is the answer d ?

6 votes

How many different $\text{BCD}$ numbers can be stored in $12$ switches ? (Assume two position or on-off switches).

- $2^{12}$
- $2^{12}-1$
- $10^{12}$
- $10^3$

17 votes

Best answer

1 Switch is equivalent to 1 bit Cz Switch store either on or off value like a bit can be 0 or 1.

So total 12 bits are there(cz there 12 switches)

Make 3 group of 12 bit . Means each group contain 4 bit.

(_ _ _ _) (_ _ _ _) (_ _ _ _)

G1. G2. G3

Here in any group max 10 numbers can be stored Cz BCD numbers we have to store.

So 10*10*10 different BCD numbers can be stored in this 12 Switches. So Ans is option D) 10^3

So total 12 bits are there(cz there 12 switches)

Make 3 group of 12 bit . Means each group contain 4 bit.

(_ _ _ _) (_ _ _ _) (_ _ _ _)

G1. G2. G3

Here in any group max 10 numbers can be stored Cz BCD numbers we have to store.

So 10*10*10 different BCD numbers can be stored in this 12 Switches. So Ans is option D) 10^3

6 votes

1

Decimal Num | Binary Num | BCD Num |
---|---|---|

0 | 0000 | 0000 |

1 | 0001 | 0001 |

2 | 0010 | 0010 |

3 | 0011 | 0011 |

4 | 0100 | 0100 |

5 | 0101 | 0101 |

6 | 0110 | 0110 |

7 | 0111 | 0111 |

8 | 1000 | 1000 |

9 | 1001 | 1001 |

10 | 1010 | 0001 0000 |

11 | 1011 | 0001 0001 |

12 | 1100 | 0001 0010 |

13 | 1101 | 0001 0011 |

14 | 1110 | 0001 0100 |

15 | 1111 | 0001 0101 |

12 Switches --> 12 bits are divided into 3 groups

as each group contains 4 bits, hence we can store maximum 10 BCD numbers in that 4 bits i.e., from 0-9

so probability of each group is 10 and total 3 groups=10*10*10

Hope this will clear your doubt.