Rangpur Higher Secondary 2011/7
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Rangpur Higher Secondary 2011/7
Problem 7:
If $A + B =1$, $B + C = 2$, $C + D = 3$, ..., $X + Y = 25$, $Y + Z = 26$, find $A – Z$.
If $A + B =1$, $B + C = 2$, $C + D = 3$, ..., $X + Y = 25$, $Y + Z = 26$, find $A – Z$.
 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Rangpur Higher Secondary 2011/7
$A+B=1$
$B+C=2$
$C+D=3$
. . .
. . .
$X+Y=25$
$Y+Z=26$
( biog ) we get
$AZ$=$123. . . .2526$
so $AZ=1(1+2+. . . .+25+26)+1$
the result is $349$.
$B+C=2$
$C+D=3$
. . .
. . .
$X+Y=25$
$Y+Z=26$
( biog ) we get
$AZ$=$123. . . .2526$
so $AZ=1(1+2+. . . .+25+26)+1$
the result is $349$.
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....

 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: Rangpur Higher Secondary 2011/7
Your formula is uncorrect.(A+B)(B+C)=AC
AC+CD=AD
(AD)+(D+E)=AE
AE+E+F=A+F
DO IT CONSECUTIVELY AND GET THE ANS 13
AC+CD=AD
(AD)+(D+E)=AE
AE+E+F=A+F
DO IT CONSECUTIVELY AND GET THE ANS 13
 Tahmid Hasan
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 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Rangpur Higher Secondary 2011/7
let's name the eqs 1,2,3,......25.then $12+3......$
get the ans
get the ans
বড় ভালবাসি তোমায়,মা
 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Rangpur Higher Secondary 2011/7
u misunderstood (may be)[\B]Mehfuj Zahir wrote:Your formula is uncorrect.(A+B)(B+C)=AC
AC+CD=AD
(AD)+(D+E)=AE
AE+E+F=A+F
DO IT CONSECUTIVELY AND GET THE ANS 13
thats a sequence from A to Z.

 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: Rangpur Higher Secondary 2011/7
By doing it at last u get Az=13
 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Rangpur Higher Secondary 2011/7
the Left side of Equation is $Az$ ok? So the right side is
$1234. . . . . . .2526$
=$1(2+3+4+5. . . .+25+26)$
=$1(1+2+3+4. . . .+25+26)+1$
=$2(27*13)$
=$2351$
=$349$
... There are 26 equations. When we substract 26 equation at a time the first and last variables will left on the left side and one right side only the value of first equation will have positive sign.
or use a CALCULATOR.
$1234. . . . . . .2526$
=$1(2+3+4+5. . . .+25+26)$
=$1(1+2+3+4. . . .+25+26)+1$
=$2(27*13)$
=$2351$
=$349$
... There are 26 equations. When we substract 26 equation at a time the first and last variables will left on the left side and one right side only the value of first equation will have positive sign.
or use a CALCULATOR.
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....

 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: Rangpur Higher Secondary 2011/7
it will like 12+34+5
 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Rangpur Higher Secondary 2011/7
i told to substract all the equation.
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....

 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: Rangpur Higher Secondary 2011/7
If u dp it u don't have Az