Math Question

[Porker]

b)

 

m = 7f

 

(m -18)/(f + 16) = 2

 

Solve the equations and get your answer

Edited April 24, 2014 by Porker

[1fast1]

  On 4/24/2014 at 11:19 AM, Porker said:

b)

 

m = 7f

 

(m -18)/(f + 16) = 2

 

Solve the equations and get your answer

 

Technically this is a pair of simultaneous equations. Teacher might quibble over the use of this technique.

 

Quite easy to do with single variable algebra.

 

Let x be the no. you're required to find (males at end).

 

So males at start = x + 18

 

And females at start = (1/7)*(x+18)

 

Females at end = (1/7)*(x+18) + 16

 

Ratio of M:F at end = 2:1, so

 

x/2 = (1/7)*(x+18) + 16

 

Cross multiply:

 

7x = 2(x+18) + 2*16*7

 

7x = 2x + 36 + 2*16*7

 

5x = 4*9 + 4*8*7

 

5x = 4*(9+56)

 

5x = 4*65

 

x = 4*13 = 52

 

The reason I arranged and factored the numbers like I did is to simplify the computation. This is a better way if they're not allowed to use a calculator.

[Rayleigh]

HI All, thanks very much

Edited April 24, 2014 by Rayleigh

[Rayleigh]

1 more question. Two different tanks A and B were filled with water. If A leaked at 10ml/hr and B leaked 5ml/h, A would still have 300ml left when B became empty. If A leak 5ml/h and B leaked 10ml/h, A would still have 750ml of water left when B was empty. How much water was there in A at first?

[1fast1]

  On 4/24/2014 at 1:51 PM, Rayleigh said:

1 more question. Two different tanks A and B were filled with water. If A leaked at 10ml/hr and B leaked 5ml/h, A would still have 300ml left when B became empty. If A leak 5ml/h and B leaked 10ml/h, A would still have 750ml of water left when B was empty. How much water was there in A at first?

Can use algebra or not? Simultaneous equations OK or not? If so problem very easy.

 

Let starting volume of tank A = x, tank B = y.

 

Scenario 1: B would drain in y/5 hours (time = amount/rate). During this time, volume leaked out of A = 10*y/5 = 2y. Remaining volume in A = x - 2y.

 

So x - 2y = 300 ---equation 1

 

Scenario 2: B would drain in y/10 hours. During this time, volume leaked out of A = 5*y/10 = (1/2)*y. Remaining volume in A = x - (1/2)y.

 

So x - (1/2)*y = 750 ---equation 2

 

Take equation 2 times 4

 

4x - 2y = 3000 ---equation 3

 

Subtract equation 1 from equation 3:

 

3x = 2700

 

x = 900

 

So tank A had 900ml at the start.

[1fast1]

Consider that solution Take 1. Here's Take 2. Much neater, but requires more insight.

 

Let Tank A have x ml at the start. In the first scenario, tank B is emptying at half the rate of scenario 2, so it takes twice as long to empty. Also, tank A is emptying at twice the rate of scenario 2, so twice as much volume in tank A would have leaked out in the same amount of time. In double the time, four times as much would've leaked out.

 

Put it together:

 

x - 300 = 4(x-750)

 

3x = 4*750 - 300

 

x = 4*250 - 100 = 1000 - 100 = 900.

 

So tank A would have started with 900ml.

Edited April 24, 2014 by Turboflat4

[Ender]

  On 4/24/2014 at 1:51 PM, Rayleigh said:

1 more question. Two different tanks A and B were filled with water. If A leaked at 10ml/hr and B leaked 5ml/h, A would still have 300ml left when B became empty. If A leak 5ml/h and B leaked 10ml/h, A would still have 750ml of water left when B was empty. How much water was there in A at first?

Let me try,

 

1st statement, A leaked at 2 twice the rate of B, and still have 300ml left, means the equation is

Volume of A = 2B +300 -----(1)

 

2nd statement, A leaked at half the rate of B and still have 750ml, means the equation is

Volume of A = (B/2)+750 -----(2)

 

Eqn (1) = Eqn(2) since they both represent the volume of A

 

2B+300= (B/2)+750

B=300

 

Subd B in eqn(1)

A=900.

[Darryn]

  On 2/8/2014 at 10:41 AM, Turboflat4 said:

 

This is essentially elimination, a well-known method of solving simultaneous equations. When the manipulations are carried out formally via matrix operations, it is known as Gaussian elimination or Gauss-Jordan elimination.

 

Whatever the name, it *is* algebra. In fact, even though people somehow seem to have developed the notion that Primary school students are not supposed to use algebra, a lot of the actual methods are indeed algebraic. It doesn't matter whether one uses blocks, symbols or the letter 'x' : as long as an unknown quantity is symbolically represented and manipulated to arrive at a solution that quantifies the unknown, it is algebra. I find that end run around "algebra" to be hypocritical and completely baffling, given that, often, highly inefficient and convoluted methods of thought are required to solve problems that would be much more directly solved by just casting them as algebraic equations from the start. It is completely ridiculous to teach students to use a trial and error method to solve a problem that is directly solvable with algebra.

 

In short, I think the way in which Math is being taught in our primary schools is *very* misguided and unsystematic. I just wanted to get that little rant out of my system.

I not smart enough to follow this. But I find the model method very powerful for understanding math

[Rayleigh]

Once again, thank you guys.

[Volvobrick]

  On 2/8/2014 at 10:41 AM, Turboflat4 said:

 

This is essentially elimination, a well-known method of solving simultaneous equations. When the manipulations are carried out formally via matrix operations, it is known as Gaussian elimination or Gauss-Jordan elimination.

 

Whatever the name, it *is* algebra. In fact, even though people somehow seem to have developed the notion that Primary school students are not supposed to use algebra, a lot of the actual methods are indeed algebraic. It doesn't matter whether one uses blocks, symbols or the letter 'x' : as long as an unknown quantity is symbolically represented and manipulated to arrive at a solution that quantifies the unknown, it is algebra. I find that end run around "algebra" to be hypocritical and completely baffling, given that, often, highly inefficient and convoluted methods of thought are required to solve problems that would be much more directly solved by just casting them as algebraic equations from the start. It is completely ridiculous to teach students to use a trial and error method to solve a problem that is directly solvable with algebra.

 

In short, I think the way in which Math is being taught in our primary schools is *very* misguided and unsystematic. I just wanted to get that little rant out of my system.

 

Exactly my views too! The current system makes our kids solve problems using certain "modeling" which can be applied only in certain situations, instead of teaching proper fundamental mathematics that has universal applications.

 

And most of the time, maths (and science) problems test more the student's English proficiency rather than maths/science.

[Ender]

  On 4/24/2014 at 3:23 PM, Ender said:

Let me try,

 

1st statement, A leaked at 2 twice the rate of B, and still have 300ml left, means the equation is

Volume of A = 2B +300 -----(1)

 

2nd statement, A leaked at half the rate of B and still have 750ml, means the equation is

Volume of A = (B/2)+750 -----(2)

 

Eqn (1) = Eqn(2) since they both represent the volume of A

 

2B+300= (B/2)+750

B=300

 

Subd B in eqn(1)

A=900.

Took sometime to do it in modelling way.

 

The Volume of A can be represented by either the two model.

 

From the above model, we can quickly determine that b/2=150, subsequently A will be 900.

Edited April 25, 2014 by Ender

[1fast1]

  On 4/25/2014 at 3:12 AM, Ender said:

Took sometime to do it in modelling way.

 

The Volume of A can be represented by either the two model.

 

From the above model, we can quickly determine that b/2=150, subsequently A will be 900.

Effectively, this is my "Take 2" way. But I like the way you drew the model out, very nice.

 

I think you'll agree with me that expecting youngsters to come up with this sort of convoluted thinking while under added time pressure in an exam is just inhumane.

 

Especially when algebra gives a no-brainer, generalisable method of solving pretty much any sort of similar problem.

[Ender]

  On 4/25/2014 at 3:18 AM, Turboflat4 said:

Effectively, this is my "Take 2" way. But I like the way you drew the model out, very nice.

 

Use powerpoint, of course nice..

 

do agree under exam stress, it's not easy to some out with this model. My immediate solution was the simul eqn. But teacher expect the student to use the the modelling way. It's looks simple once is 'already' done by someone/

[Ahgong]

shit~!

looks like i am gonna be having a hard time teaching my kid when schooling starts proper~! (facepalm)

[Volvobrick]

  On 4/25/2014 at 3:37 AM, Ahgong said:

shit~!

looks like i am gonna be having a hard time teaching my kid when schooling starts proper~! (facepalm)

 

Welcome to the club!

 

P1, P2 still OK, but P3...... :angry:

[L_club23]

Lolz MCF become Math Club Forum...

[Porker]

Can some maths genius find the area beneath the graph for SGD fluctuation against USD for the past 10 years?

[Ender]

  On 4/24/2014 at 11:10 AM, Rayleigh said:

Hi guys, need some help with a P6 math.

 

The ratio of number of male to number of female in the hall was 5:3. The number of male to number of female in auditorium was 7:1. There were twice as many people in the hall as in the auditorium. There determine the followings:

 

a. What was the ratio of the number of female in the hall to the number of male in the auditorium?

 

b. When 18 male left the auditorium and 16 female went into the auditorium, the ratio of the number of male to the number of female in the auditorium became 2:1. How many male were there in the auditorium in the end?

 

I only manage to solve a) and i hope it is correct. I'm stuck at b).

 

 

For b) please help

 

Here's the modelling method, it's more time consuming to visualise and draw the model. Algebra suggested by bro earlier is simpler and faster..

 

Therefore in the end New Male = 70-18=52 or (F+16)X2

Edited April 25, 2014 by Ender