Tutor MathsGuru: Ask me for your burning Maths questions!

[Almighty]

Dharma:

Almighty:

[quote=\"Dharma\"]
Hi Almighty,

Frankly, I find the question not clear enough. You may have gotten it from an assessment book or a school exam paper. Not to worry to much, in PSLE, the questions are checked very carefully by a panel and there is a very minimal chance of ambiguity. My advice is for maths, just use the direct interpretation. Normally it is not the intention of the setter to test your language ability.

[/quote]Seriously hope so Sir!! Thankyou for your reply.

[super star]

thank you very much

almighty is correct with the first part of the problem

blessedami:

super star:

pls hlep me with this problem
Aaron and kumar took part in a 24km marathon.Aaron ran half the distance at a speed of 10km/h and jogged the rest of the way at a speed of 8 km/h.kumar ran half of his total time at 12 km/h and jogged the rest of the time at 6 km/h. If they started the marathon at 8 am ,at what time would each of them finish?

Since Aaron ran half the distance at 10km/h and the other half at 8 km/h;

12km÷10km/h= 1 1/5 h= 1 h 12 min
12km÷8km/h=11/2 h = 1 h 30 min
Total time taken by Aaron is 2 h 50 min
He will finish at 10. 50 am

For Kumar, he ran half of his time at 12 km/h and the other half at 6 km/h
1/2 t x 12 km/h + 1/2 t x 6 km/h = 24 km
1/2 t x 18 km/ h = 24 km
1/2 t = 24 km ÷ 18 km/h = 1 h 20 min
t= 1h 20 min x 2 = 2 h 40 m
So, he will finish at 10. 40 am

[blessedami]

Almighty:

blessedami:

[quote=\"super star\"]pls hlep me with this problem

Aaron and kumar took part in a 24km marathon.Aaron ran half the distance at a speed of 10km/h and jogged the rest of the way at a speed of 8 km/h.kumar ran half of his total time at 12 km/h and jogged the rest of the time at 6 km/h. If they started the marathon at 8 am ,at what time would each of them finish?

Since Aaron ran half the distance at 10km/h and the other half at 8 km/h;

12km÷10km/h= 1 1/5 h= 1 h 12 min
12km÷8km/h=11/2 h = 1 h 30 min
Total time taken by Aaron is 2 h 50 min
He will finish at 10. 50 am

For Kumar, he ran half of his time at 12 km/h and the other half at 6 km/h
1/2 t x 12 km/h + 1/2 t x 6 km/h = 24 km
1/2 t x 18 km/ h = 24 km
1/2 t = 24 km ÷ 18 km/h = 1 h 20 min
t= 1h 20 min x 2 = 2 h 40 m
So, he will finish at 10. 40 am

Hi blessedami,
Please see your answer marked in red by me.I think there is a calcualtion error.Accordingly,
Total time taken = 1h 30 min + 1 h 12 mins = 2h 42 min
He will finfish at 10.42 am.[/quote]Thanks Almighty,
Was a calculation error.

[blessedami]

Tang:

blessedami:

[quote=\"starlight1968sg\"]
I don't understand that given solution too.

Hi,

My DS worked it out like this:

Let the boys in the order of lightest to heaviest be v,w,x,y,z

The weight of the two lightest boys, v and w is 105

So,
v+w=105 )
v+x=113 )

2 v + w + x = 105 + 113 = 218
2 v = 218 - 118 = 100
v = 100/2 = 50

v+y=115
v+z=116

So w = 105 - 50 = 55, x = 113 - 50 = 63, y = 115 - 50 = 65, z = 116 - 50 = 66


w+x=118 w + x = 55 + 63 = 118 OK
w+y=124 w + y = 55 + 65 = 120 Don't tally!
w+z=125 w + z = 55 + 66 = 121 Don't tally!

x+y=130 x + y = 63 + 65 = 128 Don't tally!
x+z=137 x + z = 63 + 66 = 129 Dpn't tally!

y+z=139 y + z = 65 + 66 = 131 Don't tally!

So in total there would be four v, four w, four x, four y and four z.

Total weight would be 1224kg
So, 4(v+w+x+y+z) --> 1224kg
v+w+x+y+z--> 1224kg/4= 306 kg

Since the two lightest weigh 105 kg and the two heaviest weigh 139 kg, the third heaviest or the last boy weighs, 306 kg- 139 kg- 105 kg= 62 kg[/i]

Hi,

See comments underlined.[/quote]Hi Tang, I will let my DS figure it out. He probably did not do reverse calculation.

[Yu Xuan]

beauty queen:

Mrs Sitoh has just enough flour to bake either 12p muffins or 9r pies. If she has already baked 40 muffins and 3r pies how many more muffins can she still bake?



please help

12p muffin => 9r pies
4p muffin => 3r pies

12p muffin - 4p muffin - 40 muffin = 8p muffin - 40 muffin
=8(p-5) muffin

[super star]

thankyou

Almighty:

super star:

please help me with these problems


(ii)How many cubes of side 2cm can be fitted into Box A of dimension 12cm by 5 cm by 7cm? my ans is 52...the ans key says 36.which is correct. pls explain

Hi Superstar,

No. of cubes in the length -------12/2= 6\t\t\t
\t\t\t\t\t\t
No. of cubes in breadth --------- 5/2 = 2.5 (Which means only 2 cubes can be fitted)\t\t\t\t\t\t\t\t\t\t
No. of cubes in height ---------- 7/2 = 3.5 (Which means only 3 cubes can be fitted )\t\t\t\t\t\t\t\t\t\t\t
No of cubes = 6 X 2 X 3\t\t\t\t\t\t\t\t\t
= 36.

[Publica]

Please help with these P5 maths question:


1. Two baskets, A and B each contain a different number of apples. After 1/7 of the apples in A are transferred to B and 1/5 of the total number of apples in B are transferred back to A, the two baskets will have the same number of apples. Find the ratio of the original number of apples in A to the original number of apples in B.

2. Two candles of the same height are lit at the same time. The first is consumed in 4 hours, the second in 3 hours. Assuming that each candle burns at a constant rate, how many hours after being lit was the height of the first candle twice the height of the second?

Thanks!

Oops! Sorry, please ignore this as I did not realise that I have started a new thread. I have posted to the correct thread.

[Publica]

Please help with these P5 maths question:


1. Two baskets, A and B each contain a different number of apples. After 1/7 of the apples in A are transferred to B and 1/5 of the total number of apples in B are transferred back to A, the two baskets will have the same number of apples. Find the ratio of the original number of apples in A to the original number of apples in B.

2. Two candles of the same height are lit at the same time. The first is consumed in 4 hours, the second in 3 hours. Assuming that each candle burns at a constant rate, how many hours after being lit was the height of the first candle twice the height of the second?

Thanks!

[CJM]

hi there,


I m trying to upload an image on a math question but unable to do so.

when i click the add image to post and upload the picture.
this was the message.
no decode delegate for this format.

can anyone pls advise how can i upload the image ?[/img]

[Muffins]

your system cannot recognise JPEG formats. Try to make it identify pictures as JPEG format, and maybe it will work???