
The number 292 in decimal system is expressed in binary system by
(A) 100001010
(B) 100010001
(C) 100100100
(D) 101010000
Answer
525.3k+ views
Hint: We are given a number 292 which is in the decimal system. We are asked in the question to express this number in a binary system. Binary system refers to the representation of any number using 0s and 1s only. That is, we will divide the given number by 2 using the long division method and we will then write the remainder after each of the computation cycle from right to left and hence, we will have the given number in binary form.
Complete step-by-step answer:
According to the given question, we are given a number 292. The number 292 is given to us in decimal system. We are asked in the question to express the given number in binary system.
Binary system is a way of representation of numbers using only 0s and 1s. It is carried out by dividing the number by 2 and writing the remainder after each division from right to left.
For example – 10 can be represented in binary system as,
\[\begin{align}
& 2\left| \!{\underline {\,
10 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
5 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
\[{{\left( 10 \right)}_{10}}={{\left( 1010 \right)}_{2}}\]
The given number we have is,
292
We will now divide the given number by 2 and we get,
\[\begin{align}
& 2\left| \!{\underline {\,
292 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
146 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
73 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
36 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
18 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
9 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
That is, we have,
\[{{\left( 292 \right)}_{10}}={{\left( 100100100 \right)}_{2}}\]
Therefore, we have the given number expressed in binary system as, \[{{\left( 292 \right)}_{10}}={{\left( 100100100 \right)}_{2}}\] and that is, option (C) is correct.
So, the correct answer is “Option C”.
Note: The division of the number carried out in the above solution is not the usual division. We divide by 2 such that the remainder is not more than 1. We keep on dividing the quotient by 2 until we get the quotient as 1. Also, while writing in the binary form, we take the right to left approach or if we see the long division in the above solution, we start from the last 1 and proceed upwards.
Complete step-by-step answer:
According to the given question, we are given a number 292. The number 292 is given to us in decimal system. We are asked in the question to express the given number in binary system.
Binary system is a way of representation of numbers using only 0s and 1s. It is carried out by dividing the number by 2 and writing the remainder after each division from right to left.
For example – 10 can be represented in binary system as,
\[\begin{align}
& 2\left| \!{\underline {\,
10 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
5 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
\[{{\left( 10 \right)}_{10}}={{\left( 1010 \right)}_{2}}\]
The given number we have is,
292
We will now divide the given number by 2 and we get,
\[\begin{align}
& 2\left| \!{\underline {\,
292 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
146 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
73 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
36 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
18 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
9 \,}} \right. \left| \!{\underline {\,
1 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \left| \!{\underline {\,
0 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
That is, we have,
\[{{\left( 292 \right)}_{10}}={{\left( 100100100 \right)}_{2}}\]
Therefore, we have the given number expressed in binary system as, \[{{\left( 292 \right)}_{10}}={{\left( 100100100 \right)}_{2}}\] and that is, option (C) is correct.
So, the correct answer is “Option C”.
Note: The division of the number carried out in the above solution is not the usual division. We divide by 2 such that the remainder is not more than 1. We keep on dividing the quotient by 2 until we get the quotient as 1. Also, while writing in the binary form, we take the right to left approach or if we see the long division in the above solution, we start from the last 1 and proceed upwards.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

