in a two digit number the sum of the digits is 7 if the number with the order of the digits are reversed is 28 greater than twice the units digit of the original number find the number
Answers
Answered by
195
Let the original number be 10(7-x)+x
Here, X is the one's digit and (7-X) is the tens digit which is why we multiply it with 10.
The reversed number would be 10x+(7-x)
We know, 10x+(7-x)=28+2x
=>10x+7-x=28+2x
=>10x-2x-x=28-7
=>7x=21
=>x=3
Hence, the original number is 10(7-3)+3= 43.
Hope it helps! xx
Here, X is the one's digit and (7-X) is the tens digit which is why we multiply it with 10.
The reversed number would be 10x+(7-x)
We know, 10x+(7-x)=28+2x
=>10x+7-x=28+2x
=>10x-2x-x=28-7
=>7x=21
=>x=3
Hence, the original number is 10(7-3)+3= 43.
Hope it helps! xx
Anonymous0071:
Thanks buddy!!
Answered by
76
Answer:
43
Step-by-step explanation
let the tens digit be x and units digit b y
the number is 10 x + y
as per question x + y = 7
and 10 y + x = 28 + 2 y
by solving x + 8 y = 28
by elimination method y - 8 y = 7 - 28
By solving y=3
x=4
the number is 43
Similar questions