leesf:

Hi, appreciate if you can help on this question from Kong Hwa practice paper.


9 fewer pupils in the lower primary than upper primary signed up for the mass run this term. 2/3 of those registered were boys. 3/7 of the lower primary pupils were girls. 1/4 of the pupils in the upper primary were girls. How many boys took part in the mass run?

How to draw model on this question?

Thanks

let number of lower pupils be 7u

Lower : [7u]
Girls to Boys : [3u][4u]

Upper: [7u+9]
Girls to Boys: [1/4(7u+9)][3/4(7u+9)]

Total Boys = 4u + 3/4(7u+9)
Total Girls = 3u + 1/4(7u+9)

If 2/3 are boys, we can write the equation:
Number of boys = 2x Number of girls
4u + 3/4(7u+9) = 2 x (3u + 1/4(7u+9))
4u + 21/4u + 27/4 = 6u + 7/2u + 9/2
16u + 21u + 27 = 24u + 14u +18
u = 9

Number of boys = 4u + 3/4(7u+9) = 4x9 + 3/4(7x9+9) = 90

There could be a easier method, will spend some time to think about this.

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:

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.

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.

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.

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!

kancheongmum:

Hi tianzhu

I remember you have posted the solution for this question before but I can't find it,pls help tks
Harry and Sally were at a party. Harry shook hands with 4 times as many boys as girls while Sally shook hands with 5 times as many boys as girls. How many boys and how many girls were at the party? (ans 25 boys, 6 girls)

Let u be the number of girls Harry shook hand with.
Harry
Girls [u]
Boys [4u]

Sally
Girls [u-1]
Boys [5 (u-1)] or [5u-5]

The number of boys that Harry shook hand with is 1 less than the boys that Sally shook hand with. We can write:

4u+1=5u-5
u=6

Therefore, there are 6 girls and 4u+1 =25 boys at the party.

tianzhu:

Hi


Another question from VC’s post(3rd Annual Mathlympics), you may wish to solve it using heuristics learnt in primary maths.

A pest inspector killed 15 spiders and cockroaches. The number of the spiders' legs is 36 more than the number of cockroaches' legs. Find the number of cockroaches killed.

Best wishes

Let number of spiders = u
Number of spiders' legs = 8u

Number of cockroaches = 15 - u
Number of cockroaches' legs = 6(15 - u)

If the number of the spiders' legs is 36 more than the number of cockroaches' legs, we can write the equation:

8u - 6(15 - u) = 36
8u - 90 + 6u = 36
14u = 126
u = 9

Therefore, number of spiders = 9 and number of cockroaches = 6

Check:
9 spiders have = 9 x 8 = 72 legs
6 cockroaches have = 6 x 6 = 36 legs
Therefore the number of the spiders' legs is 36 more than the number of cockroaches' legs

tianzhu:

Hi


An interesting question from VC’s post(3rd Annual Mathlympics) regarding diagonal of square.You may wish to try to answer it using knowledge gained from Primary Maths

The diagonal of a larger square is 2 times the diagonal of a small square. Find the percentage of the small square to the bigger square.

Here’s another one.

The inner diagonal length of the square was 38 cm. Find the area of square.

Best wishes

1) Small square is made up of two triangles of base u and height u/2.
Area of small square = 2 x 1/2 x u x u/2 = 1/2 x u^2

Large square is made up of two triangles of base 2u and height u.
Area of small square = 2 x 1/2 x 2u x u = 2 x u^2

Therefore, the percentage of the small square to the bigger square
= 1/2 x u^2 / (2 x u^2) x 100% = 25%

2) The inner diagonal length of the square was 38 cm.
This square is made up of two triangles of base 38cm and height 19cm.

Therefore, area of square = 2 x 1/2 x 38 x 19 = 722 cm2

Brenda10:

Hi all


Need help in the following question.

Elise and Joyce baked some cookies. If Elise baked another 4 more cookies, she would have baked 5/6 as many cookies as Joyce. If Joyce baked 4 more cookies, Elise would have baked 9/13 as many cookies as Joyce. How many cookies did they bake altogether?

Thank you.

If Elise baked another 4 more cookies, she would have baked 5/6 as many cookies as Joyce:
E [5u]
J [6u]

At start:
E [5u - 4]
J [6u]

If Joyce baked 4 more cookies:
E [5u - 4]
J [6u + 4]

Elise would have baked 9/13 as many cookies as Joyce, we can write the equation as:
(5u - 4) x 13 = (6u + 4) x 9
65u - 52 = 54u + 36
11u = 88
u = 8

At start, they have (5u - 4) + 6u = 5x8 - 4 + 6x8 = 84