point 36km away from the mid-point between Town A and Town B.
What was the distance between Town A and Town B?

with many thanks and cheers

Solutions:
http://singaporemathproblems.com/uploads/3/0/9/8/3098498/answer__p6-008.pdf

This is not I created page, I just helping to find the answer, pls read the note of that web, the answer key is wrong!

please note that the posted answer at provided link is incorrect.
Midpoint --> 8/16 AB
So, 8/16-7/16 = 36km, 1/16--> 36km,
Total distance betweeen AB=16 x36 = 576 km

cheers.

Hi MathIzzzFun,
Agreed with you that the link's answer is not correct. But may I ask about your approach? I hope you would be kind enough to share how you come to use 8/16 AB? What is the concept behind to using the denominator 16 and then a fraction of 8/16AB?

Thanks in advance for sharing,
Xiaohu.[/quote]using speed ratio--> car's speed : lorry's speed = 90:70 = 9u : 7u
This means that in the same time,
car travel 9u & lorry travel 7u.
when they meet,
distance travelled by car --> 9u
distance travelled by lorry --> 7u
Total distance between AB = 9u+7u=16u
Midpoint of AB = 16u / 2 = 8u
So, 9u - 8u = 36 km

cheers.

[whitety]

Hi, need help on these 2 qns.


1) George and Hamid drove at constant speeds from City A to City B. They started their journey at the same time. When George completed 3/4 of the journey, Hamid was 18km behind George. George reached City B 20min before Hamid. Find Hamid’s speed in km/h.

2) in a bag, the ratio of the number of $2 notes to the number of $10 notes was 2:5. Ten $10 notes were removed from the bag to exchange for $2 notes which were then put back into the bag. The total value of money in the bag was unchanged after the exchange. The ratio of the number of $2 notes to the number of $10 notes then became 11:5. What was the difference between the value of the $10 notes and $2 notes after the exchange?

Thanks!

[MathIzzzFun]

whitety:

Hi, need help on these 2 qns.


1) George and Hamid drove at constant speeds from City A to City B. They started their journey at the same time. When George completed 3/4 of the journey, Hamid was 18km behind George. George reached City B 20min before Hamid. Find Hamid's speed in km/h.

2) in a bag, the ratio of the number of $2 notes to the number of $10 notes was 2:5. Ten $10 notes were removed from the bag to exchange for $2 notes which were then put back into the bag. The total value of money in the bag was unchanged after the exchange. The ratio of the number of $2 notes to the number of $10 notes then became 11:5. What was the difference between the value of the $10 notes and $2 notes after the exchange?

Thanks!

1) George completed 3/4 of journey, Hamid 18km behind
So, when George completed the journey (4/4), Hamid will be 24km behind
Hamid's speed = 24km/20 x 60 km/h = 72 km/h

2) 10 x $10 = 50 x $2
At first, number of $2 notes: $10 notes --> 2u : 5u
In the end, number of $2 notes: $10 notes --> 2u+50 : 5u-10 = 11 : 5

cross multiply/equalize:
10u + 250 = 55u - 110
1u --> 8 .. can you continue from here ?

difference in value = $168 ($10 --> $300, $2--> $132)

cheers.

[Xiao Hu]

[quote="MathIzzzFun

using speed ratio–> car’s speed : lorry’s speed = 90:70 = 9u : 7u
This means that in the same time,
car travel 9u & lorry travel 7u.
when they meet,
distance travelled by car –> 9u
distance travelled by lorry –> 7u
Total distance between AB = 9u+7u=16u
Midpoint of AB = 16u / 2 = 8u
So, 9u - 8u = 36 km

cheers.[/quote]

Hi MathIzzzFun,
Perfect !! Thanks so much. I did not even begin to think about using ratio and then what you reduced them to 9u and 7u.

Have a good weekend,
Xiaohu.