in a two digit number the digit in tens place exceeds the digit in the units place by 5 if 3more than six times the sum of the digit is subtracted from the number the digit are reversed. find the original number
Answers
Answer:
Original number = 61
Step-by-step explanation:
In the two-digit number, let:
digit in 10s place = x
digit in 1s place = y
=> Number is (10x + y)
Given: digit in 10s place exceeds digit in 1s place by 5
=> x = y + 5 ........(i)
Given: If 3 more than six times the sum of the digit is subtracted from the number the digit are reversed.
6 times the sum of digits = 6(x + y)
3 more than 6 times the sum of digits = 6(x + y) + 3
This quantity has to be subtracted from the original number to get the number with digits reversed.
Number with digits reversed = 10y + x
=> 10x + y - [6(x + y) + 3] = 10y + x
=> 10x + y - 6x - 6y - 3 = 10y + x
=> 4x - 5y - 3 = 10y + x
=> 3x - 15y = 3
=> x - 5y = 1 ..........(ii)
From (i), we get:
x = y + 5
From (ii), we get:
x = 5y + 1
Equating the two, we get:
y + 5 = 5y + 1
=> 4y = 4
=> y = 1
Using this value of y in (i), we get:
x = 1 + 5 = 6
x = 6 and y = 1
=> Original number is 10(6) + 1 = 61
Verification:
1. Digit in 10s place exceeds 1s place by 5
2. 3 more than 6 times the sum of digits when substracted from original number gives
61 - [6(6+1) + 3] = 61 - (42 + 3) = 61 - 45 = 16 = original number with digits reversed.
Thus, verified
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Step-by-step explanation:
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