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Dhaka Higher Secondary 2011/4

Posted: Fri Jan 28, 2011 10:27 pm
by BdMO
\[\begin{align*}
a + b + c + d + e =& 1\\ a - b + c + d + e =& 2 \\ a + b - c + d + e =& 3\\ a + b + c - d + e =& 4\\ a + b + c + d - e =& 5\end{align*} \]
Find $a$ from the above set of equations.

Re: Dhaka Higher Secondary 2011/4

Posted: Sat Jan 29, 2011 12:22 pm
by leonardo shawon
what does form mean... Anyone please explain...

Re: Dhaka Higher Secondary 2011/4

Posted: Sat Jan 29, 2011 1:59 pm
by HandaramTheGreat
\[ a+b+c+d+e=1 \] \[b+c+d+e=1-a \]

if we add all the 5 equations, we'll get
\[ 5a+4(b+c+d+e)=15 \] \[ 5a+4(1-a)=15 \] \[ \therefore a=11 \]

Re: Dhaka Higher Secondary 2011/4

Posted: Sat Jan 29, 2011 4:00 pm
by leonardo shawon
i found different answers.....

$a+b+c+d+e=1$ its equation 1.

$a-b+c+d+e=2$ from equation 1 $ a+c+d+e=1-b$ so $b=-1/2$
by doing same in each equation, i have got $a=6$ b=-0.5 c=-1 d=-1.5 and e=-2

if im wrong, please give the correct sollution and whats the meaning question. Somebody please make it clear to me. ! ! !

Re: Dhaka Higher Secondary 2011/4

Posted: Sat Jan 29, 2011 6:25 pm
by Tahmid Hasan
i also came up with 11 :ugeek:

Re: Dhaka Higher Secondary 2011/4

Posted: Sat Jan 29, 2011 8:17 pm
by leonardo shawon
HandaramTheGreat wrote:\[ a+b+c+d+e=1 \] \[b+c+d+e=1-a \]

if we add all the 5 equations, we'll get
\[ 5a+4(b+c+d+e)=15 \] \[ 5a+4(1-a)=15 \] \[ \therefore a=11 \]
bro u made a mistake..

If we add 5 equations, we get
$5a+3(b+c+d+e)=15$
$5a-3a=15-3$
$a=6$
:)

Re: Dhaka Higher Secondary 2011/4

Posted: Sun Jan 30, 2011 10:00 am
by HandaramTheGreat
:oops: sorry...

Re: Dhaka Higher Secondary 2011/4

Posted: Sun Jan 30, 2011 11:22 am
by leonardo shawon
its okay Bro... :) dont worry.

Re: Dhaka Higher Secondary 2011/4

Posted: Tue Sep 04, 2012 11:59 am
by farhan
it was a very easy one....hohoho...bt many got perplexed..