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How many different BCD numbers can be stored in 12 switches? (Assume two position or on-off switches)

1. $2^{12}$
2. $2^{12}-1$
3. $10^{12}$
4. $10^3$

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Edit : Title : ISRO 2014 SET A Question 27

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
by Boss (23.9k points)
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With 4 bit we can represent 10 BCD no.

12 switches means 12 bit space.

Total 10*10*10 BCD no.
by Veteran (60.8k points)
+1
can you explain your solution. it is not clear to me. how it is 10*10*10.
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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.

Have a look at it : https://gateoverflow.in/13388/solve

by Boss (45.4k points)
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12 switch means 12 bit means ??
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1 Switch is equivalent to 1 bit Cz Switch store either on or off value like a bit can be 0 or 1.

BCD numbers = 4 bit.

Switch = 1 bit. (either on or off)

So, in 12 switches, we can store $\frac{12}{4}=3$ BCD numbers.

A BCD number ranges from 0 - 9. So 10 total values.

So, $10^3$

Option D

by Loyal (6.3k points)