plz solve question 9.
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Let the tens place digit be a
Ones place digit = a - 2
Number = 10 ( a) + (a - 2)
= 10a + a - 2
= 11a - 2
Reversed number = 10 ( a - 2) + a
= 10a - 20 + a
= 11a - 20
According to question,
( 11a - 2) + (11a - 20) = 66
=> 22a - 22 = 66
=> a - 1 = 3
=> a = 4
Number = 42
Similarly, if we assume the unit place digit as 'a' and tens place digit as 'a- 2' we get,
Number = 24
There are two numbers (42 and 24), that satisfy the given condition.
Ones place digit = a - 2
Number = 10 ( a) + (a - 2)
= 10a + a - 2
= 11a - 2
Reversed number = 10 ( a - 2) + a
= 10a - 20 + a
= 11a - 20
According to question,
( 11a - 2) + (11a - 20) = 66
=> 22a - 22 = 66
=> a - 1 = 3
=> a = 4
Number = 42
Similarly, if we assume the unit place digit as 'a' and tens place digit as 'a- 2' we get,
Number = 24
There are two numbers (42 and 24), that satisfy the given condition.
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