Conversion of Seconds into Minutes – Definitions, Formula, Examples | How to Convert Seconds to Minutes?

Conversion of Seconds into Minutes

Do you want to know one second is how many minutes? Then read this complete page. Here, you can learn about the seconds to minutes conversion with a detailed explanation. You will also learn the steps of converting seconds into minutes. Get to know more about what is a second, minute, and what are different units of time, and others. Refer to Seconds to Minutes Conversion Formula, Step by Step Procedure on how to convert seconds into minutes, solved examples, here.

Minutes and Seconds – Definitions

Minute, second are the different units of time. In general, hours is the highest unit, next minutes and then seconds. Minute is a period of time which is equal to sixty seconds or sixtieth of an hour. Actually, a second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom. And it is a small division of an hour.

Also, Check:

Seconds to Minutes Conversion Formula

The simple formula to convert minutes into seconds is as follows:

1 Minute = 60 Seconds

So, 1 Second = \(\frac { 1 }{ 60 } \) Minute

Number of minutes = \(\frac { 1 }{ 60 } \) x number of seconds

How to Convert Seconds to Minutes?

Check out the simple and detailed steps to convert the seconds into minutes. They are in the following fashion

  • Let us take the number of seconds you want to convert into minutes.
  • In general, 1 second = \(\frac { 1 }{ 60 } \) minute
  • Multiply the given number of seconds by \(\frac { 1 }{ 60 } \).
  • The result is called the minutes.

Also, Read

Seconds to Minutes Conversion Examples

Example 1: 

Convert 150 seconds into minutes?

Solution:

The given number of seconds = 150

Multiply 150 by (\(\frac { 1 }{ 60 } \))

So, 150 seconds = \(\frac { 150 }{ 60 } \)

Conversion of Seconds into Minutes 1

Therefore, 150 seconds = 2.5 Minutes

Example 2:

Convert 215 seconds into minutes.

Solution:

The given number of seconds = 215

Multiply 215 by \(\frac { 1 }{ 60 } \)

So, 215 seconds = \(\frac { 215 }{ 60 } \)

Conversion of Seconds into Minutes 2

Therefore, 215 seconds = 3.583 minutes.

Example 3:

Convert 180 seconds into minutes.

Solution:

The given number of seconds = 180

Multiply 180 by \(\frac { 1 }{ 60 } \)

So, 180 seconds = \(\frac { 180 }{ 60 } \) minutes

Conversion of Seconds into Minutes 3

Therefore, 180 seconds = 3 minutes

Example 4:

Convert 516 seconds into minutes.

Solution:

The given number of seconds = 516

Multiply 516 by \(\frac { 1 }{ 60 } \)

So, 516 seconds = \(\frac { 516 }{ 60 } \) minutes

Conversion of Seconds into Minutes 4

Therefore, 516 seconds = 8.6 minutes.

Example 5:

Convert 1020 seconds into minutes.

Solution:

The given number of seconds = 1020

Multiply 1020 by \(\frac { 1 }{ 60 } \)

So, 1020 seconds = \(\frac { 1020 }{ 60 } \) minutes

Conversion of Seconds into Minutes 5

Therefore, 1020 seconds = 17 minutes.

Common Conversions Facts About Time

Below are a few easy & quick conversion facts about the unit of time.

1 Hour = 60 Minutes

1 Minute = 60 seconds

1 Second = \(\frac { 1 }{ 60 } \) Minute

1 Minute = \(\frac { 1 }{ 3600 } \) Hour

1 Hour = 3600 seconds

12 hours = 1 day

12 hours = 43,200 seconds

1 day = 24 hours = 1440 minutes

1 day = 24 hours = 86,400 seconds

Seconds to Minutes Conversion Table

Here is the conversion table of Seconds (sec) to Minutes (min):

Seconds Minutes
1 Second \(\frac { 1 }{ 60 } \) Minute
2 Seconds \(\frac { 2 }{ 60 } \) Minute
10 Seconds \(\frac { 1 }{ 6 } \) Minute
25 Seconds \(\frac { 25 }{ 60 } \) Minute
50 Seconds \(\frac { 5 }{ 6 } \) Minute
60 Seconds 1 Minute
120 Seconds 2 Minutes
180 Seconds 3 Minutes
240 Seconds 4 Minutes
300 Seconds 5 Minutes

FAQ’s on Conversion of Seconds into Minutes

1. How do you convert seconds to minutes?

To convert seconds into minutes, multiply the number of seconds by \(\frac { 1 }{ 60 } \). The obtained product is the converted minutes.

2. How many minutes is 540 seconds?

540 seconds = \(\frac { 540 }{ 60 } \) = \(\frac { 54 }{ 6 } \)

540 seconds = 9 minutes.

3. How do you convert minutes into seconds?

Generally, we know that 1 minute = 60 seconds. To convert minutes into seconds, multiply the number of minutes by 60 to get the number of seconds.

Worksheet on 18 Times Table | Printable 18 Times Table Worksheet with Answers

Worksheet on 18 Times Table

Worksheet on Multiplication Table of 18 is provided here. Check out the important questions involved in the 18 Times Table Multiplication Chart. Have a look at the various methods, rules, formulas that helps to solve the questions of the 18 Times Table. After reading this page, you will be able to understand and importance of the 18 tables along with their applications. Solved example questions on the Worksheet on 18 Times Table will help you to get a piece of detailed information and also helps you to score good marks in the exam.

18 Times Multiplication Table Chart up to 25

Check out the multiplication table of 18 and remember the output to make your math-solving problems easy.

18 x 0 = 0
18 x 1 = 18
18 x 2 = 36
18 x 3 = 54
18 x 4 = 72
18 x 5 = 90
18 x 6 = 108
18 x 7 = 126
18 x 8 = 144
18 x 9 = 162
18 x 10 = 180
18 x 11 = 198
18 x 12 = 216
18 x 13 = 234
18 x 14 = 252
18 x 15 = 270
18 x 16 = 288
18 x 17 = 306
18 x 18 = 324
18 x 19 = 342
18 x 20 = 360
18 x 21 = 378
18 x 22 = 396
18 x 23 = 414
18 x 24 = 432
18 x 25 = 450

Problem 1:

Find each product using the multiplication table of 18

(i) 16 x 18

(ii) 21 x 18

(iii) 8 x 18

(iv) 5 x 18

(v) 9 x 18

Solution:

(i) The given two numbers are 16, 18

16 x 18 = 288

The product of 16 and 18 is 288.

(ii) The given two numbers are 21, 18

21 x 18 = 378

The product of 21 and 18 is 378.

(iii) The given two numbers are 8, 18

8 x 18 = 144

The product of 8 and 18 is 144.

(iv) The given two numbers are 5, 18

5 x 18 = 90

The product of 5 and 18 is 90.

(v) The given two numbers are 9, 18

9 x 18 = 162

The product of 9 and 18 is 162.


Problem 2:

Name the number just after:

(i) 7 x 18

(ii) 18 x 18

(iii) 15 x 18

(iv) 10 x 18

Solution:

(i) The given numbers are 7, 18

7 x 18 = 126

The number just after is 126 + 1 = 127.

(ii) The given numbers are 18, 18

18 x 18 = 324

The number just after is 324 + 1 = 325.

(iii) The given numbers are 15, 18

15 x 18 = 270

The number just after is 270 + 1 = 271.

(iv) The given numbers are 10, 18

10 x 18 = 180

The number just after is 180 + 1 = 181.


Problem 3:

Name the number just before:

(i) 18 x 12

(ii) 18 x 24

(iii) 18 x 17

(iv) 18 x 4

Solution:

(i) The given numbers are 18, 12

18 x 12 = 216

The number just before 216 is 216 – 1 = 215.

(ii) The given numbers are 18, 24

18 x 24 = 432

The number just before 432 is 432 – 1 = 431.

(iii) The given numbers are 18, 17

18 x 17 = 306

The number just before 306 is 306 – 1 = 305.

(iv) The given numbers are 18, 4

18 x 4 = 72

he number just before 72 is 72 – 1 = 71.


Problem 4:

The cost of a cricket bat is $18. How much will 50 such cricket bats cost?

Solution:

Given that,

The cost of one cricket bat = $18

The cost of 50 such cricket bats is obtained by multiplying one cricket bat by 50.

So, the cost of 50 cricket bats = 18 x 50 = 900

Worksheet on 18 Times Table 1

Therefore, the cost of 50 cricket bats is $900.


Problem 5:

A doll costs $15. How much will 18 such dolls cost?

Solution:

Given that,

The cost of a doll = $15

The cost for 18 dolls = 18 x 15 = 270

Worksheet on 18 Times Table 2

Therefore, $270 is the cost for 18 dolls.


Problem 6:

Complete the chart by multiplying the numbers by 18.

(i) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (ii) 52, 85, 96, 152, 638, 452

Solution:

The 18 table multiplication chart is along the lines.

(i) 18 x 1 = 18

18 x 2 = 36

18 x 3 = 54

18 x 4 = 72

18 x 5 = 90

18 x 6 = 108

18 x 7 = 126

18 x 8 = 144

18 x 9 = 162

18 x 10 = 180

(ii) 18 x 52 = 936

Worksheet on 18 Times Table 3

18 x 85 = 1530

Worksheet on 18 Times Table 4

18 x 96 = 1728

Worksheet on 18 Times Table 5

18 x 152 = 2736

Worksheet on 18 Times Table 6

18 x 638 = 11,484

Worksheet on 18 Times Table 7

16 x 452 = 8136

Worksheet on 18 Times Table 8


Problem 7:

(i) What are 18 elevens?

(ii) 18 times 9?

(iii) 12 eighteen’s?

(iv) Eighteen times 9?

(v) What are 6 times 18?

(vi) Eighteen’s 4?

(vii) 18 times 11?

(viii) Eighteen’s 7?

(ix) What is 18 multiplied by 8?

(x) 5 multiplied by 18?

Solution:

(i) 18 elevens mean the product of 18 and 11

18 x 11 = 198

So, 18 elevens is 198.

(ii) 18 times 9 means the product of 18 and 9

18 x 9 = 162

So, 18 times 9 is 162.

(iii) 12 eighteen’s means the product of 12 and 18

12 x 18 = 216

So, 12 eighteen’s is 216.

(iv) Eighteen times 9 means the product of 18 and 9

18 x 9 = 162

So, Eighteen times 9 is 162.

(v) 6 times 18 means the product of

6 x 18 = 108

So, 6 times 18 is 108

(vi) Eighteen’s 4 means the product of 18 and 4

18 x 4 = 72

So, Eighteen’s 4 is 72.

(vii) 18 times 11 means the product of 18 and 11

18 x 11 = 198

So, 18 times 11 is 198.

(viii) Eighteen’s 7 means the product of 18 and 7

18 x 7 = 126

So, eighteen’s 7 is 126.

(ix) 18 multiplied by 8 means the product of 18 and 8

18 x 8 = 144

So, 18 multiplied by 8 is 144.

(x) 5 multiplied by 18 means the product of 5 and 18

5 x 18 = 90

So, 5 multiplied by 18 is 90.


Problem 8:

The cost of a pocket Radio set is 18 dollars. What will be the cost of 7 such Radio sets?

Solution:

Given that,

The cost of a pocket Radio set = 18 dollars

The cost of 7 such Radio sets = The cost of one pocket radio set x 7

= 18 x 7

= 126

Therefore, the cost of 7 such Radio sets is 126 dollars.


Problem 9:

What does 18 × 82 mean? What number is it equal to?

Solution:

18 x 82 means the product of 18 and 82

Worksheet on 18 Times Table 9

18 x 82 = 1476

So, 18 x 82 is equal to 1476 and 18 x 82 means 1476.


Problem 10:

Solve the following using the 18 times table.

(i) How many eighteen’s in 198?

(ii) 18 times 6 minus 4

(iii) How many eighteen’s in 972?

(iv) How many eighteen’s in 1728?

(v) 18 times 16 minus 45

(vi) How many eighteen’s in 1134?

(vii) 18 times 2 plus 6

(viii) How many eighteen’s in 108?

(ix) How many eighteen’s in 1170?

(x) How many eighteen’s in 1332?

Solution:

(i) How many eighteen’s in 198 means divide 198 by 18.

Worksheet on 18 Times Table 10

198 ÷ 18 = 11

There are 11 eighteen’s in 198.

(ii) 18 times 6 minus 4

18 times 6 means the product of 18 and 6

18 x 6 = 108

108 – 4 = 104

18 times 6 minus 4 is 104.

(iii) How many eighteen’s in 972 means divide 972 by 18

Worksheet on 18 Times Table 11

972 ÷ 18 = 54

So, there are 54 eighteen’s in 972.

(iv) How many eighteen’s in 1728 means divide 1728 by 18

Worksheet on 18 Times Table 12

1728 ÷ 18 = 96

So, there are 96 eighteen’s in 1728.

(v) 18 times 16 minues 45

18 times 16 means the product of 18 and 16

18 x 16 = 288

18 times 16 minues 45 = 288 + 45 = 333

So, 18 times 16 minues 45 is 333.

(vi) How many eighteen’s in 1134 means divide 1134 by 18

Worksheet on 18 Times Table 13

1134 ÷ 18 = 63

So, there are 63 eighteen’s in 1134

(vii) 18 times 2 plus 6

18 times 2 means the product of 18 and 2

18 x 2 = 36

18 times 2 plus 6 = 36 + 6 = 42

So, 18 times 2 plus 6 is 42

(viii) How many eighteen’s in 108 means divide 108 by 18

Worksheet on 18 Times Table 14

108 ÷ 18 = 6

So, there are 6 eighteen’s in 108.

(ix) How many eighteen’s in 1170 means divide 1170 by 18

Worksheet on 18 Times Table 15

1170 ÷ 18 = 65

So, there are 65 eighteen’s in 1170

(x) How many eighteen’s in 1332 means divide1332 by 18

Worksheet on 18 Times Table 16

1332 ÷ 18 = 74

So, there are 74 eighteen’s in 1332


Examples on the Formation of Greatest and the Smallest Number | How to find the Greatest and Smallest Number?

Examples on the Formation of Greatest and the Smallest Number

Do you want to know how to form the greatest and smallest numbers using the digits? If yes, then stay tuned to this page. On this page, students can learn about the detailed steps to form the largest and smallest numbers with their definitions. We have also covered plenty of examples on the formation of the greatest and the smallest numbers in the below-mentioned sections.

What is Greatest and Smallest Number?

The greatest number is a number which is having the highest value when compared with other numbers. And we can also say that the largest number has all the digits arranged in descending order. The position of the digit at the extreme left of a number increases its place value. So the greatest digit from the given digits must be placed at the extreme left side of the number to raise its value.

The smallest number is a number that is having the lowest value compared with other numbers. In the lowest number, the digits are arranged in ascending order. If the given digits have 0, we never write 0 at the extreme left place, instead write at the second place from the left to get the smallest number.

Detailed Process on Formation of Greatest and the Smallest Number

Check out the step-by-step process of forming the greatest and smallest numbers from the given digits along the lines.

  • To form the greatest numbers from the given digits, arrange the digits in descending order. And the extreme left digit has the highest value when compared with others.
  • To form the lowest numbers from the given digits, arrange the digits in ascending order. And the extreme left digit has the lowest value compared with other digits.
  • If there is a 0, then don’t write 0 at the extreme left position instead place it at the second place from the left to obtain the lowest number.

Example:

Form the greatest and smallest number using the digits 5, 4, 8, 6.

To form the greatest number, follow these steps.

  • The smallest digit is placed at one’s place
  • The next greater digit at ten’s place and so on
  • The greatest digit is placed at the highest place of the number

To form the smallest number, follow the reverse procedure

  • The greatest digit is placed at one’s position
  • The next smaller digit is placed at ten’s position and so on.
  • So, the smallest digit is placed at the highest place of the number.

The ascending order of the numbers 4, 5, 6, 8

So, the smallest number is 4568.

The descending order of the numbers 8, 6, 5, 4

So, the greatest number is 8654.

Also, Read

Worked Out Examples on the Formation of Greatest and the Smallest Number

Example 1:

Write the greatest and smallest 4 digit numbers using the digits 1, 0, 8, 4

Solution:

The given digits are 1, 0, 8, 4.

We know that the four-digit number has four places those are thousands, hundreds, tens, and ones. If the given digits are arranged in descending order (from greatest to lowest value), we get the greatest number. If digits are arranged in ascending order (from lowest to highest), we get the smallest number.

The descending order is 8 < 4 < 1 < 0.

The ascending order is 0 < 1 < 4 < 8.

                                     Th      H       T       O

Greatest number           8       4       1       0
Lowest Number            1       0       4       8

As the given digits have 0, place zero at the second-highest position i.e hundredth position.

So, the greatest number using the digits 1, 0, 8, 4 is 8410.

The smallest number using the digits 1, 0, 8, 4 is 1048.

Example 2:

Write the greatest and smallest 5 digit number using the digits 7, 5, 6, 8, 2.

Solution:

The given digits are 7, 5, 6, 8, 2.

                           Tth      Th     H       T       O 
Greatest number    8       7       6       5       2
Smallest number    2       5       6       7       8

Arrange the given digits in the descending order 2 < 5 < 6 < 7 < 8.

To get the greatest number, the greatest digit 8 is placed at the highest valued place, i.e., ten thousand place, next smaller digit 7 at thousands place, next smaller digit 6 at hundred’s place, still smaller digit 5 is placed at ten’s place and the smallest digit 2 at one’s or units place.

Therefore, the greatest 5 digit number using the digits 7, 5, 6, 8, 2 is 87652.

Arrange the given digits in the ascending order 8 > 7 > 6 > 5 > 2.

To get the smallest number, the greatest digit 8 is placed at the lowest valued place i.e one’s place, next highest digit 7 is placed at ten’s place, next highest digit 6 is placed at hundred’s place, still greatest digit 5 is placed at thousand’s place, remaining digit 2 is placed at ten thousand place.

Therefore, the smallest 5 digit number using the digits 7, 5, 6, 8, 2 is 25678.

Example 3:

Write the greatest and smallest 5 digit number using 1, 2, 5. The digit may be repeated.

Solution:

The given digits are 1, 2, 5.
 Tth      Th     H       T       O
Greatest number    5       5       5       2       1

Smallest number    1       1       1       2       5

Arrange the given digits in the descending order 1 < 2 < 5

Since we have to make the greatest 5 digit number using 3 digits, we will repeat the greatest digit required a number of times.

To get the greatest number, the greatest digit 5 is placed at the highest valued place, i.e., ten thousand place, next smaller digit 5 at thousands place, next smaller digit 5 at hundred’s place, still smaller digit 2 is placed at ten’s place and the smallest digit 1 at one’s or units place.

Therefore, the greatest 5 digit number using the digits 1, 2, 5 is 55521.

To get the smallest number, the smallest digit 1 is placed at ten thousands-place, next greater digit 1 at thousand’s place, still greater digit 1 at hundred’s place, next greatest digit 2 is placed at ten’s place,  and greatest digit 1 at one’s or units place.

Therefore, the smallest 5 digit number using the digits 1, 2, 5 is 11125.

Example 4:

Write the greatest and smallest 4 digit number using 8, 7, 1, 4.

Solution:

The given digits are 8, 7, 1, 4.

       Th     H       T       O 
Greatest number      8       7       4       1

Smallest number       1       4       7       8

Arrange the given digits in the descending order 1 < 4 < 7 < 8

The greatest number using the digits 8, 7, 1, 4 is 8741.

To get the smallest number, we arrange the digits in ascending order.

Ascending order of 8, 7, 1, 4 is 8 > 7 > 4 > 1.

The smallest number using the digits 8, 7, 1, 4 is 1478.

FAQs on Formation of Greatest and Smallest Numbers

1. How to obtain the greatest and the smallest among the group of numbers?

The greatest number among the number of numbers is obtained by arranging the group of digits in the descending order and writing the numbers as it is. The smallest number is obtained by arranging the group of digits in the ascending order and representing them as it is.

2. Which is the greatest and smallest 4 digit number?

The greatest 4 digit number is 9999 and the smallest 4 digit number is 1000.

3. What is the ascending order?

The process of arranging the group of numbers or items of the same category from the lowest to the highest in value is called the ascending order.

Word Problems on Multiplication | Multiplication Word Problems with Solutions

Word Problems on Multiplication

Multiplication is one of the important arithmetic operations which helps to solve math problems easily. Usually, math tables are used for solving multiplication problems. So, make use of the Multiplication Tables for 0 to 25 and get the solution for your multiplication word problems quickly and easily. Students can get different types of word problems in the below sections.

What is Meant by Multiplication?

Multiplication is one of the basic arithmetic operations which gives the result of combining groups of equal sizes. Multiplication is a process of repeated addition. It is represented by the cross “x”, asterisk “*”, dot “.” symbol. When we multiply two numbers the answer is called the product.

Solved Problems on Multiplication

Question 1:

Find the product of 125 and 78.

Solution:

Given numbers are 125, 78.

125 x 78 = 9750

Word Problems on Multiplication 1

Question 2:

The product of two numbers is 175 and the multiplier is 7. Find the multiplicand?

Solution:

Given that,

The product of two numbers = 175

Multiplier = 7

Multiplicand = ?

Multiplicand * multiplier = Product

Multiplicand * 7 = 175

Multiplicand = 175/7

Word Problems on Multiplication 2

So, multiplicand = 25

Question 3:

Solve 78 x 96?

Solution:

Multiplicand = 78

Multiplier = 96

Word Problems on Multiplication 3

78 x 96 = 7488.

Also, Read

Worked out Word Problems on Multiplication

Question 1:

The weight of a rice bag is 50 kg. What will be the weight of 75 such bags?

Solution:

Given that,

Weight of a rice bag = 50 kg

Weight of 75 rice bags = 50 x 75

Word Problems on Multiplication 4

So, the total weight of 75 rice bags is 3750 kgs.

Question 2:

The price of a wooden box is $179 and a plastic box is $82. Find the cost of 40 wooden boxes and 150 plastic boxes in total?

Solution:

Given that,

The price of a wooden box = $179

The price of a plastic box = $82

The cost of 40 wooden boxes = 179 x 40 = 7160

Word Problems on Multiplication 5

The cost of 150 plastic boxes = 150 x 82 = 12300

Word Problems on Multiplication 6

The total cost of 40 wooden boxes and 150 plastic boxes = 7160 + 12300

= 19460

Word Problems on Multiplication 7

Therefore, the total cost of 40 wooden boxes and 150 plastic boxes is $19460.

 

Question 3:

A toy costs $216. How much will be paid for such 56 toys?

Solution:

Given that,

The cost of one toy = $216

The amount to be paid to buy 56 toy means add 216, 56 times

Otherwise, multiply 216 and 56 the obtained product is the amount paid.

So, the amount paid for 56 toys = 216 x 56

Word Problems on Multiplication 8

Therefore, $12096 must be paid to buy 56 toys.

Question 4:

The monthly salary of a man is $3156. What is his annual income by salary?

Solution:

Given that,

The monthly salary of a man = $3156

The annual income of the man by salary = Add man’s salary 12 times

Or, multiply the man’s monthly salary by 12 months to get the annual income in dollars.

So, the annual income of the man by salary = 3156 x 12

Word Problems on Multiplication 9

Therefore, the annual income of the man by salary is $37872.

Question 5:

Each student of class IV $75 for the flood victims. If there are 368 students in class IV, what is the total amount of money collected?

Solution:

Given that,

All the students of Class IV have given $75 for the flood victims

The total number of students in class IV = 368

The total amount of money collected from class IV students for flood victims = 75 should be added 368 times.

Otherwise, multiply the number of students by the amount each student is given.

So, the total amount of money collected from class IV students for flood victims = 75 x 368

Word Problems on Multiplication 10

Therefore, the total amount of money collected from class IV students for flood victims is $27600.

Question 6:

Prove that by adding multiplicand, multiplier times is equal to multiplicand into multiplier by solving the following question.

Ramu buys 7 pencils in a shop. The cost of each pencil is $2. Find the total amount paid to buy 7 pencils.

Solution:

Given that,

The cost of each pencil = $2

The number of pencils bought = 7

The amount paid to buy 7 pencils = cost of each pencil x total number of pencils bought

= 2 x 7

= $14

The amount paid to buy 7 pencils = Add the total number of pencils by the cost of each pencil times

= 2 + 2 + 2 + 2 + 2 + 2 + 2

= $14

Therefore, the cost of 7 pencils is $14

Hence proved.

Question 7:

A truck can carry 2,545 kg of coal per trip. How much coal will be carried if the truck makes 135 trips?

Solution:

Given that,

A truck can carry 2,545 kg of coal per trip

The amount of coal carried by 135 trips = 2545 x 135 = 343575

Word Problems on Multiplication 11

Therefore, 3,43,575 kgs of coal will be carried if the truck makes 135 trips.

Question 8:

There are 365 days in a year. How many days are there in 29 years?

Solution:

Given that,

The number of days in a year = 365

The number of days in 29 years = 365 x 29

Word Problems on Multiplication 12

Therefore, the number of days in 29 years is 10585.

Question 9:

A bookseller sells 875 books per day in the month of July. How many books will he be able to sell in 25 days of July?

Solution:

Given that,

A bookseller sells 875 books per day in the month of July

The number of books the bookseller will sell in 25 days of July = 875 x 25

Word Problems on Multiplication 13

Therefore, 21875 books will the bookseller be able to sell in 25 days of July.

Question 10:

There are 1125 students in a school. If a student pays $ 365 as fees and $ 150 as bus charge per month, how much money is collected after 8 months?

Solution:

Given that,

The number of students in a school = 1125

A student pays $ 365 as fees and $ 150 as bus charge per month

The amount of money collected from one student in a month = 365 + 150 = 515

Word Problems on Multiplication 14

The total amount of money collected from all students in a month = 515 x 1125 = $579375

Word Problems on Multiplication 15

The total amount of money collected from all students in 8 months = $579375 x 8

Word Problems on Multiplication 16

Therefore, the total amount of money collected from 1125 students of a school as fee and bus fee in 8 months is $4635000.

Nets of Solids – Definition | Geometry Shapes | Nets of a Cube, Cone, Pyramid, Prism, Cylinder

Nets of Solids

A net is a flattened out 3-dimensional solid. It is the basic skeleton outline in two dimensions, that can be folded and glued together to obtain the 3D structure. Nets are used for making 3D shapes. In the below sections, we have discussed the nets of different geometric shapes along with the examples.

Nets of Solids – Definition

A geometry net is a two-dimensional shape that can be folded to form a three-dimensional solid. When the surface of a three-dimensional figure is laid out flat showing each face of the solid, the pattern obtained is called the net. Nets are useful in finding the surface area of the solids. The following are some steps we must take to determine whether a net forms a solid:

  • Ensure that the net and a solid have the same number of faces and that there is a match between the shapes of faces of the solid and the shapes of the corresponding faces in the net.
  • Visualize how the net is to be folded to form the solid and all the sides fit together properly.

Nets of Cube

A cube is a 3-dimensional figure having 6 faces of equal length. The cube has 8 vertices, 12 edges. All the faces of a cube are in square shape. The plane angles of the cube are the right angle. The edges opposite to each other are parallel.

Nets of Solid 1

Cube Image

Nets of Solid 2

11 possible nets of a cube

Nets of Cylinder

A cylinder has two parallel bases joined by a curved surface at a fixed distance. The bases are in circular shape and the center of two bases are joined by a line segment, called the axis. The perpendicular distance between the bases is the height and the distance from the axis to the outer surface is the radius of the cylinder.

Nets of Solid 3

Cylinder Image

Nets of Solid 4

Nets of Cylinder

Nets of Rectangular Prism

A rectangular prism has six faces and each face is a rectangle. Both the bases of the prism are rectangles and other lateral faces are also rectangles. it is also called a cuboid.

Nets of solid 5

Rectangular Prism Image

nets of Solid 6

Nets of a Rectangular Prism

Nets of Triangular Prism

A triangular prism is a polyhedron having two triangular bases and three rectangular sides. Like other prisms, two bases are congruent and parallel. The prism has 5 faces, 9 sides and 6 vertices.

Nets of Solid 7

Triangular Prism Image

Nets of Solid 8

Nets of Triangular Prism

Nets of Cone

A cone is a shape formed by using a set of line segments which connects a common point, called vertex to all the points of a circular base. the distance between vertex to base of the cone is known as its height.

nets of solid 9

Cone Image

Nets of Solid 10

Nets of Cone

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Nets of Square Pyramid

A three-dimensional geometric shape with square base and four triangular faces all those faces meet at a single point is called the square pyramid. If all triangular faces have equal edges, then this pyramid is called an equilateral square pyramid.

Nets of solid 11

Square Pyramid Image

Nets of Solid 12

Nets of Square Pyramid

Solved Examples on Nets of Solid

Example 1:

Sketch the net of the solid shape given below.

Nets of Solid 13

Solution:

If the pyramid is unfolded along its edges we get the following net.

The net of the pentagonal pyramid is as follows.

Nets of Solid 14

Example 2:

Sketch the net of the solid shape given below.

Nets of Solid 16

Solution:

If the prism is unfolded along its edges we get the following net.

The net of the hexagonal prism is as follows.

Nets of solid 15

Example 3:

Sketch the net of the solid shape given below.

Nets of Solid 17

Solution:

If the prism is unfolded along its edges we get the following net.

The net of the hexagonal prism is as follows.

Nets of Solid 18

Frequently Asked Questions on Nets of Solid

1. How are nets useful in real life?

Nets are used in finding the surface area of the solids. The examples of nets are all three dimensional geometric shapes. Some of the 3 dimensional geometric shapes are square pyramid, cone, cylinder, triangular prism, rectangular pyramid, rectangular prism nad others.

2. Can a solid have different Nets?

Yes, a solid have different nets. Visualize how the net is folded to form a solid and make sure that all sides fit together properly. Make sure the solid and the net have same number of faces, shape of faces should match.

3. What is the net of a 3D shape?

The net of a 3D shape is what it looks like if it is opened out flat. A net can be folded up to ake a 3D shape. There can be several possible nets for one 3D shape. Draw a net on paper, then fold it into the shape.

4. How many nets are there for a cube?

There are exactly eleven nets that will form a cube.

Binary Addition Using 1s Complement – Definition, Examples | How to Add Binary Numbers in 1’s Complement?

Binary Addition Using 1s Complement

Are you searching for a tool that computes the addition of two binary numbers using 1’s complement method? If yes, then you have reached the correct place. Here we are giving the detailed steps on how binary addition of two numbers with 1s complement. You can also get Binary Addition Using 1s Complement definition, how to get 1s complement of a binary number, and example questions to improve the levels of understanding.

What is Binary Addition?

A binary number is a number with the base 2. Binary addition is one of the basic arithmetic operations. The binary system has only two digits 0 and 1. The binary code uses the digits 0’s and 1’s to make certain processes turn on or off. The process of the binary addition operation is very familiar to the decimal system by adjusting to the base 2.

Before attempting the binary addition, you must have complete knowledge of how the place works in the binary number system. Almost all modern digital computers and electronic circuits perform the binary operation by representing each bit as a voltage signal. The binary bit 0 means OFF state, 1 means ON state.

1’s Complement of a Binary Number

We have a simple algorithm to convert a binary number into 1’s complement. To get 1’s complement of a binary number, invert the given binary number. You can also implement a logic circuit using only NOT gate for each bit of binary number input.

Example:

Calculate the 1’s complement of a binary number 10101110

Invert each bit of a given binary number

So, 1’s complement of 10101110 is 01010001.

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Binary Addition Using 1’s Complement

Binary addition means simply performing an addition operation between two binary numbers. You can add two binary numbers digit by digit just like decimal numbers. The rules for adding two binary digits is 0 + 0 = 0, 1 + 0 = 1, 0 + 1 = 1, 1 + 1 = 10(1 is carry). At first, convert the given numbers to the 1’s complement and add those numbers using thee rules. Also, have a look at the detailed process on how to add binary numbers using 1’s complement in the following section.

How to do Binary Addition Using 1s Complement?

The process of adding binary numbers purely depends on their sign and magnitude. Here are the different cases.

Case I: Adding a positive and a negative number. If the positive number has a greater magnitude

Here, the addition of numbers is performed after taking 1’s complement of the negative number, and at the end round carry of the sum is added to the LSB (least significant bit).

Example: + 1101 – 1011

+ 1101 ⇒ 0 1 1 0 1

– 1011 ⇒ 1 0 1 0 0 (1’s complement)

⇒ 0 0 0 0 1 + 1 = 0 0 0 1 0

Hence the sum is +0010.

Case II: If the negative number has a greater magnitude

Here the addition is done in the same way as in case I but there will be non-end-around carry. The sum is obtained by taking the 1’s complement of the magnitude bits of the result and the sum is negative.

Example: 1011 – 1101

+ 1 0 1 1 ⇒ 0 1 0 1 1

– 1 1 0 1 ⇒ 1 0 0 1 0 (1’s complement)

⇒ 1 1 1 0 1

The sum is 1’s complement of 1101

Hence the sum is – 0010.

Case III: If two binary numbers are negative.

To add two negative binary numbers, 1’s complements of both the numbers are taken later addition is performed. In this case, an end-around carry will always appear. This along with a carry from the MSB will generate a 1 in the sign bit. 1’s complement of the magnitude bits of the result of an addition will give the final sum.

Example: -0110 and -0111

– 0 1 1 0 ⇒ 1 1 0 0 1 (1’s complement)

– 0 1 1 1 ⇒ 1 1 0 0 0 (1’s complement)

⇒ 1 0 0 0 1 + 1 = 1 0 0 1 0

1’s complement of 0010 is 1101 and the sign bit is 1

Hence the sum is -1101.

Binary Addition Using 1s Complement Examples

Example 1:

Calculate the sum of 0100, -1000 using the 1’s complement.

Solution:

The given binary numbers are 0100, -1000

1’s complement of -1000 is 10111

0 0 1 0 0 + 1 0 1 1 1 = 1 1 0 1 1

1’s complement of 1011 is 0100

Hence the sum is -0100.

Example 2:

Find the sum of 10000, -00111 using the 1’s complement.

Solution:

The given binary numbers are 10000, -00111

Find the 1’s complement of the negative number i.e 00111

1’s complement of 000111 = 111000

010000 + 111000 = 1 0 0 1 0 0 0

0 0 1 0 0 0 + 1 = 0 0 1 0 0 1

Hence, the sum is 001001.

Example 3:

Find the sum of -0100, -0010 using the 1’s complement method.

Solution:

The given numbers are -0100, -0010

Find the 1’s complement of the negative numbers

1’s complement of 0100 is 1011

1’s complement of 0010 is 1101

1 0 1 1 + 1 1 0 1 = 1 1 0 0 0

1 1 0 0 0 + 1 = 1 1 0 0 1

1’s complement of 1 0 01 is 0 1 1 0 and 1 is the sign bit

Hence, the sum is -0110

Example 4:

Find the sum of 1000 and -0101 using the 1’s complement.

Solution:

The given binary numbers are 1000 and -0101.

Find the 1’s complement of the negative number

1’s complement of 0101 is 11010.

0 1 0 0 0 + 1 1 0 1 0 = 1 0 0 0 1 0

1 0 0 0 1 0 + 1 = 1 0 0 0 1 1

Hence the sum is 00011.

FAQ’s on Binary Addition using 1’s Complement

1. How to find 1’s complement of a binary number?

To get the 1’s complement of binary numbers, just invert the number. Inversion means placing 1’s in place of 0’s and 0’s in place of 1’s. Otherwise, you can also use NOT logic gate to find the 1’s complement.

2. How to add two negative binary numbers using the 1’s complement?

For adding two negative binary numbers with the 1’s complement, just find the 1’s complement of both numbers. Add those 1’s complement numbers and add 1 to the obtained result. Then you will get 1 in place of the sign bit. Get the 1’s complement of the obtained sum to get the final result.

3. What’s the purpose of using complements in binary number operations?

Compared to other systems for representing the signed numbers 1’s complement, 2’s complement has the advantage that the fundamental arithmetic operations are identical to those for unsigned binary numbers.

Correct to One Decimal Place – Definition, Rules, Examples | How to Round to One Decimal Place?

Correct to One Decimal Place

Do you want to round numbers to one decimal place? If yes, then stay on this page. Here we will discuss how to correct to one decimal place. Rounding off is a type of estimation. We generally use estimation in everyday life and also in maths, physics subjects. You will also learn rounding decimals to the nearest tenth, rules, and example questions with the solutions in the following sections.

Correct to One Decimal Place – Definition

Correct to one decimal place is also known as round off to the nearest tenths. The rounding decimals technique is used to find the approximate values of a decimal number. Here the decimal numbers are rounded to the one decimal place to make them easier to read, understand instead of having lengthy string decimal places.

Rules for Rounding Decimals to Nearest Tenths or Rounding off to One Decimal Place

To correct the decimals to the nearest tenths, you have to check the below-mentioned rules.

  • Rule 1: If the digit in the hundredths place of the number is lesser than 5, then remove the following digits or substitute 0 in place of them.
  • Rule 2: If the digit in the hundredths of the number is equal to or greater than 5, then the digit in the tenths place is increased by 1 and the following digits are replaced by 0.

How to Round to One Decimal Place?

Have a look at the detailed step-by-step process on correct to one decimal place in the further sections.

  • Identify the number to which you need to round off to the nearest tenths.
  • Observe the digit in the hundredths position of the given number.
  • If the observed digit is less than 5 i.e 0, 1, 2, 3, 4 then replace the following digits with 0.
  • When the digit is more than 5 i.e 5, 6, 7, 8, 9 then add 1 to the digit in the tenths place and remove the following digits.

Also, Read

Rounding to 1 Decimal Place Examples

Example 1:

Round off the following numbers to one decimal place.

(a) 5.256

(b) 17.89

(c) 146.36

Solution:

(a) The given decimal number is 5.256

We see the digit in the hundredths place is 5 then round it to the nearest tenths which is greater than the given decimal number. Since 5 = 5 then the decimal number is rounded to 5.3.

Therefore, the solution is 5.3

(b) The given decimal number is 17.89

We can identify the digit in the hundredths place is 9 then round it to the nearest hundredths which is greater than the given decimal number. Since 9 > 5 then the decimal number is rounded to 17.80.

Therefore, the solution is 17.8.

(c) The given decimal number is 146.36

We see the digit in the hundredths place is 6 then round it to the nearest tenths which is greater than the given decimal number. Since 6 > 5 then the decimal number is rounded to 146.4.

Therefore, the solution is 146.4.

Example 2:

Round off the numbers to the nearest tenths.

(i) 14.732

(ii) 80.75

(iii) 16.54

Solution:

(i) The given decimal number is 14.732

We see the digit in the hundredths place is 3 then round it to the nearest tenths which is smaller than the given decimal number. Since 3 < 5 then the decimal number is rounded to 14.7.

(ii) The given decimal number is 80.75

The digit in the hundredths place is 5. ie equal to 5

Increase the digit in the tenths place by 1 and replace the following digits by 0.

The rounded number is 80.8.

(iii) The given decimal number is 16.54

We see the digit in the hundredths place is 4 then round it to the nearest tenths which is smaller than the given decimal number. Since 4 < 5 then the decimal number is rounded to 16.5.

Example 3:

Correct the following to one decimal place.

(i) 185.04

(ii) 77.49

(iii) 111.12

Solution:

(i) The given decimal number is 185.04

We can identify the digit in the hundredths place is 4 then round it to the nearest hundredths which is smaller than the given decimal number. Since 4 < 5 then the decimal number is rounded to 185.0.

Therefore, the solution is 185.0

(ii) The given decimal number is 77.49

We can identify the digit in the hundredths place is 9 then round it to the nearest hundredths which is greater than the given decimal number. Since 9 > 5 then the decimal number is rounded to 77.5.

Therefore, the solution is 77.5

(iii) The given decimal number is 111.12

We can identify the digit in the hundredths place is 2 then round it to the nearest hundredths which is smaller than the given decimal number. Since 2 < 5 then the decimal number is rounded to 111.1.

Therefore, the solution is 111.1.

Ratios – Definition, Formulas, Tricks, and Examples | How to Solve Ratios?

Ratios

The ratio indicates how many times a number contains another. Ratios are represented as fractions i.e a: b. The comparison or the simplified form of two quantities of the same kind is called the ratio. Interested students who want to know more about the concept of ratios can read this complete page. Here, we will discuss the basic concept of the ratio, key points, definition, and example questions.

Ratios – Definition

Ratios are an important concept in mathematics. In certain cases, the comparison of two quantities using the division method is difficult. So, at that time, we use ratio. The ratio gives us how many times one quantity is equal to another quantity.

Simply, a ratio is a number that is used to express one quantity as a fraction of another one. Two numbers in a ratio can be expressed only when they have the same unit. The sign of ratio is ‘:’. The real-life examples of a ratio are the rate of speed (distance/time), price of a material (rupees/meter, and others.

Key Points to Remember regarding Ratios

The key points to remember regarding the ratios are as follows:

  • A ratio must exist between two quantities of the same kind
  • To compare two things, their units should be the same
  • There should be significant order of terms
  • The comparison of two ratios can be performed, if the ratios are equivalent like fractions

Ratio Formulas

1. If we have two entities and you need to find the ratio of these two then the formula is defined as a: b or a/b.

Where a, b will be the entities

a is called the first term or antecedent and b is called the second term or consequent

2. If two ratios are equal, then they are proportional

a : b = c : d

d is called the fourth proportional to a, b, c

c is called third proportion to a, b

The mean proportion between a and b is √(ab)

3. If (a : b) > (c : d) = (a/b > c/d)

The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf)

4. If a: b is a ratio, then

a²: b² is a duplicate ratio

√a: √b is a sub-duplicate ratio

a³: b³ is a triplicate ratio

5. Ratio and Proportion Tricks

If a/b = x/y, then ay = bx or a/x = b/y or b/a = y/x

If a/b = x/y, then \(\frac { a + b }{ b } =\frac { x + y }{ y } \) or \(\frac { a – b }{ b } =\frac { x – y }{ y } \)

If a/b = x/y, then \(\frac { a + b }{ a – b } =\frac { x + y }{ x – y } \) this is componendo dividendo rule

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Solved Examples on Ratios

Example 1:

If x : y = 4 : 7, then find (4x – y) : (2x + 3y).

Solution:

Given ratio is x : y = 4 : 7

x = 4k, y = 7k

(4x – y) : (2x + 3y) = \(\frac { (4x – y) }{ (2x + 3y) } \) = \(\frac { (4 • 4k  – 7k) }{ (2 • 4k + 3 • 7k) } \)

= \(\frac { (16k  – 7k) }{ (8k + 21k) } \) = \(\frac { 9k }{ 29k } \)

= \(\frac { 9 }{ 29 } \)

= 9 : 29

Therefore, (4x – y) : (2x + 3y) = 9 : 29.

Example 2:

If a : b = 4 : 5, b : c = 15 : 8 then find a : c.

Solution:

Given that,

a : b = 4 : 5, b : c = 15 : 8

a : b = 4 : 5 = \(\frac { 4 }{ 5 } \) —– (i)

b : c = 15 : 8 = \(\frac { 15 }{ 8 } \) —– (ii)

By multiplying (i) and (ii), we get

\(\frac { a }{ b } \) x \(\frac { b }{ c } \) = \(\frac { 4 }{ 5 } \) x \(\frac { 15 }{ 8 } \)

\(\frac { a }{ c } \) = \(\frac { 3 }{ 2 } \)

Therefore, a : c = 3 : 2

Example 3:

If a quantity is divided in the ratio of 5: 7, the larger part is 84. Find the quantity.

Solution:

Given that,

A quantity is divided in the ratio of 5: 7

Let the quantity be x

Then the two quantities are \(\frac { 5x }{ 5 + 7 } \), \(\frac { 7x }{ 5 + 7 } \)

The larger part is 84

So, \(\frac { 7x }{ 5 + 7 } \) = 84

\(\frac { 7x }{ 12 } \) = 84

7x = 84 • 12

7x = 1008

x = \(\frac { 1008 }{ 7 } \)

x = 144

Therefore, the quantity is 144.

Example 4:

If (3a + 5b) : (7a – 4b) = 7 : 4 then find the ratio a : b.

Solution:

Given that,

(3a + 5b) : (7a – 4b) = 7 : 4

\(\frac { 3a + 5b }{ 7a – 4b } \) = \(\frac { 7 }{ 4 } \)

4(3a + 5b) = 7(7a – 4b)

12a + 20b = 49a – 28 b

20b + 28b = 49a – 12a

48b = 37a

\(\frac { 48 }{ 37 } \) = \(\frac { a }{ b } \)

So, a : b = 48 : 37

Frequently Asked Questions on Ratios

1. What are the different types of ratios?

The different types of ratios are compounded ratio, duplicate ratio, triplicate ratio, subtriplicate ratio, subduplicate ratio, the ratio of equalities, reciprocal ratio, the ratio of inequalities, the ratio of greater inequalities, and the ratio of lesser inequalities.

2. What are the 3 ways of writing ratios?

The three most used ways to write a ratio are given here. The first one is fraction 2/5. The second method is using a word to i.e 2 to 5. Finally, the third one is using the ratio colon between two numbers, 2: 5.

3. Define ratio with an example?

The ratio is a mathematical expression represented in the form of a: b, where a and b are two integers. It can also be expressed as a fraction. It is used to compare things or quantities. The example is 3: 4 = 3/4.

4. Write the differences between ratio and proportion?

The ratio is helpful to compare two things of the same unit whereas proportion is used to express the relation between two ratios. The ratio is represented using a colon: or slash / and proportion is represented using a double colon:: or equal to symbol =. The keyword to identify a ratio is “to every” and the proportion is “out of”.

Binary Addition using 2’S Complement – Definition, Examples | How to do 2’S Complement Binary Addition?

Binary Addition using 2'S Complement

A binary number is a number expressed in the base 2 numeral system which uses only two symbols 0 and 1. Each digit in the binary is called a bit or binary digit. The addition is one of the basic arithmetic operations. Binary addition is one of the binary operations. The binary addition works similarly to the base 10 decimal system. Check out more about the binary addition using 2s complement with solved example questions in the following sections of this page.

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Binary Addition

The binary addition is similar to the decimal system, but it is a base 2 system. The binary system has two digits 0 and 1. Almost all functionalities of the computer system use the binary number system. The process of the addition operation is very familiar to the decimal system by just adjusting the base 2 of the numbers.

Before attempting the binary addition process, we should have complete knowledge of how the place works in the binary number system. Most modern digital computers and electric circuits perform binary operations by representing each bit as a voltage signal. the bit 0 in the binary system denotes the “OFF” state, bit 1 in the binary system denotes the “ON” state.

2’s Complement of a Binary Number

To get the 2’s complement of a given binary number, invert the given number and add 1 to the least significant bit (LSB) of the given result. The various uses of 2’s complement of binary numbers are signed binary number representation, to perform arithmetic operations on binary numbers.

Example:

2’s complement of 101

Invert 101 = 010

Add 1 to LSB of 010 = 010 + 1

= 011

So, 2’s complement of 101 is 011.

Binary Addition using 2’s Complement

Binary Addition using 2’s Complement is similar to the normal addition of two binary numbers. When you add two positive numbers, then the result is a positive number. When you add two negative numbers, then the addition will be a negative number. The basic rules of binary addition is 1 + 1 = 10 (1 is carry), 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1. If two numbers have different signs then follow these cases.

Case I: When the positive number has a greater magnitude.

In this case, the carry obtained is discarded and the final result is the addition of two numbers

Let us take the 5-bit numbers 1011 and -0101

2’s complement of 0101 = 1011

1 0 1 1 ⇒ 0 1 0 1 1

– 0 1 0 1 ⇒ 1 1 0 1 1 (2’s Complement)

⇒ 0 0 1 1 0 (Carry 1 is discarded)

So, 1011 – 0101 = 00110

Case II: When the negative number has a greater magnitude.

If the negative number has a greater magnitude then no carry will be generated in the sign bit. The result of an addition will be negative and the final result obtained by taking 2’s complement of the magnitude bits of the result.

Let us take two numbers as + 0100 and -0111

+ 0 1 0 0 ⇒ 0 0 1 0 0

– 0 1 1 1 ⇒ 1 1 0 0 1 (2’s complement)

⇒ 1 1 1 0 1

2’s complement of 1101 is 0011.

Hence the required sum is – 0011.

Case III: When the numbers are negative.

When two negative numbers are added a carry will be generated from the sign bit which will be discarded. 2’s complement of the magnitude bits of the operation will be the final sum.

Let us take two numbers as -0111 and -0010

– 0 1 1 1 ⇒ 1 1 0 0 1 (2’s complement)

– 0 0 1 0 ⇒ 1 1 1 1 0 (2’s complement)

⇒ 1 0 1 1 1 (Carry 1 discarded)

2’s complement of 0111 is 1001

Hence the required sum is -1001.

Solved Examples on Binary Addition using 2’s Complement

Example 1:

Find the sum of -1101 and -1110 using the 2’s complement.

Solution:

Given numbers are -1101, -1110

Find the 2’s complement of the negative numbers

So, 2’s complement of 01110 is 10010 and 01101 is 10011

Add the complemet numbers

1 0 0 1 0 + 1 0 0 1 = 1 0 0 1 0 1

End carry 1 of the sum is dicarded

2’s complement of the result 00101 is 11011

So, sum of -1101, -1110 is -11011.

Example 2:

Find 10101 – 10111 using the 2’s complement concept.

Solution:

The given numbers are 10101, -10111

2’s complement of the negative number 10111 is 01001

Add first number and 2’s complement of negative number

1 0 1 0 1 + 0 1 0 0 1 = 1 1 1 1 0

2’s complement of the result 11110 is 00010.

So, 10101 – 10111 = -00010

Example 3:

Find the addition of 00110 and 01001 using the 2’s complement.

Solution:

Given numbers are 00110, 01001

0 0 1 1 0 + 0 1 0 0 1 = 0 1 1 1 1

Hence, the sum is 0 1 1 1 1.

FAQs on Binary Addition using 2’s Complement

1. What are the Rules of Binary Addition?

The four basic rules of the binary addition are 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 10.

2. How to add two binary numbers?

Add the first column if two numbers are 1 then the result is 10. Here, 1 is the carry. Continue adding till no digit is left on the left side. The result is the sum of two binary numbers.

3. How to find 2’s complement of a binary number?

Invert the given numbers and add one to the least significant bit to get the 2’s complement of the binary number.

Units of Time – SI, CGS, Other Units | Converting Units of Time(Hours, Minutes, Seconds)

Units of Time

Time is defined as the continued progress of existence in past, present, and future. Using the unit of time you can measure the existence of events. The most commonly used units of time are second, minute and hour. Time is an interesting topic and everyone is familiar with it. Want to know about the units of time then go through the following sections. Learn Converting Units of Time such as Hours, Minutes, Seconds from one unit to another by referring through solved examples available.

SI Unit of Time

The SI Unit of Time is Second and is accurately defined as the time interval equal to 9192631770 periods of radiation. Unit second is often represented as s or sec.

Different Units of Time

Some of the common and frequently used time units are minutes, hour, day, week, month, and year. If you want to measure a long duration of time, then you can use a decade which is equal to 10 years, a century which is equal to 100 years, a millennium which is equal to 1000 years, and a mega-annum which is equal to 1,000,000 years. The popular units of time are given below.

The most commonly used Units of Time are

  • Hours
  • Minutes
  • Seconds
1000 milliseconds 1 second
60 seconds 1 minute
60 minutes 1 hour
24 hours 1 day
7 days 1 week
28, 29, 30, or 31 days 1 month
365 or 366 days 1 year
12 months 1 year
10 years 1 decade
100 years 1 century
1000 years 1 millennium

How Time Became So Important?

The top ten reasons why time is so important are mentioned here.

  • Every single thing in the universe is affected by time.
  • Time is the most precious resource because you can’t get it back.
  • Because of privileges, not everyone truly has the same amount of time in a day.
  • No one knows how much time they have.
  • The only time we actually have is the present.
  • How we see time impacts happiness.
  • Managing it poorly or well has a huge impact on life.
  • Skills are impacted by how much time you invest.
  • Relationships are made or broken by how much time you invest.
  • Time is a teacher or a healer.

Also, Read

Time Conversions

To convert 1 unit of time to another, you have to know the units of time. You can use the multiplication or division operations to convert the time.

  • To convert minutes into seconds, you need to multiply each minute by 60 seconds.
  • To convert seconds into minutes, you need to divide each second by 60 minutes.

You can use these parameters for any type of time conversion.

Solved Examples on Converting Units of Time

1. Convert the following

(i) 5 hours 30 minutes into minutes

(ii)90 minutes to seconds

Solution:

(i) 5 hours 30 minutes into minutes

We know 1 hr = 60 minutes

5 hours = 5*60 minutes

= 300 minutes

5 hrs 30 minutes = 300 minutes +30 minutes

= 330 minutes

Therefore, 5 hours 30 minutes = 330 minutes

(ii)90 minutes to seconds

We know 1 minute = 60 seconds

90 minutes = 90*60 seconds

= 5400 seconds

Therefore, 90 minutes = 5400 seconds

2. Find the total time

5 hours 40 minutes and 3 hours 20 minutes

Solution:

Firstly add the hours i.e. 5 hours +3 hours

= 8 hours

Now add the minutes individually i.e. 40 minutes +20 minutes

= 60 minutes

= 1 hour

Now, add this to the hours we got in the earlier step i.e. 8 hours +1 hour

= 9 hours

Therefore, 5 hours 40 minutes and 3 hours 20 minutes is equal to 9 hours

Frequently Asked Questions on Units of Time

1. What is the SI unit of time?

The SI unit of time is seconds.

2. What are the 3 possible units of time?

The three most used units of time are seconds, minutes, and hours. 1 minute = 60 seconds, 1 hour = 60 minutes = 3600 seconds.

3. What are the different units of a second?

The various units of a second from the smallest to the largest values are along the lines:
Decisecond (1/10th of a second), centisecond (1/100th of a second), millisecond (1/1000th of a second), microsecond (one-millionth of a second), nanosecond (one-billionth of a second), picosecond (one-trillionth of a second), femtosecond (one-quadrillionth of a second), attosecond (one-quintillionth of a second), zeptosecond (one-sextillionth of a second), yoctosecond (one-septillionth of a second), and Planck time.

4. What is the largest unit of time?

The largest unit of time is the supereon. It is the combination of eons, eras, periods, epochs, and ages.

Laws of Algebra of Sets – Commutative, Associative, Distributive, Demorgan’s | Set Operations & Laws of Set Theory

Laws of Algebra of Sets

Sets are defined as a collection of well-defined elements that do not vary from person to person. It can be represented either in set-builder form or roster form. Generally, sets can be represented using curly braces {}. The different types of sets are empty set, finite set, singleton set, infinite set, equivalent set, disjoint sets, equal sets, subsets, superset, and universal sets. Get to know more about the Laws of Algebra of Sets for a better understanding of the students.

Laws of Algebra of Sets

The operations of sets are union, intersection, and complementation. The binary operations of set union, intersection satisfy many identities. The seven fundamental laws of the algebra of sets are commutative laws, associative laws, idempotent laws, distributive laws, de morgan’s laws, and other algebra laws.

1. Commutative Laws

For any two finite sets A and B

  • A U B = B U A
  • A ∩ B = B ∩ A

2. Associative Laws

For any three finite sets A, B, and C

  • (A U B) U C = A U (B U C)
  • (A ∩ B) ∩ C = A ∩ (B ∩ C)

So, union and intersection are associative.

3. Idempotent Laws

For any finite set A

  • A U A = A
  • A ∩ A = A
  • A ∩ A’ = ∅
  • ∅’ = U
  • ∅ = U’

4. Distributive Laws

For any three finite sets A, B, and C

  • A U (B ∩ C) = (A U B) ∩ (A U C)
  • A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.

5. De morgan’s Laws

For any two finite sets A and B

  • A – (B U C) = (A – B) ∩ (A – C)
  • A – (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also be written as

  • Law of union: (A U B)’ = A’ ∩ B’
  • Law of intersection: (A ∩ B)’ = A’ U B’

6. Complement Law

For any finite set A

  • A ∪ A’ = A’ ∪ A =U
  • A ∩ A’ = ∅

More laws of the algebra of sets:

7. For any two finite sets A and B;

  • A – B = A ∩ B’
  • B – A = B ∩ A’
  • A – B = A ⇔ A ∩ B = ∅
  • (A – B) U B = A U B
  • (A – B) ∩ B = ∅
  • A ⊆ B ⇔ B’ ⊆ A’
  • (A – B) U (B – A) = (A U B) – (A ∩ B)

8. For any three finite sets A, B, and C;

  • A – (B ∩ C) = (A – B) U (A – C)
  • A – (B U C) = (A – B) ∩ (A – C)
  • A ∩ (B – C) = (A ∩ B) – (A ∩ C)
  • A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

Also, Read

Solved Examples on Laws of Algebra of Sets

Example 1:

If E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7} show that

(i) (A U B)’ = A’ ∩ B’

(ii) (A U B) = B U A

(iii) A ∩ B = B ∩ A

(iv) (A ∩ B)’ = A’ U B’

Solution:

Given that

E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7}

(i) (A U B)’ = A’ ∩ B’

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

= {1, 2, 3, 4, 5, 7}

(A U B)’ = {1, 2, 3, 4, 5, 7}’

= {6}

A’ = {6, 7}

B’ = {1, 3, 4, 6}

A’ ∩ B’ = {6, 7} ∩ {1, 3, 4, 6}

= {6}

So, L.H.S = R.H.S

(ii) (A U B) = B U A

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

= {1, 2, 3, 4, 5, 7}

B U A = {2, 5, 7} U {1, 2, 3, 4, 5}

= {1, 2, 3, 4, 5, 7}

So, L.H.S = R.H.S

(iii) A ∩ B = B ∩ A

L.H.S = A ∩ B

= {1, 2, 3, 4, 5} ∩ {2, 5, 7}

= {2, 5}

R.H.S = B ∩ A

= {2, 5, 7} ∩ {1, 2, 3, 4, 5}

= {2, 5}

So, L.H.S = R.H.S

(iv) (A ∩ B)’ = A’ U B’

L.H.S = (A ∩ B)’

= {{1, 2, 3, 4, 5} ∩ {2, 5, 7}}’

= {2, 5}’

= {1, 3, 4, 6, 7}

R.H.S = A’ U B’

= {1, 2, 3, 4, 5}’ U {2, 5, 7}’

= {6, 7} U {1, 3, 4, 6}

= {1, 3, 4, 6, 7}

L.H.S = R.H.S

Hence, proved.

Example 2:

If X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e} verify that

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

Solution:

Given that,

X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e}

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

L.H.S = (X ∪ Y) ∪ Z

= ({a, b, c, d} U {b, d, f}) U {a, c, e}

= {a, b, c, d, f} U {a, c, e}

= {a, b, c, d, e, f}

R.H.S = X ∪ (Y ∪ Z)

= {a, b, c, d} U ({b, d, f} U {a, c, e})

= {a, b, c, d} U {a, b, c, d, e, f}

= {a, b, c, d, e, f}

So, L.H.S = R.H.S

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

L.H.S = (X ∩ Y) ∩ Z

= ({a, b, c, d} ∩ {b, d, f}) ∩ {a, c, e}

= {b, d} ∩ {a, c, e}

= ∅

R.H.S = X ∩ (Y ∩ Z)

= {a, b, c, d} ∩ ({b, d, f} ∩ {a, c, e})

= {a, b, c, d} ∩ ∅

= ∅

So, L.H.S = R.H.S

Hence verified.

Example 3:

If A = {p, q, r, s}, B = {u, q, s, v} find

(i) A – B

(ii) B – A

(iii) A ∩ B

Solution:

Given that,

A = {p, q, r, s}, B = {u, q, s, v}

(i)

A – B = {p, q, r, s} – {u, q, s, v}

= {p, r}

(ii)

B – A = {u, q, s, v} – {p, q, r, s}

= {u, v}

(iii)

A ∩ B = {p, q, r, s} ∩ {u, q, s, v}

= {q, s}

FAQs on Laws of Algebra of Sets

1. What is a set? Give an example?

A set is a collection of elements or objects or numbers represented using the curly brackets {}. The example is {1, 2, 3, 5} is a set of numbers.

2. What are the five basic properties of sets?

The five basic properties of sets are commutative property, identity property, associative property, complement property, and distributive property.

3. What are the 4 operations of sets?

The 4 set operations include set union, set intersection, set difference, the complement of a set, and cartesian product.

Sum of Angles of a Quadrilateral | Quadrilateral Angles Sum Property – Theorem, Proof, Examples

Sum of Angles of a Quadrilateral

A quadrilateral can be defined as a closed geometric, two-dimensional shape having 4 straight sides. It has 4 vertices and angles. The types of quadrilaterals are parallelograms, squares, rhombus, and rectangle. The sum of all interior angles of a quadrilateral is equal to 360°. The angle is formed when two line segments meet at a common point. The angle can be measured in degrees or radians. The angles of a quadrilateral are the angles formed inside the closed shape.

Sum of Angles of a Quadrilateral Theorem & Proof

The sum of interior angles of a quadrilateral is 360 degrees.

Quadrilateral

In the quadrilateral ABCD

∠ABC, ∠ADC, ∠DCB, ∠CBA are the interior angles

AC is the diagonal of the quadrilateral

AC splits the quadrilateral into two triangles ∆ABC and ∆ADC

We know that sum of angles of a quadrilateral is 360°

So, ∠ABC + ∠ADC + ∠DCB + ∠CBA = 360°

Let’s prove that sum of all interior angles of a quadrilateral is 360 degrees.

We know that the sum of angles in a triangle is 180°

In triangle ADC

∠CAD + ∠DCA + ∠D = 180° —- (i)

In the triangle ABC

∠B + ∠BAC + ∠BCA = 180° —- (ii)

Add both the equations

∠CAD + ∠DCA + ∠D + ∠B + ∠BAC + ∠BCA = 180° + 180°

∠D + (∠CAD + ∠BAC) + (∠BCA + ∠DCA) + ∠B = 360°

We can see that ∠CAD + ∠BAC = ∠DAB, ∠BCA + ∠DCA = ∠BCD

So, ∠D + ∠DAB + ∠BCD + ∠B = 360°

∠D + ∠A + ∠C + ∠B = 360°

Therefore, the sum of angles of a quadrilateral is 360°

Quadrilateral Angles Sum Propoerty

Each quadrilateral has 4 angles. The sum of its interior angles is always 360 degrees. So, we can find the angles of the quadrilateral if we know the remaining 3 angles or 2 angles or 1 angle and 4 sides. For a square or rectangle, the value of all angles is 90 degrees.

Also, Read

Examples on Quadrilateral Angles

Example 1:

Find the fourth angle of the quadrilateral if three angles are 85°, 100°, 60°?

Solution:

The given three angles of a quadrilateral are 85°, 100°, 60°

We know that the sum of angles of a quadrilateral is 360°

So, ∠A + ∠B + ∠C + ∠D = 360°

85° + 100° + 60° + x° = 360

245° + x° = 360°

x° = 360° – 245°

x° = 115°

Therefore, the fourth angle of the quadrilateral is 115°.

Example 2:

Find the measure of the missing angles in a parallelogram if ∠A = 75°?

Solution:

We know that the opposite angles of a parallelogram are equal.

So, ∠C = ∠A, ∠B = ∠D

Sum of angles is 360°

∠A + ∠B + ∠C + ∠D = 360°

75° + ∠B + 75° + ∠D = 360°

150° + ∠B + ∠D = 360°

∠B + ∠D = 360° – 150°

∠B + ∠D = 210°

∠B + ∠B = 210°

2∠B = 210°

∠B = \(\frac { 210° }{ 2 } \)

∠B = 105°

So, other angles of a parallelogram are 105°, 75°, 105°.

Example 3:

The angle of a quadrilateral are (3x + 2)°, (x – 3)°, (2x + 1)°, 2(2x + 5)° respectively. Find the value of x and the measure of each angle?

Solution:

The given angles are ∠A = (3x + 2)°, ∠B = (x – 3)°, ∠C = (2x + 1)°, ∠D = 2(2x + 5)°

We know that the sum of angles of a quadrilateral is 360°

∠A + ∠B + ∠C + ∠D = 360°

(3x + 2)° + (x – 3)° + (2x + 1)° + 2(2x + 5)° = 360°

3x + 2 + x – 3 + 2x + 1 + 4x + 10 = 360

10x + 10 = 360

10x = 360 – 10

10x = 350

x = \(\frac { 350 }{ 10 } \)

x = 35

The measurement of each angle of a quadrilateral is ∠A = (3x + 2)° = (3(35) + 2) = 105 + 2 = 107°

∠B = (x – 3)° = (35 – 3) = 32°

∠C = (2x + 1)° = (2(35) + 1) = 70 + 1 = 71°

∠D = 2(2x + 5)° = 2(2(35) + 5) = 2(70 + 5) = 2(75) = 150°

Example 4:

The three angles of a closed 4 sided geometric figure are 20.87°, 53.11°, 8.57°. Find the fourth angle?

Solution:

The given angles are ∠A = 20.87°, ∠B = 53.11°, ∠C = 8.57°

We know that the sum of angles of a quadrilateral is 360°

∠A + ∠B + ∠C + ∠D = 360°

20.87° + 53.11° + 8.57° + x° = 360°

82.55° + x° = 360°

x = 360 – 82.55

x = 277.45°

Therefore, the fourth angle of the closed 3 sided geometric figure is 277.45°.

FAQs on Sum of Angles of a Quadrilateral

1. What is the sum of the internal angles of a quadrilateral?

The sum of angles of a quadrilateral is 360 degrees.

2. What are the properties of a quadrilateral?

The three different properties of a quadrilateral are it has four sides, four vertices, four angles. And it is a closed 2-dimensional geometric figure. The sum of within angles is 360 degrees.

3. How do you prove the angle sum property of a quadrilateral?

To prove the sum property of a quadrilateral, draw a diagonal to divide it into two triangles. The sum of all interior angles of a triangle is 180 degrees. thus, the sum of angles of a quadrilateral becomes 360°.

4. What is the sum of all interior angles of a pentagon?

Draw one diagonal that should divide the pentagon into one triangle, one quadrilateral. The sum of angles of a triangle is 180 degrees, the sum of angles of a quadrilateral is 360 degrees. So, the sum of all interior angles of a pentagon is 180 + 360 = 540°.

Multiplication Table of 7 | Tips & Tricks to Remember 7 Times Table Multiplication Chart

Multiplication Table of 7

Do you need assistance in memorizing 7 table and solving the multiplication problems? Then, stay on this page. The 7 Times Table is used to find the difficult math concepts like square roots, GCF, LCM, HCF, and others. So, learning and remembering the 7 Times Table Multiplication Chart is very important. On our site, students can discover all basic information about the Math Tables like how to learn, how to read, how to write, and tips, tricks to memorize 7 Table.

Multiplication Table of Seven | 7 Times Table Chart

Multiplication Table of 7 Chart is given here in an image format for a better understanding and quick reference of students. So, you can easily download 7 Times Multiplication Chart from here and try to memorize it regularly for doing quick calculations in competitive exams. Moreover, you can also have a quick revision of the Seven Multiplication Table by taking a printed copy and pasting it on your room wall.

7 Times Multiplication Table

Importance of Learning 7 Times Table Chart

Here you will get an answer for why one should learn the multiplication chart of 7.

  • Learning Multiplication Table of 7 helps you to solve mental math problems easily. This table can be quite handy while you solve real-world problems.
  • This multiplication table saves your time while performing division and multiplication questions.

How to Read 7 Times Table Multiplication Chart?

Zero times seven is 0.

One time seven is 7.

Two times seven is 14.

Three times seven is 21.

Four times seven is 28.

Five times seven is 35.

Six times seven is 42.

Seven times seven is 49.

Eight times seven is 56.

Nine times seven is 63

Ten times seven is 70.

7 Times Multiplication Table up to 20

Multiplication Table of 7 is the easiest multiplication table to remember. Refer to the below-mentioned 7 Times Table Multiplication Chart up to 20 and understand how to write it mathematically. By seeing the Seven Multiplication Table, you can study and grasp the multiplication facts about the table easily. You can also improve your math skills and speed in answering the math problems in exams. Make use of 7 Times Tabular format and do fast calculations.

7 x 0 = 0
7 x 1 = 7
7 x 2 = 14
7 x 3 = 21
7 x 4 = 28
7 x 5 = 35
7 x 6 = 42
7 x 7 = 49
7 x 8 = 56
7 x 9 = 63
7 x 10 = 70
7 x 11 = 77
7 x 12 = 84
7 x 13 = 91
7 x 14 = 98
7 x 15 = 105
7 x 16 = 112
7 x 17 = 119
7 x 18 = 126
7 x 19 = 133
7 x 20 = 140

Check More Math Multiplication Tables

Multiplication Table of 0 Multiplication Table of 1 Multiplication Table of 2
Multiplication Table of 3 Multiplication Table of 4 Multiplication Table of 5
Multiplication Table of 6 Multiplication Table of 8 Multiplication Table of 9
Multiplication Table of 10 Multiplication Table of 11 Multiplication Table of 12
Multiplication Table of 13 Multiplication Table of 14 Multiplication Table of 15
Multiplication Table of 16 Multiplication Table of 17 Multiplication Table of 18
Multiplication Table of 19 Multiplication Table of 20 Multiplication Table of 21
Multiplication Table of 22 Multiplication Table of 23 Multiplication Table of 24
Multiplication Table of 25

Tips & Tricks to Memorize Multiplication Table of 7

  • Seven number has infinite multiples and can be multiplied by any whole number.
  • You can also get multiples of 7 by skip counting by 7.
  • As 7 is a prime number it doesn’t repeat itself unit 7 x 10.

Solved Example Questions on 7 Table

Example 1:

What does 7 x 13 mean? What number is equal to?

Solution:

7 x 13 means 7 times 13

7 x 13 = 91

91 is equal to 7 x 13.

Example 2:

Varsha has 7 glasses. She puts 7 straws in each glass. How many straws are there in all?

Solution:

As Varsha puts 7 straws in each glass,

The total number of straws = 7 x 7

= 49.

Example 3:

Using the multiplication table of 7, find the value of 7 times 7 minus 4?

Solution:

Expressing the given statement in the form of mathematical expression we get

(7 x 7) – 4 = 49 – 4

= 45.

Thousandths Place in Decimals | How to Read Decimals in Thousandths? | Decimal Place Value Examples

Thousandths Place in Decimals

A decimal number has two parts whole number part and decimal number part. The point which separates two numbers is called a decimal point. The number available after the decimal point is called the decimal number part. The first position after the decimal point is the tens place, the second position is hundreds place and the third position is called the thousands place. Willing students can learn more about Thousandths Place in Decimals in the further segments of this article in detail.

Also, Read:

Thousandths Place in Decimals

Place value is defined as something that relates to the position or place of a digit in a number. If a decimal number is represented in a general form, the position of every digit in a number is expanded. The position of digits in decimal numbers will be tens, hundreds, thousands, ten thousand, and so on. Thousandths Place in Decimals is nothing but the third digit after the point.

Rules for the Decimal Place Values

Following are the rules for the decimal place values. Everyone should remember these rules for identifying

  • The first digit available to the left of the decimal point is the one’s place
  • The first digit to the right of the decimal point is called the tenths place
  • The second digit present to the left of decimal point os tens place
  • The second digit to the right of the decimal point is called the hundredths place
  • The third digit to the left of the decimal point is hundreds place.
  • The third digit present to the right of the decimal point is called the thousandths place and so on.

How to Represent Fractions as Decimal Numbers?

  • The two parts of the fraction are the numerator and denominator.
  • If the denominator contains zero’s then add the decimal point after those many digits from the left on the numerator.
  • Otherwise, divide the numerator by denominator to get the decimal number.

Thousandths Place in Decimals Examples

Question 1:

Represent the following as decimals and identify thousandths place in decimals

(i) 125/1000

(ii)1/1000

Solution:

(i) The given fraction is 125/1000

As the denominator has 3 zero’s add decimal point after 3 digits in the numerator from the left side.

So, 125/1000 = 0.125

Thousandths place in decimals is 5. Because 5 is the third digit to the right of the decimal point.

(ii) The given fraction is 1/1000

As the denominator has 3 zero’s add decimal point after 3 digits in the numerator from the left side.

So, 1/1000 = 0.001

Thousandths place in decimals is 1. Because 1 is the third digit to the right of the decimal point.

Question 2:

Find the digit in thousandths place in decimals

(i) 1.1236

(ii) 63.5698

Solution:

(i) The given decimal number is 1.1236

Thousandths place means the digit present at the third position from the right side of decimal point.

So, the digit in the thousandths place is 3.

(ii) The given decimal number is 63.5698

Thousandths place means the digit present at the third position from the right side of the decimal point.

So, the digit in the thousandths place is 9.

Question 3:

Find the digit in thousandths place in decimals

(i) 0.2545

(ii) 0.1564

Solution:

(i) The given decimal number is 0.2545

Thousandths place means the digit present at the third position from the right side of the decimal point.

Therefore, the digit in the thousandths place is 4.

(ii) The given decimal number is 0.1564

Thousandths place means the digit present at the third position from the right side of the decimal point.

So, the digit in the thousandths place is 6.

FAQs on Decimals in Thousandths Place?

1. How many decimal places are thousandths?

Three decimal places are called the thousandths.

2. How do you place value with decimals?

The first digit after the decimal point represents tenths place, the second digit represents hundredths place, the third digit represents th thousandths place, and so on.

3. What is Place Value?

Place value tells about the place of a digit in a decimal number. Based on the position of the digit it has value varies. When you expand a decimal number, then you will use the place value concept.

Worksheet on Pythagorean Theorem | Pythagorean Theorem Worksheet with Answers

Worksheet on Pythagorean Theorem

Worksheet on Pythagorean Theorem is helpful for the students who are willing to solve the problems based on the right-angled triangle, Pythagorean theorem. If you are preparing for any exam, then you can begin preparation by referring Pythagorean Theorem Worksheet. Make use of these Pythagoras Theorem Questions during practice and perform well in the test. We have covered various types of questions that use Pythagorean Statements. So, Practice all the questions from the worksheet as many times as possible so that you will understand the concept easily.

What is Pythagorean Theorem?

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides”. The three sides of the right-angled triangle are the perpendicular side, base, and hypotenuse side, where the hypotenuse is the largest side. Pythagorean formula is hypotenuse² = base² +perpendicular_side².

c² = a² + b²

Where a is base side

b is perpendicular side

c is hypotenuse side

How to Solve Problems on Pythagorean Theorem?

Check out the manual procedure in the below sections. Follow these guidelines to solve the Pythagorean theorem.

  • Look at all terms in the question.
  • Find out which vertex has the right angle.
  • Start with those right triangles and apply the Pythagorean theorem.

Also, Read

Question 1:

In ∆ABC right angled at A. If AB = 16 m and BC = 8 cm, then find the length of AC?

Solution:

Given that,

The sides of the right-angled triangle AB = 8 cm, BC = 10 cm

As per the Pythagorean Theorem,

hypotenuse² = base² +perpendicular_side²

BC² = AC² + AB²

AC² = BC² – AB²

AC = √(BC² – AB²)

Substitute AB, BC values in the above formula

AC = √(10² – 8²)

= √(100 – 64)

= √(36)

= 6

Therefore AC = 6 cm.


Question 2:

A ladder 7 m long when set against the wall of the house just reaches a window at a height of 24 m from the ground. How far is the lower end of the ladder from the base of the wall?

Solution:

Let the ladder be AC = 7 m

Let the vertical height reached be BC = 24 m

Let the distance between the foot of the ladder and wall be AB = x meters

By Pythagoras Theorem,

AB² = AC² + BC²

x² = 7² + 24²

x² = 49 + 576

x² = 625

x = 25

Therefore, the foot of the ladder is 25 m away from the wall.


Question 3:

A carpet measures 7 feet long and has a diagonal measurement of 74 square feet. Find the width of the carpet.

Solution:

Given that,

Length of the carpet = 7 feet

Diagonal of the carpet = 74 sq ft

In a rectangle,

diagonal² = length² + width²

width² = diagonal² – length²

= 74 – 7²

= 74 – 49

= 25

width = √(25)

= 5 ft

Therefore, the width of the carpet is 5 ft.


Question 4:

Jim starts driving east for 9 miles, then takes a left turn, and then he drives north for another 40 miles. At the end of driving, what is the distance of a straight line from the starting point?

Solution:

As given in the question,

Driving for east = 9 miles

Driving for north = 40 miles

Let the straight line distance = x

The equation will be

a² + b² = c²

9² + 40² = x²

81 + 1600 = x²

1681 = x²

x = √(1681)

x = 41

Therefore, the straight line distance from the starting point is 41 miles


Question 5:

The base of an isosceles triangle is 24 cm and the two equal sides are 37 cm each. Find the altitude AD of the triangle.

Solution:

Worksheet On Pythagorean Theorem 1

Altitude AD divides the triangle ABC into two equal parts.

By using the pythagorean theorem for △ ADC, △BCD

hypotenuse² = base² + perpendicular_side²

37² = 12² + x²

1369 = 144 + x²

1369 – 144 = x²

x² = 1225

x = √(1225)

x = 35

So, the length of the altitude is 35 cm.


Question 6:

The length of a living room is 2 feet less than twice its width. If the diagonal is 2 feet more than twice the width, find the dimensions of the room?

Solution:

Let the width of the living room = x

Given that,

Length of the living room = 2 feet less than twice its width

= 2x – 2

Diagonal = 2 feet more than twice the width

= 2x + 2
From the given equations

diagonal² = length² + width²

(2x + 2)² = (2x – 2)² + x²

4x² + 8x + 4 = 4x² – 8x + 4 + x²

8x + 8x = 5x² – 4x² + 4 – 4

16x = x²

x = 16

So, the width of the living room is 16 ft

Length of the living room = 2(16) – 2 = 32 – 2 = 30 ft

Diagonal of the living room = 2(16) + 2 = 32 + 2 = 34 ft


Question 7:

A bird was sitting 9 feet from the base of an oak tree and flew 15 feet to reach the top of the tree. How tall is the tree?

Solution:

The distance between bird sitting position and tree is 9 ft

The bird flew 15 ft to reach top of the tree

According to the Pythagorean theorem

c² = a² + b²

15² = 9² + b²

225 = 81 + b²

b² = 225 – 81

b² = 144

b = √(144)

b = 12

Therefore, the height of the tree is 12 ft.


Question 8:

Mary wants to cut across a rectangular lot rather than walk around it. Of the lot is 120 feet long and 50 feet wide. Mary walks diagonally across the lot, how many feet is the short cut?

Solution:

Given that,

Length of the lot = 120 ft

Width of the lot = 50 ft

Let the shortcut distance be x

As per the Pythagorean theorem,

a² + b² = c²

50² + 120² = x²

2500 + 14400 = x²

16900 = x²

x = √(16900)

x = 130

Therefore, 130 ft is the shortcut.