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.
Here are the benefits of referring the RD Sharma textbook for practice along with its solution- RD Sharma covers different types of questions with a varying difficulty level. Practicing these questions will ensure that students have a good practice of all types of questions that can be framed in the examination. 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. Ayush Raj. Riddhi Shah. Qusai Saify. Lakshita Prajapati. The second exercise of this chapter covers problems based on inverse variation or proportion. Two values are said to be in inverse proportion if decrease in one causes increase of the other.
The chapter time and work offers a clear idea about different concerning concepts and thus help students to solve the problems based on the same easily. It comprises only 1 exercise with 27 questions in total. This chapter basically offers real-time and practical implementation of time and work. Students can download the solutions of this chapter in the PDF format for free. The chapter Percentage includes 2 exercises. The step-by-step solution of this chapter basic includes some important topics based on percentage and its various applications.
The total number of question are 54 27 each exercise. It comprises 3 exercises and covers some of the important formulas to find profit, loss, discount and value added tax. Profit, Loss, Discount and Value added tax can be calculated when a person sells something to someone. Calculating these terms in quite easy and all you need to remember their respective formulas which are mentioned in these solutions with detailed explanation. Compound Interest is one of the interesting as well as important chapters in class 8th maths.
This mainly teaches the concept of compound interest and how to obtain the same by applying formulas. There are 5 exercises in this chapter which presents different problems based on compound interest. This chapter comes under Geometry section and deals with the polygon shape and its related problems. Maths RD Sharma Solutions for Class 8th offered by us will surely help students in solving these problems easily. This chapter includes 1 exercise with 5 questions in total.
It comprises of 1 exercise which includes the basic questions related to the quadrilaterals. By solving this exercise, students will be able to improve their knowledge on quadrilaterals and its construction. A total of 3 exercises with 55 questions are given in this chapter. This chapter basically covers some special types of quadrilaterals and its associated problems.
The problems are solved by professional and experienced math experts in order to facilitate students. This chapter includes problems related to the practical geometry such as the construction of different shapes as according to the given measurement, how to write steps of construction etc. Along with this, this chapter will also teach how to use different tools of geometry for constructing different shapes.
It comprises 5 exercises in total. The chapter visualizing shapes covers the topics related to 3D shapes, scale drawing, geometrical solids etc.
There are 2 exercises in the chapter with 12 questions in total. Area of Trapezium and Polygon includes 3 exercises in total. In this chapter, students will learn how to find the area of trapezium and different polygons by using concerning formulas. There are 4 exercises in this chapter. First and second exercise comprises questions based on volume of cuboids and cube.
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