Q&A - PSLE Math

[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

[Maths Monster]

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

Hi,

What I was trying to show is that if the constants for each of the variables are different, the equation cannot be simplified to a form of D1+D2+D3 and hence solve for the total.

Apologize for not making this clear in the illustration

Let use back the previous example.

Downhill: 72 km/h
Ground level: 63 km/h
Uphill: 56 km/h
The equation derived is 2/63 D1 + 2/63 D2 + 2/63 D3 =26/3
which the constant is the same for D1, D2 and D3, hence it can be simplified to D1+D2+D3 form and it can be easily solved.

To illustrate my point that there are multiply solutions if the average of the uphill and downhill is different as the ground speed, let assume that the ground speed is 64 km/h,. The equation will be 2/63 D1 + 2/63 D2 + 2/64 D3 =26/3
The constants are different and it cannot be simplified to the D1+D2+D3 form. As this is a 3 variables in a single equation, there are infinite solutions for D1, D2 and D3 and also the total. Hope this is clear. Pm me if you need further clarification.

[tianzhu]

Q1)Jack cycles to his friend's house and then returns home by the same route. He always cycles at 4 km/h when going uphill and, 12 km/h when going downhill, and 6 km/h when on level ground. If his total cycling time is 2h 20mins, what is the distance he cycles in km?


Q2) 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?

Hi

Other than using algebra, this is another way to tackle these questions.

There are three types of terrains, uphill, downhill and level ground.
The point to note is if the cyclist moves uphill in one direction, on the way back he moves downhill. So what is the average speed for this part of the road?

Use common multiple to find the average speed. Then use the formula, Distance -----Speed*Time

Hope this helps.

Best wishes

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