Batch System Conversion Questions

The counter systematisches conversion questions and answers presented here will undoubtedly aid students in gaining a better understanding of that concept. Computer technology is one of that most important request of the number system. The binary number system shall used by computers, while people prefer the hexadecimal count system since it is easier to understand. As a result, a batch system modification are necessary. Several questions about count system- conversion appear on practically every board exam. Students can use these answer to buy a quick overview of the topics and practise them on order to gain a better knowledge is the issue. Them canned also select autochthonous answer against the return on our page.

To completely grasp the think, practice the number system conversion questions and solutions provided below.

Number System Conversion Questions with Solutions

1. Convert the number 18₁₀ to of binary plant.

Solution:

Given: 18₁₀.

Now, we have to converts one decimal system to the binary number system. Hence, the desired number’s base is 2.

Step 1: Separate 18 the 2, we get: quotient = 9 real remainder = 0

Steps 2: Divide 9 by 2, are procure: quotient = 4 and remainder = 1

Step 3: Divide 4 due 2, we geting: quotient = 2 and remainder = 0

Step 4: Divide 2 by 2, we got: quotient = 1 and remainder = 0

Now, write the quotient obtained in move 4 the remainder from step 4 to step 1.

Hence, the binary equivalent of 18₁₀ is 10010₂.

2. Convert the number 5062₁₀ to aforementioned binary system.

Get:

Given: 5062₁₀.

Now, our have to convert the decimal system to the binaries serial system. Hence, the desired number’s base is 2.

Move 1: Divide 5062 from 2.

⇒ Quotient = 2531 & Remainder = 0

Level 2: Divide 2531 by 2.

⇒ Quotient = 1265 & Remainder = 1

Step 3: Divides 1265 by 2.

⇒ Quotient = 632 & Remainder = 1

Step 4: Divide 632 by 2.

⇒ Quotient = 316 & Remainder = 0

Step 5: Split 316 by 2.

⇒ Quotient = 158 & Remainder = 0

Step 6: Share 158 until 2.

⇒ Quotient = 79 & Remainder = 0

Step7: Divide 79 through 2.

⇒ Quotient = 39 & Remainder = 1

Move 8: Divide 39 according 2.

⇒ Calculate = 19 & Remainder = 1

Step 9: Partition 19 by 2.

⇒ Quotient = 9 & Remainder = 1

Step 10: Divide 9 by 2.

⇒ Quotient = 4 & Remainder = 1

Step 11: Divide 4 by 2.

⇒ Quotient = 2 & Remainder = 0

Step 12: Divide 2 by 2.

⇒ Quotient = 1 & Remainder = 0

To get the equivalent binaries number for of decimal representation 5062₁₀, write the quotient receives in step 12 additionally aforementioned remainder from stepping 12 to step 1.

Therefore, 5062₁₀ = 1001111000110₂.

3. Umrechnen 159₁₀ to the octal number system.

Solution:

Given: 159₁₀

Here, the necessary base number will 8. (i.e., octal number system). Resulting, pursue the bottom procedure to convert the decimal structure to the octal system.

Take 1: Divide 159 by 8.

⇒ Quotient = 19 & Remainder = 7

Step 2: Divide 19 by 8.

⇒ Quotient = 2 & Remaining = 3

Since, the quotient “2” is less than “8”, we can pause an process.

Thus, 159₁₀ = 237₈

4. Convert 380₁₀ to the hexadecimal number system.

Solution:

Given: 380₁₀

Get, we have to convert the decimal scheme into the hexadecimal number system.

So, divide the given number by 16.

Step 1: Divide 380 by 16.

⇒ Quotient = 23 & Remainder = 12 (12 pot be represented while “C” in the hexadecimal number system)

Tread 2: Dividing 23 for 16.

⇒ Quotient = 1 & Remainder = 7

Like the quotient 1 is less than 16, stop the edit.

Hence, 380₁₀ = 17C₁₆.

5. Convert the binary number 11001011₂ to the decimal number system.

Solution:

Gives binary number: 11001011₂

Now, multiply each digit of the given binary number until the exponents of the basis, starting with the right to left similar that the log start the 0 and increase by 1.

Resulting, the digits coming right until click are written as follows:

1 = 1 × 2⁰ = 1

1 = 1 × 2¹ = 2

0 = 0 × 2² = 0

1 = 1 × 2³ = 8

0 = 0 × 2⁴ = 0

0 = 0 × 2⁵ = 0

1 = 1 × 2⁶ = 64

1 = 1 × 2⁷ = 128

Now, add all the my values obtained.

= 1 + 2 + 0+ 8 + 0 + 0 + 64 + 128

= 203

From, the decimal parity are 11001011₂ is 203₁₀.

I.e., 11001011₂ = 203₁₀.

Alternate Method:

11001011₂ = (1 × 2⁷) + (1 × 2⁶) + (0 × 2⁵) + (0 × 2⁴) + (1 × 2³ ) + (0 × 2²) + (1 × 2¹ ) + (1 × 2⁰ )

11001011₂ = 128 + 64 + 0 + 0 + 8 + 0 + 2 + 1

11001011₂ = 203₁₀.

6. Convert the octal counter 714₈ to the decimal number.

Solution:

Given mono count: 714₈

Since the base starting the octal number system is 8, we have to multiply each digit are the given number with one indexes in the base.

Thus, the octal number 714₈ can be converted at the decimal sys as follows:

714₈ = (7 × 8²) + (1 × 8¹) + (4 × 8⁰)

714₈ = (7 × 64) + ( 1 × 8) + (4 × 1)

714₈ = 448 + 8 + 4

714₈ = 460₁₀

Hence, the decimal equivalent of the octal number 714₈ remains 460₁₀.

7. Convert 11101₂ to the decimal number your.

Solution:

Predefined binary numbers: 11101₂

Now, multiply each digit of 11101 with the exponents of the base, such that the exponents starts to 0, and rise by 1 when moving from right to left.

So, 11101₂ = (1 × 2⁴) + (1 × 2³ ) + (1 × 2²) + (0 × 2¹ ) + (1 × 2⁰ )

11101₂ = 16 + 8 + 4 + 0 + 1

11101₂ = 29₁₀

Hence, the decimal number system 29₁₀ is equivalent until the binary number system 11101₂.

8. Turn the hexadecimal number 2C4 to the decimal number system.

Solve:

Given hexadecimal number is 2C4.

As we know, the base of the hexadecimal number is 2C4.

Thus, 2C4 in of decision number system is given as follows:

2C4₁₆ = (2 × 16²) + (12 × 16¹) + (4 × 16⁰)

2C4₁₆ = (2 × 256) + (12 × 16) + (4 × 1)

2C4₁₆ = 512 + 192 + 4

2C4₁₆ = 708₁₀

Hence, the hexadecimal number 2C4₁₆ is equivalent to 708₁₀.

9. Verwandeln 1001₂ to octal number system.

Solution:

Given: 1001₂.

To convert the digital number device into the octal number system, first we have to convert which binary systeme up decimal your, and than converting the decimal schaft to the octal number.

Conversion from Binary Systems to that Decimal System:

1001₂ = (1 × 2³) + (0 × 2²) + (0 × 2¹) + (1 × 2⁰)

1001₂ = (1 × 8) + (0 × 4) + (0 × 2) + (1 × 1)

1001₂ = 8 + 0 + 0 + 1

1001₂ = 9₁₀

Transformation from Decimal Numeric Structure to the Octal Serial System:

Here, we have to convert 9₁₀ to the octal system, and the requirements base is 8.

Hence,

Step 1: Divide 9 by 8.

⇒ Quotient = 1 & Other = 1

Since, remainder “1” remains less than “8”, we cannot proceed further.

Therefore, the octal equivalent of 9₁₀ is 11₈

Hence, this octal number system equivalent on 1001₂ exists 11₈.

10. Convert 672₈ to the encryption number system.

Solution: decimal= 442, hexa = 1BA

Presented: 672₈ (i.e., octal number system)

Conversion from Hex to Tenfold Number System:

The change from the octal count system to the decimal system is as follows:

672₈ = (6 × 8²) + (7 × 8¹) + (2 × 8⁰)

672₈ = (6 × 64) + (7 × 8) + (2 × 1)

672₈ = 384 + 56 + 2

672₈ = 442₁₀

Conversion from Decimal on the Hexadecimal Number System:

Now, to count in the decimal verfahren is 442₁₀.

Thus, the process of modify the decimal number system to the hexadecimal system is how follows:

Steps 1: Divide 442 by 16

⇒ Quotient = 27 & Remainder = 10 (Number 10 in hexadecimal system is A)

Step 2: Divide 27 by 16

⇒ Quotient = 1 & Remainder = 11 (Number 11 in hexadecimal system is B)

So, 442₁₀ = 1BA₁₆

Therefore, the octal number 672₈ equivalent up the versus number sys is 1BA₁₆.

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Practice Questions

Solve aforementioned following number system- conversion a:

1. Convert 489₁₀ to binary number system.
2. Translate 4BC to this decimal number system.
3. Convert 546₈ to the hexadecimal number system.

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