Tutor MathsGuru: Ask me for your burning Maths questions!

[biscuitqueen]

Hi there, please help me solve this Maths questn for my daughter.

I have no idea how to solve. Thanx!


Mary and Ken travelled from Town A to Town B. Mary left Town A at 0915 and took 4h to reach Town B. Ken left Town A at 1000 and took 2h 45min to reach Town B. At what time did Ken pass Mary?

[optimistforum]

optimistforum:

Hi Maths Guru


Can you look at this Clock Question, please?

Please help on the answer to the question below.

John and Tom race around a circular track, divided like the face of a clock, into 12 sections. If John gives Tom four sections start and runs half as fast again as Tom, at what point on the track will he overtake Tom? (Assume on the clock face that John and Tom start at 12 and 4, respectively).

The answer is 12, however, if John gives Tom four sections starts and is half as fast, surely does not half as fast mean that John is running at half the speed of Tom. And if he is 4 sections behind, surely Tom will win. Unless half as fast means that John runs at 1.5 times of Tom. Tearing my hair out here. Please help!!!

Hi, do all questions get answered, as I have checked on here, and my PM to see if the above question has been answered - and it has not :?

[markfch]

The qn is:


The total age of Danny and Mike is 37 yrs old. Danny is 9 yrs older than Mike. What was Mike’s age 5 yrs ago?

Of course mentally I know the ans. If I say 2y + 9 = 37, I’m afraid it’s hard for ds to follow.

How do I present the working in a simplier way so that ds can follow? (as long as I can present it systematically, I’m sure he will understand). Thanks.

[ChiefKiasu]

markfch:

The qn is:


The total age of Danny and Mike is 37 yrs old. Danny is 9 yrs older than Mike. What was Mike's age 5 yrs ago?

Of course mentally I know the ans. If I say 2y + 9 = 37, I'm afraid it's hard for ds to follow.

How do I present the working in a simplier way so that ds can follow? (as long as I can present it systematically, I'm sure he will understand). Thanks.

markfch, welcome to the world of model drawing.

Let current age of Mike be M.

So (M) and (M+9) will total 37, which means M = 14
So 5 years ago, Michael is 14-5 = 9 years old

[Dharma]

biscuitqueen:

Hi there, please help me solve this Maths questn for my daughter.

I have no idea how to solve. Thanx!


Mary and Ken travelled from Town A to Town B. Mary left Town A at 0915 and took 4h to reach Town B. Ken left Town A at 1000 and took 2h 45min to reach Town B. At what time did Ken pass Mary?

For fixed distance from Town A to Town B,

Time ratio => Mary : Ken = 240 : 165 = 48 : 33
Speed ratio => Mary : Ken = 33 : 48

At 1000hrs, Mary had travelled (33u x ¾)km = 99/4 km
Ken starts at 1000hrs and needs to catch up this distance.

Time taken by Ken to catch up with Mary
= 99u/4/(48u – 33u)
= 33/20 hrs
= 99 mins ( 1 hr 39 mins)

Ken passed Mary at 1139hrs

[Dharma]

optimistforum:

optimistforum:

Hi Maths Guru


Can you look at this Clock Question, please?

Please help on the answer to the question below.

John and Tom race around a circular track, divided like the face of a clock, into 12 sections. If John gives Tom four sections start and runs half as fast again as Tom, at what point on the track will he overtake Tom? (Assume on the clock face that John and Tom start at 12 and 4, respectively).

The answer is 12, however, if John gives Tom four sections starts and is half as fast, surely does not half as fast mean that John is running at half the speed of Tom. And if he is 4 sections behind, surely Tom will win. Unless half as fast means that John runs at 1.5 times of Tom. Tearing my hair out here. Please help!!!

Hi, do all questions get answered, as I have checked on here, and my PM to see if the above question has been answered - and it has not :?

Hi optimistforum,

Tom has a headstart of 4 units over John and Tom’s speed is 2 times John’s speed.

There is no way that John is ever gonna overtake Tom.

Anyway, if the John’s speed is 2 times Tom’s speed, then John will be able to catch up with Tom.

During the same time,(Time is fixed)
Speed ratio => John : Tom = 2 : 1
Distance ratio => John : Tom = 2 : 1
2u – 1u = 1u = 4
John will have to travel 8 units while Tom travels 4 units for them to meet at mark “8”on the clockface.

If the answer given is 12, then John’s speed must be 1.5 times Tom’s speed for them to meet at mark “12”of the clockface

Speed ratio => John : Tom = 3 : 2
Distance ratio => John : Tom = 3 : 2
3u – 2u = 4
1u = 4
John will have to travel 12 units while Tom travels 8 units for them to meet at mark “12”on the clockface.

[clblinym]

Dear Parents,

please help me solve the following Math question for my DS.
I have no idea how to solve and the answer was not given.
Thanks in advance.
Regards,
Amy

http://www.postimage.org/image.php?v=aV2Ofii

[maths6a]

In a conference, there were some men and women. In the first day, there were 80 less women than men. In the second day, the number of men who came decreased by 10%, and the number of women who had arrived increased by 20%. The total number of people who had arrived for the second day was 1542.


Let the number of
Day 1
Women be 10 units
Men be 10 units + 80

Day 2
Women will be 12 units
Men will be 9 units + 72

12units + 9 units + 72 = 1542
21 units = 1470
1 unit = 70
41 units + 80 + 72 = 3022

[markfch]

ChiefKiasu:

markfch, welcome to the world of model drawing.


Let current age of Mike be M.

So (M) and (M+9) will total 37, which means M = 14
So 5 years ago, Michael is 14-5 = 9 years old

Thanks for the tips, chief. Looks workable. Tonite I'll try out your suggestion.

[tianzhu]

Hi

Hope this helps.
http://farm5.static.flickr.com/4129/4842986868_3803e18102_b.jpg\">

Best wishes