Tutor MathsGuru: Ask me for your burning Maths questions!

[mathsguru]

super star:

Maths guru

please help me in this problem.

Alan,Ben&charles had 864 trading cards.Ben won some of the cards from alan &as a result,ben's cards increased by 50%.charles then won some cards from ben & charles'cards increased by 40%.finally charles lost some of his cards to alan & alan's cards increased by 20%.In the end ,they realised that they each had an equal number of cards.how many cards did alan have at first?

Hi Super Star,

This question was posted quite some time back by another forum member. Here's my solution anyway...trick here is to work backwards systematically.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=gxHh7s0

[Herbie]

I would like to ask one speed qn.


james and Allan were competing in a cycling race.
They cycled at a constant average speed throughout the race.
James tooke 4 hours to complete the race, by which Allan had only completed 1/5 of the race.

If allan cycled at an average of 3km/h slower than James, find the total distance of the race route.

Thanks!

[mathsguru]

Almighty:

Almighty:

[quote=\"Almighty\"]Question :5Pattern II -photo print


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

Hi MAthmonster,Dharma,Ksi & others......
Reposting...If this qt had missed yr eye...Since i posted many qt last night.
Pl.try to help me on this too..

Hi MathGuru,

This question of mine is still unanswered.Can you help?[/quote]Hi Almighty,

To be honest, I thought long and hard and tried many different ways. Unfortunately, I still cannot see the trick in this pattern... :scratchhead: Apologies for that!

Does any other experts here know? Can enlighten us please??

Regards,
MathsGuru

[mathsguru]

Herbie:

I would like to ask one speed qn.


james and Allan were competing in a cycling race.
They cycled at a constant average speed throughout the race.
James tooke 4 hours to complete the race, by which Allan had only completed 1/5 of the race.

If allan cycled at an average of 3km/h slower than James, find the total distance of the race route.

Thanks!

Hi Herbie,

Here's my solution. Hope it helps!


MathsGuru

http://www.postimage.org/image.php?v=Tsu0_60

[Almighty]

mathsguru:

Budo:

Hi need help in this question


Andrew and Brian had some money
If Andrew gave Brian $55, the ratio of Andrew to Brian amount is 3:5.
If Brian gave Andrew 20% of this money and $10, Andrew would have $108 more than Brian.
How much do Andrew and Brian have at first?

:?

Hi Budo,

Here's my solution. Hope it helps!


MathsGuru

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

Hi math monster,
I used yr algebra method.I am stuck..Pl correct me.. This is my solution:
Berfore:
A : 3m + 55
B : 5m - 55
After A gave B $55/-
A : B = 3m : 5m
B gives 20 % = (53 - 55) X 20% + 10
= m-1

So, A gets = (3m +55) + m-1
= 4m + 54
B left with = (5m -55) - m -1
= 4m -56
HOw can 4m +54 = 4m -56 n solve for m??Where did i go wrong??

If MM not active,Can MAth guru help me out? Pl.refer to P :165 for the method taught by MM.
I am not comfortable using Model...So, Pl.advise!!!

[Almighty]

mathsguru:

Almighty:

Hi All,


The figure below, not drawn to scale, is made up of a circle and 3 identical equilateral triangles. The circle has a radius of 14 cm. Find the perimeter of the shaded parts. Take ^(pie) = 22/7

http://www.postimage.org/image.php?v=Tso6CVr
My Method :
Straight line of a shaded portion is the diameter of that semi circle.
Diameter of the shaded arc = 14cm.So,
Circumference of one shaded semicircle = ( 1/2 ^ D) = 1/2 X 22/7 X14
= 22 cm
22 + 14 = 36 cm (Perimeter of one shaded portion).
So, Total perimeter of the shaded portion = 3 X 36 = 108 Cm

Book method :
Combined all the three arcs together so as to form
curved lenght of shaded parts = curved lenght of semi circle.
Ans: 86cm

MY QUESTION:
Can anyone let me know whats wrong in my method were i assumed each shaded portion as a semicircle with diameter 14cm and not 28 cm.
Hope i am clear !! If not pls let me know. Will TRY to reform the question.

Hi Almighty,

Here's my solution for your reference. The diameter should be 28cm.


MathsGuru

http://www.postimage.org/image.php?v=Pq2vyKJ

Still a bit confused...
The straight line of each shaded portion is 14cm as they form the radius of the Circle. But what i assumed is that, Each shaded semicircle has the same straight line as its diameter. Right? So, if radius of the circle is 14cm the diameter of this shaded semicircle must be 28cm.Isn't. In simple term, please expalin why the radius of the circle (Big circle) Cannot be assumed as the diameter of the shaded semicircle and be proceeded.

As Dharma said, How is it that the shaded semicircle's radius is 14cm? The whole straight line is diamter as it starts from one end to other end and doesnot end in the centre......

Plsss Dharma and MAth guru....try to get a solution for my querry.

[Dharma]

Almighty:

mathsguru:

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


The figure below, not drawn to scale, is made up of a circle and 3 identical equilateral triangles. The circle has a radius of 14 cm. Find the perimeter of the shaded parts. Take ^(pie) = 22/7

http://www.postimage.org/image.php?v=Tso6CVr
My Method :
Straight line of a shaded portion is the diameter of that semi circle.
Diameter of the shaded arc = 14cm.So,
Circumference of one shaded semicircle = ( 1/2 ^ D) = 1/2 X 22/7 X14
= 22 cm
22 + 14 = 36 cm (Perimeter of one shaded portion).
So, Total perimeter of the shaded portion = 3 X 36 = 108 Cm

Book method :
Combined all the three arcs together so as to form
curved lenght of shaded parts = curved lenght of semi circle.
Ans: 86cm

MY QUESTION:
Can anyone let me know whats wrong in my method were i assumed each shaded portion as a semicircle with diameter 14cm and not 28 cm.
Hope i am clear !! If not pls let me know. Will TRY to reform the question.

Hi Almighty,

Here's my solution for your reference. The diameter should be 28cm.


MathsGuru

http://www.postimage.org/image.php?v=Pq2vyKJ

Still a bit confused...
The straight line of each shaded portion is 14cm as they form the radius of the Circle. But what i assumed is that, Each shaded semicircle has the same straight line as its diameter. Right? So, if radius of the circle is 14cm the diameter of this shaded semicircle must be 28cm.Isn't. In simple term, please expalin why the radius of the circle (Big circle) Cannot be assumed as the diameter of the shaded semicircle and be proceeded.

As Dharma said, How is it that the shaded semicircle's radius is 14cm? The whole straight line is diamter as it starts from one end to other end and doesnot end in the centre......

Plsss Dharma and MAth guru....try to get a solution for my querry.[/quote]Hi Almighty,

I don’t really understand what you are trying to say but we know that that the radius of the circle is 14cm and the radius of 14cm is also the side of the small equilateral triangles.

Perimeter of the shaded part
= Length of the 3 sides of a triangle + Curved length of the semicircle
= ( 14 x 3 ) + ( 22/7 x ½ x 14 x 2 )
= 42 + 44
= 86

The perimeter of the shaded part = 86cm

[mathsguru]

Almighty:

mathsguru:

[quote=\"Budo\"]Hi need help in this question


Andrew and Brian had some money
If Andrew gave Brian $55, the ratio of Andrew to Brian amount is 3:5.
If Brian gave Andrew 20% of this money and $10, Andrew would have $108 more than Brian.
How much do Andrew and Brian have at first?

:?

Hi Budo,

Here's my solution. Hope it helps!


MathsGuru

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

Hi math monster,
I used yr algebra method.I am stuck..Pl correct me.. This is my solution:
Berfore:
A : 3m + 55
B : 5m - 55
After A gave B $55/-
A : B = 3m : 5m
B gives 20 % = (53 - 55) X 20% + 10
= m-1

So, A gets = (3m +55) + m-1
= 4m + 54
B left with = (5m -55) - m -1
= 4m -56
HOw can 4m +54 = 4m -56 n solve for m??Where did i go wrong??

If MM not active,Can MAth guru help me out? Pl.refer to P :165 for the method taught by MM.
I am not comfortable using Model...So, Pl.advise!!![/quote]Hi Almighty,

Think u misread the question. It says \"If Brian gave Andrew 20% of THIS money\" not \"his money\". I think it's referring to 20% of $55 and not his entire sum of money.


MathsGuru

[Budo]

mathsguru:

Budo:

Hi need help in this question


Andrew and Brian had some money
If Andrew gave Brian $55, the ratio of Andrew to Brian amount is 3:5.
If Brian gave Andrew 20% of this money and $10, Andrew would have $108 more than Brian.
How much do Andrew and Brian have at first?

:?

Hi Budo,

Here's my solution. Hope it helps!


MathsGuru

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

Thanks Maths Guru!!! $176 is the correct answer

[Tang]

Budo:

mathsguru:

[quote=\"Budo\"]Hi need help in this question


Andrew and Brian had some money
If Andrew gave Brian $55, the ratio of Andrew to Brian amount is 3:5.
If Brian gave Andrew 20% of this money and $10, Andrew would have $108 more than Brian.
How much do Andrew and Brian have at first?

:?

Hi Budo,

Here's my solution. Hope it helps!


MathsGuru

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

Thanks Maths Guru!!! $176 is the correct answer[/quote]
Hello,

There are multiple answers.

A -- 3u + $55
B -- 5u - $55

20% B -- 1/5 x (5u - $55) = 1u - $11
1u - $11 + $10 = 1u - $1

3u + $55 + 1u - $1 -- 4u + $54

5u - $55 - (1u - $1) -- 4u - $54

So condition of Difference of $108 is true regardless of value of 4u............

I believe there is a typo error - 'this' should be 'his'.