Tutor MathsGuru: Ask me for your burning Maths questions!

[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.

[Dharma]

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.

Glad u liked it. Thanks. Don't praise me so much lah, I'm just a parent like you and understand the struggles most parents go thru.

[Almighty]

Dharma:

Almighty:

[quote=\"Dharma\"]
When Kumar met Clara for the 1st time at P ; he had jogged for 80m from M.
When Kumar met Clara for the 2nd time at Q; he would have jogged another 160m (2 x 80m) from P.

So, distance from P ->N -> Q = 160m
Distance PN = 160m – 60m = 100m
Circumference of the jogging track = 2 x (80m + 100m) = 360m

Hi Dharma,
Can you please explain how did u do the thing marked in red?? :stupid:

Hi Almighty,

When Kumar met Clara for the 1st time at P, he had jogged 80m from M. When they met at P, in total they both would covered a total distance of half the circumference of the track (M to N).
Now when Kumar and Clara met for the 2nd time at Q, in total they would have covered the circumference of the track ( P to Q). -UNDERSTOOD.

M to P => 80m (Kumar) when Kumar & Clara covered ½ the circumference.UNDERSTOOD .
P to Q => 80m x 2 = 160m (Kumar) when Kumar & Clara covered the whole circumference.[/quote]Thankyou very much Dharma..I'm able to understand now ...Looks simple now..i dont know how come it dint strike me yesterday..MAy be i was a bit tensed in my preparation for my Prelim Oral...

Sorry Dharma,
Still confused...in problem it is not stated that Point P is the mid point of the circle. Then how can we assume?Still tht 80 X 2 is confusing me....Please explain :?

[Dharma]

Almighty:

Dharma:

[quote=\"Almighty\"]
Hi Dharma,
Can you please explain how did u do the thing marked in red?? :stupid:

Hi Almighty,

When Kumar met Clara for the 1st time at P, he had jogged 80m from M. When they met at P, in total they both would covered a total distance of half the circumference of the track (M to N).

Now when Kumar and Clara met for the 2nd time at Q, in total they would have covered the circumference of the track ( P to Q).

M to P => 80m (Kumar) when Kumar & Clara covered ½ the circumference
P to Q => 80m x 2 = 160m (Kumar) when Kumar & Clara covered the whole circumference.

Thankyou very much Dharma..I'm able to understand now ...Looks simple now..i dont know how come it dint strike me yesterday..MAy be i was a bit tensed in my preparation for my Prelim Oral...[/quote]No worries, Almighty.

[Almighty]

Dharma:

Almighty:

[quote=\"Dharma\"]
Hi Almighty,

When Kumar met Clara for the 1st time at P, he had jogged 80m from M. When they met at P, in total they both would covered a total distance of half the circumference of the track (M to N).

Now when Kumar and Clara met for the 2nd time at Q, in total they would have covered the circumference of the track ( P to Q).

M to P => 80m (Kumar) when Kumar & Clara covered ½ the circumference
P to Q => 80m x 2 = 160m (Kumar) when Kumar & Clara covered the whole circumference.

Thankyou very much Dharma..I'm able to understand now ...Looks simple now..i dont know how come it dint strike me yesterday..MAy be i was a bit tensed in my preparation for my Prelim Oral...

No worries, Almighty.[/quote]Sorry Dharma,
Still confused...in problem it is not stated that Point P is the mid point of the circle. Then how can we assume?Still tht 80 X 2 is confusing me....Please explain

[Dharma]

Almighty:

Dharma:

[quote=\"Almighty\"]
Thankyou very much Dharma..I'm able to understand now ...Looks simple now..i dont know how come it dint strike me yesterday..MAy be i was a bit tensed in my preparation for my Prelim Oral...

No worries, Almighty.

Sorry Dharma,
Still confused...in problem it is not stated that Point P is the mid point of the circle. Then how can we assume?Still tht 80 X 2 is confusing me....Please explain[/quote]See if can understand this ... if Kumar jogged 80m when he & Clara jogged 1/2 the circumference of track; then he must have jogged 160m when he and Clara jogged the whole circumference of the track.
Underlying assumption is the speed of Kumar and Clara do not change.
http://www.postimage.org/image.php?v=PqPN0f9

[clblinym]

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

~~~~~~~~~~~~~~~~~~
Dear Dharma

Thank you so much for your great help. I really appreciate it.

[Dharma]

clblinym:

Dharma:

[quote=\"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

~~~~~~~~~~~~~~~~~~
Dear Dharma

Thank you so much for your great help. I really appreciate it.[/quote]No worries.

[Almighty]

Dharma:

Sorry Dharma,
Still confused...in problem it is not stated that Point P is the mid point of the circle. Then how can we assume?Still tht 80 X 2 is confusing me....Please explain

See if can understand this ... if Kumar jogged 80m when he & Clara jogged 1/2 the circumference of track; then he must have jogged 160m when he and Clara jogged the whole circumference of the track.
Underlying assumption is the speed of Kumar and Clara do not change.
http://www.postimage.org/image.php?v=PqPN0f9[/quote]

Ok ok.I got it..Its based on the Principle[excuse me for the typo error]
\"When K covers 80m =1/2 of the circumference is covered
so, ? = full circumference
ie.,(80 X 2) = 160m from the point of K's journey...
Thankyou Dharma, Onceagain for a very very patient reply...I drew the same diagram and tried to understand intially..but failed to get it..My apologies for bothering you toooooo much....