Question

find the original number if the sum of the digits of two digit number is 11 and the number obtained on reversing the digits is 5 more than thrice the original number.

Answer 1

Original no. is 10x + y
sum of the no =x +y =11......(1)
Reversed no.= 10y + x
A/c
10y + x =5 +3 (10x +y)
10y + x =5+30x +3y
7y = 5 + 29x
7y - 29x =5.........(2)
Multiplying (1) by 7
=7y + 7x = 77........... (3)
Subtracting (3) by (2):
7y - 29x =5
7y + 7x =77
= - 36x = - 72
x=2
Putting value of x in (1):
2+y=11
Y=11-2
y=9
Original no. =10x + y
=10 *2+9
=20 +9
=29 (ans)

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Answer 2

hey!!!
here's ur answer
let the number be 10x+y. its is the real no.
and reversed n.o. is 10y+x
now..
a/q 10x+y=5+3(10y+x)
10x+y=5+30y+3x
7x=29y+5
and x+y is 11
so y=9 and x=2
and we got the no. which is 29