Tutor MathsGuru: Ask me for your burning Maths questions!

[OngMum]

Thank you for the solutions, Dharma and Mathsguru.

[firebird]

Dear maths guru


Good morning.

Kindly help my child’s P6 maths:

1) There are twice as many boys in P6A as in P6B.
There are four times as many girls in P6B as in P6A.
If the total number of pupils in P6A is equal to that of P6B, what is the ratio of the number of boys in P6A to the number of girls in P6B.

2) A series of numbers can be arranged into 6 colums as shown below. Under which column will the number 501 appear?

Very sorry, i could not draw the rows and columns lines.

A B C D E F
1 2 3 4 5 6
12 11 10 9 8 7
13 14 15 16 17 18
… … … … … …

3) John walks to the tuition centre from his home. If he walks at a speed of 50m/min, he will be 4 minutes late. If he walks at a speed of 70 m/min, he will reach the centre 4 minutes before the lesson starts. Find the distance between the tuition centre and John’s house.

4) There are 5970 students in School A and School B altogether. 2/5 of the students in School A and 3/5 of the students in School B are girls. There are 2769 girls altogether in the 2 schools. How many students are there in School A?

Thank you
firebird

[Dharma]

Almighty:

Dharma:

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

One more question:
http://www.postimage.org/image.php?v=TsdBLer
a) got the answer : 180Cm
For b)
My MEthod
FOr 1 glued plank ---210 cm
For 48 planks---? ie., 10080 cm

Each plank ----450 cm
?-----------10 080 cm
22.4 = 23 planks

Book method:
FOr 1 glued plank ---210 cm
for 450 cm --- 210 cm + 210 cm
No of wood needed ---- 48 /2 = 24 pieces
Am i right? If not, can anyone Please explain the book answer for (B)

1/3 of Y = ¼ of X
3u = 90cm
1u = 30cm

a)\t1 set of glued plank X + Y = 90 + 120 – 30 = 180cm
b)\tTo make 1 glued plank, Mr Tan needs 120cm length of X + 90cm length of Y
To make 1 set he needs 210cm length of plank.
1 length of plank = 4.5m = 450cm
450cm/210cm = 2.14 ( he gets 2 sets of glued plank)
So, from 1 length of plank (450cm), Mr Tan is able to make 2 pieces of glued plank.

If Mr Tan needs 48 pieces of glued plank; he needs to buy 48/2 = 24 pieces of 4.5m length plank.

Hearty thanks once again Dharma Sir. WIll go through your working..and i know u will be there to clarify my further querries (in case i have any)[/quote]No worries, Almighty. I'll try my best to answer if time permits.

Good luck and study hard.

[Dharma]

firebird:

Dear maths guru


Good morning.

Kindly help my child's P6 maths:

1) There are twice as many boys in P6A as in P6B.
There are four times as many girls in P6B as in P6A.
If the total number of pupils in P6A is equal to that of P6B, what is the ratio of the number of boys in P6A to the number of girls in P6B.

2) A series of numbers can be arranged into 6 colums as shown below. Under which column will the number 501 appear?

Very sorry, i could not draw the rows and columns lines.

A B C D E F
1 2 3 4 5 6
12 11 10 9 8 7
13 14 15 16 17 18
... ... ... ... ... ...

3) John walks to the tuition centre from his home. If he walks at a speed of 50m/min, he will be 4 minutes late. If he walks at a speed of 70 m/min, he will reach the centre 4 minutes before the lesson starts. Find the distance between the tuition centre and John's house.

4) There are 5970 students in School A and School B altogether. 2/5 of the students in School A and 3/5 of the students in School B are girls. There are 2769 girls altogether in the 2 schools. How many students are there in School A?

Thank you
firebird

Hi firebird,

Q1)
Boys
6A = 2u
6B = 1u

Girls
6A = 1p
6B = 4p

2u + 1p = 1u + 4p
1u = 3p

Boys in P6A : Girls in P6B = 2u : 4p = 6p : 4p = 3 : 2

Q2)

Divide the number by 12 and find the remainder
If Remainder is : -
0 & 1 => Column A
2 & 11 => Column B
3 & 10 => Column C
4 & 9 => Column D
5 & 8 => Column E
6 & 7 => Column F

501 divide by 12 = 41R9
Remainder 9 => Column D

Q3)

Distance is fixed (tuition centre to home)
Speed ratio : 5 : 7
Time ratio : 7 : 5

7u – 5u = (4 + 4)mins
2u = 8 mins
1u = 4 mins

7u = (7 x 4)mins = 28 mins

Distance = 50m/min x 28min = 1400m


Q4)

School A
Boys : 3u
Girls : 2u

School B
Boys : 2p
Girls : 3p

Total no of boys and girls in both schools
5u + 5p = 5970
1u + 1p = 1194


Total no of girls
= 2u + 3p = 2769
= 2u + 2p + 1p = 2769
= (1194 x 2) + 1p = 2769

1p = 2769 – 2388 = 381
1u = 1194 – 381 = 813

Total students in School A
=5 x 813
= 4065

[Nebbermind]

Dharma:

maths6a:

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





In the figure, KE = EL, AE = ED and BC is 3 times KB. Area of rect ABCD is 94 cm2 and the area of triangle KGB is 6 cm2.
Find the area of triangle KEL.

Thx for the tips. With some trial and error finally got it up. Please let me know what you think of this question. Thanks. :?

http://www.postimage.org/image.php?v=PqLd5uS

if the base of half the larger triangle is 2.5 times that of the smaller triangle, then the area is 6.25 times.

6 x 6.25 = 37.5

Area of KEL = 2 x 37.5 = 75

Is there something wrong with the question or my calculation? :?

P.S.

If BC is 3 times KB, then rect ABCD will be 9 times the area of the rect formed by 2 nos. of KGB, ie, 9 x 12 = 108. Yah? But it seems like CL is unknown

[Dharma]

Nebbermind:

Dharma:

[quote=\"maths6a\"]http://www.postimage.org/image.php?v=TsdpVqJ





In the figure, KE = EL, AE = ED and BC is 3 times KB. Area of rect ABCD is 94 cm2 and the area of triangle KGB is 6 cm2.
Find the area of triangle KEL.

Thx for the tips. With some trial and error finally got it up. Please let me know what you think of this question. Thanks. :?

http://www.postimage.org/image.php?v=PqLd5uS

if the base of half the larger triangle is 2.5 times that of the smaller triangle, then the area is 6.25 times.

6 x 6.25 = 37.5

Area of KEL = 2 x 37.5 = 75

Is there something wrong with the question or my calculation? :?

P.S.

If BC is 3 times KB, then rect ABCD will be 9 times the area of the rect formed by 2 nos. of KGB, ie, 9 x 12 = 108. Yah? But it seems like CL is unknown[/quote]Thanks for highlighting, Nebbermind.

Seems like there is a flaw in Qn. The area of rect ABCD should be 90cm2 instead of 94cm2. You're correct.


http://www.postimage.org/image.php?v=PqNP01i

[maths6a]

Is the answer 77 or 78 1/3? Which is more acceptable? Thanks....... :?:


Apparently answer from schoolS' teacherS is 77. THanks.
Oh when I use 77 to workbacks, base and height don't tally.
When I use 78 1/3, they are ok.

Who that clever person who knows to change 94 to 90?

That was what I have been told to do. But I just want to know what should the answer be if it's 94. thanks, very stubborn me.

[maths6a]

Oh another question, is it wrong to for example convert 4500m to km and write 4.500km? Is it mathematically wrong and will marks be deducted? Thanks.

[OngMum]

Hi Mathsguru,


I have two questions that need your gudiance:

1. The figure below is made up of 4 squares and a rectangle. Rectangle ABCD has a perimeter of 40cm. The total area of the 4 squares is 500cm2. Find the area of the rectangle.

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

2. The figure below, not drawn to scale, shows three triangles of differnet sizes overlapping one another. The sum of areas of Triangle A and Triangle C is 30% that of Triangle B. 12.5% of Triangle A is shaded, 5% of B is shaded and 20% of Triangle C is shaded. Express the area of Triangle A as a percentage of the area of Triangle C.

http://www.postimage.org/image.php?v=gxKY8hi

Thank you! Cheers!

[Almighty]

OngMum:

Hi Mathsguru,


I have two questions that need your gudiance:

1. The figure below is made up of 4 squares and a rectangle. Rectangle ABCD has a perimeter of 40cm. The total area of the 4 squares is 500cm2. Find the area of the rectangle.

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

2. The figure below, not drawn to scale, shows three triangles of differnet sizes overlapping one another. The sum of areas of Triangle A and Triangle C is 30% that of Triangle B. 12.5% of Triangle A is shaded, 5% of B is shaded and 20% of Triangle C is shaded. Express the area of Triangle A as a percentage of the area of Triangle C.

http://www.postimage.org/image.php?v=gxKY8hi

Thank you! Cheers!

Hi Ong mum,
Let me try,
1.Let the breadth of the rectangle be m
and its length be n
therefore, 2m + 2n = 40 \t\t\t\t
m + n = 20\t\t\t\t
given that the total area of squares = 500,ie., 2m2 + 2n2 = 500\t\t\t\t
m2 + n2 = 250 cm2\t\t\t\t
using trial n error, m = 15 and n = 5\t\t\t\t
therefore area of the rectangle = 15 X 5 = 75cm2