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Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 9:48am On Jan 29, 2015
If 4a=3b=a+b+c
find a, b, c.

Pls show ur workings clearly.
Re: Any Mathematician In The House...try Solving This Trick Question. by Nobody: 10:02am On Jan 29, 2015
Legacy44:
If 4a=3b=a+b+c
find a, b, c.

Pls show ur workings clearly.

You teach mathematics in the school?
Re: Any Mathematician In The House...try Solving This Trick Question. by derrick333(m): 10:03am On Jan 29, 2015
Shud I use almighty formula ? grin

1 Like

Re: Any Mathematician In The House...try Solving This Trick Question. by mideeniafe(f): 10:10am On Jan 29, 2015
Hmmm..lemme try

4a=3b=a+b+c
3b=a+b+c-4a
x=a+b+c-4a-3b undecided
Using Almighty Formula
x= -b +/- square root (b2 - 4ac)/2a
x= -1+/- .........


Wo....I don't know jor
Re: Any Mathematician In The House...try Solving This Trick Question. by Nobody: 10:13am On Jan 29, 2015
Mathematical your attention is needed here. Do the needful.
Re: Any Mathematician In The House...try Solving This Trick Question. by olusolaj(m): 10:26am On Jan 29, 2015
a=1
b=4/3
c=5/3

I'm too busy to type the workings for now.
I shall show it when I'm less busy

1 Like

Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 10:30am On Jan 29, 2015
check d workings in d pix posted ...
a=3, b= 4, c=5

3 Likes

Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 10:33am On Jan 29, 2015
U'all didnt get it right. Hint: a,b,c are all integers...

Yes i teach maths...nysc stuff
Re: Any Mathematician In The House...try Solving This Trick Question. by Mathematical(f): 10:35am On Jan 29, 2015
Like I understand your illogical mathe
Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 10:39am On Jan 29, 2015
Legacy44:
U'all didnt get it right. Hint: a,b,c are all integers...

Yes i teach maths...nysc stuff
enable image on your phone and see d workings I posted
Re: Any Mathematician In The House...try Solving This Trick Question. by Mathematical(f): 10:40am On Jan 29, 2015
If abc re the integers.

Then 4a = 3b
b = 4a/3

2b = 5c
b = 5c/2
Re: Any Mathematician In The House...try Solving This Trick Question. by ArchEnemy(m): 11:48am On Jan 29, 2015
mideeniafe:
Hmmm..lemme try

4a=3b=a+b+c
3b=a+b+c-4a
x=a+b+c-4a-3b undecided
Using Almighty Formula
x= -b +/- square root (b2 - 4ac)/2a
x= -1+/- .........


Wo....I don't know jor
lol
I join you in giving this up
Re: Any Mathematician In The House...try Solving This Trick Question. by ArchEnemy(m): 11:52am On Jan 29, 2015
AdeniyiA:
check d workings in d pix posted ...
a=3, b= 4, c=5
all hail the Prof!
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 12:17pm On Jan 29, 2015
AdeniyiA:
check d workings in d pix posted ...
a=3, b= 4, c=5

sorry but ur workings is not clear...it doesn't explain ur answer
Re: Any Mathematician In The House...try Solving This Trick Question. by Creamish(f): 1:17pm On Jan 29, 2015
Legacy44:


sorry but ur workings is not clear...it doesn't explain ur answer

But his answers are right... smiley

3 Likes

Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 1:23pm On Jan 29, 2015
Legacy44:


sorry but ur workings is not clear...it doesn't explain ur answer
you mean my approach is not clear to you,
it's a right approach and the answers are correct. You can request for further explanation if u wish.

1 Like

Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 1:25pm On Jan 29, 2015
ArchEnemy:
all hail the Prof!
I wan thief am...? undecided
Re: Any Mathematician In The House...try Solving This Trick Question. by ArchEnemy(m): 1:33pm On Jan 29, 2015
AdeniyiA:

I wan thief am...? undecided
ohkay Sir smiley
Re: Any Mathematician In The House...try Solving This Trick Question. by Nobody: 2:06pm On Jan 29, 2015
Legacy44:
U'all didnt get it right. Hint: a,b,c are all integers...

Yes i teach maths...nysc stuff
AdeniyiA solved it already.
Re: Any Mathematician In The House...try Solving This Trick Question. by Nobody: 2:12pm On Jan 29, 2015
AdeniyiA:
check d workings in d pix posted ...
a=3, b= 4, c=5
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 5:37pm On Jan 29, 2015
AdeniyiA:

you mean my approach is not clear to you,
it's a right approach and the answers are correct. You can request for further explanation if u wish.

yeah...i need further explanation. I understand the following deductions:
a/c=3/5, c/b=5/4, a/b=3/4.

But how did the above translate to
a=3, b=4, c=5
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 5:42pm On Jan 29, 2015
olusolaj:
a=1
b=4/3
c=5/3

I'm too busy to type the workings for now.
I shall show it when I'm less busy

Pls show ur working
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 5:45pm On Jan 29, 2015
AdeniyiA:

you mean my approach is not clear to you,
it's a right approach and the answers are correct. You can request for further explanation if u wish.

yeah...i need further explanation. I understand the following deductions:
a/c=3/5, c/b=5/4, a/b=3/4.

But how did the above translate to
a=3, b=4, c=5
because it seems like u jus guessed what should have been there...olusolaj also apparently got it 'right' cos it fitted perfectly to the question. His answer was a=1, b=4/3, C=5/3.

Pls clear workings will justify ur answer not assumptions
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 5:57pm On Jan 29, 2015
AdeniyiA:

you mean my approach is not clear to you,
it's a right approach and the answers are correct. You can request for further explanation if u wish.

yeah...i need further explanation. I understand the following deductions:
a/c=3/5, c/b=5/4, a/b=3/4.

But how did the above translate to
a=3, b=4, c=5
because it seems like u jus guessed what should have been there...olusolaj also apparently got it 'right' cos it fitted perfectly to the question. His answer was a=1, b=4/3, C=5/3.

I can also say a=3/2, b=2, c=5/2 AND THE ANSWER WILL BE JUSTIFIABLE....it still doesn't make it right.

Pls clear workings will justify ur answer not assumptions
Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 6:25pm On Jan 29, 2015
Legacy44:


yeah...i need further explanation. I understand the following deductions:
a/c=3/5, c/b=5/4, a/b=3/4.

But how did the above translate to
a=3, b=4, c=5

I neither deduce nor assume.
my answers were simply arrived at through simple ratio relation.
If u break down olusolaj 's answers, they all come down to 3,4 & 5 ,by mere multiplying through by 3. cool

1 Like

Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 6:51pm On Jan 29, 2015
AdeniyiA:


I neither deduce nor assume.
my answers were simply arrived at through simple ratio relation.
If u break down olusolaj 's answers, they all come down to 3,4 & 5 ,by mere multiplying through by 3. cool

U didnt answer my question...i said explain the last part i highlighted frm ur calculations
Re: Any Mathematician In The House...try Solving This Trick Question. by PuntGenius: 8:38pm On Jan 29, 2015
I got a=0,b=0 and c=0
Re: Any Mathematician In The House...try Solving This Trick Question. by AdeniyiA(m): 8:55pm On Jan 29, 2015
Legacy44:


U didnt answer my question...i said explain the last part i highlighted frm ur calculations
a/c=3/5, c/b=5/4, a/b=3/4.
simple ratio my dear.
Three variables nd three numbers with at least a variable appearing in two places d same .
You can even check /work it backwards.
Re: Any Mathematician In The House...try Solving This Trick Question. by tundeayo2020: 11:29pm On Jan 29, 2015
hmm solving



Modified... A=0, b=0, c=0. Or a=6, b=8, c=10. Op is it correct?
Re: Any Mathematician In The House...try Solving This Trick Question. by Legacy44(f): 12:02am On Jan 30, 2015
tundeayo2020:
hmm solving



Modified... A=0, b=0, c=0. Or a=6, b=8, c=10. Op is it correct?


yh
Re: Any Mathematician In The House...try Solving This Trick Question. by Nobody: 7:46am On Jan 30, 2015
tundeayo2020:
hmm solving



Modified... A=0, b=0, c=0. Or a=6, b=8, c=10. Op is it correct?

Legacy44:

yh


Show the working lets see.
Re: Any Mathematician In The House...try Solving This Trick Question. by PuntGenius: 8:35am On Jan 30, 2015
I got a=0, b=0 and c=0 from my workings, but 3, 4 and 5 or 6, 4 and 8vare also correct but I can't seem to work that out

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