Tutor MathsGuru: Ask me for your burning Maths questions!

[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'.

[liketoeat]

Please help me with this question. Thank you.


Najip, Kumar and Gurmit started jogging at the same time from the same starting-point round a circular track. Najip and Kumar jogged in a clockwise direction and Gurmit jogged in an anti-clockwise direction. Gurmit took 5 minutes to complete each round. Gurmit met Najip after every 3 minutes. Gurmit met Kumar after every 2 minutes. The jogging speed of each person remained the same throughout.

(a) What was the ratio of Gurmit’s speed to Najip’s speed to Kumar’s speed? (Ans: 6 : 4 : 9)
(b) When Gurmit and Najip met again at the starting-point after 15 minutes, Kumar had already jogged 3.6 km. What is the circumference of the circular track? (Ans: 0.8 km)

[liketoeat]

Here’s another question.


Each day, Mike saves $0.50 while Carol saves $0.30 more than her. Carol starts saving 6 days after Mike. She has $17.40 more than Mike now. How much does Mike save now? (Ans: $37)

[Dharma]

liketoeat:

Please help me with this question. Thank you.


Najip, Kumar and Gurmit started jogging at the same time from the same starting-point round a circular track. Najip and Kumar jogged in a clockwise direction and Gurmit jogged in an anti-clockwise direction. Gurmit took 5 minutes to complete each round. Gurmit met Najip after every 3 minutes. Gurmit met Kumar after every 2 minutes. The jogging speed of each person remained the same throughout.

(a) What was the ratio of Gurmit's speed to Najip's speed to Kumar's speed? (Ans: 6 : 4 : 9)
(b) When Gurmit and Najip met again at the starting-point after 15 minutes, Kumar had already jogged 3.6 km. What is the circumference of the circular track? (Ans: 0.8 km)

(a)
Time ratio => Najip : Gurmit = 3 : (5 – 3) = 3 : 2
Speed ratio => Najip : Gurmei = 2 : 3
Time ratio => Kumar : Gurmit = 2 : (5 – 2) = 2 : 3
Speed ratio => Kumar : Gurmit = 3 : 2

Speed ratio => Gurmit : Najip : Kumar = 6 : 4 : 9

(b)
When Gurmit and Najip met again in 15 mins (time is fixed)

Distance ratio => Gurmit :Najip : Kumar = 6 : 4 : 9
9u = 3.6km
1u = 0.4km

6u = 3 rounds ( 1 round = 5 mins; 3 round = 15 mins)
3 rounds = 6 x 0.4km = 2.4km
1 round = Circumference of track = 2.4km / 3 = 0.8km

[Dharma]

liketoeat:

Here's another question.


Each day, Mike saves $0.50 while Carol saves $0.30 more than her. Carol starts saving 6 days after Mike. She has $17.40 more than Mike now. How much does Mike save now? (Ans: $37)

Carol starts saving from Day 6 onwards.
Mike saved $ 3.00 ($0.50/day x 6 days) from Day 1 to Day 5

Day 6 to now; Carol saved $ 20.40 ($17.40 + $3.00) more than Mike.

No. of days from Day 6 to now = $20.40/ ($0.30/day) = 68 days

Amount Mike saved from Day 1 until now = $ 3.00 + ($0.50/day x 68 days) = $37.00

[Herbie]

I have one qn.


There were some boys and girls in a room.
If 24 boys leave the room, the total no. of students who remained in the room will be 7 times theno. of boys who remained the the room.
If 12 girls leave the room, the total no. of students who remained in the room will be 4 times the no. of boys.

How many students are there in the classroon?

Tq

[Dharma]

Herbie:

I have one qn.


There were some boys and girls in a room.
If 24 boys leave the room, the total no. of students who remained in the room will be 7 times theno. of boys who remained the the room.
If 12 girls leave the room, the total no. of students who remained in the room will be 4 times the no. of boys.

How many students are there in the classroon?

Tq

Boys at first = 1u + 24 = 1p ………… (1)
Girls at first = 6u = 3p + 12 ………… (2)

6u = 3(1u + 24) + 12
3u = 84
1u = 28

No. of students in classroom = 7u + 24 = 7(28 ) + 24 = 220