Morning Morgan

Nov 28th, 2012, 12:58 AM

http://i.imgur.com/u9Vb5.png

Pick your answer :oh:

Pick your answer :oh:

View Full Version : Can you do this national exam question for primary school kids?

Morning Morgan

Nov 28th, 2012, 12:58 AM

http://i.imgur.com/u9Vb5.png

Pick your answer :oh:

Pick your answer :oh:

JJ Expres

Nov 28th, 2012, 01:19 AM

28

WS=a

WX=b

XV=c

VQ=d

a*b=8

c*d=32

b+c=6

a+d=15

a/b=c/d => b*d=16, a*c=16

from a*c=16 => (15-d)*c=16 => 15*c-32=16 => c=48/15 => d=32*15/48=10

=> b= 5-48/15=42/15

area of rectangle C= b*d =10*42/15= 28

WS=a

WX=b

XV=c

VQ=d

a*b=8

c*d=32

b+c=6

a+d=15

a/b=c/d => b*d=16, a*c=16

from a*c=16 => (15-d)*c=16 => 15*c-32=16 => c=48/15 => d=32*15/48=10

=> b= 5-48/15=42/15

area of rectangle C= b*d =10*42/15= 28

JJ Expres

Nov 28th, 2012, 01:27 AM

a failed somewhere and i cant find where...

LeRoy.

Nov 28th, 2012, 01:27 AM

lol area of a triangle is 1/2 base*height.

wild.river

Nov 28th, 2012, 01:40 AM

20 :rolls:

this shouldn't take more than 10 seconds

this shouldn't take more than 10 seconds

LeRoy.

Nov 28th, 2012, 01:42 AM

Answer is 25

wild.river

Nov 28th, 2012, 01:44 AM

right. i did 16x5 :o

Joana

Nov 28th, 2012, 01:45 AM

a failed somewhere and i cant find where...

The area of the whole rectangle is 15*6=90 cm2.

The rectangle consists of two identical triangles. So, the area of one triangle is 45 cm2.

This triangle consists of triangle A, rectangle C and another triangle, identical to triangle B.

So, the area of rectangle C is: 45 cm2 - 4 cm2 (triangle A) - 16 cm2 (triangle B) = 25 cm2.

The area of the whole rectangle is 15*6=90 cm2.

The rectangle consists of two identical triangles. So, the area of one triangle is 45 cm2.

This triangle consists of triangle A, rectangle C and another triangle, identical to triangle B.

So, the area of rectangle C is: 45 cm2 - 4 cm2 (triangle A) - 16 cm2 (triangle B) = 25 cm2.

Morning Morgan

Nov 28th, 2012, 02:02 AM

No one gave a truly right answer yet...

Flavia P.

Nov 28th, 2012, 02:03 AM

Reading this thread has given me a headache.

(no offense intended at all to the OP BTW :oh:)

(no offense intended at all to the OP BTW :oh:)

Morning Morgan

Nov 28th, 2012, 02:06 AM

Reading this thread has given me a headache.

(no offense intended at all to the OP BTW :oh:)

If you're hawt and have sex with me you're forgiven. :oh:

(no offense intended at all to the OP BTW :oh:)

If you're hawt and have sex with me you're forgiven. :oh:

Flavia P.

Nov 28th, 2012, 02:18 AM

If you're hawt and have sex with me you're forgiven. :oh:

I may not be good at math but I definitely know my way around the bedroom :oh:

As for my hotness, I'll leave that for you to decide, but I don't think I'll disappoint you :angel:

I do have to ask, though.............guy or girl? :lol:

I may not be good at math but I definitely know my way around the bedroom :oh:

As for my hotness, I'll leave that for you to decide, but I don't think I'll disappoint you :angel:

I do have to ask, though.............guy or girl? :lol:

JJ Expres

Nov 28th, 2012, 02:23 AM

http://s7.postimage.org/no5o4r9ex/forum.png

after solving equations

a=2.29844

b=3.48062

c=2.51938

d=12.7016

this shitty rectangle doesn't even exists...

after solving equations

a=2.29844

b=3.48062

c=2.51938

d=12.7016

this shitty rectangle doesn't even exists...

delicatecutter

Nov 28th, 2012, 02:26 AM

I'd prefer a geography question pls. I used to be good at maths but it was never anything I actually was interested in so 99% of my knowledge is gone forever.

Morning Morgan

Nov 28th, 2012, 02:28 AM

I may not be good at math but I definitely know my way around the bedroom :oh:

As for my hotness, I'll leave that for you to decide, but I don't think I'll disappoint you :angel:

I do have to ask, though.............guy or girl? :lol:

Guy :angel:

As for my hotness, I'll leave that for you to decide, but I don't think I'll disappoint you :angel:

I do have to ask, though.............guy or girl? :lol:

Guy :angel:

JJ Expres

Nov 28th, 2012, 02:32 AM

first solution b*d=44.20944299

and there is even one more solution where is b*d=5.7777. the thing is that that with this numeric data you can't have picture like that. diagonal is not gonna be straight at all, this question is lame and i want my sleep time back.

and there is even one more solution where is b*d=5.7777. the thing is that that with this numeric data you can't have picture like that. diagonal is not gonna be straight at all, this question is lame and i want my sleep time back.

Joana

Nov 28th, 2012, 02:35 AM

Of course the numbers don't add up. This is a question for primary school kids. The only formula you need to know is a*b.

delicatecutter

Nov 28th, 2012, 02:46 AM

I picked all of the answers because I think it's funny when ppl make multiple choice polls that aren't, well, multiple choice.

égalité

Nov 28th, 2012, 02:50 AM

None of those are right. :spit:

Solve the system of equations:

xy=8

x(15-y)+y(6-x)=50

You get some gross shit involving sqrt(41) and there are two possible solutions.

After looking at it for 2 second I voted for 20 though. :oh:

Solve the system of equations:

xy=8

x(15-y)+y(6-x)=50

You get some gross shit involving sqrt(41) and there are two possible solutions.

After looking at it for 2 second I voted for 20 though. :oh:

Flavia P.

Nov 28th, 2012, 02:55 AM

Guy :angel:

Good news :angel:

Good news :angel:

LeRoy.

Nov 28th, 2012, 03:08 AM

If you HAVE to pick ONE answer out of the FOUR given options, then (3) 25 is CORRECT.

Cajka

Nov 28th, 2012, 03:22 AM

I think it's 2 by 11. The triangle is probably 2*4, that's the only way to get 4cm2 as an area of it. If I remember correct, the formula for a triangle is ab/2. :unsure:

wild.river

Nov 28th, 2012, 03:29 AM

None of those are right. :spit:

Solve the system of equations:

xy=8

x(15-y)+y(6-x)=50

You get some gross shit involving sqrt(41) and there are two possible solutions.

After looking at it for 2 second I voted for 20 though. :oh:

what??? why?

it has to be 25.

Solve the system of equations:

xy=8

x(15-y)+y(6-x)=50

You get some gross shit involving sqrt(41) and there are two possible solutions.

After looking at it for 2 second I voted for 20 though. :oh:

what??? why?

it has to be 25.

Mynarco

Nov 28th, 2012, 03:34 AM

20. The rectangle is bisected diagonally. Hence the area should be 40 each. Do the subtraction and you get the answer

ETA: Crap I read 16x5. It should be 25 then

ETA: Crap I read 16x5. It should be 25 then

égalité

Nov 28th, 2012, 03:38 AM

what??? why?

it has to be 25.

The "diagonal" is not a straight line. The drawing is not to scale.

it has to be 25.

The "diagonal" is not a straight line. The drawing is not to scale.

Cajka

Nov 28th, 2012, 03:39 AM

20. The rectangle is bisected diagonally. Hence the area should be 40 each. Do the subtraction and you get the answer

15*6 is no 80, it's 90. And the OP said that 25 is not the correct answer.

15*6 is no 80, it's 90. And the OP said that 25 is not the correct answer.

LeRoy.

Nov 28th, 2012, 03:40 AM

25. The rectangle is bisected diagonally. Hence the area should be 45 each. Do the subtraction and you get the answer

Corrected.

Corrected.

égalité

Nov 28th, 2012, 03:41 AM

Here is the answer:

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

Mynarco

Nov 28th, 2012, 03:41 AM

15*6 is no 80, it's 90. And the OP said that 25 is not the correct answer.

I edited it. But how was 25 not the answer wuuuut?

I edited it. But how was 25 not the answer wuuuut?

LeRoy.

Nov 28th, 2012, 03:41 AM

The "diagonal" is not a straight line. The drawing is not to scale. None of the possible answers work if the diagonal is actually a straight line.

And you are getting a PhD in Math ?:confused:

And you are getting a PhD in Math ?:confused:

égalité

Nov 28th, 2012, 03:42 AM

And you are getting a PhD in Math ?:confused:

Yes, what's your point? :weirdo:

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

Yes, what's your point? :weirdo:

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

wild.river

Nov 28th, 2012, 03:42 AM

The "diagonal" is not a straight line. The drawing is not to scale. None of the possible answers work if the diagonal is actually a straight line.

ok, well i'm pretty sure it's ok to assume that the diagonal is a straight line and the drawing isn't to scale :p

ok, well i'm pretty sure it's ok to assume that the diagonal is a straight line and the drawing isn't to scale :p

LeRoy.

Nov 28th, 2012, 03:44 AM

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

There should be a 5th option then. :)

As stated, the correct answer to the problem is 25.

There should be a 5th option then. :)

As stated, the correct answer to the problem is 25.

wild.river

Nov 28th, 2012, 03:46 AM

Yes, what's your point? :weirdo:

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

holy shit.

i'm gonna take my bachelors in physics and quietly slink away :worship:

http://www.aaasurvey.com/tutor_html/bottom_no13detail.html

holy shit.

i'm gonna take my bachelors in physics and quietly slink away :worship:

LeRoy.

Nov 28th, 2012, 03:48 AM

Yes, what's your point? :weirdo:

If you ever teach primary school students, may god help them.

If you ever teach primary school students, may god help them.

~{X}~

Nov 28th, 2012, 03:48 AM

Fuck this. :o

I just had a flashback of years of torture in math classes because of this. LORD I almost cried. :lol:

I just had a flashback of years of torture in math classes because of this. LORD I almost cried. :lol:

Cajka

Nov 28th, 2012, 03:49 AM

Can OP give us an answer? :lol:

Lin Lin

Nov 28th, 2012, 03:51 AM

a piece of cake:lol:

égalité

Nov 28th, 2012, 03:51 AM

If you ever teach primary school students, may god help them.

I don't understand what is bothering you so much. I said nothing wrong. :hysteric:

But thankfully for all the primary school students out there, I'm only teaching college courses at the moment and get stellar reviews from my students every semester.

Also, the link I provided you goes into detail about how 25 is an impossible answer, whether the "diagonal" in the picture is a straight line or not. I honestly have no idea what you're talking about and what I said to rile you up.

I don't understand what is bothering you so much. I said nothing wrong. :hysteric:

But thankfully for all the primary school students out there, I'm only teaching college courses at the moment and get stellar reviews from my students every semester.

Also, the link I provided you goes into detail about how 25 is an impossible answer, whether the "diagonal" in the picture is a straight line or not. I honestly have no idea what you're talking about and what I said to rile you up.

LeRoy.

Nov 28th, 2012, 03:54 AM

I don't understand what is bothering you so much. :hysteric:

Much like your fave, your total lack of practical reasoning.

....(and a formal education) :haha:

Much like your fave, your total lack of practical reasoning.

....(and a formal education) :haha:

égalité

Nov 28th, 2012, 03:56 AM

Much like your fave, your total lack of practical reasoning.

....(and a formal education) :haha:

Elaborate. And I suppose you're not going to address the rest of the post. :lol:

All I said is that the drawing is not to scale, which is the truth. :spit: The two correct answers look like this:

http://www.aaasurvey.com/gif/math456/psle2005c.gifhttp://www.aaasurvey.com/gif/math456/psle2005f.gif

....(and a formal education) :haha:

Elaborate. And I suppose you're not going to address the rest of the post. :lol:

All I said is that the drawing is not to scale, which is the truth. :spit: The two correct answers look like this:

http://www.aaasurvey.com/gif/math456/psle2005c.gifhttp://www.aaasurvey.com/gif/math456/psle2005f.gif

wild.river

Nov 28th, 2012, 03:56 AM

wait...the link assumes VW and TU are lines and area VXUQ is 32 and TSWX 8. maybe they aren't!

then this is unsolvable.

then this is unsolvable.

Morning Morgan

Nov 28th, 2012, 03:57 AM

Well, I won't give the answer yet. :angel: But I will post the following:

This problem makes us want to assume the following:

1) SQ is a straight line.

2) The rectangle is 15 by 6.

3) Area of triangle A is 4.

4) Area of triangle B is 16.

Stare at these 4 assumptions and see what you get. :angel:

And egalite, I'm really sorry, this must be so frustrating for you. :oh:

This problem makes us want to assume the following:

1) SQ is a straight line.

2) The rectangle is 15 by 6.

3) Area of triangle A is 4.

4) Area of triangle B is 16.

Stare at these 4 assumptions and see what you get. :angel:

And egalite, I'm really sorry, this must be so frustrating for you. :oh:

LeRoy.

Nov 28th, 2012, 03:57 AM

Also, the link I provided you goes into detail about how 25 is an impossible answer, whether the "diagonal" in the picture is a straight line.....

Area of Rectangle C = 45 - 4 - 16 = 25 cm2

The above calculation is correct provided RST is a straight line.

Area of Rectangle C = 45 - 4 - 16 = 25 cm2

The above calculation is correct provided RST is a straight line.

Cajka

Nov 28th, 2012, 03:58 AM

I don't understand what is bothering you so much. I said nothing wrong. :hysteric:

But thankfully for all the primary school students out there, I'm only teaching college courses at the moment and get stellar reviews from my students every semester.

Also, the link I provided you goes into detail about how 25 is an impossible answer, whether the "diagonal" in the picture is a straight line or not. I honestly have no idea what you're talking about and what I said to rile you up.

Since you're here, I have a question. How is the area of B 16cm2? :sobbing: For me, that's the part that makes it all confusing.

But thankfully for all the primary school students out there, I'm only teaching college courses at the moment and get stellar reviews from my students every semester.

Also, the link I provided you goes into detail about how 25 is an impossible answer, whether the "diagonal" in the picture is a straight line or not. I honestly have no idea what you're talking about and what I said to rile you up.

Since you're here, I have a question. How is the area of B 16cm2? :sobbing: For me, that's the part that makes it all confusing.

LeRoy.

Nov 28th, 2012, 04:01 AM

Elaborate. And I suppose you're not going to address the rest of the post. :lol:

This is a multiple choice problem for primary school students who HAVE to pick ONE answer.

"Most pupils gave the 'correct' answer of 25 cm2 based on the incorrect figure. If you start with an incorrect figure, you are not going to get a correct answer. "

And I did. :awww:

This is a multiple choice problem for primary school students who HAVE to pick ONE answer.

"Most pupils gave the 'correct' answer of 25 cm2 based on the incorrect figure. If you start with an incorrect figure, you are not going to get a correct answer. "

And I did. :awww:

égalité

Nov 28th, 2012, 04:05 AM

Area of Rectangle C = 45 - 4 - 16 = 25 cm2

The above calculation is correct provided RST is a straight line.

Okay, fine, this is vacuously true. So what? That's like saying "If the capital of Finland is Madrid, then the answer is 25." The fact is that the diagonal is not a straight line. It is impossible for the diagonal to be a straight line given that triangles A and B have areas 4 and 16.

Also, through equally valid reasoning "assuming RST is a straight line," you can get 20 as an answer.

Since the area of B is 4 times the area of A, the sides of B are twice as long as the sides of A. This is true because RST being a straight line implies A and B are similar. So TX=5, XU=10, and WX=2, XV=4. Therefore rectangle C has area 2x10=20.

The above calculation is correct provided RST is a straight line.

Okay, fine, this is vacuously true. So what? That's like saying "If the capital of Finland is Madrid, then the answer is 25." The fact is that the diagonal is not a straight line. It is impossible for the diagonal to be a straight line given that triangles A and B have areas 4 and 16.

Also, through equally valid reasoning "assuming RST is a straight line," you can get 20 as an answer.

Since the area of B is 4 times the area of A, the sides of B are twice as long as the sides of A. This is true because RST being a straight line implies A and B are similar. So TX=5, XU=10, and WX=2, XV=4. Therefore rectangle C has area 2x10=20.

LeRoy.

Nov 28th, 2012, 04:09 AM

Okay, fine, this is vacuously true....

Thank you !

http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif

Now, sashay away.

http://24.media.tumblr.com/tumblr_ly6f2sOSdN1rn95k2o1_500.gif

Thank you !

http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif

Now, sashay away.

http://24.media.tumblr.com/tumblr_ly6f2sOSdN1rn95k2o1_500.gif

égalité

Nov 28th, 2012, 04:11 AM

Thank you !

Now, sashay away.

http://24.media.tumblr.com/tumblr_ly6f2sOSdN1rn95k2o1_500.gif

Gurl no. You can read the rest of the post. You're not right. Get over it. 20 is just as "correct" an answer as the one you're fapping over.

Now, sashay away.

http://24.media.tumblr.com/tumblr_ly6f2sOSdN1rn95k2o1_500.gif

Gurl no. You can read the rest of the post. You're not right. Get over it. 20 is just as "correct" an answer as the one you're fapping over.

Novichok

Nov 28th, 2012, 04:12 AM

Gurl no. You can read the rest of the post. You're not right. Get over it.

LOL.

LOL.

LeRoy.

Nov 28th, 2012, 04:13 AM

FOR A PRIMARY SCHOOL KID, the correct answer to the problem stated is 25.

http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif)http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif)

http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif)http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif)

Cajka

Nov 28th, 2012, 04:15 AM

If we would agree that the answer is 25, wouldn't that mean that the area of the unnamed square (let's call it "D"), would also be 25? :unsure:

Novichok

Nov 28th, 2012, 04:15 AM

FOR A PRIMARY SCHOOL KID, the correct answer to the problem stated is 25.

http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif)http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif)

egalite, are the truths of mathematics contingent?

http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif37.gif)http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif36.gif)

egalite, are the truths of mathematics contingent?

LeRoy.

Nov 28th, 2012, 04:17 AM

egalite, are the truths of mathematics contingent?

Wipe the cum out of your eyes and quote and/or address the right person ?

Wipe the cum out of your eyes and quote and/or address the right person ?

égalité

Nov 28th, 2012, 04:18 AM

egalite, are the truths of mathematics contingent?

I die at this thread. :hysteric:

No they are not. Thank you.

If we would agree that the answer is 25, wouldn't that mean that the are of the unnamed square (let's call it "D"), would also be 25? :unsure:

Yes, which would mean that triangles A and B also should be the same area. Which is why 25 is wrong, even FOR A PRIMARY SCHOOL KID. :hysteric: :hysteric: :hysteric:

I die at this thread. :hysteric:

No they are not. Thank you.

If we would agree that the answer is 25, wouldn't that mean that the are of the unnamed square (let's call it "D"), would also be 25? :unsure:

Yes, which would mean that triangles A and B also should be the same area. Which is why 25 is wrong, even FOR A PRIMARY SCHOOL KID. :hysteric: :hysteric: :hysteric:

Novichok

Nov 28th, 2012, 04:21 AM

Wipe the cum out of your eyes and quote and/or address the right person ?

It's okay. You can't be right all the time. :hug:

It's okay. You can't be right all the time. :hug:

Cajka

Nov 28th, 2012, 04:23 AM

Yes, which would mean that triangles A and B also should be the same area. Which is why 25 is wrong, even FOR PRIMARY SCHOOL STUDENTS. :hysteric: :hysteric: :hysteric:

My idea is this: B is not the triangle at all. :oh: B is actually the unnamed square above the triangle A. So, B is 16 = 4*4. So, my theory is correct, C is 22. :oh:

My idea is this: B is not the triangle at all. :oh: B is actually the unnamed square above the triangle A. So, B is 16 = 4*4. So, my theory is correct, C is 22. :oh:

LeRoy.

Nov 28th, 2012, 04:26 AM

It's okay. You can't be right all the time. :hug:

I assume this is for egalite then ?

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

I assume this is for egalite then ?

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

égalité

Nov 28th, 2012, 04:28 AM

I assume this is for egalite then ?

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

This means about as much to me as being told by GoDokic that I'm rarely right. It's probably a sign that you're just not keeping up.

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

This means about as much to me as being told by GoDokic that I'm rarely right. It's probably a sign that you're just not keeping up.

Novichok

Nov 28th, 2012, 04:28 AM

I assume this is for egalite then ?

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

His argument for the order of operations wasn't right. :oh:

He's right this time. But you aren't. :awww:

He is rarely right. :D

http://media.tumblr.com/tumblr_lbe3m1zHEp1qcde8x.gif

His argument for the order of operations wasn't right. :oh:

He's right this time. But you aren't. :awww:

LeRoy.

Nov 28th, 2012, 04:31 AM

http://blog.vh1.com/files//2010/04/dr2_9_252gif17.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif17.gif)http://blog.vh1.com/files//2010/04/dr2_9_252gif17.gif (http://blog.vh1.com/files//2010/04/dr2_9_252gif17.gif)

http://25.media.tumblr.com/tumblr_lpbxj1hayi1r0zbwoo1_400.gif

http://25.media.tumblr.com/tumblr_lpbxj1hayi1r0zbwoo1_400.gif

Cajka

Nov 28th, 2012, 04:33 AM

My "the wrong B" theory totally works. :sobbing: 15-4 is 11, 6-4 is 2. 2 by 11 rectangle it is. :oh:

Novichok

Nov 28th, 2012, 04:37 AM

My "the wrong B" theory totally works. :sobbing: 15-4 is 11, 6-4 is 2. 2 by 11 rectangle it is. :oh:

I don't think it does. :oh:

I don't think it does. :oh:

Cajka

Nov 28th, 2012, 04:39 AM

I don't think it does. :oh:

I know it doesn't, 16 is not 25, but who cares. :oh:

Oh no, in that case the new unnamed triangle D is 25. :oh:

I know it doesn't, 16 is not 25, but who cares. :oh:

Oh no, in that case the new unnamed triangle D is 25. :oh:

Morning Morgan

Nov 28th, 2012, 04:40 AM

LOL this thread is turning into the hot mess I was predicting it would be

Cajka

Nov 28th, 2012, 04:44 AM

LOL this thread is turning into the hot mess I was predicting it would be

Explain this whole mess, please. :lol:

Explain this whole mess, please. :lol:

Lin Lin

Nov 28th, 2012, 05:05 AM

It's 25:cool:

Lin Lin

Nov 28th, 2012, 05:09 AM

I think people shoudl solve this in less than 30 seconds,it's 25:shrug:

Dominic

Nov 28th, 2012, 05:13 AM

25 and that's not primary level.

Flavia P.

Nov 28th, 2012, 05:13 AM

LOL this thread is turning into the hot mess I was predicting it would be

At least we got to talk :oh:

At least we got to talk :oh:

Lin Lin

Nov 28th, 2012, 05:18 AM

WS=a

WX=b

XV=c

VQ=d

a/b=d/c(similar triangles)=>ac=bd(means the aera of C(m) = the aera of rectangle PVXT(n).)

m+n=15*6-2A-2B

2*m=90-2*4-2*16=50

m=25.

In less than 30 seconds.:p

WX=b

XV=c

VQ=d

a/b=d/c(similar triangles)=>ac=bd(means the aera of C(m) = the aera of rectangle PVXT(n).)

m+n=15*6-2A-2B

2*m=90-2*4-2*16=50

m=25.

In less than 30 seconds.:p

égalité

Nov 28th, 2012, 05:19 AM

WS=a

WX=b

XV=c

VQ=d

a/b=d/c(similar triangles)=>ac=bd(means the aera of C(x1) = the aera of rectangle PVXT(x2).)

X1+x2=15*6-2A-AB

2*X1=90-2*4-2*16=50

X1=25.

In less than 30 seconds.:p

You should read the rest of the thread :hysteric:

WX=b

XV=c

VQ=d

a/b=d/c(similar triangles)=>ac=bd(means the aera of C(x1) = the aera of rectangle PVXT(x2).)

X1+x2=15*6-2A-AB

2*X1=90-2*4-2*16=50

X1=25.

In less than 30 seconds.:p

You should read the rest of the thread :hysteric:

Lin Lin

Nov 28th, 2012, 05:50 AM

:unsure:

Cajka

Nov 28th, 2012, 05:53 AM

:unsure:

long story short - try to draw the figure with the numbers you got as a result and you'll see it.

long story short - try to draw the figure with the numbers you got as a result and you'll see it.

Lin Lin

Nov 28th, 2012, 06:04 AM

ok:unsure:

JJ Expres

Nov 28th, 2012, 10:15 AM

Egalite is right( i wrote the same shit at the beginning.)

@MOrgan Diagonal is straight line (ofc :rolleyes:), the thing egalite wants to say is that SXQ(SXQ != SQ) in not a straight line so it's not dividing big rectangle in 2 equal areas but wte...there are 2 answers...

@MOrgan Diagonal is straight line (ofc :rolleyes:), the thing egalite wants to say is that SXQ(SXQ != SQ) in not a straight line so it's not dividing big rectangle in 2 equal areas but wte...there are 2 answers...

JJ Expres

Nov 28th, 2012, 10:17 AM

and my mistake in post 1 is that i used Similar triangles to solve it and obviously there are not similar because of the diagonal...

Kəv.

Nov 28th, 2012, 10:33 AM

Wally picking all 4 :hysteric:

Super Dave

Nov 28th, 2012, 02:24 PM

http://cdn.memegenerator.net/instances/400x/23705197.jpg

Not enough information given, not drawn to scale, etc.

Not enough information given, not drawn to scale, etc.

Gagsquet

Nov 28th, 2012, 02:31 PM

I don't know. Wonder why I studied maths during 14 years.

Kon.

Nov 28th, 2012, 03:40 PM

It's no 6÷2（2+1）but this thread is still funny. :sobbing:

Igorche

Nov 28th, 2012, 04:32 PM

rectangle SRQP = 15 * 6 = 90

triangle SRQ = 90 / 2 = 45

A = 4

triangle XUQ = B = 16

triangle SRQ = A + B + C => C = triangle SRQ - A - B = 45 - 4 -16 = 25

triangle SRQ = 90 / 2 = 45

A = 4

triangle XUQ = B = 16

triangle SRQ = A + B + C => C = triangle SRQ - A - B = 45 - 4 -16 = 25

Sally Struthers

Nov 28th, 2012, 08:30 PM

let's just agree that it is a poorly constructed question

GoofyDuck

Nov 28th, 2012, 08:35 PM

I can't :sad:

vBulletin® v3.6.8, Copyright ©2000-2014, Jelsoft Enterprises Ltd.