Q&A - PSLE Math

[MathIzzzFun]

Chan09:

Hi Tianzhu,

Apologies. Yes shaded area is 81cm2? Rest of question is correct.

Can u advise how to solve, thank you.

Hi
unshaded square : unshaded triangle --> 4u : u
square --> 4u + 81
triangle --> u + 81
square = 3 x triangle
so, 4u + 81 = 3u + 3 x 81
1u = 2 x 81
square --> 4u + 81 = 4x2x81 + 81 = 9 x 81
each side = 3 x 9 = 27

cheers.

[tianzhu]

Chan09:

Hi Tianzhu,

Apologies. Yes shaded area is 81cm2? Rest of question is correct.

Can u advise how to solve, thank you.

Hi

Good Morning.

I hope I’ve drawn the diagram correctly.

Remember the “age question” which appears regularly in primary school maths, you may apply the same concept to solve your question.

Carry out a simple demonstration at home with your kid by cutting out one square and one triangle to show that the difference in total areas between the square and the triangle is the same as the difference in unshaded areas between the square and the triangle.

Hope this helps.

Best wishes

http://farm8.staticflickr.com/7081/7040442625_6f64fc677d_z.jpg\">

[tianzhu]

MathIzzzFun:

Hi

using speed & time ratio...

Time taken for whole journey at V km/h : 1.2V km/h = 6:5 = 48u : 40u
8u --> 80 min, 1u --> 10 min
30 min --> 3u
Time taken for 2nd part of journey at V km/h : 1.25V km/h = 5 : 4 = 15u : 12u
So, time taken for 275km --> 48u - 15u = 33u
Total time taken for whole journey at V km/h --> 48 u
48 /33 x 275 = 400km --> total distance between X and Y = 400 km

cheers.

Hi

Good Morning.

Just curious, could you help to explain why you used 48u: 40u and 15u: 12u in your solution.

Why not using other equivalent ratios like 60:50, 25:20 or others?

I observe that using different ratios will lead to different answers.

Best wishes

[MathIzzzFun]

tianzhu:

MathIzzzFun:

Hi

using speed & time ratio...

Time taken for whole journey at V km/h : 1.2V km/h = 6:5 = 48u : 40u
8u --> 80 min, 1u --> 10 min
30 min --> 3u
Time taken for 2nd part of journey at V km/h : 1.25V km/h = 5 : 4 = 15u : 12u
So, time taken for 275km --> 48u - 15u = 33u
Total time taken for whole journey at V km/h --> 48 u
48 /33 x 275 = 400km --> total distance between X and Y = 400 km

cheers.

Hi

Good Morning.

Just curious, could you help to explain why you used 48u: 40u and 15u: 12u in your solution.

Why not using other equivalent ratios like 60:50, 25:20 or others?

I observe that using different ratios will lead to different answers.

Best wishes

Hi
48u & 40u is selected because the difference in time is 80 min, so I want to get a whole number.
The underlying concept is time taken to travel the whole journey at V km/h is the same for both scenarios ie in this case, 48u.
So we take reference from this total time.
In the first scenario, we get 1u --> 10 min
In the second scenario, I am comparing the bus travelling at V km/h for whole journey (A) vs the bus travelling V km/h for 275 km and then 1.25 V km/h for rest of journey (B). The time difference in this scenario will come in the 2nd part of the journey, which is 30 min --> this translate to 3u using the same 48u time taken at V km/h.
Since for the 2nd part of journey, the time ratio is 5:4, I have to get an equivalent ratio with a difference of 3 ie 15u : 12u. This means that for the second scenario, time taken for (A) will be 33u + 15u, and for (B) is 33u + 12u.

It would easier see with a diagram - I will post later.

cheers.

[tianzhu]

Hi MathIzzzFun


Thank you for the clarification.

Best wishes

[MathIzzzFun]

tianzhu:

Hi MathIzzzFun


Thank you for the clarification.

Best wishes

Hi tianzhu,

u r welcome

http://i44.tinypic.com/2d6p6ba.jpg\">

have a great day !

cheers.

[YumYum]

Dear Tianzhu and MathIzzzun,


Thank you for yr explanation. Will pass it to my ds to digest.

[lousymum]

pls help


kumar had 50% more bookmarks than leon.
max had 75% as many bookmarks as kumar.
kumar and leon gave max a number of bookmarks in the ratio 3:1.
as a result, maz had twice as many bookmakrs as before, and leon had 16 bookmarks more than kumar.
how many bookmarks did kumar give to max?

tia

[Winx5015]

Need help for the following questions.


(1). Alex left Town P at 7.30am and travelled towards Town Q. Jia Hao left Town Q at 8am and travelled towards Town P along the same road. Jia Hao’s average speed was 10 1/2 km/h faster than Alex’s average speed and he reached Town P at 12 noon while Alex was still 6km from Town Q.
(a). What was the distance between the 2 towns?
(b). At what time did Alex reach Town Q?

(2). Candle A and Candle B are of the same length. Candle A,which is broader, can burn for 5h while Candle B, the thinner candle, can burn for 4h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long left as Candle B?

Thanks!

[MathIzzzFun]

Winx5015:

Need help for the following questions.


(1). Alex left Town P at 7.30am and travelled towards Town Q. Jia Hao left Town Q at 8am and travelled towards Town P along the same road. Jia Hao's average speed was 10 1/2 km/h faster than Alex's average speed and he reached Town P at 12 noon while Alex was still 6km from Town Q.
(a). What was the distance between the 2 towns?
(b). At what time did Alex reach Town Q?

(2). Candle A and Candle B are of the same length. Candle A,which is broader, can burn for 5h while Candle B, the thinner candle, can burn for 4h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long left as Candle B?

Thanks!

Hi

Q2 .. I think this was discussed some time ago.. this is a speed question in disguise

In the same period of time, the length burned off Candle A to length burned off Candle B --> 4 : 5

http://www.flickr.com/photos/62167097@N02/6895397206/



cheers.