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Number Systems

Number Systems

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Roman Numerals
Conversion to Base 10
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Number Systems 1
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Number Systems : Level 1
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CAT 2025 Lesson : Number Systems - Past Questions

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6. Past Questions

Question 1

In a number system the product of 444444 and 111111 is 103410341034. The number 311131113111 of this system, when converted to the decimal number system, becomes
[CAT 2001]

406406406
108610861086
213213213
691691691

Observation/Strategy
1) We can compare the product with that in decimal system and deduce the base.
2) Product of 444444 and 111111 in base nnn is 103410341034, which is more than what is in base 101010. ∴ nnn is less than 101010.



Looking at the tens digit, we note that
888 in base 101010 is probably 131313 or 232323 in base nnn system.

If
(8)10=(13)n(8)_{10} = (13)_{n}(8)10​=(13)n​
⇒
8=n+38 = n + 38=n+3
⇒
n=5n = 5n=5

If
(8)10=(23)n(8)_{10} = (23)_{n}(8)10​=(23)n​
⇒
8=2n+38 = 2n + 38=2n+3
⇒
n=2.5n = 2.5n=2.5(not possible)

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

Answer: (1)
406406406


Question 2

The symbol 252525b represents a two-digit number in base b. If the number 525252b is double the number 252525b, then b is :
[FMS 2010]

777
888
999
111111

52b=2×25b52_{b} = 2 \times 25_{b}52b​=2×25b​
⇒
5b+2=2×(2b+5)5b + 2 = 2 \times (2b + 5)5b+2=2×(2b+5)
⇒
b=10−2=8b = 10 - 2 = 8b=10−2=8

Answer: (2)
888


Question 3

Two numbers in the base system B are 2111B2111_{\text{B}}2111B​ and 1101B1101_{\text{B}}1101B​. The sum of these two numbers in decimal system is 432432432. Find the value of 1010B1010_{\text{B}}1010B​ in decimal system.
[XAT 2016]

110
120
130
140
150

Observation/Strategy
1) Checking 1010B1010_{\text{B}}1010B​ amongst the options is easier than the equation formed by adding 2111B2111_{\text{B}}2111B​ and 1101B1101_{\text{B}}1101B​
2)
1010B1010_{\text{B}}1010B​ should be between 110110110 and 150150150. We can work backwards to verify.

(1010)B(1010)_{\text{B}}(1010)B​ = B3+B\text{B}^{3} + \text{B}B3+B should be between 110110110 and 150150150

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

Verifying the result

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

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

Answer: (3)
130130130

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