Get 5 TISSNET 2023 Mocks At just Rs. 299 <\/a><\/p>\n Question 1:\u00a0<\/b>If a – b : b – c : c – d = 1 : 2 : 3, then what is the value of (a + d) : c?<\/p>\n a)\u00a01 : 2<\/p>\n b)\u00a02 : 1<\/p>\n c)\u00a04 : 1<\/p>\n d)\u00a03 : 1<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given = $(a-b):(b-c):(c-d)=1:2:3$<\/p>\n => $\\frac{a-b}{b-c}=\\frac{1}{2}$<\/p>\n => $2a-2b=b-c$<\/p>\n => $2a+c=3b$ ———–(i)<\/p>\n Similarly, $\\frac{b-c}{c-d}=\\frac{2}{3}$<\/p>\n => $3b-3c=2c-2d$<\/p>\n => $3b=5c-2d$ ———-(ii)<\/p>\n Substituting above value in equation (i), we get\u00a0:<\/p>\n => $2a+c=5c-2d$<\/p>\n => $2a+2d=5c-c$<\/p>\n => $2(a+d)=4c$<\/p>\n => $\\frac{a+d}{c}=\\frac{4}{2}=2$<\/p>\n $\\therefore$ $(a+d):c=2:1$<\/p>\n => Ans – (B)<\/p>\n Question 2:\u00a0<\/b>What is the value of\u00a0$\\frac{(0.7)^{3}-(0.4)^{3}}{(0.7)^{2}+\\ 0.7\\times0.4+(0.4)^{2}}\\ $?<\/p>\n a)\u00a00.3<\/p>\n b)\u00a00.4<\/p>\n c)\u00a00.7<\/p>\n d)\u00a01.1<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression =\u00a0$\\frac{(0.7)^{3}-(0.4)^{3}}{(0.7)^{2}+\\ 0.7\\times0.4+(0.4)^{2}}\\ $<\/p>\n Let $x=0.7$ and $y=0.4$<\/p>\n = $\\frac{x^3-y^3}{x^2+xy+y^2}$<\/p>\n =\u00a0$\\frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2}$<\/p>\n = $x-y=0.7-0.4=0.3$<\/p>\n => Ans – (A)<\/p>\n Question 3:\u00a0<\/b>What is the value of\u00a0$3^{2}+7^{2}+13^{2}+17^{2}-1^{2}-5^{2}-9^{2}-11^{2}-15^{2}$?<\/p>\n a)\u00a05<\/p>\n b)\u00a072<\/p>\n c)\u00a092<\/p>\n d)\u00a063<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression =\u00a0$3^{2}+7^{2}+13^{2}+17^{2}-1^{2}-5^{2}-9^{2}-11^{2}-15^{2}$<\/p>\n = $(9+49+169+289)-(1+25+81+121+225)$<\/p>\n = $516-453=63$<\/p>\n => Ans – (D)<\/p>\n Question 4:\u00a0<\/b>If P = (\u221a7 – \u221a6)\/(\u221a7 + \u221a6), then what is the value of P + (1\/P)?<\/p>\n a)\u00a012<\/p>\n b)\u00a013<\/p>\n c)\u00a024<\/p>\n d)\u00a026<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given\u00a0:\u00a0$P=\\frac{\\sqrt7-\\sqrt6}{\\sqrt7+\\sqrt6}$ —————(i)<\/p>\n =>\u00a0$\\frac{1}{P}=\\frac{\\sqrt7+\\sqrt6}{\\sqrt7-\\sqrt6}$ ———-(ii)<\/p>\n Adding equations (i) and (ii),<\/p>\n => $P+\\frac{1}{P}=(\\frac{\\sqrt7-\\sqrt6}{\\sqrt7+\\sqrt6})+(\\frac{\\sqrt7+\\sqrt6}{\\sqrt7-\\sqrt6})$<\/p>\n = $\\frac{(\\sqrt7-\\sqrt6)^2+(\\sqrt7+\\sqrt6)^2}{(\\sqrt7+\\sqrt6)(\\sqrt7-\\sqrt6)}$<\/p>\n = $\\frac{(13-2\\sqrt{42})+(13+2\\sqrt{42})}{7-6}$<\/p>\n = $13+13=26$<\/p>\n => Ans – (D)<\/p>\n Question 5:\u00a0<\/b>$If\\ \\frac{cos\\ \\theta}{1+sin\\ \\theta}+ \\frac{cos\\ \\theta}{1-sin\\ \\theta}=2\\sqrt{2}\\ and\\ \\theta\\ $is acute, then what is the value (in degrees) of $\\theta$?<\/p>\n a)\u00a030<\/p>\n b)\u00a045<\/p>\n c)\u00a060<\/p>\n d)\u00a090<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n $ \\frac{cos\\ \\theta}{1+sin\\ \\theta}+ \\frac{cos\\ \\theta}{1-sin\\ \\theta}=2\\sqrt{2}\\ $<\/p>\n $\\cos\\theta(\\frac{1}{1+sin\\ \\theta}+ \\frac{1}{1-sin\\ \\theta})=2\\sqrt{2}\\ $<\/p>\n $\\cos\\theta (\\frac{1 – \\sin\\theta + 1 + \\sin\\theta}{1^{2} – \\sin^{2}\\theta}) = 2\\sqrt{2}$<\/p>\n $\\cos\\theta (\\frac{2}{\\cos^{2}\\theta}) = 2\\sqrt{2}$<\/p>\n $\\cos\\theta = \\frac{1}{\\sqrt{2}}$<\/p>\n $\\theta = 45^{\\circ}$<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 6:\u00a0<\/b>If cot\u00a0$\\theta=\\sqrt{11}\\ and\\ \\theta\\ $is acute, then what is the value of $(\\frac{cosec^{2}\\ \\theta\\ +\\ sec^{2}\\ \\theta}{cosec^{2}\\ \\theta\\ -\\ sec^{2}\\ \\theta})$?<\/p>\n a)\u00a02\/3<\/p>\n b)\u00a06\/5<\/p>\n c)\u00a03\/4<\/p>\n d)\u00a07\/6<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, cot\u00a0$\\theta=\\sqrt{11}$<\/p>\n $(\\frac{cosec^{2}\\ \\theta\\ +\\ sec^{2}\\ \\theta}{cosec^{2}\\ \\theta\\ -\\ sec^{2}\\ \\theta})$ =\u00a0$(\\frac{1 + cot^{2}\\ \\theta\\ +\\ 1 + tan^{2}\\ \\theta}{1 + cot^{2}\\ \\theta\\ -\\ 1 + tan^{2}\\ \\theta})$<\/p>\n $\\Rightarrow (\\frac{1 + 11\\ +\\ 1 + \\frac{1}{11}}{1 + 11\\ -\\ 1 + \\frac{1}{11}})$\u00a0= $(\\frac{\\frac{144}{11}}{\\frac{120}{11}})$\u00a0= $\\frac{144}{20}$<\/p>\n $\\Rightarrow \\frac{6}{5}$<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 7:\u00a0<\/b>If sin \u03b8 + $sin^{2}$ \u03b8 = 1, then what is the value of $(cos^{12}\\ \\theta\\ + 3\\ cos^{10}\\ \\theta+3\\ cos^{8}\\ \\theta\\ + cos^{6}\\ \\theta -1)$<\/p>\n a)\u00a0-1<\/p>\n b)\u00a00<\/p>\n c)\u00a01<\/p>\n d)\u00a02<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given $sin \\theta + sin^{2} \\theta = 1$<\/p>\n $\\Rightarrow sin$ \u03b8 = 1 – $sin^{2}$ \u03b8 $\\Rightarrow$ $sin$ \u03b8 = $cos^{2}$ \u03b8 …..(1)<\/p>\n Now, $(cos^{12}\\ \\theta\\ + 3\\ cos^{10}\\ \\theta+3\\ cos^{8}\\ \\theta\\ + cos^{6}\\ \\theta -1)$<\/p>\n $(cos^{4} \\theta$\u00a0+ $cos^{2} \\theta)^{3}$ – 1<\/p>\n Substitute equation (1) in the above equation<\/p>\n $(sin^{2} \\theta$\u00a0+ $cos^{2} \\theta)^{2}$ – 1<\/p>\n 1 – 1 = 0<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 8:\u00a0<\/b>If\u00a0$(1+ tan^{2} \\theta) = \\frac{625}{49}\\ and\\ \\theta\\ $is acute, then what is the value of $\\ (\\sqrt{sin \\theta+cos\\ \\theta})\\ $?<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\frac{5}{4}\\sqrt{\\frac{31}{42}}$<\/p>\n c)\u00a0$\\frac{\\sqrt{31}}{5}$<\/p>\n d)\u00a05\/7<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n $(1+ tan^{2}\\theta)$ = $\\frac{625}{49}$\u00a0 \u00a0(or) $sec^{2}\\theta$ =\u00a0$\\frac{625}{49}$<\/p>\n $sec\\ \\theta$ =\u00a0$\\frac{25}{7}$<\/p>\n $Cos\\ \\theta = \\frac{7}{25}$ and $sin\\ \\theta = \\frac{24}{25}$<\/p>\n $\\ \\sqrt{sin \\theta+cos\\ \\theta}\\ = \\sqrt{\\frac{7 + 24}{25}}$ = $\\sqrt{\\frac{31}{25}}$\u00a0 =\u00a0 $\\frac{\\sqrt{31}}{5}$<\/p>\n Hence, option C is the correct answer.<\/p>\n Question 9:\u00a0<\/b>If sec$\\theta=\\frac{13}{12}\\ and\\ \\theta\\ $is acute, then what is the value of\u00a0$(\\sqrt{cot\\theta+tan \\theta})$?<\/p>\n a)\u00a0$\\frac{13}{2\\sqrt{15}}$<\/p>\n b)\u00a0$\\frac{12}{2\\sqrt{13}}$<\/p>\n c)\u00a0$\\frac{13}{2\\sqrt{5}}$<\/p>\n d)\u00a0$\\frac{2}{13}$<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n sec$\\theta = \\frac{13}{12}$<\/p>\n Then, cot$\\ \\theta = \\frac{12}{5}$ and tan$\\ \\theta = \\frac{5}{12}$<\/p>\n $(\\sqrt{cot\\theta+tan \\theta})$ =\u00a0$(\\sqrt{\\frac{12}{5}+\\frac{5}{12}})$ = \u00a0$(\\sqrt{\\frac{169}{60}})$ =\u00a0$\\frac{13}{2\\sqrt{15}}$<\/p>\n Hence, option A is the correct answer.<\/p>\n Question 10:\u00a0<\/b>In \u0394PQR, \u2220P : \u2220Q : \u2220R = 1 : 3 : 5. What is the value (in degrees) of \u2220R – \u2220P?<\/p>\n a)\u00a030<\/p>\n b)\u00a080<\/p>\n c)\u00a045<\/p>\n d)\u00a060<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let,\u00a0\u2220P = 1x, \u2220Q = 3x, \u2220R = 5x<\/p>\n Sum of interior angles of a triangle = 180 degrees i.e<\/p>\n \u2220P + \u2220Q + \u2220R = 180 degrees<\/p>\n $\\Rightarrow$ 9x = 180 (or) x = 20<\/p>\n \u2220R – \u2220P = 5x – 1x<\/p>\n \u2220R – \u2220P = 4x (or) 4 x 20<\/p>\n \u2220R – \u2220P = 80<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 11:\u00a0<\/b>If x + y + z = 0, then what is the value of\u00a0$\\frac{x^{2}}{yz}+\\frac{y^{2}}{xz}+\\frac{z^{2}}{xy}$?<\/p>\n a)\u00a00<\/p>\n b)\u00a01\/3<\/p>\n c)\u00a01<\/p>\n d)\u00a03<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given x + y + z = 0. So, x + y = -z, y + z = -x, z + x = -y ———–(1)<\/p>\n $\\frac{x^{2}}{yz}+\\frac{y^{2}}{xz}+\\frac{z^{2}}{xy}$<\/p>\n $\\frac{x^{3} + y^{3} + z^{3}}{xyz}$ ———–(2)<\/p>\n We know that, $a^{3} + b^{3} + c^{3} = (a+b+c)^{3} – 3ab(a+b) – 3bc(b+c) – 3ac(a+c) – 6abc$<\/p>\n Hence, equation (2) can be written as,<\/p>\n $\\frac{(x+y+z)^{3} – 3xy(x+y) – 3yz(y+z) – 3zx(x+z) – 6xyz}{xyz}$<\/p>\n Now substitute equation (1) in the above equation,<\/p>\n $\\frac{(0)^{3} + 3xy(z) + 3yz(x) + 3zx(y) – 6xyz}{xyz}$<\/p>\n $\\frac{3xyz}{xyz}$ = 3<\/p>\n Hence, option D is the correct answer.<\/p>\n Question 12:\u00a0<\/b>If$\\ x^{2}+\\frac{1}{x^{2}}=\\frac{7}{4}$ for x > 0, then what is the value of x +\u00a0$\\frac{1}{x}$?<\/p>\n a)\u00a02If<\/p>\n b)\u00a0\u221a15\/2<\/p>\n c)\u00a0\u221a5<\/p>\n d)\u00a0\u221a3<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let,<\/p>\n $(x + \\frac{1}{x})^{2}$ =\u00a0$\\ x^{2}+\\frac{1}{x^{2}} + 2(x)(\\frac{1}{x})$<\/p>\n $(x + \\frac{1}{x})^{2}$ =\u00a0$\\frac{7}{4} + 2$<\/p>\n $(x + \\frac{1}{x})^{2}$ =\u00a0$\\frac{15}{4}$<\/p>\n $(x + \\frac{1}{x})$ =\u00a0$\\frac{\\sqrt{15}}{2}$<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 13:\u00a0<\/b>If$\\frac{1}{x+2}=\\frac{3}{y+3}=\\frac{1331}{z+1331}=\\frac{1}{3}$, then what is the value of\u00a0$\\frac{x}{x+1}+\\frac{4}{y+2}+\\frac{z}{z+2662}?$<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a03\/2<\/p>\n d)\u00a03<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given\u00a0$\\frac{1}{x+2}=\\frac{1}{3}$<\/p>\n We get x = 1 …..(1)<\/p>\n $\\frac{3}{y+3}=\\frac{1}{3}$<\/p>\n we get y = 6 …..(2)<\/p>\n $\\frac{1331}{z+1331}=\\frac{1}{3}$<\/p>\n we get z = 2662 ….(3)<\/p>\n We need to find $\\frac{x}{x+1}+\\frac{3}{y+3}+\\frac{z}{z+2662}$<\/p>\n Substitute equations (1), (2) and (3) in the above equation<\/p>\n = $\\frac{1}{1+1}+\\frac{3}{6+3}+\\frac{2662}{2662+2662}$<\/p>\n = $\\frac{1}{2}+\\frac{3}{9}+\\frac{1}{2}$<\/p>\n = $\\frac{3}{2}$<\/p>\n Hence, option C is the correct answer.<\/p>\n Question 14:\u00a0<\/b>When a = 61, b = 63 and c = 65, then what is the value of a$^{3}$ + b$^{3}$ + c$^{3}$ – 3abc?<\/p>\n a)\u00a01456<\/p>\n b)\u00a02268<\/p>\n c)\u00a04536<\/p>\n d)\u00a05460<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that $(a + b + c)^{3} – 3abc= \\frac{1}{2}(a + b + c)((a – b)^{2} + (b – c)^{2} + (c – a)^{2})$<\/p>\n $\\Rightarrow \\frac{1}{2} (61 + 63 + 65)((61 – 63)^{2}$ + $(63 – 65)^{2}$ + $(65 – 61)^{2})$<\/p>\n $\\Rightarrow \\frac{1}{2} (189)(2^{2} + 2^{2} + 4^{2})$ =\u00a0$ \\frac{1}{2} (189)(24) = 2268$<\/p>\n Hence, option B is the correct answer.<\/p>\n Instructions<\/b><\/p>\n The bar graph given below represents the revenue of a firm for 8 years. All the revenue figures have been shown in terms of Rs crores.<\/p>\n Question 15:\u00a0<\/b>What is the total value of revenue of the firm (in crores Rs) in years 2010, 2011 and 2012?<\/p>\n a)\u00a0910<\/p>\n b)\u00a0930<\/p>\n c)\u00a0950<\/p>\n d)\u00a01020<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total value of revenue of the firm in years 2010, 2011 and 2012 = 260 + 350 + 320 = 930<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 16:\u00a0<\/b>A certain sum becomes Rs 1020 in 5 years and Rs 1200 in 8 years at simple interest. What is the value of principal?<\/p>\n a)\u00a0820<\/p>\n b)\u00a0780<\/p>\n c)\u00a0700<\/p>\n d)\u00a0720<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Simple interest for 3 years = 1200 – 1020 = 180 (or 60 for 1 year)<\/p>\n Then simple interest for 5 years = 300<\/p>\n Sum becomes 1020 in 5 years, hence, principal amount = 1020 – 300 = 720.<\/p>\n Hence, option D is the correct answer.<\/p>\n Question 17:\u00a0<\/b>The price of motorcycle depreciates every year by 8%. If the value of the motorcycle after 2 years will be Rs 84640, then what is the present value (in\u00a0Rs) of the motorcycle?<\/p>\n a)\u00a090000<\/p>\n b)\u00a0102000<\/p>\n c)\u00a0110000<\/p>\n d)\u00a0100000<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that price of motorcycle depreciates every year by 8% then after 2 years,<\/p>\n 92% of 92% of $x = 84,640$<\/p>\n $\\frac{92}{100}$ x $\\frac{92}{100}$ x $x = 84,640$<\/p>\n $x = 100000$<\/p>\n Hence, option D is the correct answer.<\/p>\n Question 18:\u00a0<\/b>P is 20% more than Q and 40% less than R. If value of Q is Rs 150, then what is the value of R (in Rs)?<\/p>\n a)\u00a0300<\/p>\n b)\u00a0320<\/p>\n c)\u00a0220<\/p>\n d)\u00a0250<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given P is 20% more than Q,<\/p>\n P = 120% of Q (or) P = (6\/5) x 150 (Q = 150)<\/p>\n P = 180<\/p>\n P is 40% less than R,<\/p>\n P = 60% of R (or) 180 = (3\/5) x R<\/p>\n R = 300<\/p>\n Hence, option A is the correct answer.<\/p>\n Question 19:\u00a0<\/b>If 2A = 3B, then what is the value of (A + B)\/A?<\/p>\n a)\u00a05\/4<\/p>\n b)\u00a02\/3<\/p>\n c)\u00a05\/2<\/p>\n d)\u00a05\/3<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, 2A = 3B i.e<\/p>\n $\\frac{A}{B} = \\frac{3}{2}$\u00a0(or) $\\frac{B}{A} = \\frac{2}{3}$<\/p>\n Now, $\\frac{A+B}{A} \\Rightarrow (\\frac{A}{A} + \\frac{B}{A}$) $\\Rightarrow (1 + \\frac{2}{3}) = \\frac{5}{3}$<\/p>\n Hence, option D is the correct answer.<\/p>\n Question 20:\u00a0<\/b>What is the value of $\\frac{(0.5^{3})-(0.1)^{3}}{(0.5)^{2}+0.5 \\times 0.1+(0.1)^{2}}$?<\/p>\n a)\u00a00.1<\/p>\n b)\u00a00.4<\/p>\n c)\u00a00.5<\/p>\n d)\u00a00.6<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given $\\frac{(0.5^{3})-(0.1)^{3}}{(0.5)^{2}+0.5 \\times 0.1+(0.1)^{2}}$……(1)<\/p>\n we know that $a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})$……(2)<\/p>\n Substitute (2) in (1)<\/p>\n $\\frac{(0.5-0.1)(0.5)^{2} + 0.5 \\times 0.1 + (0.1)^{2}}{(0.5)^{2}+0.5 \\times 0.1+(0.1)^{2}}$<\/p>\n $0.5 – 0.1 = 0.4$<\/p>\n Hence, option B is the correct answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/41567_IxCqU07.PNG\")
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