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प्रश्न
The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
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उत्तर
Let the digits at units and tens place of the given number be x and y respectively.
Thus, the number is 10y + x.
The two digits of the number are differing by 3. Thus, we have x - y = -3
After interchanging the digits, the number becomes 10x + y.
The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have
(10x + y) + (10y + x) = 99
⇒ 10x + y + 10y + x = 99
⇒ 11x + 11y = 99
⇒ 11(x + y) = 99
⇒ x + y = `99/11`
⇒ x + y = 9
So, we have two systems of simultaneous equations
x - y = 3
x + y = 9
x - y = -3
x + y = 9
Here, x and y are unknowns. We have to solve the above systems of equations for x and y.
(i) First, we solve the system
x - y = 3
x + y = 9
Adding the two equations, we have
(x - y) + (x + y) = 3 + 9
⇒ x - y + x + y = 12
⇒ 2x = 12
⇒ x = `12/2`
⇒ x = 6
Substituting the value of x in the first equation, we have
x - y = 3
⇒ y = 6 - 3
⇒ y = 3
Hence, the number is 10 × 3 + 6 = 36.
(ii) Now, we solve the system
x - y = -3
x + y = 9
Adding the two equations, we have
(x - y) + (x + y) = -3 + 9
⇒ x - y + x + y = 6
⇒ 2x = 6
⇒ x = `6/2`
⇒ x = 3
Substituting the value of x in the first equation, we have
3 - y = -3
⇒ y = 3 + 3
⇒ y = 6
Hence, the number is 10 × 6 + 3 = 63.
Note that there are two such numbers.
