Tutor MathsGuru: Ask me for your burning Maths questions!

[tianzhu]

Hi moderator


Please delete post.

Thanks

[clblinym]

Dear Dharma & Mathsguru,


Thank you so much for the clear explanation.
It’s great to have people like u guys to help selflessly.

Regards,
Amy

[clblinym]

Please help me solve the following Question.


There is no answer because it was one of the 30 questions in NUSHS mathematical Olympiad this year. I guess it is $37, but no confident at all.

Thanks in advance.




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

[Dharma]

clblinym:

Please help me solve the following Question.


There is no answer because it was one of the 30 questions in NUSHS mathematical Olympiad this year. I guess it is $37, but no confident at all.

Thanks in advance.




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

$37 is the lowest amount of money Mr Tan
will needs to spend.

By spending
1) 1 set of (11 chocs + 9 candies) => $15
2) 2 sets of (7 chocs + 9 candies) => $11x 2 = $22

Total chocs = 11 + 14 = 25
Total candies = 9 + 18 = 27
Each student will get 1 piece of chocolate and 1 piece of candy.( 2 pieces of candy left for Mr Tan to enjoy)

Minimum Amt spent = $15 + $22 = $37

[hapie_always]

Hi Mathsguru,


I have a math problem. Hope you can help me with it.

3 boys shared a box of oranges. Ted took 1/3 and 6 of the oranges. Bob took 1/2 of the remainder and 9 of the oranges. Ken took the last remaining 5 oranges. How many oranges were in the box at first?

Thanks in advance!

[Dharma]

hapie_always:

Hi Mathsguru,


I have a math problem. Hope you can help me with it.

3 boys shared a box of oranges. Ted took 1/3 and 6 of the oranges. Bob took 1/2 of the remainder and 9 of the oranges. Ken took the last remaining 5 oranges. How many oranges were in the box at first?

Thanks in advance!

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

[hapie_always]

Dharma:

hapie_always:

Hi Mathsguru,


I have a math problem. Hope you can help me with it.

3 boys shared a box of oranges. Ted took 1/3 and 6 of the oranges. Bob took 1/2 of the remainder and 9 of the oranges. Ken took the last remaining 5 oranges. How many oranges were in the box at first?

Thanks in advance!

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

Thanks Dharma!

[tiger262]

Dear Dharma,


In the problem about Tan buying chocolates and candies for his 25 students, you have worked out as follows:

$37 is the lowest amount of money Mr Tan will needs to spend.

By spending
1) 1 set of (11 chocs + 9 candies) => $15
2) 2 sets of (7 chocs + 9 candies) => $11x 2 = $22

Total chocs = 11 + 14 = 25
Total candies = 9 + 18 = 27
Each student will get 1 piece of chocolate and 1 piece of candy.( 2 pieces of candy left for Mr Tan to enjoy)

Minimum Amt spent = $15 + $22 = $37

My question is:

In exam, we will try several combinations of getting 25 chocolates plus 25 candies and calculate the price of each combination and only then choose the $ 37 combination which you have given above. How did you arrive at this final answer so quickly? Is there any short-cut or heuristic involved to reach the final answer in one go? Can you please explain?

Because it looks like a 'guess & check' problem. But you have neatly solved it without any false guesses. How? Kindly educate me.

Thanks a lot.

[Dharma]

tiger262:

Dear Dharma,


In the problem about Tan buying chocolates and candies for his 25 students, you have worked out as follows:

$37 is the lowest amount of money Mr Tan will needs to spend.

By spending
1) 1 set of (11 chocs + 9 candies) => $15
2) 2 sets of (7 chocs + 9 candies) => $11x 2 = $22

Total chocs = 11 + 14 = 25
Total candies = 9 + 18 = 27
Each student will get 1 piece of chocolate and 1 piece of candy.( 2 pieces of candy left for Mr Tan to enjoy)

Minimum Amt spent = $15 + $22 = $37

My question is:

In exam, we will try several combinations of getting 25 chocolates plus 25 candies and calculate the price of each combination and only then choose the $ 37 combination which you have given above. How did you arrive at this final answer so quickly? Is there any short-cut or heuristic involved to reach the final answer in one go? Can you please explain?

Because it looks like a 'guess & check' problem. But you have neatly solved it without any false guesses. How? Kindly educate me.

Thanks a lot.

Hi tiger262,

Actually I checked Amy’s answer and found it to be the lowest. You’re right …mainly Guess and Check and I did some combination mentally though I didn’t post them. You can actually try look at the 3 packages and find out most economic package in terms of unit rate. Then start buying from the cheapest.

3 packages
1. 11 chocs + 9 candies = $15
2. 7 chocs + 9 candies = $11
3. 3 chocs + 2 candies = $5

I tried to find out the unit cost of sweets (combination of chocs and candies) for each of the packages.

Package 1 => Unit cost = $15/20 sweets = $0.75/sweet
Package 2 => Unit cost = $11/16 sweets = $0.6875/sweet [approx. $0.69/sweet]
Package 3 => Units cost = $5/5 sweets = $1.00/sweet

If Mr Tan wants to get 25 chocs and 25 candies at the cheapest price; he should go for Package 2 first followed by Package 1 and finally Package 3

If he buys Package 2 only => he needs to buy 4 sets which will cost him ($44)
If he buys 3 sets of Package 2 + 2 sets of Package 3 = $43
If he buys 2 sets of Package 2 + 1 set of Package 1 = $37 (Best combi)

You can try other combinations ....

[tiger262]

Dear Dharma,


Thanks for explaining in such a detailed way, that too within an hour or so of my posting my question. Really a splendid response.

I had once before mentioned in this forum - "I wish every Maths teacher in our Primary schools undergo an orientation program under you (under Dharma)". I reaffirm that again. You are a great Maths teacher.

God bless you.