the sum of two 2 digit number and the number obtained by reversing its digits is 121 find the number if its unit place digit is greater than the tens place digit by 7
Answers
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
Step-by-step explanation:
ORIGINAL NUMBER IS 29
HOPE IT HELP
