Tutor MathsGuru: Ask me for your burning Maths questions!

[Brenda10]

sachiko:

Dear all


I would appreciate if you could enlighten me on the following Maths Qns:

1. A box contained some blue pens and red pens in the ratio of 11:3. After 28 blue pens were removed from the box and 7 red pens were added in the box, the ratio of the number of blue pens to the number of red pens in the box was 7:4. How many pens were there in the box at first?

Thanks in advance.

Hi, dd answer:

B : R
11 : 3 (Before : 14u)
-28 +7
7 : 4 (after)


R 21u 7*7 + 28*4
B 44u

23u =49+ 112
23u = 161
1u = 7

14u = 7 * 14
= 98

Therefore there are 98 pens at first.

*Please note that our answer are different from the answer sheet : 28 pens

[Vanilla Cake]

sachiko:

1. A box contained some blue pens and red pens in the ratio of 11:3. After 28 blue pens were removed from the box and 7 red pens were added in the box, the ratio of the number of blue pens to the number of red pens in the box was 7:4. How many pens were there in the box at first?

This 4-mark question was from http://www.orlesson.org/orp/09Ma/2009-Math-SA2-Nanyang.pdf set on 29 Oct 2009.
In addition to solutions provided by Brenda10 and YLH88, here's another solution by http://www.onsponge.com/forum/35-thinkingmathonsponge/5065-p5-nanyang-sa2-2009-external-changed.html using External Changed concept / Total Change concept to solve this P5 Problem sum.

VC's mum

[sachiko]

Vanilla Cake:

sachiko:

1. A box contained some blue pens and red pens in the ratio of 11:3. After 28 blue pens were removed from the box and 7 red pens were added in the box, the ratio of the number of blue pens to the number of red pens in the box was 7:4. How many pens were there in the box at first?

This 4-mark question was from http://www.orlesson.org/orp/09Ma/2009-Math-SA2-Nanyang.pdf set on 29 Oct 2009.
In addition to solutions provided by Brenda10 and YLH88, here's another solution by http://www.onsponge.com/forum/35-thinkingmathonsponge/5065-p5-nanyang-sa2-2009-external-changed.html using External Changed concept / Total Change concept to solve this P5 Problem sum.

VC's mum

Dear Vanilla Cake, Brenda10 & YLH88,
Thanks for your help.

[acehkr3009]

Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

[Brenda10]

acehkr3009:

Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

Hi,

According to dd, the height is 5cm.

Area of 1/2 of the triangle : 1/2 x 5 x 10 = 25CM²
Total : 25CM² x 2 = 50CM²


Hope the answer is correct.

[YLH88]

Post deleted

[amberzbon]

Please help to solve these problems. thanks in advance

\t
Q1) Ali, Ben and Col each had a certain number of cards. Ali gave to Ben and Col some cards which is equal to the amount Ben and Col each originally had. Next, Ben gave to Ali an Col some cards which is equal to the amount Ali and Col each already had and finally Col gave to Ali and Ben in the similar manner. At the end, each had 32 cards each. Find the number of cards each had originally.


Q5) A box contains both pink and blue pens. 1/2 of the pink pens and 1/3 of the blue pens make up a total of 13 pens. 1/3 of the pink pens and 1/2 of the blue pens make up a total of 12 pens. How many pens are there of each colour?

[Dharma]

Brenda10:

acehkr3009:

Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

Hi,

According to dd, the height is 5cm.

Area of 1/2 of the triangle : 1/2 x 5 x 10 = 25CM²
Total : 25CM² x 2 = 50CM²


Hope the answer is correct.

Perpendicular height is 8cm

[YLH88]

Dharma:

Brenda10:

[quote=\"acehkr3009\"]Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

Hi,

According to dd, the height is 5cm.

Area of 1/2 of the triangle : 1/2 x 5 x 10 = 25CM²
Total : 25CM² x 2 = 50CM²


Hope the answer is correct.

Perpendicular height is 8cm[/quote]Hi Dharma,

How do you get the height = 8 cm ?

[Dharma]

YLH88:

acehkr3009:

Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

If you join 2 of the same isosceles triangles together, as shown in the diagram below, it will become a square.

http://postimage.org/image/uh05zzus/

area of the square = 10 x 10 = 100 cm2
area of the triangle ABC = 1/2 x 100 = 50 cm2

Hope the answer is correct.

Triangle ABC is not a right angle isoceles triangle . The 2 triangles put together as shown will not form a square.