Tutor MathsGuru: Ask me for your burning Maths questions!
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Dharma:
Hi Dharma Sir,
Q1)Almighty:
Hi all,
Can anyone help on the following qts:
1)48% of the pupils go to school by car.50% of those who donot go to school by car are girls.\t\t\t\t
25% of those who go to school by car are boys. How many percent of the total number of boys\tin the school go to school by car?\t\t
\t\t\t\t\t
\t\t\t\t\t\t
My method:\t\t\t\t\t\t
\t\t\t\t\t
\t\t\t\t\t\t
Boys :Girls\t (Car ---48%) \tBoys : Girls (Not by car -----52%\t)\t\t
12% : 36% (by car) --------\t26% :26% (Not by car)\t\t
total number of boys in the school = 12 % + 26% = 38%\t\t\t\t\t\t
total number of boys in the school go to school by car\t\t\t\t\t\t= 12% / 38% X 100
\t\t\t\t\t\t= 31 11/19%
Book answer = 12%\t\t\t\t\t\t
Can anyone clarify how can it just be 12%\t\t\t\t\t\t
2)The ratio of number of cars to the number of vans sold in a certain country\t\t\t\t\t\t
was 3:2 last year. This year. A total of 9600 more vans and cars are sold and there are twice as many vans and thrice as many cars as last year.How many cars are sold this year? \t\t
\t\t\t\t\t\t
Last year - cars sold\tThis year cars sold\t\t
C : V : Total\t\tC : V : Total\t\t
3 : 2 : 5\t\t\t 9 : 4 : 13\t\t
\t\t\t\t\t\t
\t8 units = 9600\t\t\t\t\t
\t1 unit = 1200\t\t\t\t\t
\t
So, we must find for 9 units to find the number of cars sold this year.\t
\t
ie., 9 units = 10800 cars My answer(forgot to write this?)
Book method\t
6 units ( 9 - 3) ----- ? 1200 X 6\t
\t = 7200 cars
Can anyone clarify why should we do 9 - 3 ???\t
3)In a ballot school, 60% of the pupils are girls and the rest are boys.
How many percent more girls than boys are there in the ballet school?\t
\t
Girls ---60% \tBoys ----40%
\t\t\t
My method : ans : 50%\t
\t\t\t
\t20 / 40 X 100% = 50%\t\t
Work backwards:\t\t\t
50% X 40% = 20% ie , 40% + 20% = 60% \t\t\t
\t\t\t
Book method:\t\t\t
\t\t\t
Ans : 20% ie 60 % - 40%\t\t\t
\t\t\t
Work backwards:\t\t\t
20% X 40% = 8% ie., 40% + 8% = 48% not equal to 60%\t
\t\t\t
\t\t\t
My method: interpretating the question as:
How many percent more girls than boys are there in the ballet school
Book method : interpretating the question as
How many percent more girls are there than boys in the ballet school
Which is right??Can anyone please clarify on this?? :?:
http://www.postimage.org/image.php?v=TsfKTo0
Q2)
Last year
Cars : 3u
Vans : 2u
This year
Cars : 9u
Vans : 4u
13u – 5u = 8u = 9600
1u = 1200
No. of cars sold this year = 9u = 9 x 1200 = 10800
Q3)
Your method is correct
There was a spacing mistake (as indicated in violet) for qt 2.
According to your answer, i think i have got qt 2 n 3 right .
Will go thru yr working for qt 1.
Thankyou very much for the effort taken by you to clarify my doubts.
Thankyou others too... -
Dharma:
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)
1/3 of Y = ¼ of XAlmighty:
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)
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. -
OngMum:
Dear OngMum,I have two questions that need guidance :-
1. Edison cycled at a speed of 10m/min while Damien cycled at a speed of 15m/min. Edison started cycling from Sleepy Town and Damien started at the same time from Happy Town. They cycled towards each other. They first met at Point Q and then continued to cycle. Upon reaching their destination, they made a U-turn and continued cycling towards each other again. This time, they met at Point R. The distance between Point Q and Point R was 324m.
(a) What was the distance between Sleepy Town and Happy Town?
(b) How long would Edison have cycled when he met Damien at Point R?
2. 2 identical tins, X and Y, were half filled with paint. Tin X contained only white paint while Tin Y contained only black paint. 1/2 of the white paint in Tin X was poured into Tin Y. After stirring the contents in Tin Y thoroughly, 1/2 of the mixed paint in Tin Y was poured into Tin X. The contents in Tin X ws mixed thoroughly and 1/2 of the mixture was then poured into Tin Y. If 7.2 litres of black paint was in the mixture in Tin X, how much white paint would there be in the mixture in Tin X?
Thank you. Cheers!
Here are my solutions. Hope they are clear enough!
Cheers,
MathsGuru
http://www.postimage.org/image.php?v=TshsJX9
http://www.postimage.org/image.php?v=TshsMri -
Thank you for the solutions, Dharma and Mathsguru.
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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 -
Almighty:
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.
1/3 of Y = ¼ of XDharma:
[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)
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.
Good luck and study hard. -
firebird:
Hi 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
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 -
Dharma:
if the base of half the larger triangle is 2.5 times that of the smaller triangle, then the area is 6.25 times.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
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 -
Nebbermind:
if the base of half the larger triangle is 2.5 times that of the smaller triangle, then the area is 6.25 times.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
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 -
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.
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