The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Solution:
Let the digits at tens place and units place in the first number be x and y respectively.
A number can be expressed in the expanded form as 10(x) + y.
On reversing the digits, x is the units digit and y is the tens digit. The expanded notation for the second number be 10(y) + x
As per the question,
(10x + y) + (10y + x) = 66
⇒ 11(x + y) = 66
⇒ x + y = 6 be equation (1)
Also, it is given that the difference between the two digits is 2.
x - y = 2 be equation (2) or y - x = 2 be equation (3)
When x - y = 2
Subtracting equation 2 from equation 1.
(x + y = 6) - (x - y = 2)
2y = 4
y = 2 and x = 4.
∴ The two digit number is 10x + y = 40 + 2 = 42.
When y - x = 2
Subtracting equation 3 from equation 2.
(x + y = 6) - (y - x = 2)
2x = 4
x = 2 and y = 4
∴ The two digit number is 10y + x = 20 + 4 = 24
Thus, the two digits are 42 and 24
☛ Check: NCERT Solutions for Class 10 Maths Chapter 3
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Summary:
The two digit numbers are 42 and 24, when the sum of a two-digit number and the number obtained by reversing the digits is 66
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