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
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.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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