Rd sharma class 8 pdf download

Class 8 is the basic foundational class for the students and it is essential for them to have a good understanding of the concepts rather than memorizing them. Maths can be a nightmare for many students. One of the major reason for this is due to the lack of understanding of the concepts.

So, it is essentially required from the basic foundational class that students focus more on the understanding of the concepts and building up the basics rather than mugging things up. This can be easily achieved by referring to the world-class and renowned book, like RD Sharma.

Referring to RD Sharma textbook has various benefits for the user, as it covers various types of questions, which students can get in their final examination. While solving questions from the RD Sharma textbook, students can come across difficult questions, thus we have provided the solution for all the questions from the RD Sharma textbook, which they can refer at any time, and get to know the various ways of solving a question.

Thus it is essential for every student to refer to the RD Sharma solution for class 8 to build their basics as well as prepare themselves for the classes ahead.

How many were thet in each row? It is given that after arranging men in the form of a square, 9 men W' over. Hence, there were men in each row The product of two numbers is and their quotient is 2 Find the numbers. Let one of the two numbers be a. Since the product of numbers is Find the smallest square number divisible by each one of the numbers 8, 9 and The smallest number divisible by each one of the numbers 8, 9, 10 is their LCM.

Clearly, the L. Find the smallest number by which must be multiplied so that. Which : i ii iti iv 2, Find the square root of each of the following by prime factorization i ii iii iv v vi vii viii ix x xi xii xiii xiv xv , Find the smallest number by which must be multiplied so that it : becomes perfect square. Also, find the square root of the perfect square so obtained, a ie it becomes, fi perfect square.

Also, find the square root of the number so obtained. Find the smallest number by which must be divided so that it becomes g Pettey 7 square. Also, find the square root of the resulting number.

The product of two numbers is If one number is 16 times the other, find 4, numbers. A welfare association collected Rs as donation from the residents. If each pa as many rupees as there were residents, find the number of residents. A society collected Rs Each member collected as many paise as there wen members. How many members were there and how much did each contribute? A school collected Rs as fees from its students.

If each student paid as man paise as there were students in the school, how many students were there in th school? The area of a square field is m?. A rectangular field, whose length is twice is breadth has its perimeter equal to the perimeter of the square field. Find the area i the rectangular field. Find the least square number, exactly divisible by each one of the numbers: i 6,9, 15 and 20 ii 8, 12, 15 and 20 Find the square roots of and by the method of repeated subtraction.

Write the prime factorization of the following numbers and hence find their sq" roots. T7rows 3. Stepll Place a bar over every pair of digits starting with the units digit. Each pair and remaining one digit if any on the extreme left is called a period. For example i will be written as 29 09, In this 28 is called the first period and 09 is called the second period.

Here, 3 is the first period, 92 is the second period and 04 is the third period. Step Ill Count the number of bars. The number of bars is the number of digits in the square root of the given number. For example, the square root of has two digits and the square root of has three digits. Find the number of digits in the square roots of each of the following perfect squares: i ii iii Solution Placing a bar over pair of digits starting with the units digits, we have i , So, the square root of has three digits.

So, has four digits. It follows from the above discussion that: If the number of digits in a square number is n, then i the number of digits in its square root is - when n is even. Stepll Place bars over every pair of digits starting with the units digit Also, py bar on one digit if any not forming a pair on the extreme left. Each a te the remaining one digit if any on the extreme left is called a period, MF On Step — Think of the largest number whose square is less than or e: period.

Take this number as the divisor and the quotient. StepIV Put the quotient above the period and write the product of divisor and quq just below the first period. Step VIL Think of a digit, to fill the blank in step VI, in such a way that the product of ny, divisor and this digit is equal to or just less than the new dividend obtaineg , step V. Step VIM Subtract the product of the digit chosen in step VII and the new divisor from th, dividend obtained in step V and bring down the next period to the right of, remainder.

This becomes new dividend. Step X Obtain the quotient as the square root of the given number. Following examples will illustrate the above procedure. Clearly, the largest number whose square is just less than 5 is 2.

So, we take? Bringing down the next period i. For the Ge ad ae the quotient 2 i. So, take 43 as the new divisor and 23 as the new quotient. Now, subtract from to get 18 as the remainder. Bring down the next period 56 to the right of 18 to get as the new dividend For a ew a an the quotient 23 to get Take 46 as the left two digits of the new divisor.

Solution — Let us work out the process of finding the square root of by long division method. This means that ifaoy subtracted from the given number, the remainder will be zero and the ey number will be a perfect square. Let us first work out the Process of finding the square root by the division method: 5 25 , , It is evident from the above working that ?

Thus, will be a perfect square. Hence, the required least number is Find the greatest number of six digits which is a perfect square. We know that the greatest number of six digits is In order to find th Greatest number of six digits which is a perfect square, we must first find tht smallest number that must be subtracted from to make it a perfect square. For this, we work out the process of finding the square root of by long division method as given below.

What is that fraction which when multiplied by itself gives The area of a square playground is Find the length of one side of the playground. What is the fraction which when multiplied by itself gives 0. Simplify, Find the value of We use the following stepwise procedure to do the same.

Determine the number of decimal places to which the square root of the number is to be computed. Suppose the square root of the given number is to be computed correct to n places of decimal. Count the number of digits in the decimal part. Solution Since we have to find the square root of 2 correct to three places of decimal, we shall first find the square root of 2 upto four places of decimal. For this purpose, we affix 8 zeros to the right of the decimal point. For yet Purpose, we must add 8 zeros to the right of the decimal point.

Thus, we writ,. Solution Since we have to find the square root of Therefore, we first find the square root upto four decimal places, For this, we require 8 digits in the decimal part.

So, we affix 5 zeros to the right of decimal part. That is, we write For this reason, tables have been prepared which provide the approximate values of square roots of different numbers correct to a certain decimal place. With the help of these tables the square roots of most of the numbers can be written down The following table giv es values of square roots of all natural numbers from 1 to Mathematics for Classy, y ir ve x ve fa 1.

Look at the row containing 7. We find that the entry in the column of Vx is 2. Ay oe Nat Gi. For the difference of 0. However, cube roots are , cube root of a negative number is 4. In other words, a natural number n is a perfect cube if there exists a natural number m whose cube isn i.

So, 27, 64 and are perfect cubes of natural numbers 3, 4 nq. Obtain the natural number. Step] Express the given natural number as a product of prime factors. Step Ll Group the factors in triples in such a way that all the three factors in each trip , are equal Step IV. If no factor is left over in grouping in step III, then the number is a perfect cub, otherwise not.

To find the natural number whose cube is the given number, take one factor fron each triple and multiply them. The cube of the number so obtained will be th, given number.

Example2 Is a perfect cube? What is that number whose cube is ? What is the number whose cube is ? Therefore, is perfect cube. To determine the number whose cube is , we collect one factor from each group.

What is the smallest number by which product is a perfect cube? Thus, if we multiply by 7, 7 will also occur as a prime factor thrice and the product will be 2x2x 2x7x7x7, which is a perfect cube. What is the smallest number by which must be divided so that 3 the quotient is a perfect cube? Therefore, quotient is a perfect cube. Prove that if a number is doubled, then its cube is eight times the cube of the given number. Let b denote the double of a i. Evaluate tho following wy feat ere iy dio" atl!

That is, odd, PHI on esas gy nyt Property 4 Cubes ofthe numbers ending in digite 1, 4,5, 6 and 9 are the numbers ending" the same digit. The cube of 2 ends in 8 and the cube of 8 ends in 2. Similarly, the cube of 3 ends in 7 and the cube of 7 ends in 3. Also, if a number ends in 0, then its cube will end in three zeros, 4. In this section, we will discuss column method for finding the cubes of two digit natural numbers. The remaining procedure is exactly indentical to the method of finding the square of a two digit natural numbers, Following examples will illustrate the procedure.

Thus, - is also a perfect cube. Similarly, we define the cube of a rational number which is not an integer as given below.

Also, find that rational number. Find the cubes of ii 12 iii Which of the following numbers are cubes of negative integers i iii iv v - 3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer. Find which of the following numbers are cubes of rational numbers: 27 : da eee iii 0. Gill, iv, , In other words, the cube root of a number n is that number m whose cube gives Nn.

The cube root of a number n is denoted by Yn. Subtract 1 from it. If you get zero 1. Otherwise go to next step. Do you get 0 as the result, If yes, the cube root of the given number is 3. Do you get 0-as the vesult. If yes, the cube root of the given number is 4. Otherwise, go to next step. Therefore, unite digit of its cube root is After striking out the last three digits from the right, the number left is Using the method of suecessive subtraction examine whether or not the following numbers are perfect cubes: i ii 3.

Find the smallest number that mi question 2 which are not perfect cubes, to m corresponding cube roots? Find the cube root of each of the following natural numbers: i ii iii , iv v yi vii viii ix x xi xii 5. Find the smallest number which when multiplied with will make the Produ perfect cube.

Further, find the cube root of the product. Also, find the cube root of the quotient so obtained. Three numbers are in the ratio 1 : 2: 3. The sum of their cubes is Find y, numbers. Find the side of the cube. Solution We have, 1. Find the length of a side of the box. Solution Let the length of a side of the box be x metres.

Then, its volume is x cubic meters. But, the volume is given as Find the cube roots of each of the a ee ant i ii - iii - Show that. TxIs Id x18 8. The volume of a cubi i aay ical box is Find the length of each side Three numbers are t numbers. The sum of their cubes is 0. Fin4" In fact, there are only ten numbers between 1 and which are perfect cubes. The remaining natural numbers are not perfect cubes. Consequently, their cube roots are not whole numbers and they cannot be found exactly.

These cube a and are therefore irrational numbers, Only approximate values of the cube roots of these numbers can be found. The third column gives multiplied by So, to find the cube root of 62, we look at the row containing 62 in the column of x.

The following are some illustrations for the above assertions. Therefore, their sum when divided by!! Let us now take a three digit number abe, By changi its digits i "i order, we obtain numbers bea and cab. Gili 37 iv 3. Interchanging its ones and hundreds digits, we get the number cba. Therefore, the difference between these two numb. Thus, the difference of a three digit number abc and the number obtained by interchangi its ones and hundreds digits ie.

Without performing actual addition and division write the quotient when the sum of 69 and 96 is divided by a 2. Without performing actual computations, 9 Gi 5 3. If sum of the number and two other numbers obtained by arranging the digits of in eyelic order is divided by , 22 and 37 respectively. Find the quotient in each. Find the quotient when the differen: Gi 15 find the quotient when 94 - 49 is divided by. Mathematics for Clas. Also, problems on divisibility of the above mentioned divisors.

Clearly, 10 is a multiple of 5. Therefore, 10a is also a multiple of 5. Since the sum of any two multiples of 5 is a multiple of 5. Thus, an integer is divisible by 5, if its units digit is a multiple of 5. That is its units digit is either 0 or 5. So, we have the following test of divisibility by 5. It also follows from the above discussion that if the units digit of a number is not 0 or 5, then it is not divisible by 5.

Let n be any natural number. Thus, the remainder when an integer is divided by 5 is equal to the remainder when its units digit is divided by 5. For example, if is divided by 5, the remainder is 1. Solution i If-n is divided by 5, then the remainder is equal to the remainder when its ones digit is divided by 5.

Consequently, the units digit of n must be 3 or 8. So, the units digit of n is 1 or 6. Since 10a is an even number and the sum of two even numbers is an even number and sum ofan even number and an odd number is an odd number.

A number is divisible by 2, if its units digit isan even digit, , , or Since 10a is divisible by 2. If the what might be the units digit of n?

So, n must be an odd natnr. Hence, its units digit can be 1, 3, 5, 7, or 9. Hence, its units digit can be 0, 2,4, 6, or 8, 2, the remainder is zero if n is even, otherwise the remainder is 1.

Wh must be the units digit of n? Solution It is given that the division of m by 5 leaves a remainder of 4. Therefore, h: division of units digit of n by 5 must leave a remainder of 4. So, the units digi of nis either 4 or 9.

It is also given that the division of n by 2 leaves a remainder of 1. So, its units digit can be 1, 3, 5, 7 or 9. Clearly, 9 is the common value of units digit in two cases, Hence, the units digit of n is 9. Uptill now, we have studied three tests of divisibility, In all the th is decided just by the units digit. So, we have used only the units digit of the given number without even bothering about the rest of the number, This has happened because 10, 5 and 2 are divisors of 10, which is the key number in our place value system.

Consider now a three digit number abe. Thus, we have following test of divisibility by 9. It should be noted that a number is not divisible by 9 when the sum of its digits is not divisible by 9. It is evident from the above discussion that the remainder obtained by dividing a number by 9 is equal to the remainder when the sum of its digits is divided by 9. So, n is divisible by 9.

So, m is not divisible by 9. Let us now discuss more illustrations on divisibility by 9. Solution Since 34q is divisible by 9. Therefore, the sum of its digits is a multiple , Since 21 ie.

But, y is a digit. Solution It is given that the number 2a25 is a multiple of 9. Therefore, the sum of digits is a multiple of 9. But, a is a digit. So,a can take values 0, J, 2, Since a multiple of 9 is also a multi i i : 4 multiple of 3, So, a natura visible by ifthe sum ofits digits is also divisible by 3. Nilesh Gupta. Shalih Abdul Qodir Qodir. Jaikumar Periyannan.

Swaran Kanta. Bhishma Pandya. Nehal Goel. Gissele Abolucion. Ramkumar Sundaram. Ritu Mittal. Suresh Solanki.