Question

in a 2 didit number the digit in the tens place is one less than the digit in the ones place . the sum ofthe number and the number obtained by reversing its digits is 121 . assuming the digit in the units place to be x . Find the number

Answer 1

Answer:

the original no. is 29

Step-by-step explanation:

let tens place digit =x

let ones place digit=y

the original no.=10x+y(any numbers expanded form looks like this for example 23=2*10+3)

the reversed no.is =10y+x

according to the question,

10x+y+10y+x=121-------1

x+7=y------2

from equation2 we get the value of y(ones place digit)in terms of x)

which is y=x+7

now we will place it in the first equation like this,but before we do this we should simplify the equation1

10x+y+10y+x=121

10x+x+10y+y=121

11x+11y=121

11(x+y)=121

x+y=121/11

x+y=11

we will now put the value of y in this equation,

x+(x+7)=11

x+x+7=11

2x=11-7

2x=4

x=4/2

x=2

now we know the value of x so we can easily find the value of y,

in equation 2 by putting the value of x we can get the value of y

x+7=y

2+7=y

y=9

thus the original no. is 29