the sum of the digits of a two-digit number is 10 if the number formed by reversing the digit is less than the original number by 18 find the original number
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Answered by
8
Let the two digit number is 10X+Y
Given sum of two digits=10
X+Y=10 ...........eq-1
number formed by reversing the digits is 18 less than the original.
10Y+X+18=10X+Y
=>9X-9Y=18 ..........eq-2
solving eq-1 and 2
X=6 and Y=4
Given sum of two digits=10
X+Y=10 ...........eq-1
number formed by reversing the digits is 18 less than the original.
10Y+X+18=10X+Y
=>9X-9Y=18 ..........eq-2
solving eq-1 and 2
X=6 and Y=4
RanjanaJha:
How to solve equation 2
Answered by
1
Answer:
Let the two digit number is 10x + y
Given
Sum of two digits = 10
x + y = 10 -------- (i)
Number formed by reversing the digits is 18 less than the original.
10y + x + 18 = 10x + y
= 9x - 9y = 18 -------(ii)
Solving equation -(i) & (ii)
x = 6 and y = 4
So the original number is
64
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