CAT 2025 Lesson : Number Systems - Past Questions

Observation/Strategy
1) We can compare the product with that in decimal system and deduce the base.
2) Product of $44$ and $11$ in base $n$ is $1034$, which is more than what is in base $10$. ∴ $n$ is less than $10$.



Looking at the tens digit, we note that $8$ in base $10$ is probably $13$ or $23$ in base $n$ system.

If $(8)_{10} = (13)_{n}$
⇒ $8 = n + 3$
⇒ $n = 5$

If $(8)_{10} = (23)_{n}$
⇒ $8 = 2n + 3$
⇒ $n = 2.5$(not possible)

$(3111)_{5} = 3 \times 125 + 1 \times 25 + 1 \times 5 + 1$
$= 375 + 25 + 5 + 1$
$= 406$

Answer: (1) $406$

$52_{b} = 2 \times 25_{b}$
⇒ $5b + 2 = 2 \times (2b + 5)$
⇒ $b = 10 - 2 = 8$

Answer: (2) $8$

Observation/Strategy
1) Checking $1010_{\text{B}}$ amongst the options is easier than the equation formed by adding $2111_{\text{B}}$ and $1101_{\text{B}}$
2) $1010_{\text{B}}$ should be between $110$ and $150$. We can work backwards to verify.

$(1010)_{\text{B}}$ = $\text{B}^{3} + \text{B}$ should be between $110$ and $150$

Only B $= 5$ results in $\bm{1010_{\text{B}} = 130}$

Verifying the result

$(2111)_{5} = 250 + 25 + 5 + 1 = 281$
$(1101)_{5} = 125 + 25 + 1 = 151$

$(2111)_{5} + (1101)_{5} = 281 + 151 = \bm{432}$ (Verified)

Answer: (3) $130$