The sum of the digits of a two digit number is 10. If 18 be subtracted...
x = 10's digit of a two digit number(Let)
y = units digit(Let)
n=Resulting equal digits(Let)
Condition 1-the sum of a two digit number is 10
x + y = 10
further simplifying
y = (10-x)
Condition 2-if 18 be subtracted from it the digits in the resulting number will be equal
10x + y - 18 = 10n + n
10x + y = 11n+ 18
Substitutituting values found above
10x + (10-x) = 11n + 18
10x - x = 11n + 18 - 10
9x = 11n+ 8
If we solve it we will found the values as - x = 7, y = 3, z = 5 and finally we can say the no. is 73.
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The sum of the digits of a two digit number is 10. If 18 be subtracted...
Solution:
Let the ten's digit be x and unit's digit be y.
According to the question,
x + y = 10 ---(1)
(10x + y) - 18 = (10y + x)
9x - 9y = 18
x - y = 2 ---(2)
Adding equations (1) and (2), we get
2x = 12
x = 6
Putting the value of x in equation (1), we get
y = 4
Therefore, the number is 64.
When we subtract 18 from 64, we get 46 whose digits are equal.
Hence, the correct answer is option B) 73.
The sum of the digits of a two digit number is 10. If 18 be subtracted...
73,
Because 7+3=10
and, 73-18=55