Download Algebra Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $a^{2}+b^{2}-c^{2}=0$, then the value of $\\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$ is:<\/p>\n a)\u00a03<\/p>\n b)\u00a01<\/p>\n c)\u00a00<\/p>\n d)\u00a02<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n If a + b + c\u00a0 = 0 then\u00a0$a^3 + b^3 + c^3 = 3abc$ so,<\/p>\n $a^{6}+b^{6}-c^{6} = 3a^2b^2c^2$<\/p>\n $\\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$<\/p>\n =\u00a0$\\frac{2(3a^{2}b^{2}c^{2})}{3a^{2}b^{2}c^{2}}$ = 2<\/p>\n Question 2:\u00a0<\/b>If $a+\\frac{1}{a}=5$ then $a^{3} + \\frac{1}{a^{3}}$ is:<\/p>\n a)\u00a0110<\/p>\n b)\u00a010<\/p>\n c)\u00a080<\/p>\n d)\u00a0140<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $a^{3} + \\frac{1}{a^{3}}$<\/p>\n = $($a+\\frac{1}{a})^3 – 3(a+\\frac{1}{a})$<\/p>\n ($\\because (a + b)^3 = a^3 + b^3 + 3ab(a + b))$<\/p>\n = $5^3 – 3(5)$ = 110<\/p>\n Question 3:\u00a0<\/b>The coefficient of x in $(x – 3y)^{3}$ is :<\/p>\n a)\u00a0$3 y^{2}$<\/p>\n b)\u00a0$27 y^{2}$<\/p>\n c)\u00a0$-27 y^{2}$<\/p>\n d)\u00a0$-3 y^{2}$<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(x – 3y)^{3} = x^3 – (3y)^3 – 3x.3y(x – 3y)$<\/p>\n ($(a – b)^3 = a^3 – b^3 – 3ab(a – b)$)<\/p>\n = $x^3 – 27y^3 – 9xy(x – 3y)$<\/p>\n = $x^3 – 27y^3 – 9x^2y – 27xy^2$<\/p>\n The coefficient of x = 27$y^2$<\/p>\n Question 4:\u00a0<\/b>Expand $\\left(\\frac{x}{3} + \\frac{y}{5}\\right)^3$<\/p>\n a)\u00a0$\\frac{x^3}{27} + \\frac{x^2y}{25} + \\frac{xy^2}{25} + \\frac{y^3}{125}$<\/p>\n b)\u00a0$\\frac{x^3}{25} + \\frac{x^2y}{15} + \\frac{xy^2}{25} + \\frac{y^3}{125}$<\/p>\n c)\u00a0$\\frac{x^3}{27} + \\frac{xy}{15} + \\frac{xy^2}{25} + \\frac{y^3}{125}$<\/p>\n d)\u00a0$\\frac{x^3}{27} + \\frac{x^2y}{15} + \\frac{xy^2}{25} + \\frac{y^3}{125}$<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left(\\frac{x}{3} + \\frac{y}{5}\\right)^3$<\/p>\n ($\\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$)<\/p>\n =\u00a0 $(\\frac{x}{3})^3 + (\\frac{y}{5})^3 + 3(\\frac{x}{3})(\\frac{y}{5})(\\frac{x}{3} + \\frac{y}{5})\u00a0$<\/p>\n = $\\frac{x^3}{27} + \\frac{y^3}{125} + \\frac{xy}{5} (\\frac{x}{3} + \\frac{y}{5}) $<\/p>\n = $\\frac{x^3}{27} + \\frac{y^3}{125} + \\frac{x^2y}{15} + \\frac{xy^2}{25} $<\/p>\n Question 5:\u00a0<\/b>If $a^2 + b^2 + c^2 = 300$ and $ab + bc + ca = 50$, then what is the value of $a + b + c$ ? (Given that a, b and c are all positive.)<\/p>\n a)\u00a022<\/p>\n b)\u00a020<\/p>\n c)\u00a025<\/p>\n d)\u00a015<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(a + b + c)^2 =\u00a0a^2 + b^2 + c^2 +\u00a02(ab + bc + ca)$<\/p>\n $(a + b + c)^2 = 300 + 2(50)$<\/p>\n $(a + b + c)^2 = 400$<\/p>\n a + b + c = 20<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>If x + y + z = 10 and xy + yz + zx = 15, then find the value of $x^3 + y^3 + z^3 \u2014 3xyz$.<\/p>\n a)\u00a0660<\/p>\n b)\u00a0525<\/p>\n c)\u00a0550<\/p>\n d)\u00a0575<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^3 + y^3 + z^3 \u2014 3xyz = (x + y + z)(x^2 + y^2 +\u00a0z^2 – xy – yz – xz)$<\/p>\n x + y + z = 10<\/p>\n Taking square on both sides,<\/p>\n $(x + y + z)^2 = 100$<\/p>\n $x^2 + y^2 + z^2 + 2(xy + yz + xz) = 100$<\/p>\n $x^2 + y^2 + z^2 = 100 – 2\\times 15 = 00 – 30 = 70$<\/p>\n $x^3 + y^3 + z^3 \u2014 3xyz = (10)(70 – 15) = 10 \\times 55 = 550$<\/p>\n Question 7:\u00a0<\/b>If $x^{2} – 4x+4=0 $, then the value of 16$(x^{4} – \\frac{1}{x^{4}})$ is<\/p>\n a)\u00a0127<\/p>\n b)\u00a0255<\/p>\n c)\u00a0$\\frac{127}{16}$<\/p>\n d)\u00a0$\\frac{255}{16}$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^{2} \u20144x+4=0 $<\/p>\n $x^{2} \u20142x – 2x+4=0 $<\/p>\n $x(x \u2014 2) – 2(x \u2014 2)=0 $<\/p>\n $(x \u2014 2)(x \u2014 2)=0 $<\/p>\n x = 2<\/p>\n now,<\/p>\n 16$(x^{4}-\\frac{1}{x^{4}})$<\/p>\n =\u00a016$(2^{4}-\\frac{1}{2^{4}})$<\/p>\n = 16$(16-\\frac{1}{16})$<\/p>\n = $16^2 – 1$ = 255<\/p>\n Question 8:\u00a0<\/b>If $a^{3}+\\frac{1}{a^{3}} = 52$ then the value of $2\\left(a + \\frac{1}{a}\\right)$ is :<\/p>\n a)\u00a08<\/p>\n b)\u00a02<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $a^{3}+\\frac{1}{a^{3}} = 52$<\/p>\n $(a + \\frac{1}{a})^3 – 3.a.\\frac{1}{a}(a + \\frac{1}{a}) = 52$<\/p>\n $(\\because a^3 + b^3 = (a + b)^3 – 3ab(a + b))$<\/p>\n $(a + \\frac{1}{a})^3 – 3(a + \\frac{1}{a}) = 52$<\/p>\n From the option A) –<\/p>\n Put the value of $2(a + \\frac{1}{a}) = 8$,<\/p>\n $(a + \\frac{1}{a}) = 4$<\/p>\n L.H.S.,<\/p>\n $4^3 – 3 \\times 4$ = 52<\/p>\n = R.H.S.<\/p>\n $\\therefore$ The value of $2\\left(a + \\frac{1}{a}\\right)$ is 8.<\/p>\n Question 9:\u00a0<\/b>If $b + c = ax, c + a = by, a + b = cz$, then the value $\\frac{1}{9}\\left[\\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}\\right]$ is:<\/p>\n a)\u00a0$\\frac{1}{9}$<\/p>\n b)\u00a01<\/p>\n c)\u00a00<\/p>\n d)\u00a0$\\frac{1}{3}$<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $b + c = ax, c + a = by, a + b = cz$<\/p>\n x =$\\frac{b + c}{a}$<\/p>\n y =$\\frac{c + a}{b}$<\/p>\n z =$\\frac{a + b}{c}$<\/p>\n Now,<\/p>\n $\\frac{1}{9}\\left[\\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}\\right]$<\/p>\n x + 1 =\u00a0$\\frac{b + c}{a}$ + 1 = $\\frac{a + b + c}{a}$<\/p>\n y + 1 = $\\frac{c + a}{b}$ + 1 = $\\frac{a + b + c}{b}$<\/p>\n z + 1 = $\\frac{a + b}{c}$ + 1 = $\\frac{a + b + c}{c}$<\/p>\n =\u00a0$\\frac{1}{9}\\left[\\frac{1}{\\frac{a + b + c}{a}}+\\frac{1}{\\frac{a + b + c}{b}}+\\frac{1}{\\frac{a + b + c}{c}}\\right]$<\/p>\n $\\frac{1}{9}\\left[\\frac{a}{a + b + c}+\\frac{b}{a + b + c}+\\frac{c}{a + b + c}\\right]$<\/p>\n $\\frac{1}{9}[\\frac{a + b + c}{a + b + c}]$ = 1\/9<\/p>\n Question 10:\u00a0<\/b>Find the product of $(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)$<\/p>\n a)\u00a0$a^{3} + b^{3} + 8c^{3} – 2abc$<\/p>\n b)\u00a0$a^{3} + b^{3} + 8c^{3} – abc$<\/p>\n c)\u00a0$a^{3} + b^{3} + 6c^{3} – 6abc$<\/p>\n d)\u00a0$a^{3} + b^{3} + 8c^{3} – 6abc$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)$<\/p>\n $= a^3 + b^3 + (2c)^3 – 3 \\times a \\times b \\times 2c$<\/p>\n $(\\because\u00a0a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca))$<\/p>\n $=\u00a0a^3 + b^3 + 8c^3 – 6abc$<\/p>\n Question 11:\u00a0<\/b>$25a^{2}-9$ is factored as<\/p>\n a)\u00a0$(5a + 3)(5a\u00a0–\u00a03)$<\/p>\n b)\u00a0$(5a + 1)(5a –\u00a09)$<\/p>\n c)\u00a0$(5a – 3)^{2}$<\/p>\n d)\u00a0$(25a +\u00a01)(a -9)$<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $25a^{2}-9$<\/p>\n = $(5a)^2 – (3)^2$ = (5a + 3)(5a – 3)<\/p>\n Question 12:\u00a0<\/b>If $a^{4} + \\frac{1}{a^{4}} = 50$, then find the value of $a^{3} + \\frac{1}{a^{3}}$<\/p>\n a)\u00a0$\\sqrt{2(1+\\sqrt{3})}+(-1+2\\sqrt{13})$<\/p>\n b)\u00a0$\\sqrt{2(1+\\sqrt{3})}(3-2\\sqrt{13})$<\/p>\n c)\u00a0$\\sqrt{2(\\sqrt{13}+1)}(3+2\\sqrt{13})$<\/p>\n d)\u00a0$\\sqrt{2(1-\\sqrt{3})}(-1+2\\sqrt{13})$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $a^{4} + \\frac{1}{a^{4}} = 50$<\/p>\n $a^{4} + \\frac{1}{a^{4}} + 2= 50 + 2$<\/p>\n $(a^2+\\frac{1}{a^2})^2=52$<\/p>\n $(a^2+\\frac{1}{a^2})=\\sqrt{52}$<\/p>\n $a^2+\\frac{1}{a^2} + 2\u00a0= \\sqrt{52} + 2$<\/p>\n $(a + \\frac{1}{a})^2 =\u00a0\\sqrt{52} + 2$<\/p>\n $(a + \\frac{1}{a})\u00a0= \\sqrt{\\sqrt{52} + 2}$<\/p>\n $a^{3} + \\frac{1}{a^{3}} = (a + b)^3 + 3ab(a + b)$<\/p>\n =$(\\sqrt{\\sqrt{52} + 2})^3 +\u00a0\\sqrt{\\sqrt{52} + 2}$<\/p>\n =$(\\sqrt{2\\sqrt{13} + 2})^3 + \\sqrt{2\\sqrt{13} + 2}$<\/p>\n =$\\sqrt{2\\sqrt{13} + 2}(1 +\u00a0(\\sqrt{2\\sqrt{13} + 2})^2)$<\/p>\n =$\\sqrt{2\\sqrt{13} + 2}(1 + {2\\sqrt{13} + 2})$<\/p>\n =$\\sqrt{2(\\sqrt{13} + 1})(3 + {2\\sqrt{13}})$<\/p>\n Question 13:\u00a0<\/b>$(a + b – c + d)^2 – (a – b + c – d)^2 = ?$<\/p>\n a)\u00a0$4a(b + d – c)$<\/p>\n b)\u00a0$2a(a + b – c)$<\/p>\n c)\u00a0$2a(b + c – d)$<\/p>\n d)\u00a0$4a(b – d + c)$<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(a + b – c + d)^2 – (a – b + c – d)^2$<\/p>\n = [(a + b – c + d) +\u00a0(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 = (2a)(2b-2c + 2d)<\/p>\n = 4a(b – c + d)<\/p>\n Question 14:\u00a0<\/b>The value of $27a^3 – 2\\sqrt{2}b^3$ is equal to:<\/p>\n a)\u00a0$(3a – \\sqrt{2}b)(9a^2 – 2b^2 + 6\\sqrt{2}ab)$<\/p>\n b)\u00a0$(3a – \\sqrt{2}b)(9a^2 + 2b^2 + 6\\sqrt{2}ab)$<\/p>\n c)\u00a0$(3a – \\sqrt{2}b)(9a^2 + 2b^2 + 3\\sqrt{2}ab)$<\/p>\n d)\u00a0$(3a – \\sqrt{2}b)(9a^2 – 2b^2 – 3\\sqrt{2}ab)$<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $27a^3 – 2\\sqrt{2}b^3 = (3a – \\sqrt{2}b)(9a^2 + 2b^2 + 6\\sqrt{2}ab)$<\/p>\n ($\\because a^3 – b^3 = (a – b)(a^2 +\u00a0ab + b^2)$)<\/p>\n here,<\/p>\n a = 3a<\/p>\n b = $\\sqrt{2}b$<\/p>\n Question 15:\u00a0<\/b>If $x+3y+2=0$ then value of $x^{3}+27y^{3}+8-18xy$ is:<\/p>\n a)\u00a0-2<\/p>\n b)\u00a02<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x+3y+2=0$<\/p>\n x + 3y = -2<\/p>\n Taking cube both sides,<\/p>\n $(x + 3y)^3 = -8$<\/p>\n $x^3 + 27y^3 + 3x.3y(x +\u00a03y) = -8$<\/p>\n $x^3 + 27y^3 + 9xy(-2) = -8 $<\/p>\n $x^{3}+27y^{3} -18xy = -8$<\/p>\n $x^{3}+27y^{3}+8-18xy$ = 0<\/p>\n Question 16:\u00a0<\/b>If $p+q=7$ and $pq=5$, then the value of $p^{3}+q^{3}$ is:<\/p>\n a)\u00a034<\/p>\n b)\u00a0238<\/p>\n c)\u00a0448<\/p>\n d)\u00a064<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $p^{3}+q^{3} = (p + q)^3 – 3pq(p\u00a0+ q)$<\/p>\n =$7^3 – 3 \\times 5(7)$<\/p>\n = 343 – 105 = 238<\/p>\n Question 17:\u00a0<\/b>If $30x^2 – 15x + 1 = 0$, then what is the value of $25x^2 + (36x^2)^{-1}$?<\/p>\n a)\u00a0$\\frac{9}{2}$<\/p>\n b)\u00a0$6\\frac{1}{4}$<\/p>\n c)\u00a0$\\frac{65}{12}$<\/p>\n d)\u00a0$\\frac{55}{12}$<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $30x^2 – 15x + 1 = 0$<\/p>\n Dividing by x,<\/p>\n $30x – 15 + \\frac{1}{x} = 0$<\/p>\n $5x – 15\/6 +\u00a0\\frac{1}{6x} = 0$<\/p>\n $5x\u00a0 +\u00a0\\frac{1}{6x}\u00a0= 5\/2$<\/p>\n taking square both side,<\/p>\n $(5x + \\frac{1}{6x})^2 = 25\/4$<\/p>\n $25x^2 + \\frac{1}{36x^2} + 2 \\times 5x \\times\u00a0\\frac{1}{6x} = 25\/4$<\/p>\n $25x^2 + \\frac{1}{36x^2} = 25\/4 – 5\/3$<\/p>\n $25x^2 + \\frac{1}{36x^2} = \\frac{55}{12} $<\/p>\n Question 18:\u00a0<\/b>If a + b + c = 7 and ab + bc + ca = -6, then the value of $a^3 + b^3 + c^3 – 3abc$ is:<\/p>\n a)\u00a0469<\/p>\n b)\u00a0472<\/p>\n c)\u00a0463<\/p>\n d)\u00a0479<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that,<\/p>\n $a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 +\u00a0b^2 +\u00a0c^2 – (ab + bc + ac))$<\/p>\n a + b + c = 7<\/p>\n Squaring both sides,<\/p>\n $(a + b + c)^2 = 49$<\/p>\n $a^2 + b^2 + c^2 + 2(ab + bc + ac) = 49$<\/p>\n $a^2 + b^2 + c^2 = 49 + 12 = 61$<\/p>\n $a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – (ab + bc + ac))$<\/p>\n = 7(61 – (-6)) = 7 $\\times$ 67 = 469<\/p>\n Question 19:\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 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Question 20:\u00a0<\/b>If $P = \\frac{x^4 – 8x}{x^3 – x^2 – 2x}, Q = \\frac{x^2 + 2x + 1}{x^2 – 4x – 5}$ and $R = \\frac{2x^2 + 4x + 8}{x – 5}$, then $(P \\times Q) \\div R$ is equal to:<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a04<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $P = \\frac{x^4 – 8x}{x^3 – x^2 – 2x}$<\/p>\n $Q = \\frac{x^2 + 2x + 1}{x^2 – 4x – 5}$<\/p>\n =\u00a0$\\frac{(x +\u00a01)^2}{x^2 – 4x – 5 + 9 – 9}$<\/p>\n $(P \\times Q) \\div R$<\/p>\n = $(\\frac{x^4 – 8x}{x^3 – x^2 – 2x} \\times \\frac{x^2 + 2x + 1}{x^2 – 4x – 5}) \\div \\frac{2x^2 + 4x + 8}{x – 5}$<\/p>\n = $\\frac{x^4 – 8x}{x^3 – x^2 – 2x} \\times \\frac{x^2 + 2x + 1}{x^2 – 4x – 5} \\times \\frac{x – 5}{2x^2 + 4x + 8}$<\/p>\n = $\\frac{x(x^3 – 8)}{x^3 – x^2 – 2x} \\times \\frac{x^2 + 2x + 1}{x^2 – 4x – 5} \\times \\frac{x – 5}{2(x^2 + 2x + 4)}$<\/p>\n = $\\frac{x(x – 2)(x^2 + 2x – 4)}{x(x^2 – x – 2)} \\times \\frac{(x + 1)^2}{x^2 – 5x + x – 5} \\times \\frac{x – 5}{2(x^2 + 2x + 4)}$<\/p>\n = $\\frac{(x – 2)}{(x^2 – 2x + x- 2)} \\times \\frac{(x + 1)^2}{(x +1)(x – 5)} \\times \\frac{x – 5}{2}$<\/p>\n = $\\frac{(x – 2)}{(x – 2)(x + 1)} \\times \\frac{(x + 1)}{2}$<\/p>\n = $\\frac{1}{2}$<\/p>\n
\n= ($\\because a^2 – b^2 = (a + b)(a – b)$)<\/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