Download Algebra Questions for CMAT PDF \u2013 CMAT Algebra questions PDF by Cracku. Practice CMAT solved Algebra Questions paper tests, and these are the practice question to have a firm grasp on the Algebra topic in the XAT exam. Top 20 very Important Algebra Questions for XAT based on asked questions in previous exam papers. \u00a0The CMAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Algebra Questions for CMAT<\/a><\/p>\n Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $a=2b=8c$ and $a+b+c=13$ then the value of $\\frac{\\sqrt{a^{2}+b^{2}+c^{2}}}{2c}$ is:<\/p>\n a)\u00a0$\\frac{9}{2}$<\/p>\n b)\u00a0$-\\frac{5}{6}$<\/p>\n c)\u00a0$-\\frac{9}{2}$<\/p>\n d)\u00a0$\\frac{5}{6}$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $a+b+c=13$<\/p>\n Put the value of a and b,<\/p>\n 8c + 4c + 8c = 13<\/p>\n c = 13\/20 $\\frac{\\sqrt{(8c)^{2}+(4c)^{2}+c^{2}}}{2c}$<\/p>\n $\\frac{\\sqrt{81c^{2}}}{2c}$ = $\\frac{9c}{2c}$ = $\\frac{9}{2}$<\/p>\n Question 2:\u00a0<\/b>If x,y,z are three numbers such that $x+y=13,y+z=15$ and $z+x=16$, the a)\u00a0$\\frac{5}{36}$<\/p>\n b)\u00a0$\\frac{36}{5}$<\/p>\n c)\u00a0$\\frac{18}{5}$<\/p>\n d)\u00a0$\\frac{5}{18}$<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n x+y=13 —(1) By (1) +\u00a0(2) + (3),<\/p>\n 2(x + y + z) = 13 + 15 + 16<\/p>\n x + y + z = 44\/2 = 22<\/p>\n put the value from eq(1),<\/p>\n 13 + z = 22<\/p>\n z = 9<\/strong><\/p>\n From eq(3),<\/p>\n 9 + x =16<\/p>\n x = 7<\/strong><\/p>\n From eq(3),<\/p>\n 7 + y = 13<\/p>\n y = 6<\/strong><\/p>\n Now,<\/p>\n $\\frac{xy+xz}{xyz}$<\/p>\n = $\\frac{(7)(6)+(7)(9)}{(7)(6)(9)}$<\/p>\n =\u00a0$\\frac{6 + 9}{6 \\times\u00a09}$<\/p>\n = $\\frac{5}{18}$<\/p>\n Question 3:\u00a0<\/b>If $x – \\frac{1}{x} = 11$, then $x^3 – \\frac{1}{x^3}$ is:<\/p>\n a)\u00a01474<\/p>\n b)\u00a01364<\/p>\n c)\u00a01188<\/p>\n d)\u00a01298<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x – \\frac{1}{x} = 11$<\/p>\n $x^3 – \\frac{1}{x^3}$\u00a0= $(x – \\frac{1}{x})^3 – 3(x –\u00a0\\frac{1}{x})$<\/p>\n ($\\because (a – b)^3 = a^3 – b^3 -3ab(a – b)$)<\/p>\n = $(11)^3 – 3(11)$<\/p>\n = 1331 – 33 = 1298<\/p>\n Question 4:\u00a0<\/b>The coefficient of $x^2$ in $(2x + y)^3$ is:<\/p>\n a)\u00a08<\/p>\n b)\u00a0$12 y^2$<\/p>\n c)\u00a0$12 y$<\/p>\n d)\u00a012<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(2x + y)^3$<\/p>\n = $(2x)^3 + y^3 + 3.2x.y(2x + y)$<\/p>\n ($\\because (a + b)^3 = a^3 +\u00a0b^3 + 3ab(a + b)$)<\/p>\n = $8x^3 + y^3 + 6xy(2x + y)$<\/p>\n = $8x^3 + y^3 + 12x^2y + 6xy^2$<\/p>\n The coefficient of $x^2$ = 12y<\/p>\n Question 5:\u00a0<\/b>$(a + b + c – d)^2 – (a – b – c + d)^2 = ?$<\/p>\n a)\u00a0$2a(b + c – d)$<\/p>\n b)\u00a0$4a(b + c – d)$<\/p>\n c)\u00a0$2a(b + c + d)$<\/p>\n d)\u00a0$4a(b + c + d)$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(a + b + c – d)^2 – (a – b – c + d)^2$<\/p>\n [(a + b + c – d)+(a – b – c + d)][(a + b + c – d) – (a – b – c + d)]<\/p>\n ($\\because a^2 – b^2 = (a + b)(a – b))$<\/p>\n =\u00a0(2a)(2b + 2c – 2d)<\/p>\n = 4a(b + c – d)<\/p>\n Question 6:\u00a0<\/b>Expand: $(4a+3b+2c)^{2}$<\/p>\n a)\u00a0$16a^{2}-9b^{2}+4c^{2}-24ab+12bc-16ca$<\/p>\n b)\u00a0$16a^{2}+9b^{2}+4c^{2}+24ab+12bc+16ca$<\/p>\n c)\u00a0$4a^{2}+3b^{2}+2c^{2}+24ab+12bc+16ca$<\/p>\n d)\u00a0$16a^{2}+9b^{2}+4c^{2}-24ab-12bc-16ca$<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(4a+3b+2c)^{2}$<\/p>\n = $(4a)^2 + (3b)^2 +\u00a0(2c)^2 + 2(4a.3b +\u00a03b.2c +\u00a02c.4a)$<\/p>\n ($\\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(a + b + c))$<\/p>\n = $16a^2 +\u00a09b^2 + 4c^2 + 2(12ab + 6bc + 8ac)$<\/p>\n = $16a^2 + 9b^2 + 4c^2 + 24ab + 12bc + 16ac$<\/p>\n Question 7:\u00a0<\/b>$(3a-4b)^{3}$ is equal to:<\/p>\n a)\u00a0$9a^{2}-16b^{2}$<\/p>\n b)\u00a0$27a^{3}-64b^{3}-108a^{2}b+144ab^{2}$<\/p>\n c)\u00a0$27a^{3}-64b^{3}$<\/p>\n d)\u00a0$9a^{2}-24ab+16b^{2}$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(3a-4b)^{3}$<\/p>\n = $(3a)^3 – (4b)^3 – 3(3a.4b)(3a-4b)$<\/p>\n ($\\because (a – b)^3 = a^3 – b^3 – 3ab(a – b)$)<\/p>\n = $27a^3 – 64b^3 – 36ab(3a-4b)$<\/p>\n =\u00a0$27a^3 – 64b^3 – 108a^2b + 144ab^2$<\/p>\n Question 8:\u00a0<\/b>If $A+B=12$ and $AB=17$, What is the value of $A^{3}+B^{3}$ ?<\/p>\n a)\u00a01116<\/p>\n b)\u00a01166<\/p>\n c)\u00a01106<\/p>\n d)\u00a01213<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $A^{3}+B^{3}$<\/p>\n = $(A + B)^3 -3AB(A+B)$<\/p>\n = $(12)^3 – 3 \\times 17(12)$<\/p>\n = 1728 – 612 = 1116<\/p>\n Question 9:\u00a0<\/b>If $16a^4 + 36a^2b^2 + 81b^4 = 91$ and $4a^2 + 9b^2 – 6ab = 13$, then what is the value of $3ab$?<\/p>\n a)\u00a0$\\frac{3}{2}$<\/p>\n b)\u00a0-3<\/p>\n c)\u00a0$-\\frac{3}{2}$<\/p>\n d)\u00a05<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $4a^2 + 9b^2 – 6ab = 13$<\/p>\n $(4a^2 + 9b^2 – 6ab)^2 = (13)^2$<\/p>\n $(\\because(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac))$<\/p>\n $(4a^2)^2 + (9b^2)^2 + (6ab)^2 +2(4a^2.9b^2 –\u00a09b^2.6ab –\u00a06ab.4a^2) = 169$<\/p>\n $16a^4 + 36a^2b^2 + 81b^4 + 2(36a^2b^2 – 54ab^3 – 24a^3b) = 169$<\/p>\n $91 + 2(36a^2b^2 – 54ab^3 – 24a^3b) = 169$<\/p>\n $36a^2b^2 – 54ab^3 – 24a^3b = \\frac{169 – 91}{2}$<\/p>\n $36a^2b^2 – 54ab^3 – 24a^3b = 39$<\/p>\n $6ab(6ab – 9b^2 – 4a^2) = 39$<\/p>\n $6ab(-13) = 39$<\/p>\n 6ab = -3<\/p>\n 3ab = -3\/2<\/p>\n Question 10:\u00a0<\/b>If $P = \\frac{x^3 + y^3}{(x – y)^2 + 3xy}, Q = \\frac{(x + y)^2 – 3xy}{x^3 – y^3}$ and $R = \\frac{(x + y)^2 + (x – y)^2}{x^2 – y^2}$, then what is the value of $(P \\div Q) \\times R$?<\/p>\n a)\u00a0$2(x^2 + y^2)$<\/p>\n b)\u00a0$x^2 + y^2$<\/p>\n c)\u00a0$4xy$<\/p>\n d)\u00a0$2xy$<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(P \\div Q) \\times R$<\/p>\n =\u00a0$(\\frac{x^3 + y^3}{(x – y)^2 + 3xy} \\div \\frac{(x + y)^2 – 3xy}{x^3 – y^3}) \\times \\frac{(x + y)^2 + (x – y)^2}{x^2 – y^2}$<\/p>\n = $\\frac{x^3 + y^3}{(x – y)^2 + 3xy} \\times\u00a0\\frac{x^3 – y^3}{(x + y)^2 – 3xy} \\times \\frac{(x + y)^2 + (x – y)^2}{x^2 – y^2}$<\/p>\n = $\\frac{(x + y)(x^2 – xy + y^2)}{x^2 + y^2 – 2xy\u00a0+ 3xy} \\times \\frac{(x – y)(x^2 + xy + y^2)}{x^2 + y^2 + 2xy – 3xy} \\times \\frac{x^2 + y^2 + 2xy\u00a0+ x^2 + y^2 – 2xy}{(x + y)(x – y)}$<\/p>\n = $\\frac{(x^2 – xy + y^2)}{x^2 + xy + y^2} \\times \\frac{(x^2 + xy + y^2)}{x^2 – xy + y^2} \\times 2(x^2 + y^2) $<\/p>\n = $ 2(x^2 + y^2) $<\/p>\n Question 11:\u00a0<\/b>If $x^2 – 2\\sqrt{5}x + 1 = 0$, then what is the value of $x^5 + \\frac{1}{x^5}$?<\/p>\n a)\u00a0$610\\sqrt{5}$<\/p>\n b)\u00a0$406\\sqrt{5}$<\/p>\n c)\u00a0$408\\sqrt{5}$<\/p>\n d)\u00a0$612\\sqrt{5}$<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^2 – 2\\sqrt{5}x + 1 = 0$<\/p>\n Divide by x,<\/p>\n $x – 2\\sqrt{5} + \\frac{1}{x} = 0$<\/p>\n $x +\u00a0 \\frac{1}{x} =\u00a02\\sqrt{5}$ —(1)<\/p>\n $(x + \\frac{1}{x})^2 = (2\\sqrt{5})^2$<\/p>\n $(x + \\frac{1}{x})^2 = 20$<\/p>\n $x^2 +\u00a0(\\frac{1}{x})^2 + 2 = 20$<\/p>\n $x^2 + (\\frac{1}{x})^2 = 18$\u00a0—-(2)<\/p>\n From eq(1),<\/p>\n $(x + \\frac{1}{x})^3 = (2\\sqrt{5})^3$<\/p>\n $x^3 +\u00a0(\\frac{1}{x})^3 + 3(x + \\frac{1}{x}) = 40\\sqrt{5}$<\/p>\n $x^3 + (\\frac{1}{x})^3 =\u00a040\\sqrt{5} –\u00a03(2\\sqrt{5}) =\u00a034\\sqrt{5}$ —(3)<\/p>\n From eq(2) and (3),<\/p>\n $(x^2 + (\\frac{1}{x})^2)(x^3 + (\\frac{1}{x})^3) = (18)(34\\sqrt{5})$<\/p>\n $(x^5 + \\frac{1}{x} + x + (\\frac{1}{x})^5) = 612\\sqrt{5}$<\/p>\n $x^5 +\u00a0 \\frac{1}{x^5} =612\\sqrt{5} –\u00a02\\sqrt{5}$ = $610\\sqrt{5}$<\/p>\n Question 12:\u00a0<\/b>If $2x^2 + y^2 + 8z^2 – 2\\sqrt{2}xy + 4\\sqrt{2}yz – 8zx = (Ax + y + Bz)^2,$ then the value of $(A^2 + B^2 – AB)$ is:<\/p>\n a)\u00a016<\/p>\n b)\u00a014<\/p>\n c)\u00a06<\/p>\n d)\u00a018<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2x^2 + y^2 + 8z^2 – 2\\sqrt{2}xy + 4\\sqrt{2}yz – 8zx = (Ax + y + Bz)^2$<\/p>\n =\u00a0$2x^2 + y^2 + 8z^2 + 2(-\\sqrt{2}xy – 2\\sqrt{2}yz + 4zx) = (Ax + y + Bz)^2,$<\/p>\n = $2x^2 + y^2 + 8z^2 + 2(-\\sqrt{2}x.y – y.2\\sqrt{2}z + \\sqrt{2}x.2\\sqrt{2}z) = (Ax + y + Bz)^2$<\/p>\n =\u00a0$(+\\sqrt{2}x – y +\u00a02\\sqrt{2}z)^2 =\u00a0(Ax + y + Bz)^2$<\/p>\n Ax = $\\sqrt{2}x$<\/p>\n A = $\\sqrt{2}$<\/p>\n y = -1<\/p>\n B = $2\\sqrt{2}$<\/p>\n Now,<\/p>\n $(A^2 + B^2 – AB)$<\/p>\n $((\\sqrt{2})^2 + (2\\sqrt{2})^2 – \\sqrt{2}.2\\sqrt{2})$<\/p>\n = 2 + 8 – 4 = 6<\/p>\n Question 13:\u00a0<\/b>If $12x^2 – 21x + 1 = 0, $ then what is the value of $9x^2 + (16x^2)^{-1}$?<\/p>\n a)\u00a0$\\frac{429}{8}$<\/p>\n b)\u00a0$\\frac{465}{16}$<\/p>\n c)\u00a0$\\frac{417}{16}$<\/p>\n d)\u00a0$\\frac{453}{8}$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $12x^2 – 21x + 1 = 0 $<\/p>\n Divide by 4x,<\/p>\n $3x – 21\/4 + 1\/4x = 0 $<\/p>\n $3x + \\frac{1}{4x} = \\frac{21}{4}$<\/p>\n On taking square,<\/p>\n $(3x + \\frac{1}{4x})^2 = (\\frac{21}{4})^2$<\/p>\n $9x^2 + \\frac{1}{16x^2} + 2.3x.\\frac{1}{4x} = \\frac{441}{16}$<\/p>\n $9x^2 + \\frac{1}{16x^2} =\u00a0\\frac{441}{16} – \\frac{3}{2}$<\/p>\n $9x^2 + \\frac{1}{16x^2} = \\frac{417}{16}$<\/p>\n Question 14:\u00a0<\/b>If x + y + z = 3, and $x^2 + y^2 + z^2 = 101$, then what is the value of $\\sqrt{x^3 + y^3 + z^3 – 3xyz}$?<\/p>\n a)\u00a019<\/p>\n b)\u00a021<\/p>\n c)\u00a024<\/p>\n d)\u00a028<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n x\u00b3 + y\u00b3 + z\u00b3 – 3xyz = (x+y+z) [(x + y + z)\u00b2 – 3(xy + yz + zx)] —(1)<\/p>\n And,<\/p>\n (x + y + z)\u00b2 = x\u00b2 + y\u00b2 + z\u00b2 + 2(xy + yz + zx)<\/p>\n On put the values,<\/p>\n $3^2 = 101 +\u00a02(xy + yz + zx)$<\/p>\n (xy + yz + zx) = -92\/2 = -46<\/p>\n From the eq(1),<\/p>\n x\u00b3 + y\u00b3 + z\u00b3 – 3xyz =\u00a0(3) [(3)\u00b2 – 3(-46)] = 3 $\\times 147 = 441<\/p>\n $\\sqrt{x^3 + y^3 + z^3 – 3xyz} = \\sqrt{441}$ = 21<\/p>\n Question 15:\u00a0<\/b>If $x^2 – 3x + 1 = 0$, then what is the value of $x^6 + \\frac{1}{x^6}$?<\/p>\n a)\u00a0324<\/p>\n b)\u00a0322<\/p>\n c)\u00a0318<\/p>\n d)\u00a0327<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^2 – 3x + 1 = 0$<\/p>\n Divide both sides by x,<\/p>\n $x – 3 + \\frac{1}{x}\u00a0= 0$<\/p>\n $x + \\frac{1}{x} =\u00a03$<\/p>\n On taking both sides cube,<\/p>\n $(x + \\frac{1}{x})^3 = 3^3$<\/p>\n $x^3 +\u00a0\\frac{1}{x^3} + 3(x + \\frac{1}{x}) = 27$<\/p>\n $(\\because (a + b)^3 = a^3 + b^3 + 3ab(a + b))$<\/p>\n $x^3 + \\frac{1}{x^3} + 3(3) = 27$<\/p>\n $x^3 + \\frac{1}{x^3} = 18$<\/p>\n on taking square both sides,<\/p>\n $(x^3 + \\frac{1}{x^3})^2 = 18^2$<\/p>\n $x^6 +\u00a0\\frac{1}{x^6} + 2 = 324$<\/p>\n $(\\because (a + b)^2 = a^2 + b^2 + 2ab)$<\/p>\n $x^6 + \\frac{1}{x^6} = 322$<\/p>\n Question 16:\u00a0<\/b>If $x^4 + x^2y^2 + y^4 = 21$ and $x^2 + xy + y^2 = 7$, then the value of $\\left(\\frac{1}{x^2} + \\frac{1}{y^2}\\right)$ is:<\/p>\n a)\u00a0$\\frac{5}{2}$<\/p>\n b)\u00a0$\\frac{7}{4}$<\/p>\n c)\u00a0$\\frac{5}{4}$<\/p>\n d)\u00a0$\\frac{7}{3}$<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^4 + x^2y^2 + y^4 = 21$<\/p>\n =\u00a0$x^4 + x^2y^2 + y^4 +\u00a0x^2y^2\u00a0–\u00a0x^2y^2= 21$<\/p>\n = $x^4 + 2x^2y^2 + y^4 – x^2y^2= 21$<\/p>\n $(\\because (a + b)^2 = a^2 + 2ab +\u00a0b^2)$<\/p>\n = $(x^2 +\u00a0y^2)^2 = 21 +\u00a0x^2y^2$ —-(1)<\/p>\n $x^2 + xy + y^2 = 7$<\/p>\n $x^2 + y^2 = 7 – xy$ —(2)<\/p>\n From eq(1) and (2),<\/p>\n $(7 – xy)^2 = 21 + x^2y^2$<\/p>\n $49 +\u00a0x^2y^2 – 14xy = 21 +\u00a0x^2y^2$<\/p>\n 14xy = 49 – 21<\/p>\n xy = 28\/14 = 2<\/strong><\/p>\n From eq(2),<\/p>\n $x^2 + y^2 = 7 – 2$ $x^2 + y^2 = 5$<\/strong><\/p>\n Now,<\/p>\n $\\left(\\frac{1}{x^2} + \\frac{1}{y^2}\\right)$<\/p>\n = $\\frac{x^2 + y^2}{x^2y^2}$<\/p>\n = $\\frac{5}{2^2}$<\/p>\n = $\\frac{5}{4}$<\/p>\n Question 17:\u00a0<\/b>The expression $(a + b – c)^3 + (a – b + c)^3 – 8a^3$is equal to:<\/p>\n a)\u00a0$6a(a – b + c)(c – a – b)$<\/p>\n b)\u00a0$3a(a + b – c)(a – b + c)$<\/p>\n c)\u00a0$6a(a + b – c)(a – b + c)$<\/p>\n d)\u00a0$3a(a – b + c)(c – a – b)$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u00a0$(a + b – c) = A$,\u00a0\u00a0$(a – b + c) = B$ and\u00a0\u00a0$2a^3$ = C<\/p>\n $(a + b – c)^3 + (a – b + c)^3 – 8a^3$<\/p>\n =\u00a0$(A)^3 + (B)^3 – C^3$<\/p>\n $(a^3 +\u00a0b^3 + c^3 = 3abc$ when a + b + c = 0)<\/p>\n So,<\/p>\n = -3ABC<\/p>\n =$ -3(a + b – c)(a – b + c).2a^3$<\/p>\n = $6a^3(a + b – c)(c – a – b)$<\/p>\n Question 18:\u00a0<\/b>If a + b + c = 11, ab + bc + ca = 3 and abc = \u2014135, then what is the value of $a^3 + b^3 + c^3$?<\/p>\n a)\u00a0827<\/p>\n b)\u00a0929<\/p>\n c)\u00a0823<\/p>\n d)\u00a0925<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $a^3 + b^3 + c^3$<\/p>\n = $(a\u00a0+ b + c)[(a + b + c)^2 – 3(ab + bc +\u00a0ac)] +\u00a03abc$<\/p>\n =\u00a0$(11)[(11)^2 – 3(3)] + 3 \\times (-135)\u00a0$<\/p>\n = $(11)[112] – 405$<\/p>\n = 1132 – 405 = 827<\/p>\n Question 19:\u00a0<\/b>If $5x + \\frac{1}{3x} = 4$, then whatis the value of $9x^2 + \\frac{1}{25x^2}$?<\/p>\n a)\u00a0$\\frac{174}{125}$<\/p>\n b)\u00a0$\\frac{119}{25}$<\/p>\n c)\u00a0$\\frac{144}{125}$<\/p>\n d)\u00a0$\\frac{114}{25}$<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $5x + \\frac{1}{3x} = 4$<\/p>\n $\\frac{3}{5}(5x + \\frac{1}{3x}) =\u00a0\\frac{3}{5} \\times\u00a04$<\/p>\n $(3x + \\frac{1}{5x}) = \\frac{12}{5}$<\/p>\n On square both sides,<\/p>\n $(3x + \\frac{1}{5x})^2 = (\\frac{12}{5})^2$<\/p>\n $9x^2 +\u00a0\\frac{1}{25x^2} + 2.3x.\\frac{1}{5x} = \\frac{144}{25}$<\/p>\n $((a+b)^2 = a^2 + b^2 + 2ab)$<\/p>\n $9x^2 + \\frac{1}{25x^2} =\u00a0\\frac{144}{25} –\u00a0\\frac{6}{5} = 0$<\/p>\n $9x^2 + \\frac{1}{25x^2} =\u00a0\\frac{114}{25}$<\/p>\n Question 20:\u00a0<\/b>On simplification, $\\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \\div \\frac{y[(x – y)^2 + 3xy]}{x^3 + y^3} \\times \\frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$ is equal to:<\/p>\n a)\u00a04<\/p>\n b)\u00a01<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$\\frac{1}{4}$<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \\div \\frac{y[(x – y)^2 + 3xy]}{x^3 + y^3} \\times \\frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$<\/p>\n = $\\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \\times \\frac{x^3 + y^3}{y[(x – y)^2 + 3xy]} \\times \\frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$<\/p>\n = $\\frac{(x – y)(x^2 + xy + y^2)}{x[(x + y)^2 – 3xy]} \\times \\frac{(x + y)(x^2 – xy + y^2)}{y[(x – y)^2 + 3xy]} \\times \\frac{(x^2 + y^2 + 2xy) – (x^2 + y^2 – 2xy)}{(x +\u00a0y)(x – y)}$<\/p>\n = $\\frac{x^2 + xy + y^2}{x[x^2 – xy + y^2]} \\times \\frac{x^2 – xy + y^2}{y[x^2 + xy + y^2]} \\times 4xy$ = 4<\/p>\n
\nOn put the value of a, b and c,<\/p>\n
\nvalue of $\\frac{xy+xz}{xyz}$ is:<\/p>\n
\ny+z=15 —(2)
\nz+x=16$\u00a0—(3)<\/p>\n
\n<\/strong><\/p>\n