Monday, February 2, 2009

More maths problems by request

I've received a couple of maths problems on my blog. While I'm flattered that you have such faith in my maths abilities, I can't promise that I'll entertain all future requests (or be able to solve them all!) since this is not a dedicated maths blog. Do click on my mathematics label to see the workings of other maths problems I've done.

But for now, these are the two I received. Thanks for your patience!

By wo wen tian:

Hi, how to solve this problem without using Algebra? At first, two shops A & B had a total of 1,040 sacks of rice. After Shop A has cleared 3/4 of its stock and Shop B has cleared 3/5 of its stock. Shop B now has 52 more sacks of rice than Shop A. How many sacks of rice does each shop have at first?


It's quite easy to solve this problem using models, you just need to work backwards. First, you draw the model for the final scenario, ie Shop B has 52 more sacks of rice than Shop A (right pic).

Then you add in the initial stock of rice. Shop A now has only 1/4 of its original stock so you need to draw in another 3 units (to make up the 3/4). Shaded parts depict stock of rice that was cleared.

Shop B now has only 2/5 of its original stock, meaning 1 unit + 52 sacks = 2/5 of original stock. To add in the original 3/5, you need to draw another 1 unit + 52 sacks (2/5) and ½ unit + 26 sacks (1/5).

Remember one of my rules for models is every unknown unit has to be equal in value, so since you have a half unit, you need to cut every other unknown unit into half (including those for Shop A - just slice horizontally across). Now, you can see (below) that you have 13 units + 52 + 52 + 26 and all that is equivalent to 1,040 sacks of rice.

13 units = 1,040 - 52 - 52 - 26 = 910
1 unit = 910 ÷ 13 = 70

Shop A is 8 units, so 8 x 70 = 560
Shop B is 5 units + 52 + 52 + 26 = 5 x 70 + 130 = 480

Answer: Shop A has 560 sacks of rice at first and Shop B has 480.


Help needed said...

There are 600 children in Team A and 30% of them are boys. There are 400 children in Team B and 60% of them are boys. After some children are transferred from Team B to Team A , 40% of the children in Team A and 60% of the children in Team B are boys. How many children are transferred from Team B to Team A?

This question is tricky - I couldn't solve it using models, I used a combination of ratio, percentage and algebra. First, we are given the number of children in each team, so we can work out how many boys and girls there were in each team originally.

600 x 30% = 180, so Team A originally had 180 boys and 420 girls (600 - 180)
400 x 60% = 240, so Team B originally had 240 boys and 160 girls (400 - 240)

Next, we use ratio. The ratio of girls to boys was:

At first: Team A - 7 : 3 Team B - 4 : 6
After : Team A - 6 : 4 Team B - 4 : 6

Notice that the ratio of girls to boys for Team B remained the same, even after the transfer. This means that the proportion of girls and boys transferred out of Team B was also 4 : 6 (or 2 : 3), ie 2 units of girls and 3 units of boys were transferred out.

Using algebra, the number of children in Team A after the transfer can be expressed as:

(Girls) 420 + 2 units = 60%
(Boys) 180 + 3 units = 40%

The lowest common multiple of 60 and 40 is 120, so convert both equations to = 120 and you can combine both equations.

2 (420 + 2 units) = 3 (180 + 3 units)
840 + 4 units = 540 + 9 units
9 units - 4 units = 840 - 540
5 units = 300

Remember a total of 5 units of children were transferred out of Team B (2 units of girls and 3 units of boys), so you don't even need to find out how many children is represented by 1 unit.

Answer: 300 children were transferred from Team B to Team A.

If anyone can solve this problem using models, do let me know.

8 comments:

eunice said...

Once again, I kow tow to you and can I say (again) " I'll be sending Sean to you for tutoring!"

monlim said...

no lah, but can i take it that you're coming back to SG??

Lilian said...

Hey Mon, I have the second question in my notebook and above the question I had written "Model?", cos couldn't do it by model. Brian and I did it by algebra, but slightly differently. I could do the first one by model but he still resorted to algebra; getting lazy after I told him he could use algebra for back-up. Now he's using it as first, not last, resort.

monlim said...

I think Brian has got algebra down pat in which case, no need to worry about model lah! I don't think Lesley-Anne would be able to do Q2.

eunice said...

Monica, not going back to Sin but the next time I do go back I definitely want to meet up with you.It's funny that we have never met!!!

monlim said...

Eunice: I know, it's so funny to "know" someone that you've never met so well, such is the internet age! Would love to meet up the next time you're back.

Great Hope said...

Hi

Another one, but this time there is transfer from Hall A to Hall B as well as from Hall B to Hall A

In Hall A, 30% of the 800 people were men. In Hall B, 40% of the 400 people were women and children. After some of the people in both halls had switched hall, 25% of the people in Hall A and 75% of those in Hall B were men. How many people were there in Hall B after the change?

Great Hope said...

Hi

Can this problem be solved with Model Drawing?

9 fewer students in the lower primary than upper primary competed in the story telling this year. 2/3 of all those who competed were boys. 3/7 of the lower primary participants were girls. 1/4 of the participants in the upper primary were girls. How many boys took part in the competition?

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