The sum of the digits of a 2-digits number is 11. The number obtained by adding 4 to this number 41 less than the reserved number.. Find orijinal number
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38 will be answer.
10x+y+4=10y+x-41
45=9y-9x
y-x=5 & x+y=11
so y=8 & x=3
10x+y+4=10y+x-41
45=9y-9x
y-x=5 & x+y=11
so y=8 & x=3
Answered by
0
solution:
Let the 2-Digit Number is 10a + b (where a, b ∈ (0, 1..., 9) & a ≠ 0.
A/q: a + b = 11 --------------------------( i )
and 10a + b + 4 = 10b + a - 41
=> 9b - 9a = 45
=> b - a = 5 ------------- ( ii )
Solving ( i ) & ( ii ), we will get a = 3 & b = 8.
Required number is : 10a + b = 10x3 + 8 = 38.
Hope it helps!
Let the 2-Digit Number is 10a + b (where a, b ∈ (0, 1..., 9) & a ≠ 0.
A/q: a + b = 11 --------------------------( i )
and 10a + b + 4 = 10b + a - 41
=> 9b - 9a = 45
=> b - a = 5 ------------- ( ii )
Solving ( i ) & ( ii ), we will get a = 3 & b = 8.
Required number is : 10a + b = 10x3 + 8 = 38.
Hope it helps!
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