solve this if you got the brains

Typing...

iOluwaMichael:
Typing...

waiting

This simple thing? Where's this primary four boy, come and see o

(3.5) + (4.5)=8,
(9.5) - (3.5)=6.
.......................
Now add 3.5 and 9.5 you have 13.
Then add 4.5 and 3.5 you have 8.

Simple!

Hmmmmmm

3.5 + 4.5 =8 9.5 - 3.5= 6 3.5+9.5 =13 4.5+3.5=8

chai some people go don solve mathematics full plain sheet...where are the copy and paste crew it's your turn to answer the question

I see it has already been slaughtered

veekid:
This simple thing? Where's this primary four boy, come and see o

lol...you fit carry mouth cut down Iroko tree...you get mouth but answer you no give oh

aragon4realz:

lol...you fit carry mouth cut down Iroko tree...you get mouth but answer you no give oh

Swear say you no know the answer

aragon4realz:

waiting

Wait no more. Ademasta has slaughtered it. This is too easy na.... *winks*...

let a+b=8 a+c=13 b+d=8 c-d=6
a+b=(13-c)+(8-d)=8 c=6+d ; solving this way,we get (d=3.5),(c=9.5),(b=8-3.5=4.5),(a=8-4.5=3.5)

The onez above me used Google search

Doesn't take a genius jare, it's easy. The guys above got it, but here are my workings.

There are 4 unknowns and 4 equations. Let the unknowns a, b, c, and d. So the equations become:
a + b = 8 .................. (1)
b + d = 8 .................. (2)
a + c = 13 .................. (3)
c - d = 6 .................. (4)

Comparing (1) and (2), we notice that
a = d

And comparing (1) and (3), we also notice that
c = b + 5

Armed with this new info, let's play with (4). Substitute a with d and c with b + 5. We'd have:
b + 5 - a = 6

Subtract 5 from both sides to eliminate 5 from the left hand side. It becomes:
b - a = 1
-a + b = 1 .................. (5)
Recall
a + b = 8 .................. (1)
Now we see that (5) has the same variables as (1) so eliminate one variable. If you add both equations, you'd eliminate a, but if you subtract, b goes. Let's add.
-a + a + b + b = 1 + 8
0 + 2b = 9
b = 4.5

but a + b = 8
Therefore
a = 8 - b = 8 - 4.5 = 3.5

d = a = 3.5
Then
c - d = 6
c - 3.5 = 6
c = 6 + 3.5 = 9.5

So the original equations are
3.5 + 4.5 = 8 .................. (1)
4.5 + 3.5 = 8 .................. (2)
3.5 + 9.5 = 13 .................. (3)
9.5 - 3.5 = 6 .................. (4)


Don't mind me, there are shorter routes. To type long story dey hungry me.

Oahray:
Doesn't take a genius jare, it's easy. The guys above got it, but here are my workings.

There are 4 unknowns and 4 equations. Let the unknowns a, b, c, and d. So the equations become:
a + b = 8 .................. (1)
b + d = 8 .................. (2)
a + c = 13 .................. (3)
c - d = 6 .................. (4)

Comparing (1) and (2), we notice that
a = d

And comparing (1) and (3), we also notice that
c = b + 5

Armed with this new info, let's play with (4). Substitute a with d and c with b + 5. We'd have:
b + 5 - a = 6

Subtract 5 from both sides to eliminate 5 from the left hand side. It becomes:
b - a = 1
-a + b = 1 .................. (5)
Recall
a + b = 8 .................. (1)
Now we see that (5) has the same variables as (1) so eliminate one variable. If you add both equations, you'd eliminate a, but if you subtract, b goes. Let's add.
-a + a + b + b = 1 + 8
0 + 2b = 9
b = 4.5

but a + b = 8
Therefore
a = 8 - b = 8 - 4.5 = 3.5

d = a = 3.5
Then
c - d = 6
c - 3.5 = 6
c = 6 + 3.5 = 9.5

So the original equations are
3.5 + 4.5 = 8 .................. (1)
4.5 + 3.5 = 8 .................. (2)
3.5 + 9.5 = 13 .................. (3)
9.5 - 3.5 = 6 .................. (4)


Don't mind me, there are shorter routes. To type long story dey hungry me.

you just proved not to belong to the copy and paste crew good work

iOluwaMichael:
Wait no more. Ademasta has slaughtered it. This is too easy na.... *winks*...

Slaughtered like Christmas chicken

Hi guys ,

I thought i could do it easily using Brute Force SQL but it gave a tuff time

First i tried with whole numbers but it didnt work but then i found this site (I got the puzzle in a whatsapp group). Once here i saw the answer in decimals.

Of course then i solved it easily with Decimal brute force

use dba

drop table tableA
drop table TableAandC


create table tableA (a bigint)



declare @icount int
set @icount = -50000
while @icount < 50000
begin

insert into Tablea select @icount
set @icount = @icount +1
end

select * From tablea a, tablea b, tablea c , tablea d
where a.a + b.a = 8 and a.a + c.a = 13 and b.a + d.a = 8 and c.a- d.a = 6


--TRY to do it in parts first find all combinations that can have a+b = 8 and a+ c = 13 then check for the rest two


select * From tablea a, tablea c
where a.a + c.a = 13





select A.a AS A , C.A as C into TableAandC From tablea a, tablea c
where a.a + c.a = 13


--Nothing found with whole numbers
-- So lets try with decimals


/**SECOND APPROACH ***/
use dba

drop table tableA


create table tableA (a money , b money,c money,d money)



declare @icount money
set @icount = -10.00
while @icount < 100.00
begin

insert into Tablea select @icount , 8-@icount , 13-@icount , 13-@icount - 6
set @icount = @icount +0.1
end

select * From tablea a
where a.a + b = 8 and a + c = 13 and b+ d = 8 and c- d = 6

The above query gives correct answer. The code will work in SQL SERVER

funooni:
Hi guys ,

I thought i could do it easily using Brute Force SQL but it gave a tuff time

First i tried with whole numbers but it didnt work but then i found this site (I got the puzzle in a whatsapp group). Once here i saw the answer in decimals.

Of course then i solved it easily with Decimal brute force

use dba

drop table tableA
drop table TableAandC


create table tableA (a bigint)



declare @icount int
set @icount = -50000
while @icount < 50000
begin

insert into Tablea select @icount
set @icount = @icount +1
end

select * From tablea a, tablea b, tablea c , tablea d
where a.a + b.a = 8 and a.a + c.a = 13 and b.a + d.a = 8 and c.a- d.a = 6


--TRY to do it in parts first find all combinations that can have a+b = 8 and a+ c = 13 then check for the rest two


select * From tablea a, tablea c
where a.a + c.a = 13





select A.a AS A , C.A as C into TableAandC From tablea a, tablea c
where a.a + c.a = 13


--Nothing found with whole numbers
-- So lets try with decimals


/**SECOND APPROACH ***/
use dba

drop table tableA


create table tableA (a money , b money,c money,d money)



declare @icount money
set @icount = -10.00
while @icount < 100.00
begin

insert into Tablea select @icount , 8-@icount , 13-@icount , 13-@icount - 6
set @icount = @icount +0.1
end

select * From tablea a
where a.a + b = 8 and a + c = 13 and b+ d = 8 and c- d = 6

The above query gives correct answer. The code will work in SQL SERVER

lol alright maybe someday you will teach me data base management or sq language