A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.
Let the tens place digit be a and the units place digit be b.
Then, number is (10a + b).
According to the question:
4(a + b) + 3 = (10 a + b)
4a + 4b + 3 = 10a + b
6a - 3b = 3
3(2a - b) = 3
2a - b =1 ... (1)
Given:
If 18 is added to the number, its digits are reversed.
The reverse of the number is (10b + a).
∴ (10a + b) + 18 = 10b + a
10a - a + b -10b = -18
9a - 9b = -18
9(a - b) = -18
a - b = -2 ... (2)
Subtracting equation (2) from equation (1):
Using a = 3 in equation (1):
2(3) - b = 1
6 - b = 1
∴ b = 5
Number = 10a+b = 10 × 3 + 5 = 35
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