CAT Exam Model Paper 1 with solutions for free online practice

Directions (Q.Nos. 98-99) A function is defined as follows

\f ( a _ { 1 } a _ { 2 } a _ { 3 } , ...a_ { n } ) =

a _ { 1 } 2 ^ { n - 1 } + a _ { 2 } 2 ^ { n - 2 } + a _ { 3 } 2 ^ { n - 3 } + ... a _ { n } 2°

The above function is repeated until the value of function reduces to a single digit number

Question : 99 of 100

Marks: +1, -0

What is the value of f[f(888222) + f( 113113)]?

Solution:

\f ( 888222 ) = 8 . 2 ^ { 5 } + 8 . 2 ^ { 4 } + 8 . 2 ^ { 3 } + 2 . 2 ^ { 2 } + 2 . 2 ^ { 1 } + 2 .2 ^ { 0 }

= 2 ^ { 8 } + 2 ^ { 7 } + 2 ^ { 6 } + 2 ^ { 3 } + 2 ^ { 2 } + 2 ^ { 1 } = 2 ^ { 6 } ( 7 ) + 14

= 448 + 14 = 462

\f ( 462 ) = 4 . 2 ^ { 2 } + 6 .2 ^ { 1 } + 2 . 2 ^ { 0 } = 30

\f ( 30 ) = 3 . 2 ^ { 1 } + 0 .2 ^ { 0 } = 6

\Again \f(113113) = 1 . 2 ^ { 5 } + 1 .2 ^ { 4 } + 3 . 2 ^ { 3 } + 1 .2 ^ { 2 } + 1 . 2 ^ { 1 } + 3 . 2 ^ { 0 }

= 32 + 16 + 24 + 4 + 2 + 3 = 81

\f ( 81 ) = 8.2 ^ { 1 } + 1 .2 ^ { 0 } = 16 + 1 = 17

\f ( 17 ) = 1 . 2 ^ { 1 } + 7 . 2 ^ { 0 } = 2 + 7 = 9

∴f[f(888222) + f(113113)]= f(6 + 9)

= \f(15) =1.2 ^ { 1 } + 5 . 2 ^ { 0 }

= 2 + 5= 7

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