Any Mathematician In The House...try Solving This Trick Question. - Forum Games

Any Mathematician In The House...try Solving This Trick Question. - Forum Games - Nairaland

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Any Mathematician In The House...try Solving This Trick Question. by Legacy44(op): 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(op): 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
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(op): 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...

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...?

ohkay Sir

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(op): 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(op): 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(op): 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(op): 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(op): 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.

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.