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Maths problem


Sosaria
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Turbocharged

Maths give me a headache now, even though I use to score well in school.

 

Very strenuous on the brain cells.

 

Physics is the law of the universe while Mathematics is the language of the universe! You cannot describe Physics without Mathematics.

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Physics is the law of the universe while Mathematics is the language of the universe! You cannot describe Physics without Mathematics.

 

and you are gainfully employed in the world's oldest profession. :D

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Hi guys. Need help!

 

There were 500 beads in 4 jars P Q R and S. If the number of beads in P was halved, the number of beads in Q was increased by 18, the number of beads in R was tripled and the number of beads in S was decreased by 24, they would all have the same nukber of beads. What was the difference in the humber of beads between P and Q at first?

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Hi guys. Need help!

 

There were 500 beads in 4 jars P Q R and S. If the number of beads in P was halved, the number of beads in Q was increased by 18, the number of beads in R was tripled and the number of beads in S was decreased by 24, they would all have the same nukber of beads. What was the difference in the humber of beads between P and Q at first?

 

Let P start out with 6 units. Then R has to start out with 1 unit, Q with 3 units minus 18 beads and S with 3 units plus 24 beads. This is so that they all end up with the same number:

 

P: 6 units -> half -> 3 units

 

Q: 3 units - 18 beads -> add 18 beads -> 3 units

 

R: 1 unit -> triple -> 3 units

 

S: 3 units + 24 beads-> subtract 24 beads - >3 units

 

The initial sum was 500 beads, so:

 

(6+3+1+3) units + 6 beads = 500 beads

 

13 units = 494 beads.

 

1 unit = 494/13 = 38 beads.

 

Hence P started with 6*38 = 228 beads and Q with 3 units - 18 beads = 3*38 -18 = 96 beads.

 

The required difference is therefore 228 - 96 = 132 beads.

 

Explanatory note: the reason I started out by saying let P have 6 units is because you halve P and triple R. The least common multiple of 2 and 3 is 6.

Edited by Turboflat4
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After seeing all these so-called "simple heuristic" solutions, I have to ask: why the f**k are our educators so bloody dumb?

 

These problems are handled much more efficiently with simple algebra. So why expect people to come up with convoluted methods to solve the same problem in the name of a fad?

 

To screw something in, we use a screwdriver, not a hammer. But our educators seem to be the sort who would teach our kids to use a hammer "just because".

 

Right now Singapore is still ranked highly in academic terms. Keep in mind that a lot of this is from past performance and traditional curricula. But if these jokers are allowed to carry on with these new-fangled fads, I see our general mathematical literacy dropping relative to the other nations.

i second that.....i taught my son using algebra when he had a problem in sec 2.....nb the teacher said wrong as he didnt use model method. Got the answer right though.....but my son was a bit angry but i told him that was how i was thought in school.......hahaha
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[laugh]

 

Enough math and math modelling experts here to start a math tuition centre... or even some "gifted" type of math training centre that can charge high fees! [laugh]

Edited by Sosaria
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[laugh]

 

Enough math and math modelling experts here to start a math tuition centre... or even some "gifted" type of math training centre that can charge high fees! [laugh]

 

Problem is even if can get the answer but different method from the one taught in school, still wrong <_<

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Problem is even if can get the answer but different method from the one taught in school, still wrong <_<

I just learned a NEW word from the teacher.

"Flute". Correct answer wrong method.

Knnb....dont know whether the teacber know or not.

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I just learned a NEW word from the teacher.

"Flute". Correct answer wrong method.

Knnb....dont know whether the teacber know or not.

 

Flute is musical instrument lah.

 

Should be fluke [scholar] Tell that teacher to go back to school [rolleyes]

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Post my son's solution here, in case this may be the more approved one from the school.. He told me there isn't a need to draw the 2nd model, which could save some precious exam time. Straight away from the 1st model, the student should be able to derived 13R=494.

I included 2nd model in for the sake of complete understanding..

 

152f43m.gif

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Hi TB4 and others: i need help to solve the question below:



There are two tanks A and B of different capacities. If A is fill at a rate of 3L/min and B is fill at a rate of 5L/min, when A is completely filled, 5 litres of water flowed out from tank B.



If A is filled at a rate of 4L/min and B at a rate of 3L/min, when A is completely filled, B is only half filled. What is the capacity of B.


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You should get these 2 equations

 

B = 5/3A - 5 ---------(1)

B = 3/4A X 2 ---------(2)

 

Eqn(1) - Eqn(2) --- >

6/4A - 5/3A-5 = 0

1/6A=5

A=30L

 

And B= 45L

 

Will dig further if there's any moe approve solution.

Edited by Ender
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Just figured this out

 

ratio of amount of water flow

1st statement A is filled at 3L/min while B is 5L/min. of the amount of water

The amount of water from A:B is 3:5. Take note this is amount of water flow during the same period of time when A is cmpletely filled, but the NOT capacity of B

 

Similarly from statement 2, by the time A is filled up, the ratio of water flowed during this period for A:B is 4:3

 

These are the two ratio derived

3:5 (take note capacity of B is -5L from 5units)

4:3 (take note capacity of B is twice of 3 units)

 

We have common A in both of the above ratio, The 3 and 4 in the above 2 ratio represent the capacity of A. So we can common multiple A. 12 is the closest common mulitple.

 

12:20 We know cap of B is 20Units - 5

12:9 We know cap of B is two times of 9units, i.e 18units

 

20units - 5 =18units (they are both cap of B)

2 units = 5

1 unit=2.5

 

B= 18units = 45L

A= 12units=30L

 

 

Edited by Ender
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Just figured this out

 

ratio of amount of water flow

1st statement A is filled at 3L/min while B is 5L/min. of the amount of water

The amount of water from A:B is 3:5. Take note this is amount of water flow during the same period of time when A is cmpletely filled, but the NOT capacity of B

 

Similarly from statement 2, by the time A is filled up, the ratio of water flowed during this period for A:B is 4:3

 

These are the two ratio derived

3:5 (take note capacity of B is -5L from 5units)

4:3 (take note capacity of B is twice of 3 units)

 

We have common A in both of the above ratio, The 3 and 4 in the above 2 ratio represent the capacity of A. So we can common multiple A. 12 is the closest common mulitple.

 

12:20 We know cap of B is 20Units - 5

12:9 We know cap of B is two times of 9units, i.e 18units

 

20units - 5 =18units (they are both cap of B)

2 units = 5

1 unit=2.5

 

B= 18units = 45L

A= 12units=30L

 

 

 

Thank you Ender. The solution has been understood clearly

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Hi TB4 and others: i need help to solve the question below:

There are two tanks A and B of different capacities. If A is fill at a rate of 3L/min and B is fill at a rate of 5L/min, when A is completely filled, 5 litres of water flowed out from tank B.

If A is filled at a rate of 4L/min and B at a rate of 3L/min, when A is completely filled, B is only half filled. What is the capacity of B.

Not sure if my method is acceptable by school

 

Say t = time taken to fill up A

 

Capacity of A = 3 x t = 3t

Capacity of B = 5t - 5

(Less 5 since 5L flow out)

 

Base on 2nd scenario:

Rate of flow = volume/time

 

Now say T = time to fill A

 

T = 3t(capacity of A)/ 4 (rate of flow)

 

Therefore,

Capacity of B/ 2 = 3 x T

 

(5t - 5)/ 2 = 3 x 3t/4

5/2 t - 5/2 = 9/4 t

t = 10

 

Since capacity of B = 5t-5

 

Capacity of B = 45 L

 

Cheers

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Thank you Ender. The solution has been understood clearly

No prob. Keep the questions coming, I'm also benefiting from these questions, as I need more of them to test on my kids.

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