Algebra - Number System - Previous Year CAT/MBA Questions

Answer: Option D

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Text Explanation :

Of the four given options 63 and 75 are multiples of 3. Their remainder cannot be 1.

∴ The last digit cannot be 1. Thus, 63 and 75 are eliminated.

For the options 31 and 91, the remainder is 1. Thus, the last digit is 1.

31 = (11111)₂ = (1011)₃ = (111)5

91 = (1011011)₂ = (10101)₃ = (331)₅

91 has 1 as the first digit in only 2 of the notations.

Hence, option (d).

Answer: Option A

Video Explanation

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Text Explanation :

Let the base be b.

∴ (4b + 4)(b + 1) = b³ + 3b + 4

∴ b³ − 4b² − 5b = 0

Solving we get, b = 0, –1, 5

∵ Base cannot be zero or negative, so base is 5.

∴ (3111)₅ = 3 × 125 + 25 + 5 + 1 = 406

Hence, option (a).

Alternatively,

Let the required base be x.

We know that the answer of 44 × 11 in the required base is 1034.

∴ As 4 occurs in the product, we can say that the base is greater than 4.

Also, 44 × 11 = 484 in base 10.

As 1034 > 484, the base is lesser than 10.

So we can represent the multiplication as follows:

44 × 11 = 44(x + 1) = 44x + 44 = 1034

4 + 4 → 3 or 13 or 23 …

As the base is less than 10, 4 + 4, which is 8 in base 10 cannot be expressed as 3 in the required base.

∴ 4 + 4 → 13 or 23…

4 + 4 → 13

∴ 8 → 13

∴ 8 → x + 3

∴ x = 5

If 4 + 4 → 23

∴ 8 → 23

∴ 8 → x + 13

∴ x = –5, which is not possible.

∴ (3111)₅ = 1 × 5⁰ + 1 × 5¹ + 1 × 5² + 3 × 5³  = (406)^₁₀

Hence, option (a).

Answer: Option C

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Text Explanation :

To convert 1982 from base 10 to 12:

Divide 1982 by 12, write the quotient in the column below 1982 and the remainder in the next column. Repeat the process till the quotient is 0.

After that, write the remainders in the reverse order i.e. upward direction.

1982₁₀ = 1192₁₂

Hence, option (c).