Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>If $\\sqrt{x}=\\sqrt{3}-\\sqrt{5}$, then the value of $x^2-16x+6$ is:<\/p>\n a)\u00a04<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0-2<\/p>\n Question 2:\u00a0<\/b>If $x=\\frac{\\sqrt{3}}{2}$, then the value of $\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}$ is equal to:<\/p>\n a)\u00a0$\\sqrt 2$<\/p>\n b)\u00a0$\\sqrt 3$<\/p>\n c)\u00a03<\/p>\n d)\u00a02<\/p>\n Question 3:\u00a0<\/b>If $a^3+b^3=62$ and a + b = 2, then the value of ab is:<\/p>\n a)\u00a0-6<\/p>\n b)\u00a09<\/p>\n c)\u00a06<\/p>\n d)\u00a0-9<\/p>\n Question 4:\u00a0<\/b>If $a-b=18$ and $a^3-b^3=324$, then find ab.<\/p>\n a)\u00a0105<\/p>\n b)\u00a0-102<\/p>\n c)\u00a0-104<\/p>\n d)\u00a0103<\/p>\n Question 5:\u00a0<\/b>If $a^2+\\frac{2}{a^2}=16$, then find the value of $\\frac{72a^2}{a^4+2+8a^2}$<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a01<\/p>\n d)\u00a03<\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:<\/p>\n a)\u00a09<\/p>\n b)\u00a08<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n Question 7:\u00a0<\/b>If $x^3 \u2014 6x^2 + ax + b$ is divisible by $(x^2 \u2014 3x + 2)$, then the values of a and b are:<\/p>\n a)\u00a0a = -6 and b = -11<\/p>\n b)\u00a0a = -11 and b = 6<\/p>\n c)\u00a0a = 11 and b = -6<\/p>\n d)\u00a0a = 6 and b = 11<\/p>\n Question 8:\u00a0<\/b>If a + b + c + d = 2, then the maximum value of (1 + a)(1 + b)(1 + c)(1 + d) is<\/p>\n a)\u00a0$\\frac{91}{9}$<\/p>\n b)\u00a0$\\frac{63}{22}$<\/p>\n c)\u00a0$\\frac{54}{13}$<\/p>\n d)\u00a0$\\frac{81}{16}$<\/p>\n Question 9:\u00a0<\/b>If $x + \\frac{1}{x} = 4,$ then the value of $x^4 + \\frac{1}{x^4} $ is :<\/p>\n a)\u00a016<\/p>\n b)\u00a0196<\/p>\n c)\u00a0194<\/p>\n d)\u00a014<\/p>\n Question 10:\u00a0<\/b>If x – y = 13 and xy = 25, then the value of $x^2 – y^2$ = ?<\/p>\n a)\u00a0$13 \\sqrt 240$<\/p>\n b)\u00a0$13 \\sqrt 229$<\/p>\n c)\u00a0$13 \\sqrt 269$<\/p>\n d)\u00a0$13 \\sqrt 210$<\/p>\n Question 11:\u00a0<\/b>2x \u2014 3y is a factor of:<\/p>\n a)\u00a0$4x^2 + 2x – 3y + 9y^2 – 12xy$<\/p>\n b)\u00a0$4x^2 + 9y^2 + 12xy$<\/p>\n c)\u00a0$8x^3 + 27y^3$<\/p>\n d)\u00a0$4x^2 + 2x – 3y + 36y^2 + 12xy$<\/p>\n Question 12:\u00a0<\/b>If a and b are two positive real numbers such that a + b = 20 and ab = 4, then the value of $a^3 + b^3$ is:<\/p>\n a)\u00a07760<\/p>\n b)\u00a08000<\/p>\n c)\u00a08240<\/p>\n d)\u00a0240<\/p>\n Question 13:\u00a0<\/b>If x + y = 15 and xy = 14, then the value of x \u2014 y is:<\/p>\n a)\u00a013<\/p>\n b)\u00a012<\/p>\n c)\u00a011<\/p>\n d)\u00a014<\/p>\n Question 14:\u00a0<\/b>If a = 355, b = 356, c = 357,then find the value of $a^3 + b^3 + c^3 – 3abc$.<\/p>\n a)\u00a03204<\/p>\n b)\u00a03206<\/p>\n c)\u00a03202<\/p>\n d)\u00a03208<\/p>\n Question 15:\u00a0<\/b>If $x + y = 14; x^3 + y^3 = 1064$, then the value of $(x – y)^2$ is:<\/p>\n a)\u00a036<\/p>\n b)\u00a064<\/p>\n c)\u00a081<\/p>\n d)\u00a0100<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Question 16:\u00a0<\/b>If $a + b = 8$ and $a + a^2b + b + ab^2 = 128$ then the positive value of $a^3 + b^3$ is:<\/p>\n a)\u00a0344<\/p>\n b)\u00a096<\/p>\n c)\u00a0224<\/p>\n d)\u00a0152<\/p>\n Question 17:\u00a0<\/b>The coefficient of y in the expansion of $(2y – 5)^3$, is:<\/p>\n a)\u00a0150<\/p>\n b)\u00a050<\/p>\n c)\u00a0-30<\/p>\n d)\u00a0-150<\/p>\n Question 18:\u00a0<\/b>The given table represents the revenue (in \u20b9 crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows. a)\u00a044.5<\/p>\n b)\u00a031.2<\/p>\n c)\u00a043.6<\/p>\n d)\u00a045.4<\/p>\n Question 19:\u00a0<\/b>If $a + b + c = 19, ab + bc + ca = 120$, then what is the value of $a^3 + b^3 + c^3 – 3abc$?<\/p>\n a)\u00a018<\/p>\n b)\u00a023<\/p>\n c)\u00a031<\/p>\n d)\u00a019<\/p>\n Question 20:\u00a0<\/b>Solve the following: a)\u00a0(a + b)(b + c)(c – a)<\/p>\n b)\u00a0(a + b)(b – c)(c + a)<\/p>\n c)\u00a0(a + b)(b + c)(c + a)<\/p>\n d)\u00a0(a – b)(b – c)(c – a)<\/p>\n Question 21:\u00a0<\/b>If $x^6 – 512 y^6 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$, then what is the value of $(A + B – C)$?<\/p>\n a)\u00a0-80<\/p>\n b)\u00a0-72<\/p>\n c)\u00a072<\/p>\n d)\u00a048<\/p>\n Question 22:\u00a0<\/b>If $x, y, z$ are three integers such that $x + y = 8, y + z = 13$ and $z + x = 17$, then the value of $\\frac{x^2}{yz}$ is:<\/p>\n a)\u00a0$\\frac{7}{5}$<\/p>\n b)\u00a0$\\frac{18}{11}$<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n Question 23:\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 Question 24:\u00a0<\/b>If $\\left(x^3 + \\frac{1}{x^3} – k\\right)^2 + \\left(x + \\frac{1}{x} – p\\right)^2 = 0$\u00a0where k and p are real numbers and x\u00a0\u2260 0, then $\\frac{k}{p}$ is equal to:<\/p>\n a)\u00a0$P^2+1$<\/p>\n b)\u00a0$P^2+3$<\/p>\n c)\u00a0$P^2-1$<\/p>\n d)\u00a0$P^2-3$<\/p>\n Question 25:\u00a0<\/b>For real a, b, c if $a^2 + b^2 + c^2 = ab + bc + ca$, the value of $\\frac{(a+b)}{c}$<\/p>\n a)\u00a03<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a00<\/p>\n Enroll to CAT 2022 Complete Course<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given,\u00a0 $\\sqrt{x}=\\sqrt{3}-\\sqrt{5}$<\/p>\n $\\Rightarrow$ \u00a0$x=\\left(\\sqrt{3}-\\sqrt{5}\\right)^2$<\/p>\n $\\Rightarrow$ \u00a0$x=3+5-2\\sqrt{15}$<\/p>\n $\\Rightarrow$ \u00a0$x=8-2\\sqrt{15}$ ……………(1)<\/p>\n $\\Rightarrow$ \u00a0$x^2=\\left(8-2\\sqrt{15}\\right)^2$<\/p>\n $\\Rightarrow$ \u00a0$x^2=64+60-32\\sqrt{15}$<\/p>\n $\\Rightarrow$ \u00a0$x^2=124-32\\sqrt{15}$ ………..(2)<\/p>\n $\\therefore\\ $ $x^2-16x+6=124-32\\sqrt{15}-16\\left(8-2\\sqrt{15}\\right)+6$<\/p>\n $=124-32\\sqrt{15}-128+32\\sqrt{15}+6$<\/p>\n $=130-128$<\/p>\n $=2$<\/p>\n Hence, the correct answer is Option C<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given,\u00a0$x=\\frac{\\sqrt{3}}{2}$<\/p>\n $\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}=\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}\\times\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}+\\sqrt{1-x}}$<\/p>\n $=\\frac{1+x+1-x+2\\left(\\sqrt{1+x}\\right)\\left(\\sqrt{1-x}\\right)}{1+x-\\left(1-x\\right)}$<\/p>\n $=\\frac{2+2\\left(\\sqrt{1-x^2}\\right)}{2x}$<\/p>\n $=\\frac{1+\\sqrt{1-x^2}}{x}$<\/p>\n $=\\frac{1+\\sqrt{1-\\left(\\frac{\\sqrt{3}}{2}\\right)^2}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{1+\\sqrt{1-\\frac{3}{4}}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{1+\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{3}{2}\\times\\frac{2}{\\sqrt{3}}$<\/p>\n $=\\sqrt{3}$<\/p>\n Hence, the correct answer is Option B<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,\u00a0 $a+b=2$<\/p>\n $a^3+b^3=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+b\\right)\\left(a^2-ab+b^2\\right)=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(2\\right)\\left(a^2+2ab+b^2-3ab\\right)=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+b\\right)^2-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$\\left(2\\right)^2-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$4-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$3ab=-27$<\/p>\n $\\Rightarrow$ \u00a0$ab=-9$<\/p>\n Hence, the correct answer is Option D<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given,\u00a0 $a-b=18$ and<\/p>\n $a^3-b^3=324$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-b\\right)\\left(a^2+ab+b^2\\right)=324$<\/p>\n $\\Rightarrow$\u00a0 $\\left(18\\right)\\left(a^2-2ab+b^2+3ab\\right)=324$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-b\\right)^2+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$\\left(18\\right)^2+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$324+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$3ab=-306$<\/p>\n $\\Rightarrow$ \u00a0$ab=-102$<\/p>\n Hence, the correct answer is Option B<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, \u00a0$a^2+\\frac{2}{a^2}=16$<\/p>\n $\\frac{72a^2}{a^4+2+8a^2}=\\frac{72a^2}{a^2\\left(a^2+\\frac{2}{a^2}+8\\right)}$<\/p>\n $=\\frac{72}{a^2+\\frac{2}{a^2}+8}$<\/p>\n $=\\frac{72}{16+8}$<\/p>\n $=\\frac{72}{24}$<\/p>\n $=3$<\/p>\n Hence, the correct answer is Option D<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given,\u00a0$a+3b=13$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+3b\\right)^2=12^2$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+6ab=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+6\\left(9\\right)=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+54-6ab+6ab=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2-6ab+6ab=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+6ab=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+6\\left(9\\right)=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+54=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2=36$<\/p>\n $\\Rightarrow$ \u00a0$a-3b=6$<\/p>\n Hence, the correct answer is Option C<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given,\u00a0$x^3\u20146x^2+ax+b$ is divisible by $(x^2 \u2014 3x + 2)$<\/p>\n Let the quotient when\u00a0$x^3 \u2014 6x^2 + ax + b$ is divisible by $(x^2 \u2014 3x + 2)$ be $x-p$<\/p>\n $\\Rightarrow$ $(x^2\u20143x+2)\\left(x-p\\right)=x^3\u20146x^2+ax+b$<\/p>\n $\\Rightarrow$\u00a0 $x^3-3x^2+2x-px^2+3px-2p=x^3\u20146x^2+ax+b$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\left(3+p\\right)x^2+\\left(2+3p\\right)x-2p=x^3\u20146x^2+ax+b$<\/p>\n Comparing both sides,<\/p>\n $-\\left(3+p\\right)=-6$<\/p>\n $\\Rightarrow$ \u00a0$p=3$<\/p>\n $2+3p=a$<\/p>\n $\\Rightarrow$ \u00a0$2+3\\left(3\\right)=a$<\/p>\n $\\Rightarrow$ \u00a0$a=11$<\/p>\n $-2p=b$<\/p>\n $\\Rightarrow$ \u00a0$-2\\left(3\\right)=b$<\/p>\n $\\Rightarrow$ \u00a0$b=-6$<\/p>\n $\\therefore\\ $ a = 11 and b = -6<\/p>\n Hence, the correct answer is Option C<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,\u00a0a + b + c + d = 2<\/p>\n We know that, AM\u00a0$\\ge\\ $ GM<\/p>\n $\\Rightarrow$ Arithmetic mean of\u00a0(1 + a),(1 + b),(1 + c),(1 + d)\u00a0$\\ge\\ $ Geometric mean of\u00a0(1 + a),(1 + b),(1 + c),(1 + d)<\/p>\n $\\Rightarrow$\u00a0$\\frac{\\left(1+a\\right)+\\left(1+b\\right)+\\left(1+c\\right)+\\left(1+d\\right)}{4}\\ge\\left[\\ \\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\right]^{\\frac{1}{4}}$<\/p>\n $\\Rightarrow$\u00a0$\\frac{4+a+b+c+d}{4}\\ge\\left[\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\right]^{\\frac{1}{4}}$<\/p>\n $\\Rightarrow$\u00a0$\\frac{4+2}{4}\\ge\\left[\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\right]^{\\frac{1}{4}}$<\/p>\n $\\Rightarrow$\u00a0$\\frac{6}{4}\\ge\\left[\\ \\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\right]^{\\frac{1}{4}}$<\/p>\n $\\Rightarrow$\u00a0$\\left[\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\right]^{\\frac{1}{4}}\\le\\ \\frac{3}{2}$<\/p>\n $\\Rightarrow$\u00a0$\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\le\\ \\left(\\frac{3}{2}\\right)^4$<\/p>\n $\\Rightarrow$ $\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)\\left(1+d\\right)\\le\\ \\frac{81}{16}$<\/p>\n $\\therefore\\ $Maximum value of\u00a0(1 + a)(1 + b)(1 + c)(1 + d) =\u00a0$\\frac{81}{16}$<\/p>\n Hence, the correct answer is Option D<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, \u00a0$x + \\frac{1}{x} = 4$<\/p>\n $=$> \u00a0$\\left(x+\\frac{1}{x}\\right)^2=4^2$<\/p>\n $=$> \u00a0$x^2+\\frac{1}{x^2}+2.x.\\frac{\\ 1}{x}=16$<\/p>\n $=$> \u00a0$x^2+\\frac{1}{x^2}+2=16$<\/p>\n $=$> \u00a0$x^2+\\frac{1}{x^2}=14$<\/p>\n $=$> \u00a0$\\left(x^2+\\frac{1}{x^2}\\right)^2=14^2$<\/p>\n $=$> \u00a0$x^4+\\frac{1}{x^4}+2.x^2.\\frac{\\ 1}{x^2}=196$<\/p>\n $=$> \u00a0$x^4+\\frac{1}{x^4}+2=196$<\/p>\n $=$> \u00a0$x^4+\\frac{1}{x^4}=196-2$<\/p>\n $=$> \u00a0$x^4+\\left(\\frac{1}{x}\\right)^4=194$<\/p>\n Hence, the correct answer is Option C<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given,<\/p>\n $x – y = 13$ and\u00a0 $xy = 25$<\/p>\n $=$> \u00a0$\\left(x-y\\right)^2=13^2$<\/p>\n $=$> \u00a0$x^2+y^2-2xy=169$<\/p>\n $=$> \u00a0$x^2+y^2+2xy-4xy=169$<\/p>\n $=$> \u00a0$\\left(x+y\\right)^2-4xy=169$<\/p>\n $=$> \u00a0$\\left(x+y\\right)^2-4\\left(25\\right)=169$<\/p>\n $=$> \u00a0$\\left(x+y\\right)^2-100=169$<\/p>\n $=$> \u00a0$\\left(x+y\\right)^2=269$<\/p>\n $=$> \u00a0$x+y=\\sqrt{269}$<\/p>\n $\\therefore\\ $ $x^2-y^2=\\left(x+y\\right)\\left(x-y\\right)=\\left(\\sqrt{269}\\right)\\left(13\\right)=13\\sqrt{269}$<\/p>\n Hence, the correct answer is Option C<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $4x^2 + 2x – 3y + 9y^2 – 12xy=4x^2+9y^2-12xy+2x-3y$<\/p>\n $=\\left(2x-3y\\right)^2+2x-3y$<\/p>\n $=\\left(2x-3y\\right)\\left(2x-3y+1\\right)$<\/p>\n $\\therefore\\ $ $2x+3y$ is factor of $4x^2 + 2x – 3y + 9y^2 – 12xy$<\/p>\n Hence, the correct answer is Option A<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given,\u00a0 $a+b=20$\u00a0 and\u00a0 $ab=4$<\/p>\n $=$> \u00a0$\\left(a+b\\right)^3=20^3$<\/p>\n $=$> \u00a0$a^3+b^3+3ab\\left(a+b\\right)=8000$<\/p>\n $=$> \u00a0$a^3+b^3+3\\left(4\\right)\\left(20\\right)=8000$<\/p>\n $=$> \u00a0$a^3+b^3+240=8000$<\/p>\n $=$> \u00a0$a^3+b^3=8000-240$<\/p>\n $=$> \u00a0$a^3+b^3=7760$<\/p>\n Hence, the correct answer is Option A<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given,\u00a0$x+y\\ =15$\u00a0 and\u00a0$xy=14$<\/p>\n $=$>\u00a0$\\left(x+y\\right)^2=15^2$<\/p>\n $=$>\u00a0$x^2+y^2+2xy=225$<\/p>\n $=$>\u00a0$x^2+y^2+2xy-2xy+2xy=225$<\/p>\n $=$>\u00a0$x^2+y^2-2xy+4xy=225$<\/p>\n $=$>\u00a0$\\left(x-y\\right)^2+4\\left(14\\right)=225$<\/p>\n $=$>\u00a0$\\left(x-y\\right)^2+=225-56$<\/p>\n $=$>\u00a0$\\left(x-y\\right)^2+=169$<\/p>\n $=$>\u00a0 $x-y=13$<\/p>\n Hence, the correct answer is Option A<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, $a$ = 355, $b$ = 356, $c$ = 357<\/p>\n $a^3+b^3+c^3\u20143abc=\\left(a+b+c\\right)\\left(a^2+b^2+c^2-ab-bc-ca\\right)$<\/p>\n $=\\frac{1}{2}\\left(a+b+c\\right)\\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\\right)$<\/p>\n $=\\frac{\\left(a+b+c\\right)}{2}\\left[\\left(a-b\\right)^2+\\left(b-c\\right)^2+\\left(c-a\\right)^2\\right]$<\/p>\n $=\\frac{\\left(355+356+357\\right)}{2}\\left[\\left(355-356\\right)^2+\\left(356-357\\right)^2+\\left(357-355\\right)^2\\right]$<\/p>\n $=\\frac{1068}{2}\\left[1+1+4\\right]$<\/p>\n $=\\frac{1068}{2}\\left[6\\right]$<\/p>\n $=3204$<\/p>\n Hence, the correct answer is Option A<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given,\u00a0$x + y = 14$<\/p>\n $x^3 + y^3 = 1064$<\/p>\n $=$> \u00a0$\\left(x+y\\right)\\left(x^2+y^2-xy\\right)=1064$<\/p>\n $=$> \u00a0$14\\left(x^2+y^2+2xy-3xy\\right)=1064$<\/p>\n $=$> \u00a0$\\left(x+y\\right)^2-3xy=76$<\/p>\n $=$> \u00a0$\\left(14\\right)^2-3xy=76$<\/p>\n $=$> \u00a0$196-3xy=76$<\/p>\n $=$> \u00a0$3xy=120$<\/p>\n $=$> \u00a0$xy=40$<\/p>\n $\\therefore\\ $ $(x – y)^2=x^2+y^2-2xy$<\/p>\n $=x^2+y^2+2xy-4xy$<\/p>\n $=\\left(x+y\\right)^2-4xy$<\/p>\n $=\\left(14\\right)^2-4\\left(40\\right)$<\/p>\n $=196-160$<\/p>\n $=36$<\/p>\n Hence, the correct answer is Option A<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,\u00a0$a + b = 8$<\/p>\n $a + a^2b + b + ab^2 = 128$<\/p>\n $=$> \u00a0$a\\left(1+ab\\right)+b\\left(1+ab\\right)=128$<\/p>\n $=$> \u00a0$\\left(1+ab\\right)\\left(a+b\\right)=128$<\/p>\n $=$> \u00a0$\\left(1+ab\\right)8=128$<\/p>\n $=$> \u00a0$ 1+ab =\\frac{128}{8}$<\/p>\n $=$> \u00a0$1+ab=16$<\/p>\n $=$> \u00a0$ab=15$<\/p>\n $\\therefore\\ $ $a^3+b^3=\\left(a+b\\right)\\left(a^2+b^2-ab\\right)$<\/p>\n $= 8 (a^2+b^2+2ab-3ab)$<\/p>\n $=8 \\left(\\left(a+b\\right)^2-3ab\\right)$<\/p>\n $=8 \\left(8^2-3\\left(15\\right)\\right)$<\/p>\n $=8 \\left(64-45\\right)$<\/p>\n $=152$<\/p>\n Hence, the correct answer is Option D<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $(2y – 5)^3$<\/p>\n = $8y^3 – 125 +150y – 60y^2$<\/p>\n $(\\because(a-b)^3 = a^3 – b^3 – 3a^2b + 3ab^2)$<\/p>\n The coefficient of y = 150<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Total revenue of the company from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538<\/p>\n Total revenue from the sale of product B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370<\/p>\n Required percentage = $\\frac{538 – 370}{370} \\times 100$ = 45.4%<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $(a + b + c)^2 = a^2 +b^2 + c^2 + 2ab +\u00a02bc +\u00a02ca$<\/p>\n $ 19^2 =\u00a0a^2 +b^2 + c^2 + 2(120)$<\/p>\n $a^2 +b^2 + c^2 = 361 – 240 = 121$<\/p>\n $a^3 + b^3 + c^3 – 3abc$<\/p>\n =$ (a + b + c)(a^2 + b^2 + c^2 -ab -bc – ca)$<\/p>\n = 19(121$ \\times -120) = 19$<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n (a + b + c)(ab + bc + ca) – abc<\/p>\n = $a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc$<\/p>\n = $a^2(b +c)+bc(b+c)+a(b^2+c^2+2bc)$<\/p>\n = $a^2(b+c)+bc(b+c)+a(b+c)^2$<\/p>\n =$(b+c)(a^2+bc+ab+ac)$<\/p>\n = (b + c)[a(a + b) + c(a + b)]<\/p>\n = (a + b)(b + c)(c + a)<\/p>\n 21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $x^6 – 512 y^6 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$<\/p>\n $(x^2)^3 +\u00a0\u00a0(-8y^2)^3 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$<\/p>\n $(\\because a^3 +\u00a0b^3 = (a + b)(a^2 – ab + b^2))$<\/p>\n by comparison –<\/p>\n $-8y^2 = Ay^2$<\/p>\n A = -8<\/p>\n $Bx^2y^2 = x^2 \\times -8y^2$<\/p>\n B = -8<\/p>\n $(-8y^2)^2 =\u00a0Cy^4$<\/p>\n C = 64<\/p>\n The value of $(A + B – C)$ = -8 – 8 – 64 = -80<\/p>\n 22)\u00a0Answer\u00a0(B)<\/strong><\/p>\n x + y = 8 —-(1)<\/p>\n y + z = 13\u00a0—-(2)<\/p>\n z + x = 17\u00a0—-(3)<\/p>\n Eq (1) + (2) + (3),<\/p>\n 2(x + y + z) = 8 + 13 +\u00a017<\/p>\n x + y + z = 38\/2 = 19 —-(4)<\/p>\n From eq (3) and (4),<\/p>\n x + 13 = 19<\/p>\n x = 6<\/strong><\/p>\n From eq(3),<\/p>\n z + x = 17<\/p>\n z + 6 = 17<\/p>\n z = 11<\/strong><\/p>\n From eq(2),<\/p>\n y + z = 13<\/p>\n y + 11 = 13<\/p>\n y = 2<\/strong><\/p>\n $\\frac{x^2}{yz}$<\/p>\n =\u00a0$\\frac{6^2}{2 \\times 11}$<\/p>\n = 36\/22 = 18\/11<\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/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 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\left(x^3 + \\frac{1}{x^3} – k\\right)^2 + \\left(x + \\frac{1}{x} – p\\right)^2 = 0$<\/p>\n It will be zero, if the individual terms will be zero.<\/p>\n So, $(x^3 + \\frac{1}{x^3} – k)^2=0$ and $(x + \\frac{1}{x} – p)^2$<\/p>\n So, $k=x^3 + \\frac{1}{x^3}$ and\u00a0$(x + \\frac{1}{x} – p)^2=0$<\/p>\n $k=x^3 + \\frac{1}{x^3}$ and $(x + \\frac{1}{x} )=p$<\/p>\n Now, $(x + \\frac{1}{x} )=p$ taking cube of both side,<\/p>\n $\\Rightarrow (x + \\frac{1}{x} )^3=p^3$<\/p>\n $\\Rightarrow x^3+\\dfrac{1}{x^3}+3(x+\\dfrac{1}{x})=p^3$<\/p>\n Now substituting the values in the above,<\/p>\n $\\Rightarrow k+3p=p^3$<\/p>\n $\\Rightarrow k=p^3-3p$<\/p>\n $\\Rightarrow \\dfrac{k}{p}=p^2-3$<\/p>\n 25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, \u00a0 \u00a0$a^2+b^2+c^2=ab+bc+ca$<\/p>\n Multiplying both sides by “2”, it becomes<\/p>\n $2\\left(a^2+b^2+c^2\\right)=2\\left(ab+bc+ca\\right)$<\/p>\n $2a^2+2b^2+2c^2=2ab+2bc+2ca$<\/p>\n $a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0$<\/p>\n $\\left(a^2+b^2-2ab\\right)+\\left(b^2+c^2-2ca\\right)+\\left(c^2+a^2-2ca\\right)=0$<\/p>\n $\\left(a-b\\right)^2+\\left(b-c\\right)^2+\\left(c-a\\right)^2=0$<\/p>\n Sum of squares is zero so each term should be zero<\/p>\n $=$>\u00a0\u00a0 \u00a0$\\left(a-b\\right)^2=0$, $\\left(b-c\\right)^2=0$, $\\left(c-a\\right)^2=0$<\/p>\n $=$> \u00a0 \u00a0 \u00a0$a-b=0$, \u00a0 \u00a0 \u00a0 $b-c=0$, \u00a0 \u00a0 \u00a0 $c-a=0,$<\/p>\n $=$> \u00a0 \u00a0\u00a0 \u00a0 \u00a0$a=b$,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0$b=c$, \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0$c=a$<\/p>\n $=$>\u00a0 \u00a0\u00a0 \u00a0 \u00a0 $a=b=c$<\/p>\n Therefore \u00a0 \u00a0\u00a0$\\ \\frac{\\ a+b}{c}=\\ \\frac{\\ a+a}{a}=\\ \\frac{\\ 2a}{a}=2$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/295231_HVVEkWJ.png\")
By what percentage is the total revenue of the company from the sale of products A, B and D in 2012 and 2013 more than the total revenue from the sale of product B in 2013 to 2016?(Correct to one decimal place)<\/p>\n
\n(a + b + c)(ab + bc + ca) – abc = ?<\/p>\n