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The sum of a two - digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.
Answer :
Let the two - digit number be xy (i.e. 10x + y).
After reversing the digits of the number xy, the new number becomes yx (i.e. 10y + x).
According to question -
(10x + y) + (10y + x) = 121
⇒ 11x + 11y = 121
⇒ x + y = 11.....(1)
and,
x - y = 3 or y - x = 3
[as we don't know which digit is greater out of x and y]
⇒ x - y = ±3.....(2)
Adding Equation (1) and (2), we get -
2x = 14 or 8
⇒ x = 7 or 4
Case 1. when x = 7
y = 4 [from equation (1)]
Case 2. when x = 4
y = 7 [from equation (1)]
Thus, the possible numbers are 47 or 74.
Answer :
Let the two - digit number be xy (i.e. 10x + y).
After reversing the digits of the number xy, the new number becomes yx (i.e. 10y + x).
According to question -
(10x + y) + (10y + x) = 121
⇒ 11x + 11y = 121
⇒ x + y = 11.....(1)
and,
x - y = 3 or y - x = 3
[as we don't know which digit is greater out of x and y]
⇒ x - y = ±3.....(2)
Adding Equation (1) and (2), we get -
2x = 14 or 8
⇒ x = 7 or 4
Case 1. when x = 7
y = 4 [from equation (1)]
Case 2. when x = 4
y = 7 [from equation (1)]
Thus, the possible numbers are 47 or 74.
cutipie13:
which sum?
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sorry disturb kiya to..
bt aise block ni karte
Maine koi galat baat koi ki thi...
byee gud 9i8
chalo, meri vajah se kisi ka to bhala hua....
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