The ones digit of a two-digit number is 4 more than the tens digit. Find the number, if the sum of that number and the number reversed is 110.
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hey mate here is your sollution.. sorry for bad handwriting...
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Let ones digit be "x" and tens digit be "y".
According to the question,
x=y+4
Then, x-y=4........1
The two digit number is 10y+x, then number formed by reversing the order of digits is 10x+y.
According to the question,
10y+x+10x+y=110
11x+11y=110
11(x+y)=110
x+y=110/11
x+y=10..........2
Solving 1 and 2,
x-y=4
x+y=10
2x=14
Therefore x=7.
Substitute x=7 in ......1
7-y=4
Therefore y=3.
Therefore the two digit number is 10y+x=(10×3)+7=30+7=37.
According to the question,
x=y+4
Then, x-y=4........1
The two digit number is 10y+x, then number formed by reversing the order of digits is 10x+y.
According to the question,
10y+x+10x+y=110
11x+11y=110
11(x+y)=110
x+y=110/11
x+y=10..........2
Solving 1 and 2,
x-y=4
x+y=10
2x=14
Therefore x=7.
Substitute x=7 in ......1
7-y=4
Therefore y=3.
Therefore the two digit number is 10y+x=(10×3)+7=30+7=37.
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