{"id":27376,"date":"2020-06-29T23:28:30","date_gmt":"2020-06-29T17:58:30","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27376"},"modified":"2021-07-02T15:07:10","modified_gmt":"2021-07-02T09:37:10","slug":"kseeb-solutions-for-class-8-maths-chapter-2-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-2-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 2 Algebraic Expressions Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions<\/h2>\n

Question 1.
\nChoose the correct answer
\n(a) Terms having the same literal factor with the same exponents are called
\n(A) exponents
\n(B) like terms
\n(C) factors
\n(D) unlike term
\nAnswer:
\n(B) like terms<\/p>\n

(b) The coefficient of ab in 2ab is
\n(A) ab
\n(B) 2
\n(C) 2a
\n(D) 2b
\nAnswer:
\n(B) 2<\/p>\n

(c) The exponential form of a x a x a is
\n(A) 3a
\n(B) 3 + a
\n(C) a3<\/sup>
\n(D) 3 – a
\nAnswer:
\n(C) a3<\/sup><\/p>\n

(d) Sum of two negative integers is
\n(A) negative
\n(B) positive
\n(C) zero
\n(D) infinite
\nAnswer:
\n(A) negative<\/p>\n

(e) What should be added to a2<\/sup> + 2ab to make it a complete square?
\n(A) b2<\/sup>
\n(B) 2ab
\n(C) ab
\n(D) 2a
\nAnswer:
\n(A) b2<\/sup><\/p>\n

\"KSEEB<\/p>\n

(f) What is the product of (x + 2) (x – 3)?
\n(A) 2x – 6
\n(B) 3x – 2
\n(C) x2<\/sup> – x – 6
\n(D) x2<\/sup> – 6x
\nAnswer:
\n(C) x2<\/sup> – x – 6<\/p>\n

(g) The value of (7.2)2<\/sup> is (use an iden-tity to expand)
\n(A) 49.4
\n(B) 14.4
\n(C) 51.84
\n(D) 49.04
\nAnswer:
\n(C) 51.84<\/p>\n

(h) The expansion of (2x – 3y)2<\/sup>is
\n(A) 2x2<\/sup> + 3y2<\/sup>+ 6xy
\n(B) 4x2<\/sup> + 9y2<\/sup> – 12xy
\n(C) 2x2<\/sup> +3y2<\/sup> – 6xy
\n(D) 4x2<\/sup> +9y2<\/sup> + 12xy
\nAnswer:
\n(B) 4x2<\/sup> + 9y2<\/sup> – 12xy<\/p>\n

(i) The product of 58 \u00d7 62 is (use an identity)
\n(A) 4596
\n(B) 2596
\n(C) 3596
\n(D) 6596
\nAnswer:
\n(C) 3596<\/p>\n

Question 2.
\nTake away 8x – 7y – 8p + 10q from 10x + 10y – 7p + 9q
\nAnswer:
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nExpand
\n(i) (4x + 3)2<\/sup>
\nAnswer:
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ‘
\nHere a = 4x, b = 3
\n(4x + 3)2<\/sup> = (4x)2<\/sup> + 2.4×3 .+ 32<\/sup>
\n= 16x2<\/sup> + 24x + 9<\/p>\n

(ii) (x + 2y)2<\/sup>
\nAnswer:
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\nHere a = x, b = 2y
\n(x + 2y)2<\/sup> = x2<\/sup> + 2.x.2y + (2y)2<\/sup>
\n= x2<\/sup> + 4xy + 4y2<\/sup><\/p>\n

(iii) \\(\\left(x+\\frac{1}{x}\\right)^{2}\\)
\nAnswer:
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\nHere a = x, b = 1\/x
\n\"KSEEB<\/p>\n

Question 4.
\nExpand
\n(i) (2t + 5) (2t – 5).
\nAnswer:
\n(a + b) (a – b) = a2<\/sup> – b2<\/sup>
\nHere = a = 2t, b = 5 ]
\n(2t + 5) (2t – 5) = (2t)2<\/sup> – 52<\/sup>
\n= 4t2<\/sup> – 25<\/p>\n

(ii) (xy + 8) (xy – 8)
\nAnswer:
\n(a + b) (a -b) = a2<\/sup> – b2<\/sup>
\nHere a = xy, b = 8
\n(xy + 8) (xy – 8) = (xy)2<\/sup> – 82<\/sup>
\n= x2<\/sup>y2<\/sup> – 64<\/p>\n

(ii) (2x + 3y) (2x – 3y)
\nAnswer:
\n(a + b) (a -b) = a2<\/sup> – b2<\/sup>
\nHere = a = 2x, b = 3y
\n(2x + 3y) (2x – 3y) = (2X)2<\/sup> – (3y)2<\/sup> = 4x2<\/sup> – 9y2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nExpand
\n(i) (n – 1) (n + 1) (n2<\/sup> + 1)
\nAnswer:
\n(n – 1) (n + 1) (n2<\/sup> + 1)
\n= (n2<\/sup> – l2<\/sup>) (n2<\/sup> + 1)
\n= (n2<\/sup> – 1) (n2<\/sup> + 1)
\n= (n2<\/sup>)2<\/sup> – l2
\n= n4<\/sup> – 1<\/p>\n

(ii)\\(\\left(n-\\frac{1}{n}\\right)\\left(n+\\frac{1}{n}\\right)\\left(n^{2}+\\frac{1}{n^{2}}\\right)\\)
\nAnswer:
\n\"KSEEB<\/p>\n

(iii) (x – 1) (x + 1) (x2<\/sup> + 1) (x4<\/sup> + 1)
\nAnswer:
\n(x – 1) (x + 1) (x2<\/sup> + 1) (x4<\/sup> + 1)
\n= (x2<\/sup> – l2<\/sup>) (x2<\/sup> + 1) (x4<\/sup> + 1)
\n= (x2<\/sup> – 1) (x2<\/sup> + 1) (x4<\/sup> + 1)
\n= [(x2<\/sup>)2<\/sup> – l2<\/sup>)] (x4<\/sup> + 1)
\n= (x4<\/sup> – 1) (x4<\/sup> + 1)
\n= (x4<\/sup>)2<\/sup> – l2<\/sup>
\n= X8<\/sup> – 1<\/p>\n

(iv) (2x – y) (2x + y) (4x2<\/sup> + y2<\/sup>)
\nAnswer:
\n(2x – y) (2x + y) (4x2<\/sup> + y2<\/sup>)
\n= [(2x)2<\/sup> – y2<\/sup>) (4x2<\/sup> + y2<\/sup>)
\n= (4x2<\/sup> – y2<\/sup>) (4x2<\/sup> + y2<\/sup>)
\n= [(4x2<\/sup>)2<\/sup> – (y2<\/sup>)2<\/sup>)]
\n=16x4<\/sup> – y4<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nUse appropriate formulae and compute
\n(i) (103)2<\/sup>
\nAnswer:
\n1032<\/sup> = (100 + 3)2<\/sup>
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\nHere a = 100, b = 3
\n(100 + 3)2<\/sup> = 1002<\/sup> + 2.100.3 + 32<\/sup>
\n= 10000 + 600 + 9
\n1032<\/sup> = 10609<\/p>\n

(ii) (96)3<\/sup>
\nAnswer:
\n962<\/sup> = (100 – 4)2<\/sup>
\n(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
\nHere a = 100, b = 4
\n(100 + 3)2<\/sup> = 1002<\/sup> + 2.100.4 + 42
\n= 10000 – 800 + 16
\n1032<\/sup> = 9216<\/p>\n

(iii) 107 x 93
\nAnswer:
\n107 x 93 = (100 + 7)(100 – 7)
\n(a + b) (a – b) = a2<\/sup> – b2<\/sup>
\nHere a = 100, b = 7
\n(100 + 7) (100 – 7) = 1002<\/sup> – 72<\/sup>
\n= 10000 – 49 = 9951<\/p>\n

(iv) 1008 x 992
\nAnswer:
\n1008 x 992 = (1000 + 8)(100 – 8)
\n(a + b) (a – b) = a2<\/sup> – b2<\/sup>
\nHere a = 1000, b = 8
\n(1000 + 8) (100 – 8) = (1000)2<\/sup> – 82<\/sup>
\n= 1000000 – 64
\n1008 x 992 = 999936<\/p>\n

(v) 1852<\/sup> – 1152<\/sup>
\nAnswer:
\n1852<\/sup> – 1152<\/sup>
\n(a – b)2<\/sup> = (a + b) (a – b)
\nHere a = 185, b = 115
\n1852<\/sup> – 1152<\/sup> = (185 + 115) (185 – 115)
\n= (300) (70)
\n1852<\/sup> – 1152<\/sup> = 21000<\/p>\n

Question 7.
\nIf x + y = 7 and xy = 12 find x2<\/sup> + y2<\/sup>
\nAnswer:
\n(x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup>
\n(x + y)2<\/sup> = x2<\/sup> + y2<\/sup> + 2xy
\n72<\/sup> = x2<\/sup> + y2<\/sup> + 2(12)
\n[Substituting for x +y and xy]
\n49 = x2<\/sup> + y2<\/sup> +24
\nx2<\/sup> + y2<\/sup> = 49 24
\nx2<\/sup> + y2<\/sup> = 25<\/p>\n

Question 8.
\nIf x + y = 12 and xy = 32 find x2<\/sup> + y2<\/sup>
\nAnswer:
\n(x + y)2<\/sup> = x2<\/sup> + y2<\/sup> + 2xy
\n122<\/sup> = x2<\/sup> + y2<\/sup> + 2(32)
\n[Substituting for x +y and xy]
\n144 = x2<\/sup> + y2<\/sup> + 64
\nx2<\/sup> + y2<\/sup> = 144 – 64
\nx2<\/sup> + y2<\/sup> = 80<\/p>\n

Question 9.
\nIf 4x2<\/sup> + y2<\/sup> = 40 and xy = 6 find 2x + y
\nAnswer:
\n4x2<\/sup> + y2<\/sup> = 40
\n(2x)2<\/sup> + y2<\/sup> = 40
\n(2x + y)2<\/sup> = (2x)2<\/sup> + y2<\/sup> + 2.2x.y
\n(2x + y)2<\/sup> = 4x2<\/sup> + y2<\/sup> + 4xy
\n= 40 + 4 x 6
\n= 40 + 24
\n(2x + y)2<\/sup> = 64
\n2x + y = \u00b1\u221a64
\n2x + y = \u00b1\u221a8<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf x – y = 3 and xy = 10 find x2<\/sup> + y2<\/sup>
\nAnswer:
\n(x – y)2<\/sup> = x2 + y2 – 2xy
\n32<\/sup> = x2<\/sup> + y2<\/sup> + 2(10)
\n[Substituting for x – y and xy]
\n9 = x2<\/sup> + y2<\/sup> – 20
\n9 + 20 = x2<\/sup> + y2<\/sup>
\n\u2234 29 = x2<\/sup> + y2<\/sup>
\nx2<\/sup> + y2<\/sup> = 29<\/p>\n

Question 11.
\nIf \\(x+\\frac{1}{x}\\) = 3 find \\(x^{2}+\\frac{1}{x^{2}}\\) and \\(x^{3}+\\frac{1}{x^{3}}\\)
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\n\\(x+\\left(\\frac{1}{x}\\right)=6\\) find \\(x^{2}+\\frac{1}{x^{2}}\\) and \\(x^{4}+\\frac{1}{x^{4}}\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question 13.
\nSimplify (i) (x + y)2<\/sup> + (x – y)2<\/sup>
\nAnswer:
\n(x + y)2<\/sup> + (x – y)2<\/sup>
\n= x2<\/sup> + y2<\/sup> + 2xy + x2<\/sup> + y2<\/sup> – 2xy
\n= x2<\/sup> + y2<\/sup> + x2<\/sup> + y2<\/sup> = 2x2<\/sup> + 2y2<\/sup> ]
\n= 2(x2<\/sup> + y2<\/sup>)<\/p>\n

(ii) (x + y)2<\/sup> x (x – y)2<\/sup>
\nAnswer:
\n(x + y)2<\/sup> x (x – y)2<\/sup>
\n= (x2<\/sup> + y2<\/sup> + 2xy) (x2<\/sup> + y2<\/sup> – 2xy)
\n= x2<\/sup> (x2<\/sup> + y2<\/sup> – 2xy) + y2<\/sup> (x2<\/sup> + y2<\/sup> – 2xy) + 2xy (x2<\/sup> + y2<\/sup> – 2xy)
\n= x4<\/sup> + x2<\/sup>y2<\/sup> – 2x3<\/sup>y + x2<\/sup>y2<\/sup> + y4<\/sup> – 2xy3<\/sup> + 2x3<\/sup>y + 2xy3<\/sup> – 4x2<\/sup>y2<\/sup>
\n= X4<\/sup> + 2x2<\/sup>y2<\/sup> – 4x2<\/sup>y2<\/sup> + y4<\/sup>
\n= x4<\/sup> – 2x2<\/sup>y2<\/sup> + y4<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nExpress the following as difference of two squares
\n(i) (x + 2z) (2x + z)
\nAnswer:
\nx(2x + z) + 2z (2x + z)
\n= 2x2<\/sup> + xz + 4xz + 2z2<\/sup>
\n= 2x2<\/sup> + 2z2<\/sup> + 4xz + xz
\n= 2 (x2<\/sup> + z2<\/sup> + 2xz) + xz
\n= 2(x + z)2<\/sup> + xz
\n\"KSEEB
\n\"KSEEB<\/p>\n

(iii) 4(x + 2y) (2x + y)
\nAnswer:
\n4(x + 2y) (2x + y)
\n= 4(2x2<\/sup> + xy + 4xy + 2y2<\/sup>)
\n= 4(2x2<\/sup> + 2y2<\/sup> + 4xy + xy)
\n=4 (2 (x2<\/sup> + y2<\/sup> + 2xy) + xy)
\n= 4 (2 (x + y)2<\/sup> + xy)
\n= 8 (x + y)2<\/sup> + 4xy
\n= 8 (x + y)2<\/sup> + (x + y)2<\/sup> – (x – y)2<\/sup>
\n[\u2235 4xy = (x + y)2<\/sup> – (x – y)2<\/sup>] = 9 (x + y)x^{4}+\\frac{1}{x^{4}} – (x – y)2 = [3 (x + y)]2<\/sup> – (x – y)2<\/sup><\/p>\n

(iii) (x + 98)( (x+ 102)
\nAnswer:
\n(x + 98)( (x + 102)
\n= (x + 100 – 2) (x + 100 + 2)
\n= (x + 100)2 – 22
\n[\u2235(a + b) (a – b) = a2<\/sup> – b2<\/sup>]<\/p>\n

(iv) 505 x 495
\nAnswer:
\n505 x 495 = (500 + 5) (500 – 5)
\n= (500)2<\/sup> – 52<\/sup><\/p>\n

Question 15.
\nIf a = 3x – 5y, b = 6x + 3y and c = 2y – 4x find
\n(i) a + b – c
\nAnswer:
\na + b – c = 3x – 5y + 6x + 3y – (2y – 4x)
\n= 9x – 2y – 2y + 4x
\n= 13x – 4y<\/p>\n

(ii) 2a – 3b + 4c = 2(3x – 5y) – 3 (6x + 3y) + 4 (2y – 4x)
\n= 6x – 10y – 18x – 9y + 8y – 16 x
\n= 6x – 18x – 16 x 10y – 9y + 8y
\n= 6x – 34x – 19y + 8y = – 28x – 11y
\n= – (28x + 11y)<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nThe perimeter of a triangle is 15×2 – 23x + 9 and two of its sides are 5×2 + 8x – 1 and 6×2 – 9x + 4. Find the third side
\nAnswer:
\nLet the sides of the triangle be a, b and c
\nLet a = 5x2<\/sup> + 8x – 1,
\nb = 6x2<\/sup> – 9x + 4
\nPerimeter = 15x2<\/sup> – 23x + 9
\na + b + c = 15x2<\/sup> – -23x + 9 5x2<\/sup> + 8x – 1 + 6x2<\/sup> – 9x + 4 + c
\n= 15x2<\/sup> – 23x + 9
\nllx2<\/sup> – x + 3 + c <= 15x2<\/sup> – 23x + 9
\nc = 15x2<\/sup> – 23x + 9 – (llx2<\/sup> – x + 3)
\n= 15x2<\/sup> – 23x + 9 – llx2<\/sup> + x – 3
\nc = 4x2<\/sup> 22x + 6
\n\u2234 The third side is 4x2<\/sup> – 22x + 6<\/p>\n

Question 17.
\nTwo adjacent sides of a rectangle are 2x2<\/sup> – 5xy + 3z2<\/sup> and 4xy – x2<\/sup> – z2<\/sup>. Find its perimeter.
\nAnswer:
\nLet l = 2x2<\/sup> – 5xy + 3z2<\/sup> and b = 4xy – x2<\/sup> – z2<\/sup>
\nPerimeter of a rectangel = 2(l + b)
\n= 2 [2x2<\/sup> – 5xy + 3z2<\/sup> + 4xy – x2<\/sup> – z2<\/sup>]
\n= 4x2<\/sup> – lOxy + 6z2<\/sup> + 8xy – 2x2<\/sup>– 2z2<\/sup>]
\n= 2x2<\/sup> + 4z2<\/sup> – 2xy<\/p>\n

Question 18.
\nThe base and altitutde of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
\nAnswer:
\nArea of a triangle = \\(\\frac{1}{2}\\) x base x height
\n= \\(\\frac{1}{2}\\) x(3x-4y)(5x + 5y)
\n= \\(\\frac{1}{2}\\)[(3x(6x + 5y) – 4y(6x + 5y) ]
\n= \\(\\frac{1}{2}\\)[18x2<\/sup> +15y)-24y-20y2<\/sup>]
\n= \\(\\frac{1}{2}\\)[18x2<\/sup> – 9xy – 20y2<\/sup>]<\/p>\n

Question 19.
\nThe sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed. What is the area of the remaining region.
\nAnswer:
\nArea of the remaining region
\n= Area of rectangle – Area of sq\u00fbare
\n= length x breadth – (side)2<\/sup>
\n= (2x + 3y) (3x + 2y) – (x + y)2<\/sup>
\n=2x(3x + 2y) + 3y(3x + 2y) – [x2<\/sup> + y2<\/sup> + 2xy]
\n= 6x2<\/sup> + 4xy + 9xy + 6y2<\/sup> – [x2<\/sup> + y2<\/sup> + 2xy]
\n= 6x2<\/sup> + 13xy + 6y2<\/sup> – x2<\/sup> – y2<\/sup> – 2xy = 5x2<\/sup> + 11xy + 5y2<\/sup><\/p>\n

Question 20.
\nIf a, b, c are rotational numbers such that a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – be – ca = 0 prove that a = b = c
\nAnswer:
\na2<\/sup> +b2<\/sup> + c2<\/sup> – ab – bc – ca = 0 Multiplying both the sides by 2
\n2(a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – be – ca) = 2 x 0
\n2a2<\/sup> + 2b2<\/sup> + 2c2<\/sup> – 2ab – 2bc – 2ca = 0
\na2<\/sup> + a2<\/sup> + b2<\/sup> + b2<\/sup> + c2<\/sup> + c2<\/sup> – 2ab – 2bc – 2ca = 0
\na2<\/sup> – 2ab + b2<\/sup> + b2<\/sup> – 2bc + c2<\/sup> + c2<\/sup> – 2ca + a2<\/sup> = 0
\n(a – b)2<\/sup> + (b – c)2<\/sup> + (c – a)2<\/sup> =0
\n(a – b)2<\/sup> = 0, (b – c)2<\/sup> = 0, (c – a)2<\/sup> = 0 [v If the sum of positive quantities is equal to zero then each quantity is equal to zero]
\n\u2234 (a-b)2<\/sup> = 0 \u21d2 a – b = 0 \u21d2 a = b
\n(b-c)2<\/sup> = 0 \u21d2 b – c = 0 \u21d2 b = c
\n(c-a)2<\/sup> = 0 \u21d2 c – a = 0 \u21d2 c = a
\n\u2234 a = b = c.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Algebraic Expressions Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions Question 1. Choose the correct answer (a) …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27376"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27376"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27376\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27376"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27376"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27376"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}