Tutor MathsGuru: Ask me for your burning Maths questions!

[starlight1968sg]

Dharma:

315 is correct.

Red : 12u
Yellow : 6u – 1p = 4u
Blue : 3u – 15 = 2u

3u – 2u = 15
1u = 15
Total no. of marbles at first = 12u + 6u + 3u = 21u =21 x 15 = 315

Thanks.
I solved using the model diagram.

Is it convincing enough that the number of Yellow marbles taken is 15, so that the ratio for Yellow to Blue remains 2:1 ?

[Dharma]

starlight1968sg:

Dharma:

315 is correct.

Red : 12u
Yellow : 6u – 1p = 4u
Blue : 3u – 15 = 2u

3u – 2u = 15
1u = 15
Total no. of marbles at first = 12u + 6u + 3u = 21u =21 x 15 = 315

Thanks.
I solved using the model diagram.

Is it convincing enough that the number of Yellow marbles taken is 15, so that the ratio for Yellow to Blue remains 2:1 ?

To maintain the ratio of yellow to blue marbles of 2 : 1 but ratio of red to yellow marbles of 3 : 1, you need to remove 2 x 15 = 30 yellow marbles.

[starlight1968sg]

Dharma:

To maintain the ratio of yellow to blue marbles of 2 : 1 but ratio of red to yellow marbles of 3 : 1, you need to remove 2 x 15 = 30 yellow marbles.

YES!

I take each block (ie blue marbles) as \"remaining plus 15\". So for yellow block, it is \"remaing plus 15\" and \"remaing plus 15\" ie remove 15*2 = 30 yellow marbles.

Since the question says \"some yellow marbles\" are removed and to maintain the ratio of Yellow to Blue at 2:1, the number of yellow marbles removed has to be 15*2 = 30.

Though somehow am not very convinced about this part. Ok, maybe am just a bit slow to understand.

[Dharma]

starlight1968sg:

Dharma:

To maintain the ratio of yellow to blue marbles of 2 : 1 but ratio of red to yellow marbles of 3 : 1, you need to remove 2 x 15 = 30 yellow marbles.

YES!

I take each block (ie blue marbles) as \"remaining plus 15\". So for yellow block, it is \"remaing plus 15\" and \"remaing plus 15\" ie remove 15*2 = 30 yellow marbles.

Since the question says \"some yellow marbles\" are removed and to maintain the ratio of Yellow to Blue at 2:1, the number of yellow marbles removed has to be 15*2 = 30.

Though somehow am not very convinced about this part. Ok, maybe am just a bit slow to understand.

I’m no expert at models but I am able to explain to u what I have done.

Actually, the number of yellow marbles removed is irrelevant for answering this question, so don’t worry too much about it.

1.\tThe initial condition is ratio of red : yellow : blue = 4 : 2 : 1
2.\tFinal condition is ratio of red : yellow : blue = 6 : 2 : 1
3.\tNo red marbles are given out, so the no. of red marbles DO NOT change.
4.\tNow u have initial ratio of 12: 6 : 3 and final ratio of 12 : 4 : 2
5.\tFrom (4) u can see for every 1 unit reduction of blue marbles, there will be 2 units of reduction in yellow marbles, based on the ratios above.

[starlight1968sg]

Dharma:

I’m no expert at models but I am able to explain to u what I have done.


1.\tThe initial condition is ratio of red : yellow : blue = 4 : 2 : 1
2.\tFinal condition is ratio of red : yellow : blue = 6 : 2 : 1
3.\tNo red marbles are given out, so the no. of red marbles DO NOT change.
4.\tNow u have initial ratio of 12: 6 : 3 and final ratio of 12 : 4 : 2
5.\tFrom (4) u can see for every 1 unit reduction of blue marbles, there will be 2 units of reduction in yellow marbles, based on the ratios above.

Thanks.
Your solution is so short and clear. Earlier, I was struggling with the model diagram.
Myself is lousy in model & ratio. So I can't expect my dd to do well too. :stupid:

[KSP]

I need help on the following questions.


1. Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles. How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red? Ans: 350


2. In a math competition, participants can obtain 4 possible award: Gold, Silver, Bronze and Participation. 3/7 of the participants obtained Gold awards, 1/4 of the them obtained Silver awards, and 1/6 of them obtained Bronze awards. Given that there were less than 100 participants taking the competition,
a) How many participants obtained the awards for Participation? Ans: 13
b) How many more participants obtained Gold awards then Bronze awards? Ans: 22

[Dharma]

starlight1968sg:

Dharma:

I’m no expert at models but I am able to explain to u what I have done.


1.\tThe initial condition is ratio of red : yellow : blue = 4 : 2 : 1
2.\tFinal condition is ratio of red : yellow : blue = 6 : 2 : 1
3.\tNo red marbles are given out, so the no. of red marbles DO NOT change.
4.\tNow u have initial ratio of 12: 6 : 3 and final ratio of 12 : 4 : 2
5.\tFrom (4) u can see for every 1 unit reduction of blue marbles, there will be 2 units of reduction in yellow marbles, based on the ratios above.

Thanks.
Your solution is so short and clear. Earlier, I was struggling with the model diagram.
Myself is lousy in model & ratio. So I can't expect my dd to do well too. :stupid:

I personally never liked models … find it time consuming and u don’t get marks for it.

Ratio ... your child needs to understand and master ...cause it is a very powerful tool to solve most PSLE problem solving qns.

[mathsguru]

P5G:

Pl help.


Paper 1 Q13.
http://www.orlesson.org/orp/09Ma/2009-Math-SA1-MGS.pdf


Worksheet answer is (4) 135 [not likely].


Please advise how to get (3) 125 as Benny's score?


TIA.

Hi P5G,

Sorry for the late reply. Here's my answer.


MathsGuru

http://www.postimage.org/image.php?v=aVjUVjJ

[KSP]

I need help on the 2 questions below.


1. The figure below is not drawn to scale. Given that the area of WXYZ is 40cm sq, find the area of the shaded part. Ans: 68cm sq

http://www.postimage.org/image.php?v=aVjVKHJ



2. The figure below is made up of two identical right-angled triangles, a small square and a big square. The parimeter of the shaded region is 50cm, and the total area of the two unshaded squares is 254.5cm sq. Find the total area of the two shaded right-angled triangles. Ans: 114.75cm sq

http://www.postimage.org/image.php?v=gxNVMm9

[mathsguru]

mmmouse:

Anyone knows how to solve this Pr. 3 Maths question?


5 erasers + 3 rulers = S$5.05
3 erasers + 6 rulers = S$3.95
2 erasers + 2 rulers = ???

Thanks!

Hi Mmmouse,

Think there's probably a typo error in the question, cos I can't seem to get nice numbers for the answers and I think for P3, they haven't learnt rounding off yet.

In any case, here's my solution. You can follow the steps. It's the same kind of steps to solve such questions.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=aVjVXar