Grade 08 Maths Revision Notes PDF

#Science  Grade 08: Maths Exam Important Questions Direct and Inverse Proportions 93  Grade 08 Maths Chapter Notes 'S JU BY  Grade 08: Maths Exam Important Questions  Grade 08 Maths Chapter Notes 'S JU BY  Grade 08 Maths...

#Science  Grade 08: Maths Exam Important Questions Direct and Inverse Proportions 93  Grade 08 Maths Chapter Notes 'S JU BY  Grade 08: Maths Exam Important Questions  Grade 08 Maths Chapter Notes 'S JU BY  Grade 08 Maths Chapter Notes 'S JU BY  General Instructions: 1) This question paper contains five sections, Section A to E. 2) Section A has 10 questions carrying 01 mark each. 3) Section B has 09 questions carrying 02 marks each. 4) Section C has 08 questions along with an extra question carrying 03 marks each. 5) Section D has 05 questions along with an extra question carrying 04 marks each. 6) Section E has 02 case study questions carrying 04 marks each. SECTION - A Q.1Do as directed 10M 1)If 9the value of is _____. 2)The common factor of and is _____. 3)If 6 boys can do a piece of work in 4 days , one boy can do it in ___. 4)The same number cannot be added to or subtracted from both sides of an equation. (True / False) 5)The number of terms in the expression 6)The factors of is __________ 7)What will be the cube of 0.2 ? 8)What is the coefficient of in the term 9)The ones digit in the square of 132 will be 4. (True /False) 10)The Product of two rational number always a _______. SECTION - B Q.2Solve the following questions. 118M 1)Solve the equation and check your result. 2)Show that 1728 is a perfect cube. 3)Subtract 4)Factorize : 5)Add : 6)An aeroplane flies 4800 Km in 4 hours. Flying at the same speed, in how many hours will it cover 7200 Km ? 7)Factorizeby the splitting method. 8)What is the sum of Multiplicative Inverse and Additive Inverse of 9)Find the square root of 3481 by Long Division Method. SECTION - C Q.3Attempt the following questions.(Any 8) 24M 1)Simplify and solve the linear equation : 2)Find the length of the side of a square whose area 441 3)Simplify : 4)Factorize using appropriate identities. (a) (b) 5)Find : 6)Divide by the given binomial. (a) (b) 7)At a camp there is sufficient food for 300 scouts for 24 days. If 150 more scouts join the camp, how long will the food last ? (a)8)Find the cube root of 54872 by Prime factorisation method. 9)Find the smallest square number which is divisible by each of the numbers 6 , 9 and 15. SECTION - D Q.4Solve the following questions.(Any 5) 24M 1)Find the least number which must be added to 525 so as to get a perfect square. Also find the square root of the perfect square so obtained. 2)Solve : 3)Factorize : i) ii) 4)Find the smallest number by which 704 must be divided to obtain a perfect cube 5)Simplify : and find its value for. 6)Divide  i) ii) Section - E Q.5Case Study. 04M 1) Priya and Aditi are very good friends. They decided to play a game, Priya asked Aditi to think of a number and subtract from it. Then she ask multiply the result by 6 , she asked to add 8 in the result. Now Aditi said, “The number I obtained is 8 times the same number I thought of”. Based upon this information , answer the following questions: i) Formulate Aditi equation from the above passage. ii) Find the number Aditi thought of? iii) The shifting of a number from one side of an equation to other is called ____ iv) In a linear equations the highest power of the variable is ____.  Chapter Notes Algebraic Expressions and Identities 'S JU Grade 08 BY 80  BY JU 'S 81  BY JU 'S 82  BY JU 'S 83  BY JU 'S 84  BY JU 'S 85  BY JU 'S 86  BY JU 'S 87  BY JU 'S 88  Revision Notes Class 8 Maths Chapter 14 – Factorisation Factorisation: Factorisation of a number or an algebraic expression is writing it as a product of numbers, variables or expressions. These numbers, variables or expressions are called factors. Factorisation of numbers – A number can be broken down as a product of two or more factors depending upon the number given. For example, the number 15 can be written as the product of 1 and 15 or the product of 3 and 5 or the product of 1,3 and 5. So, 1,15,3 and 5 are the factors of number 30. Here, if the factors are prime numbers, then they are called the prime factors. Factorisation of algebraic expressions – The factors can be a number, variable or an expression. For example, factors of 5ab are 5,a and b. Similarly, the expression 3ab  6a can be written as 3ab  6a  3  a  (b  2) , where 3,a and (b  2) are the factors. Method of common factors: In this method of finding the factors of an algebraic expression, we first write each term of an expression as a product of irreducible factors. Then by using distributive law, we take the common factors and thus determine the factors of that algebraic expression. For example, let us find the factors of the expression 4x  2xy. First find the factors of each term. So, 4x can be written as 4x  2  2  x and 2xy can be written as 2  x  y. Thus, we can write 4x  2xy  (2  2  x)  (2  x  y) , where it can be noticed that 2 and x are common in both terms. So, 4x  2xy  2  x  (2  y). Thus, the common factors are 2,x and (2  y). Factorisation by regrouping terms: When the terms of a given expression do not have any single common factor, then we rearrange the terms to group them to lead to factorisation. This method of Class VII Maths www.vedantu.com 1 forming groups is called regrouping. There can be one or more ways of regrouping for a given expression. For example, take the expression z  7  7xy  xyz , here the terms do not have any common factors, so we will rearrange the terms as z  xyz  7  7xy and then taking common z and 7 from the first two and last two terms respectively. We get; z  xyz  7  7xy  z  1  xy   7  1  xy    z  7   (1  xy) Hence, the factors are  z  7  and (1  xy). Factorisation using identities: The expressions directly or indirectly given in certain forms can be factored using standard identities. Some common identities and their expressions have been tabulated below; Expressions of the form Identity for factorisation a 2  2ab  b 2 a  b 2 a 2  2ab  b2 a  b 2 a 2  b2 (a  b)(a  b) a 2  b2  c2  2ab  2bc  2ac a  b  c 2 a 3  3a 2b  3ab2  b3 a  b 3 a 2  (x  y)a  xy  a  x  (a  y) Apart from the above identities, many other identities are also used. The signs of the terms should be specially taken care of while using the identities. Division of algebraic expressions: The division operation in case of algebraic expressions unlike numbers is a bit different and factorisation plays an important role in it. The division is carried out keeping factorisation as a key player, which will see in following cases. Note that the examples considered upon division give 0 as their remainder. Division of a monomial by another monomial – Since, monomials have only one term, it is written as products of its irreducible factors and then the Class VII Maths www.vedantu.com 2  common factors in two monomials are then cancelled. Let us suppose we need to divide 14x 2 yz 3 by 7xyz. We can write this as; 14x 2 yz3 2  7  x  x  y  z  z  z  7xyz 7 x yz  2xz 2 Hence, the quotient is 2xz2. Division of a polynomial by a monomial – In this case, each term of the polynomial is separately divided by the monomial. Hence, each term is written into its factor form and then divided separately. For example, let us divide the polynomial 4a 3  2a 2  6a by monomial 2a. So, first we can write the polynomial as; 4a 3  2a 2  6a  (2  2  a  a  a)   2  a  a    2  3  a  and the monomial as  2  a . 4a 3  2a 2  6a (2  2  a  a  a)   2  a  a    3  2  a  Hence,  2a 2  a    2  2  a  a  a    2  a  a   3  2  a  2  a 2  a 2  a   2  a  a    a    3  2a 2  a  3 Division of a polynomial by another polynomial – In this case also, both numerator and denominator are factored and the factors common in both are cancelled. For example, let us divide  7x 2  14x  by  x  2  , then we can write;  7x 2  14x   7  x  x   2  7  x   x  2  x  2 Take 7x common from the two terms of numerator, 7  x   x  2   7x  x  2 Hence the quotient is 7x. Class VII Maths www.vedantu.com 3 Common errors while solving exercises: There might be terms in an algebraic expression without any coefficient, such terms have 1 as their coefficient. So, we need to take care about that while doing addition or subtraction. While substituting negative values in an expression, always use brackets so as to avoid confusion in final signs. When we multiply an expression enclosed in a bracket by a constant or a variable or an expression, we should multiply it with each term of the expression that is within brackets. Whenever we raise the power of a monomial, the power of both constant and variable is needed to be raised. Class VII Maths www.vedantu.com 4  Chapter Notes Linear Equations in One Variable 'S JU Grade 08 BY 10  BY JU 'S 11  BY JU 'S 12  BY JU 'S 13  BY JU 'S 14  BY JU 'S 15  BY JU 'S 16  Grade 08: Maths Exam Important Questions Algebraic Expressions and Identities 69 70 71 72 73  Chapter Notes Direct and Inverse Proportions 'S JU Grade 08 BY 112  BY JU 'S 113  BY JU 'S 114  BY JU 'S 115  BY JU 'S 116 Linear Equations in One Variable 10 11 12 13 14 15 16 17 18 19  Important questions for Class 8 Maths Chapter 14 – Factorisation Very Short Answer Questions 1 Mark 1. Factorise 12a2b+15ab2 (a) 3ab(4a+5b) (b) 3ab (c) 4a+5b (d) 3ab(5a+4b) Ans: 12a 2b = 2  2  3  a  a  b 15ab2 = 3  5  a  b  b Taking common factor from each term, 12a 2 b  15ab 2  3ab(4a  5b) Hence the correct option is (a). 2. When we factorise an expression, we write it as a ______ of factors. Ans: When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. 3. Factorise 6xy – 4y + 6 – 9x (a) (3x – 2) (b) (3x – 2)(2y – 3) (c) (2y – 3) (d) (2x – 3)(3y – 2) Ans: 6xy  4y  6  9x  2  3  x  y  2  2  y  2  3  3  3  x  2y(3x  2)  3(3x  2)  (3x  2)(2y  3) Hence the correct option is (b). 4. Find the common factors of 12x, 36 (a) 12 (b) 36 (c) x Class VIII Maths www.vedantu.com 1 (d) 12x Ans: Here, 12x = 2  2  3  x 36 = 2  2  3  3 So, common factor = 2  2  3  4  3  12 Hence the correct option is (a). 5. Solve 7x 2 y 2 z 2 ÷14xyz (a) 2 (b) 4x (c) 3y xyz (d) 2 Ans: 7x 2 y 2 z 2 ÷14xyz = 7  x  x  y  y  z  z  2  7  x  y  z = x yz  2 xyz = 2 Hence the correct option is (d). 6. Factorise 49p 2 -36 (a) (7p+6)(7p+6) (b) (7p-6)(7p-6) (c) (7p-6)(7p+6) (d) (6p-7)(7p-6) Ans: Using the identity, x 2  y 2  (x  y)(x  y) 49p 2 -36 = (7p)2 - 62 = (7p+6)(7p-6) Hence the correct option is (c). Short Answer Questions 2 Marks 7. Factorise ax 2 +by 2 +bx 2 +ay 2 Ans: ax 2 +by 2 +bx 2 +ay 2  ax 2 +bx 2 +ay 2 +by 2 = a  x  x+b  x  x+a  y  y+b  y  y = x 2 (a  b)  y 2 (a  b) Class VIII Maths www.vedantu.com 2 = (a  b)(x 2  y 2 ) 8. Factorise using identity x2 +10x+25 Ans: x 2 +10x+25 = x 2 +2  5  x+52 Using the identity, (a  b) 2  a 2  2ab  b 2 = x 2 +2  5  x+52 = (x+5) 2 = (x+5)(x+5) 9. Factorise (a+b)2 - (a-b)2 Ans: (a  b) 2  a 2  2ab  b 2 (a  b) 2  a 2  2ab  b 2 (a+b) 2 - (a-b) 2  (a 2 +2ab+b 2 ) - (a 2 -2ab+b 2 )  a 2 +2ab+b2 - a 2 +2ab-b2  4ab 10. Simplify 102 -18×10+81 Ans: 102 -18×10+81  102  2  9  10  92  Using the identity, (a  b) 2  a 2  2ab  b 2  102  2  9  10  92   10  9 2  1 2 1 25 11. Factorise 16a 2 - 4a 2 2 25  5  Ans:16a  2  (4a) 2    2 4a  2a  Using the identity, x  y  (x  y)(x  y) 2 2 2  5   (4a)    2  2a  Class VIII Maths www.vedantu.com 3   5  5    4a   4a    2a  2a  4m 2 -169n 2 12. Simplify 2m+13n 4m -169n 2 (2m) 2 -(13n) 2 2 Ans: = 2m+13n (2m+13n) Using the identity, x 2  y 2  (x  y)(x  y) (2m)2 - (13n)2 = (2m+13n) (2m-13n) - (2m+13n) = (2m+13n) =(2m-13n) Long Answer Questions 3 Marks 13. Simplify  p 2 +11p+28   p+4  Ans: Here, the numerator can be further factorised as follow (p 2 +11p+28) = (p2 +7p+4p+28)  p(p  7)  4(p  7)  (p  7)(p  4) Now,  p +11p+28  (p  7)(p  4) 2  p+4  (p  4)  (p  7) (4y 2 -4z 2 ) 14. Find x if x(y-z)= (y+z) Ans: Taking 4 common from the numerator we get 4(y 2 - z 2 ) x(y-z)= (y+z) Using the identity, x 2  y 2  (x  y)(x  y) in the numerator Class VIII Maths www.vedantu.com 4  4(y-z)(y+z) x(y-z)= (y+z) x(y-z) = 4(y-z) On comparing both sides we can see that x=4. 15. Find x and y that xy = 72, x + y = 17. Ans: Here we have been given xy and x + y To find out the value of x and y we can the identity (x  y) 2  x 2  2xy  y 2 (x  y) 2  x 2  2xy  y 2 (17) 2  x 2  2(72)  y 2 (17) 2  144  x 2  y 2 x 2  y 2  289  144 x 2  y 2  145 Now, putting the values in identity (x  y) 2  x 2  2xy  y 2 (x  y) 2  x 2  2xy  y 2 (x  y) 2  145  2(72) (x  y) 2  145  144 x  y 1 Adding both the equations i.e., x + y = 17 and x – y = 1 (x + y)+(x - y) = 17+1 x + y + x – y = 18 2x = 18 x=9 Putting the value of x in equation x + y = 17 x + y = 17 9 + y = 17 y = 17 – 9 y=8 Hence the value of x is 9 and y is 8. 16. Find the remainder in the following (x4 -a4 )÷(x 2 +a 2 ). Ans: For dividing both the equation first we have to simplify it. So, (x 4  a 4 ) can be written as (x 2 ) 2  (a 2 ) 2 Class VIII Maths www.vedantu.com 5 Now by applying the identity a 2  b 2  (a  b)(a  b) (x 2 ) 2  (a 2 ) 2  (x 2  a 2 )(x 2  a 2 ) Now,  x -a  = (x +a )(x -a ) 4 4 2 2 2 2 (x 2 +a 2 ) (x 2 +a 2 )  x -a  =(x -a ) 4 4 2 2 2 2 (x +a ) Hence the remainder is (x 2 -a 2 ). (0.87)2 -(0.13)2 17. Simplify (0.87-0.13) Ans: By applying the identity a 2  b 2  (a  b)(a  b) in the numerator. (0.87)2 -(0.13)2 (0.87+0.13)-(0.87-0.13)2  (0.87-0.13) (0.87-0.13) (0.87)2 -(0.13)2  (0.87+0.13) (0.87-0.13) (0.87)2 -(0.13)2 1 (0.87-0.13) 18. Using identity (a-b)2 =a 2 -2ab+b 2 find the value of (98)2. Ans: (98) 2 can be further written as (100 - 2) 2 By comparing the above equation with the identity, we get a = 100 and b = 2 (98)2  (100)2  2(100)(2)  (2)2  10000  400  4  9604 19. Using identity a 2 -b 2 =(a+b)(a-b). Find (1.02)2  (0.98)2. Ans: Using identity a 2 -b 2 =(a+b)(a-b) (1.02) 2 -(0.98)2  (1.02  0.98)(1.02  0.98)  (2)(0.04)  (0.08) Class VIII Maths www.vedantu.com 6  (a 2 -16) (a 2 -2a-8) 20. Simplify 2 ÷ (a -25) (a 2 +10a+25) Ans: Identities to be used in the question are; a 2  b 2  (a  b)(a  b) (a  b) 2  a 2  2ab  b 2 (a 2 -16) (a 2 -2a-8) (a 2 -42 ) (a 2 -2a-8) ÷  ÷ (a 2 -25) (a 2 +10a+25) (a 2 -52 ) (a 2 +2  5  a+52 ) (a+4)(a-4) (a 2 -4a+2a-8)  ÷ (a+5)(a-5) (a 2 +2  5  a+52 ) (a+4)(a-4) a(a-4)+2(a-4)  ÷ (a+5)(a-5) (a+5) 2 (a+4)(a-4) (a+5)(a+5)   (a+5)(a-5) (a+2)(a-4) (a+5)(a+4)  (a-5)(a+2) Long Answer Questions 5 Marks 2 2 1. Simplify (xy+yz) -(xy-yz) Ans: Using the identities (a  b) 2  a 2  2ab  b 2 and (a  b) 2  a 2  2ab  b 2 =(xy) 2 +(yz) 2 +2(xy)(yz) - (xy) 2  (yz) 2  2(xy)(yz)  =(xy) 2 +(yz) 2 +2(xy)(yz) - (xy) 2  (yz) 2  2(xy)(yz) =2(xy 2 z)+2(xy 2 z) =4xy 2 z 2. The area of a rectangle is 6a2  36a and its width is 36a. Find its length. Ans: Let the length of the rectangle is x Breath = 36a Area of rectangle = Length ×Breath 6a 2 +36a = x  36a 6a 2 +36a =x 36a Class VIII Maths www.vedantu.com 7  6a(a+6) x 36a (a+6) x 6 (a+6) Hence the length of rectangle is 6 3. Match the following: a) 25p 2 -q 2 1) x 2 -2x-35 b)  x-1 -36  5p-q  5p+q  2 2) e5  1 e4  c) dj+hj+dq+hq 3)  +  m 2r 5  m 2 r 4  e5 e9 d) 4 5 + 2 9 4)  d+h  q+j mr mr 12p 2 +132p e) 5)  p+11  12p Ans: a) Using the identity: a 2  b 2  (a  b)(a  b) 25p 2 -q 2 =  5p  -q 2 2   5p-q  5p+q  ⇒ Option (2). b)  x-1 -36 =  x 2  2x  1  36 2  x 2 - 2x - 35 ⇒ Option (1). c) dj+hj+dq+hq = d  j  q   h( j  q)   d+h  q+j ⇒ Option (4). e5 e9 e5  1 e 4  d) 4 5 + 2 9 = 2 5  2 + 4  ⇒ Option (3). mr mr m r m r  12p2 +132p 12p  p  11 e) = ⇒ Option (5). 12p 12p Class VIII Maths www.vedantu.com 8 4. The combined area of two squares is 20cm2. Each side of one square is twice as long as a side of the other square. Find the length of the sides of each square. Ans:Let the side of the smaller square be S and that of the bigger square be 2S. Combined area of the two squares = 20cm2 S2  (2S) 2  20 S2  4S2  20 5S2  20 S2  4 S  2cm Hence the side of the smaller square is 2cm and that of the bigger square is 4cm. 5. Find the factors of 25x 2 -4y 2 +28yz-49z 2. Ans: 25x 2 -4y 2 +28yz-49z 2  25x 2 -(4y 2 -28yz+49z 2 )  25x 2 -[(2y) 2 -2  2y  7z+(7z) 2 ] Using the identity (a  b) 2  a 2  2ab  b 2 (5x) 2  (2y  7z) 2 Now using the identity a 2  b 2  (a  b)(a  b) (5x  2y  7z)(5x  2y  7y) Class VIII Maths www.vedantu.com 9 94 95 96 97 98 99 100 101 102