Tutor MathsGuru: Ask me for your burning Maths questions!

[Andrew Lee]

Andrew Lee:

MadScientist:

Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

Can help me solve Q20 :?
http://i53.tinypic.com/2ld831i.jpg\">

[whywhy]

Tin A contains some 20-cents coins and Tin B contains some 50-cents coins.

There are 9 more coins in Tin A than in Tin B.
The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50-cents coins are there in Tin B ?

Please help, Thank u

[Andrew Lee]

Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm[/quote]

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:[/quote]

Can help me solve this question, Q3 ? tks :?
http://i55.tinypic.com/2jg0x2c.jpg\">

[Andrew Lee]

Andrew Lee:

Andrew Lee:

[quote=\"MadScientist\"]Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

[/quote]

[tdes]

The ratio of the number of girls to the number of boys in a school is 5:8. After 538 girls and 538 boys left the school, the ratio of girls to boys in the school is now 3:7.


How many students were there in the school at the beginning?

Thanks!

[blitz]

Help please.

Tracy counted the number of 10cents coins, 20cent coins and 50cents coins in her piggy bank. The number of 10c coins was 24 more than the number of 50c coins. After spending 3/5 of the 10c coins, 1/4 of the 20c coins and 18 50c coins, she had an equal number of 10c and 20c coins. The number of 50c coins now formed 40% of the remaining number of coins.

(a) How many 10c coins did Tracy have at first?
(b) How much did she spend in all?

[MathIzzzFun]

whywhy:

Tin A contains some 20-cents coins and Tin B contains some 50-cents coins.

There are 9 more coins in Tin A than in Tin B.
The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50-cents coins are there in Tin B ?

Please help, Thank u

Hi

this was discussed before - http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=27&t=149&start=1220

cheers.

[MathIzzzFun]

Andrew Lee:

Andrew Lee:

[quote=\"MadScientist\"]Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

Can help me solve Q20 :?
http://i53.tinypic.com/2ld831i.jpg\">[/quote]Hi

angle BOA = 284 deg - this refers to the reflex angle ie the \"bigger\" of the two angles BOA, so the \"smaller\" or the acute angle BOA = 360 - 284 = 76 deg

angle AOC = 118 deg, angle BOC = 118 - 76 = 42 deg

cheers.

[MathIzzzFun]

Andrew Lee:

Andrew Lee:

[quote=\"MadScientist\"]Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

[/quote]Hi

I thought it would be easier to solve with these two clues:
- Area of triangle DEC is 1/2 of rectangle ABCD

- AE=EB, so area of triangle EBC = 1/2 of triangle DEC

here goes....

Shaded area : Area of triangle DEC = 40u : 100u
Area of rectangle = 2 x area of triangle DEC = 200u
Shaded area : Area of rectangle = 40u : 200u
Percentage of rectangle shaded = 40/200 x 100% = 20%
a) 20% of rectangle ABCD is shaded.

AE=EB, so area of triangle ECB
= ½ of area of triangle DEC
= ¼ of area of rectangle
Shaded area = 5 / (5+9) x 168 cm² = 60 cm²= 20% of area of rectangle
Area of rectangle = 100/20 x 60 cm² = 300cm²
b) Area of triangle ECB = ¼ x 300cm² = 75 cm²

cheers.

[MathIzzzFun]

Andrew Lee:

Can help me solve this question, Q3 ? tks :?
http://i55.tinypic.com/2jg0x2c.jpg\">

Hi

When Xavier and Yoke first meet at C, the total distance that they walked = 1/2 circumference of circular track.

When Xavier and Yoke left C and then meet again at D, the total distance they walked = 1 x circumference of circular track.

So, the total distance that they walked ie Xavier from A to D, and Yoke from B to D in opposite direction = 1.5 times circumference of circular track.

So, circumference of circular track = (500+340)/1.5 = 560 m

cheers.