Question

the sum of the digits of a two digit number is 12. if the digits are reversed,the new number becomes 4/7 times the original number find the original number

Answer 1

here you go ☆☆

▪let digit at ones place = x
and at tens place= y

▪so, no. is 10y + x

▪given, y+x= 12------《1》

▪if no. is reversed , then

▪10x+ y = 4/7(10y +x)

▪10x-4/7x + y -40/7y

66/7x-33/7y = 0

▪multiply term with 7 , we get ,

▪66x -33y = 0

taking 33 as common,

▪2x -y =0------《2》

▪by elimination method,
multiply 1 by 2 and subtract 2 from 1 ,we get,

▪2x +2y =24
▪2x - y = 0
- + -
▪3y = 24
▪y = 8

▪so x= 4

▪so original no.= 84

hope it helps you....

Answer 2

Let the digit at tens place be 'x' and the digit at the once place be 'y'
Therefore the number must be: 10x+y
According to the question
= (x+y) =12 - - - - - - - - - eqn (1)
Again case II
= 10y+x = 4/7(10x+y)
= 10y+x= 40x/7+4y/7 ( TALKING LCM)
= 10Y+X= (40x+4y)/7 (Transposing) 7 to other side, we obtain
= 70y+7x=40x+4y
= 70y-4y+7x-40x
= 66y-33x= 0 ( it can be written as below)
= 33x-66y= 0
= 3x-6y=0---------------eqn (2)
NOW USING elimination method and eliminating x by multiplying equation 1 with 3 and equation 2 with 1 we get
= 9y= 36
= y= 4
NOW putting the value of 'y' in eqn (1)we get
= x+y=12
= x+4=12
= x= 8
Therefore the original number is (10x+y)
= 10*8+4
= 84 is the answer..
HOPE THIS HELPS :)