Tutor MathsGuru: Ask me for your burning Maths questions!

[Dharma]

Almighty:

Dharma:

[quote=\"tiger262\"]Can anyone post a solution for this speed question?


Kumar and Clara started jogging along a circular track.
Kumar started at Point M while Clara started at Point N where the line MN formed the diameter of the circle.
Kumar and Clara jogged toward each other along the circular track from their respective starting point and first met at Point P which was 80 m from Point M.
After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Q which was 60 m from Point N. Find the distance of the circular track.

Thank you so much.

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:[/quote]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.

[clblinym]

Dear parents,


My son has two questions to ask. We tried to solve them, but failed.
Hope parents here can help us get solutions. Thanks in advance.

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

http://www.postimage.org/image.php?v=PqN6UjA

Amy

[Dharma]

clblinym:

Dear parents,


My son has two questions to ask. We tried to solve them, but failed.
Hope parents here can help us get solutions. Thanks in advance.

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

http://www.postimage.org/image.php?v=PqN6UjA

Amy

Hi Amy,

Q1

Area of quadrlateral ABCD
= Area of triangle ABC + Area of triangle ADC
= (1/2 x 3cm x 4cm) + (1/2 x 2cm x 2cm)
= 6cm2 + 2cm2
= 8cm2


Q2

Shaded area
= (Area of triangle GBE + Area of triangle HBD) - Area of quad. ABCD
= (1/2 x 10cm x 10cm) - 7cm2
= 50cm2 - 7cm2
= 43cm2

[mathsguru]

super star:

pls help me in solving this problem

Mr.Ong had 5kg rice.after he sold 5/12 of it ,he repacked the remaining rice into smaller packets of 2/3 kg each.how much rice had he left?

Hi Super Star,

Here's my solution for your reference!


MathsGuru

http://www.postimage.org/image.php?v=gxB_e8r

[mathsguru]

clblinym:

Dear parents,


My son has two questions to ask. We tried to solve them, but failed.
Hope parents here can help us get solutions. Thanks in advance.

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

http://www.postimage.org/image.php?v=PqN6UjA

Amy

Hi Amy,

Here are my solutions.


MathsGuru


http://www.postimage.org/image.php?v=PqNlX9S

http://www.postimage.org/image.php?v=PqNlZF0

[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!