For converting to base 10, we first divide the given number by the base and then get the remainder. Then, repeat this process by ultimately dividing the quotient of the first step by the new obtained base. We then repeat this process unless and until the value of the quotient happens to be comparatively less than the base.
Before you proceed ahead with this concept, check out the basics of conversion to various bases. In this article, we will take a look at the Conversion to Base 10 according to the GATE Syllabus for CSE (Computer Science Engineering). Read ahead to learn more.
Table of Contents
How to Perform Conversion to Base 10?
In the case of number systems, it is very crucial that one has a thorough knowledge of how one can convert various numbers from a given base to another one. In this article, we will learn how we can convert some numbers from any available base to base 10.
We can convert a number from a base to base 10 with the help of the expansion method. Now, according to the expansion method, if mno.pq is a given number in base x, its value in base 10 will be like this:
(mno.pq)x = (mx2 + nx + o + px-1 + qx-2)10
Explanation
If we want to use the expansion method for the process of conversion, then:
- Assign the position number to every digit of the available number.
- The digits available at the left of the decimal are numbered from 0.
- The digits available at the right of the decimal are numbered from -1.
- Write a certain term for every digit as digit x (which is the base of the given number), the digit’s position number.
- Add all the terms so as to get the number in base 10.
- We can expand this formula for any given number of digits.
Practice Problems on Conversion to Base 10
1. Convert the given numbers to base 10-
1.1. (10010)2
Answer – We use the expansion method here and we get-
(10010)2
= ( 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 0 x 20 )10
= ( 16 + 0 + 0 + 2 + 0 )10
= ( 18 )10
1.2. (254)8
Answer – We use the expansion method here and we get-
(254)8
= ( 2 x 82 + 5 x 81 + 4 x 80 )10
= ( 128 + 40 + 4 )10
= ( 172 )10
1.3. (AC)16
Answer – We use the expansion method here and we get-
(AC)16
= ( A x 161 + C x 160 )10
= ( 10 x 16 + 12 x 1 )10
= ( 160 + 12 )10
= ( 172 )10
1.4. (10010.101)2
Answer – We use the expansion method here and we get-
(10010.101)2
= ( 1 x 24 + 0 x 23 + 0 x 22 + 1 x 21 + 0 x 20 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3 )10
= ( 16 + 0 + 0 + 2 + 0 + 0.5 + 0.125 )10
= ( 18.625 )10
1.5. (254.7014)8
Answer – We use the expansion method here and we get-
(254.7014)8
= ( 2 x 82 + 5 x 81 + 4 x 80 + 7 x 8-1 + 0 x 8-2 + 1 x 8-3 + 4 x 8-4 )10
= ( 128 + 40 + 4 + 0.875 + 0.0019 + 0.0009 )10
= ( 172.8778 )10
1.6. (AC.FBA5)16
Answer – We use the expansion method here and we get-
(AC.FBA5)16
= ( A x 161 + C x 160 + F x 16-1 + B x 16-2 + A x 16-3 + 5 x 16-4 )10
= ( 10 x 16 + 12 x 1 + 15 x 16-1 + 11 x 16-2 + 10 x 16-3 + 5 x 16-4 )10
= ( 160 + 12 + 0.9375 + 0.0429 + 0.0024 + 0.0001 )10
= ( 172.9829 )10
1.7. (0.1402)8
Answer – We use the expansion method here and we get-
(0.1402)8
= ( 0 x 80 + 1 x 8-1 + 4 x 8-2 + 0 x 8-3 + 2 x 8-4 )10
= ( 0 + 0.125 + 0.0625 + 0 + 0.0005 )10
= ( 0.188 )10
1.8. (0.ABDF)16
Answer – We use the expansion method here and we get-
(0.ABDF)16
= ( 0 x 160 + A x 16-1 + B x 16-2 + D x 16-3 + F x 16-4 )10
= ( 0 x 1 + 10 x 16-1 + 11 x 16-2 + 13 x 16-3 + 15 x 16-4 )10
= ( 0 + 0.625 + 0.0429 + 0.0032 + 0.0002 )10
= ( 0.6713 )10
Keep learning and stay tuned to get the latest updates on GATE Exam along with GATE Eligibility Criteria, GATE 2023, GATE Admit Card, GATE Syllabus, GATE Previous Year Question Paper, and more.
Also Explore,
- Combinational Circuits
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- Various Implicants in K-Map
- Representation of Boolean Functions
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- Flip-Flop Types, Conversion and Applications
- The Base of Number System
- Number System Notes
- Decimal to Binary Conversion
- Decimal to Hexadecimal Conversion
- Decimal to Octal Conversion
- Minimization of Boolean Functions
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