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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}, \ldots a_{n}) =

a_{1} 2^{n-1} + a_{2} 2^{n-2} + a_{3} 2^{n-3} + \ldots a_{n} 2^{\circ}

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 \cdot 2^{5} + 8 \cdot 2^{4} + 8 \cdot 2^{3} + 2 \cdot 2^{2} + 2 \cdot 2^{1} + 2 \cdot 2^{0}

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

= 448 + 14 = 462

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

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

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

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

f(81) = 8 \cdot 2^{1} + 1 \cdot 2^{0} = 16 + 1 = 17

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

∴f[f(888222) + f(113113)]= f(6 + 9)

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

= 2 + 5= 7

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