Q&A - PSLE Math

[Maths Monster]

Vanilla Cake:

Just to share an interesting 5-mark speed question from 3rd Annual Mathlympics - 2010 Prelim round Q28 (Not in exact words). It's not a typical speed question (in my humble opinion) and do you have any alternative solutions to share?


Downhill: 72 km/h
Ground level: 63 km/h
Uphill: 56 km/h
Car travels from Town A to Town B in 4 hours. It returns to Town A using the same route in 4 h 40 min. What is the distance between Town A and Town B?

D1 - Uphill, D2 - Downhill, D3 - Ground level
Time from Town A to Town B --> D1/56 + D2/72 + D3/63

D3 - Ground level, D2 - Uphill, D1 - Downhill
Time from Town B to Town A --> D3/63 + D2/56 + D1/72

Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/63 + 1/63)
= 2/63 D1 + 2/63 D2 + 2/63 D3
= 2/63 D (Distance D = D1+D2+D3)

2/63 D -> 26/3
D -> 26/3x63/2 = 273 km

Distance between Town A and Town B = 273 km

A similar http://psle2010a.blogspot.com/2010/08/speed-p6.html posted on 27 Aug 2010 can be found in Uncle Observer's blog.

To share my thoughts on this.... this question is only applicable if the average speed of the uphill and downhill is equal to the ground level. In this case:

(1/56 + 1/72) / 2 = 1/63

Or in the other question:

(1/12 + 1/4) / 2 = 1/6

Otherwise, there will be multiply answers.

Note to calculate average speed. Example a car travels at 10 km/h from town A to B and takes 20 km/h to travel back from town B to A. What is the average speed?

Average speed = 1 / ((1/10 + 1/20) / 2) = 13 1/3 km/h

It is not = (10 + 20) / 2 = 15 km/h........this is a common mistake student make!

[tianzhu]

Maths Monster:

Otherwise, there will be multiply answers.

Hi

For the benefits of members,please help to provide more details.

Best wishes

[kancheongmum]

Hi all thank you for the solutions on the shook hand question. I have come across something interesting while teaching my ds. I felt that I have to share but I hope it will not cause any confusion. Tianzhu, Dharam and all the maths guru please contribute.


Tony had 15%more 50cts coins than 20cts coins. If there were 6 more 50cts coins than 20cts coins, how much did Tony have?

My solution:
100% — 20cts coins
115% — 50cts coins
15% ---- 6 coins
1% — 6/15
100% ---- 6/15 x 100 — 40 coins (no. of 20cts coins)
total money Tony have — (40 x 20cts) + (46 x 50cts) — $31.00

My ds did the above and was marked wrong. This is my understanding of the question.

The school teacher solution which I think is the answer key. This question is from My pals Maths Test book:

100% - 15% — 85%
85%/2 — 42.5% (20cts coins)
42.5% + 15% — 57.5% (50cts coins)
15% — 6 coins
42.5% — 6/15 x 42.5 — 17 20cts coins
57.5% — 6/15 x 57.5 — 23 50cts coins
total money Tony have — (17 x 20cts) + (23 x 50cts) — $14.90

Very confuse now please help. Thanks

[Maths Monster]

tianzhu:

Maths Monster:

Otherwise, there will be multiply answers.

Hi

For the benefits of members,please help to provide more details.

Best wishes

Sure, if we use the same question but now we change the ground speed to say 64km/h instead of 63km/h.

The equation will be:

Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/64 + 1/64)
= 2/63 D1 + 2/63 D2 + 2/64 D3
= 2/63 (D1 + D2) + 2/64 D3 = 26/3

This is one equation with 3 variables that means there are infinitely many solutions to this equation.

[Maths Monster]

kancheongmum:

Hi all thank you for the solutions on the shook hand question. I have come across something interesting while teaching my ds. I felt that I have to share but I hope it will not cause any confusion. Tianzhu, Dharam and all the maths guru please contribute.


Tony had 15%more 50cts coins than 20cts coins. If there were 6 more 50cts coins than 20cts coins, how much did Tony have?

My solution:
100% --- 20cts coins
115% --- 50cts coins
15% ---- 6 coins
1% --- 6/15
100% ---- 6/15 x 100 --- 40 coins (no. of 20cts coins)
total money Tony have --- (40 x 20cts) + (46 x 50cts) --- $31.00

My ds did the above and was marked wrong. This is my understanding of the question.

The school teacher solution which I think is the answer key. This question is from My pals Maths Test book:

100% - 15% --- 85%
85%/2 --- 42.5% (20cts coins)
42.5% + 15% --- 57.5% (50cts coins)
15% --- 6 coins
42.5% --- 6/15 x 42.5 --- 17 20cts coins
57.5% --- 6/15 x 57.5 --- 23 50cts coins
total money Tony have --- (17 x 20cts) + (23 x 50cts) --- $14.90

Very confuse now please help. Thanks

I would agree with the school teacher solution if the question is phased as \"Tony had more 50cts coins than 20cts coins and the different between 50cts coins and 20cts coins is 15% of the total number of coins.\", or something similar.

Perhaps others can comment.

[teachingmum]

The teacher’s solution is incorrect. Since the no of 50 cents is compared to the no of 20 cents. the no of 20 cents is taken as 100%.

[tianzhu]

Maths Monster:

Sure, if we use the same question but now we change the ground speed to say 64km/h instead of 63km/h.

The equation will be:

Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/64 + 1/64)
= 2/63 D1 + 2/63 D2 + 2/64 D3
= 2/63 (D1 + D2) + 2/64 D3 = 26/3

This is one equation with 3 variables that means there are infinitely many solutions to this equation.

Hi Maths Monster

Good Morning.

Thank you for your reply.

The point to note is that the question asks for the sum of D1+D2+D3 which the distance between the two towns.

Even though there may be multiple answers for D1, D2 and D3 individually, they add up to give the same figure which is 273.

Best wishes

[tianzhu]

kancheongmum:

Tony had 15%more 50cts coins than 20cts coins. If there were 6 more 50cts coins than 20cts coins, how much did Tony have?

Hi kancheongmum

Good Morning.

Your answer is correct.

You may wish to check with the teacher again.Or, you may write to the publisher(MC) for clarification.

Best wishes

[Maths Monster]

tianzhu:

Maths Monster:

Sure, if we use the same question but now we change the ground speed to say 64km/h instead of 63km/h.

The equation will be:

Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/64 + 1/64)
= 2/63 D1 + 2/63 D2 + 2/64 D3
= 2/63 (D1 + D2) + 2/64 D3 = 26/3

This is one equation with 3 variables that means there are infinitely many solutions to this equation.

Hi Maths Monster

Good Morning.

Thank you for your reply.

The point to note is that the question asks for the sum of D1+D2+D3 which the distance between the two towns.

Even though there may be multiple answers for D1, D2 and D3 individually, they add up to give the same figure which is 273.

Best wishes

Hi,

No. The sum D1 + D2 + D3 will be different if the average speed of the uphill and downhill is not equal to ground level speed. For illustration purpose, say D1 = D2 and the equation is

D1 + D2 + 2D3 = 10

D1 or D2 = 0, D3 = 5 and Total = 5
D1 or D2 = 1, D3 = 4 and Total = 6
D1 or D2 = 2, D3 = 3 and Total = 7
D1 or D2 = 3, D3 = 2 and Total = 8
and so on....

There are multiple answers for D1, D2 and D3 and they add up to give the different answers.

D1 + D2 + D3 will give one answer if the average of the uphill and downhill is equal to the ground level speed.

[tianzhu]

Maths Monster:

Hi,

No. The sum D1 + D2 + D3 will be different if the average speed of the uphill and downhill is not equal to ground level speed. For illustration purpose, say D1 = D2 and the equation is

D1 + D2 + 2D3 = 10

D1 or D2 = 0, D3 = 5 and Total = 5
D1 or D2 = 1, D3 = 4 and Total = 6
D1 or D2 = 2, D3 = 3 and Total = 7
D1 or D2 = 3, D3 = 2 and Total = 8
and so on....

There are multiple answers for D1, D2 and D3 and they add up to give the different answers.

D1 + D2 + D3 will give one answer if the average of the uphill and downhill is equal to the ground level speed.

Hi

D1 + D2 + 2D3 = 10

Please explain how you arrive at D1=D2(not indicated in question) and 2D3.

(D1 - Uphill, D2 - Downhill, D3 - Ground level)

Best wishes