Let ten's place digit be a and unit's place digit be a2
Original number = 10 x a + 1 x a2 = 10a + a2
The number formed by interchanging the digits,
New number = 10 x a 2 + 1 x a = 10a 2 + a
According to the question
(10a2 + a) - ( 10a + a2) = 54
? 10a2 + a - 10a - a 2 = 54
? 9a2 - 9a = 54
? 9( a2 - a) = 54
? ( a2 - a) = 54/9
? ( a2 - a) = 6
? a2 - a - 6 = 0
? a2 - 3a + 2a - 6 = 0
? a (a - 3) + 2 (a - 3) = 0
? (a - 3) (a + 2) = 0
? a = 3, - 2
? Ten,s digit = a = 3
Unit's digit = a2 = 32 = 9
Original number = 39
? Required number = 39 x 40/100 = 15.6
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
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