SSC CHSL Algebra Set-2 Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra Set-2 questions for SSC CHSL Exam.<\/p>\n
Download Algebra Questions For SSC CHSL Set-2<\/a><\/p>\n\n Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Find the area $(in\u00a0 cm^2)$. If the circumference of a circle is $110 cm$?<\/p>\n a)\u00a0962.5<\/p>\n b)\u00a01024.5<\/p>\n c)\u00a0866.5<\/p>\n d)\u00a01925<\/p>\n Question 2:\u00a0<\/b>Classify the triangle as a type of triangle if the sides of the triangle are 8, 11 and 17 units.<\/p>\n a)\u00a0Isoceles triangle<\/p>\n b)\u00a0Acute angled triangle<\/p>\n c)\u00a0Obtuse angled triangle<\/p>\n d)\u00a0Right angled triangle<\/p>\n Question 3:\u00a0<\/b>Find the\u00a0area$(in\u00a0 cm^2)$\u00a0of a square whose perimeter is 17 cm.<\/p>\n a)\u00a020.0625<\/p>\n b)\u00a014.0625<\/p>\n c)\u00a018.0625<\/p>\n d)\u00a016.0625<\/p>\n Question 4:\u00a0<\/b>find the\u00a0roots of the equation\u00a0$x^2 + 16x – 57 = 0$<\/p>\n a)\u00a0-24, 6<\/p>\n b)\u00a0-14, -2<\/p>\n c)\u00a014, 2<\/p>\n d)\u00a0-19, 3<\/p>\n Question 5:\u00a0<\/b>If $p + \\frac{1}{p} = 3$, then find the value of $p^6 + \\large\\frac{1}{p^6}$.<\/p>\n a)\u00a0322<\/p>\n b)\u00a0348<\/p>\n c)\u00a0329<\/p>\n d)\u00a0342<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>If $px^3 + qx^2 – 19x – 30 = 0$ is completely divided by $x^2 + 5x + 6$, find the value of $(p, q)$.<\/p>\n a)\u00a0(1,0)<\/p>\n b)\u00a0(2,1)<\/p>\n c)\u00a0(1,0)<\/p>\n d)\u00a0(4,1)<\/p>\n Question 7:\u00a0<\/b>If $l^{3} + m^{3}$ = -218 and $lm = -35$ , then what is the value of $l + m$?<\/p>\n a)\u00a0-6<\/p>\n b)\u00a0-2<\/p>\n c)\u00a0-4<\/p>\n d)\u00a0-5<\/p>\n Question 8:\u00a0<\/b>When a number is increased by 138, it becomes 123% of itself. Find the number?<\/p>\n a)\u00a0600<\/p>\n b)\u00a0450<\/p>\n c)\u00a0750<\/p>\n d)\u00a0900<\/p>\n Question 9:\u00a0<\/b>If $l^{3} + m^{3}$ = -316 and $l + m = -4$, then what is the value of $l \\times m$?<\/p>\n a)\u00a0-20<\/p>\n b)\u00a0-21<\/p>\n c)\u00a0-22<\/p>\n d)\u00a0-23<\/p>\n Question 10:\u00a0<\/b>If $3[\\large\\frac{5p}{6} – \\frac{7}{3}\u00a0] $ $ = 2[\\large\\frac{3p}{4} – $ $9]\u00a0$, find $p$ ?<\/p>\n a)\u00a0-13<\/p>\n b)\u00a0-19<\/p>\n c)\u00a0-17<\/p>\n d)\u00a0-11<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>If $p + q = -7$ and $p \\times q = 5$, then what is the value of $p^{3} + q^{3}$ ?<\/p>\n a)\u00a0448<\/p>\n b)\u00a0238<\/p>\n c)\u00a0-448<\/p>\n d)\u00a0-238<\/p>\n Question 12:\u00a0<\/b>Sum of ten times a fraction and its reciprocal is 7. What is the fraction? {It is known that the fraction is less than 0.5}<\/p>\n a)\u00a0$\\large\\frac{1}{3}$<\/p>\n b)\u00a0$\\large\\frac{1}{5}$<\/p>\n c)\u00a0$\\large\\frac{1}{2}$<\/p>\n d)\u00a0$\\large\\frac{1}{7}$<\/p>\n Question 13:\u00a0<\/b>If $\\large\\frac{12p}{7} $ $ – 4= \\large\\frac{4}{7}-\\frac{10p}{14}$ , then find $p$ ?<\/p>\n a)\u00a0$\\large\\frac{32}{17}$<\/p>\n b)\u00a0$\\large\\frac{39}{17}$<\/p>\n c)\u00a0$\\large\\frac{28}{17}$<\/p>\n d)\u00a0$\\large\\frac{27}{17}$<\/p>\n Question 14:\u00a0<\/b>If the roots of the equation $x^{2} + 10x \u00ad-119=0$ are p & q, then find the difference between the roots.<\/p>\n a)\u00a0$-10$<\/p>\n b)\u00a0$10$<\/p>\n c)\u00a0$27$<\/p>\n d)\u00a0$24$<\/p>\n Question 15:\u00a0<\/b>Find the value of $(36)^\\frac{3}{2}$ $\\times$ $(1024)^\\frac{2}{5}$ $\\times$ $(343)^\\frac{1}{3}$<\/p>\n a)\u00a024192<\/p>\n b)\u00a024882<\/p>\n c)\u00a024912<\/p>\n d)\u00a024921<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>The square root of $27-4\\sqrt{35}$ is :<\/p>\n a)\u00a0$\\pm(\\sqrt{5}+2\\sqrt{7})$<\/p>\n b)\u00a0$\\pm(\\sqrt{5}-2\\sqrt{7})$<\/p>\n c)\u00a0$\\pm(\\sqrt{7}-2\\sqrt{5})$<\/p>\n d)\u00a0$\\pm(\\sqrt{7}+2\\sqrt{5})$<\/p>\n Question 17:\u00a0<\/b>If $n-\\large\\frac{1}{n}$ $=7$, find the value of $\\large\\frac{n^{2}}{n^{4}+49n^2 +1}$ is<\/p>\n a)\u00a0$\\large\\frac{1}{60}$<\/p>\n b)\u00a0$\\large\\frac{1}{70}$<\/p>\n c)\u00a0$\\large\\frac{1}{100}$<\/p>\n d)\u00a0$\\large\\frac{1}{50}$<\/p>\n Question 18:\u00a0<\/b>If $n-5=2\\sqrt{6}$, then the value of $\\large\\frac{n^{4}+1}{n^{2}}$ is<\/p>\n a)\u00a098<\/p>\n b)\u00a047<\/p>\n c)\u00a023<\/p>\n d)\u00a079<\/p>\n Question 19:\u00a0<\/b>If $9^{0.20} \\times 3^{1.2} \\times 9^{0.4} \\times 3^{0.6} = 81^{n}$. Then find $n$.<\/p>\n a)\u00a00.4<\/p>\n b)\u00a00.6<\/p>\n c)\u00a00.75<\/p>\n d)\u00a00.45<\/p>\n Question 20:\u00a0<\/b>If $\\large\\frac{p}{q}+\\frac{q}{p}$ $=-1$, then the value of $p^{3}-q^{3}$ is<\/p>\n a)\u00a064<\/p>\n b)\u00a027<\/p>\n c)\u00a0100<\/p>\n d)\u00a00<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n SSC CHSL Important Q&A PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, the circumference of a circle is 110 cm<\/p>\n Let radius of circle = $r$ cm<\/p>\n => Circumference = $2\\pi r=110$<\/p>\n => $2\\times\\large\\frac{22}{7}\\times$ $ r=110$<\/p>\n => $r=110\\times\\large\\frac{7}{44}$<\/p>\n => $r=\\large\\large\\frac{35}{2}$ $=17.5$ cm<\/p>\n $\\therefore$ Area of circle = $\\pi r^2$<\/p>\n = $\\large\\frac{22}{7}$ $\\times(17.5)^2$\u00a0$cm^2$<\/p>\n =\u00a0 $962.5$<\/p>\n => Ans – (A)<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the sides of $\\triangle$ ABC be $a,b,c$, where the largest side = $’c’$<\/p>\n If $c^2=a^2+b^2$, then the angle at $C$ is right angle.<\/p>\n If $c^2<a^2+b^2$, then the angle at $C$ is acute angle.<\/p>\n If $c^2>a^2+b^2$, then the angle at $C$ is obtuse angle.<\/p>\n Given, sides of the triangle are 8, 11 and 17\u00a0units<\/p>\n Now, according to ques, => $17^2=289$<\/p>\n and $8^2+11^2=64+121=185$<\/p>\n $\\therefore c^2>a^2+b^2$, hence it is an obtuse angled triangle.<\/p>\n => Ans – (C)<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let side of square = $s$ cm<\/p>\n =>Given, Perimeter = $4s=17$<\/p>\n => $s=\\large\\frac{17}{4}$ $=4.25$ cm<\/p>\n $\\therefore$ Area = $(4.25)^2=18.0625$ $cm^2$<\/p>\n => Ans –\u00a0(C)<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, equation\u00a0$x^2 + 16x – 57 = 0$<\/p>\n This can be written as,<\/p>\n $x^2 + 19x – 3x – 57 = 0$<\/p>\n (x+19)(x-3)=0<\/p>\n x = -19 or x = 3<\/p>\n Hence, option A is the right answer.<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, Squaring both sides,<\/p>\n $p^2$ $+$ $\\large\\frac{1}{p^2}$ $+$ $2 \\times p \\times \\large\\frac{1}{p}$ $=$ $9$<\/p>\n $p^2 + \\large\\frac{1}{p^2}$ $=$ $7$<\/p>\n Cubing on both sides,<\/p>\n $p^6 + \\large\\frac{1}{p^6}$ $+ 3\u00a0\\times p^2 \\times\u00a0\\large\\frac{1}{p^2}(p^2 + \\large\\frac{1}{p^2}) $ $= 343$<\/p>\n $p^6 + \\large\\frac{1}{p^6}$ $+ 3 (p^2 + \\large\\frac{1}{p^2})$ $=343$<\/p>\n $p^6 + \\large\\frac{1}{p^6}$ $+ 3 (7)$= $343$<\/p>\n $p^6 + \\large\\frac{1}{p^6}$\u00a0 $=$ $343-21$<\/p>\n $p^6 + \\large\\frac{1}{p^6}$\u00a0 $=$ $322$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, $px^3 + qx^2 – 19x – 30 = 0$ is divisible by $x^2 + 5x + 6$<\/p>\n $x^2 + 5x + 6$ can be factorized as $(x + 2)(x + 3)$<\/p>\n x = -2 and x = -3 should satisfy\u00a0$px^3 + qx^2 – 19x – 30 = 0$.<\/p>\n Substituting the values,<\/p>\n When, $x = -2$<\/p>\n $-8p + 4q + 8 = 0$ or $2p – q – 2 = 0$<\/p>\n When, $x = -3$<\/p>\n $-27p + 9q + 27 = 0$ or $3p – q – 3 = 0$<\/p>\n Subtracting both, $p = 1$<\/p>\n Substituting the value of $p = 1$, we get $q = 0$<\/p>\n Hence,\u00a0$(p, q)$=\u00a0$(1, 0)$.<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, $l^{3} + m^{3}$ = -218 and $lm = -35$<\/p>\n $(l+m)^{3} = l^{3}+ m^{3} + 3lm(l+m)$<\/p>\n $(l+m)^{3} = -218 + 3(-35)(l+m)$<\/p>\n $(l+m)^{3} = -218 -105(l+m)$<\/p>\n we need to solve 3rd degree equation<\/p>\n so,without solving, by verification, we get $l+m = -2$,<\/p>\n so the answer is option B.<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n When the number is increased by $138$, it becomes $123$% of itself. Let the number be ‘n’<\/p>\n n + $138$ = $123$% of n<\/p>\n $138$ = $\\large\\frac{123}{100}$n – n<\/p>\n $138$ = n$(\\large\\frac{123}{100} – 1)$<\/p>\n $132$ = $\\large\\frac{23}{100}$<\/p>\n x = $600$.<\/p>\n Hence, option A is the correct answer.<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given,<\/p>\n $l^{3} + m^{3}$ = -316 and $l + m = -4$<\/p>\n We know that, $l^{3}+m^{3}=(l+m)^{3}-3lm(l+m)$<\/p>\n $-316=(-4)^{3}-3lm(-4)$<\/p>\n $-316=-64+12lm$<\/p>\n $12lm=-252$<\/p>\n $lm=-21$<\/p>\n So the answer is option B.<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,<\/p>\n $3[\\large\\frac{5p}{6} –\u00a0\\frac{7}{3}\u00a0] $ $ = 2[\\large\\frac{3p}{4} -$ $ 9]\u00a0$<\/p>\n $\\large\\frac{5p}{2} – $ $7 $ $ = [\\large\\frac{3p}{2} -$ $ 18]\u00a0$<\/p>\n $\\large\\frac{5p}{2} – \\frac{3p}{2} $ $ = 7 – 18 $<\/p>\n $\\large\\frac{5p-3p}{2}\u00a0$ $ = -11 $<\/p>\n $p = -11 $<\/p>\n GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, $p + q = -7$ and $p \\times q = 5$<\/p>\n $p^{3} + q^{3} = (p+q)^{3} – 3pq(p+q)$<\/p>\n $ = (-7)^{3} – 3(5)(-7)$<\/p>\n $ = -343 + 105 = -238$<\/p>\n So the answer is option D.<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let that fraction be $\\large\\frac{1}{f}$<\/p>\n $ 10(\\large\\frac{1}{f})$ $ + f = 7$<\/p>\n $\\Rightarrow (10+f^{2}) = 7 \\times f$<\/p>\n $\\Rightarrow f^{2}-7f+10 = 0$<\/p>\n $\\Rightarrow(f-2)(f-5)=0$<\/p>\n $\\Rightarrow f = 2$ or $5$<\/p>\n $\\Rightarrow \\text{fraction} = \\large\\frac{1}{x}$ $ = 1\/2 \\text{ or } 1\/5$<\/p>\n {Since, It is known that the fraction is less than 0.5} i.e. $ f<\\large\\frac{1}{2}$<\/p>\n $\\therefore f =\u00a0\\large\\frac{1}{5}$<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\large\\frac{12p}{7} $ $ – 4=\u00a0\\large\\frac{4}{7}-\\frac{10p}{14}$<\/p>\n $\\large\\frac{12p}{7} $ $ – 4=\u00a0\\large\\frac{4}{7}-\\frac{5p}{7}$<\/p>\n $\\frac{12p}{7} $ $ +\\large\\frac{5p}{7}=\u00a0\\frac{4}{7} + 4$<\/p>\n $\\large\\frac{12p+5p}{7} $ $ =\u00a0\\large\\frac{4+28}{7} $<\/p>\n $\\large\\frac{17p}{7} $ $ =\\large\u00a0\\frac{32}{7} $<\/p>\n $17p = 32$<\/p>\n $p=\\large\\frac{32}{17}$<\/p>\n <\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given Expression: $x^2 + 10x – 119$<\/p>\n = $x^2 + 17x – 7x – 119$<\/p>\n = $x(x + 17) – 7(x + 17)$<\/p>\n = $(x + 17)(x – 7)$<\/p>\n = $(x = -17 or 7)$<\/p>\n Therefore the roots are 7 & -17<\/p>\n The\u00a0difference between the roots is 24(p ~ q)<\/p>\n => Ans – (D)<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $(36)^\\frac{3}{2} \\times (1024)^\\frac{2}{5}\\times (343)^\\frac{1}{3}$<\/p>\n = $(6)^{2\\times\\large\\frac{3}{2}} \\times$ $ (2)^{10\\times\\large\\frac{2}{5}}$ $\\times (7)^{3\\times\\large\\frac{1}{3}}$<\/p>\n = $216\\times16\\times7$<\/p>\n = $24192$<\/p>\n <\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We have to find $27-4\\sqrt{35}$<\/p>\n We can write it as :<\/p>\n = $\\sqrt{{27} – 2\u00a0\\times 2\\times \\sqrt{5}\u00a0\\times \\sqrt{7}}$<\/p>\n Since, $(a^2 + b^2 – 2ab) = (a-b)^2$<\/p>\n = $\\sqrt{(2\\sqrt{5})^2 + (\\sqrt{7})^2 – 2\\times2\\sqrt{5}\\times\\sqrt{7}}$<\/p>\n = $\\sqrt{(\\sqrt{7} – 2\\sqrt{5})^2}$<\/p>\n = $\\pm(\\sqrt{7} – 2\\sqrt{5})$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\large\\frac{n^{2}}{n^{4}+49n^2 +1}$<\/p>\n Given,\u00a0$n-\\large\\frac{1}{n}$ $=7$<\/p>\n Squaring on both sides,<\/p>\n $(n-\\large\\frac{1}{n})^2$ $ =7^2$<\/p>\n $(n^2 – 2 +\\large\\frac{1}{n^2})$ $ =7^2$<\/p>\n $(n^2 +\\large\\frac{1}{n^2})$ $ = 49+2$<\/p>\n $(n^2 +\\large\\frac{1}{n^2})$ $ = 51$<\/p>\n $(n^4 +1)$ $ = 51n^2$<\/p>\n Adding $49n^2$ on both sides<\/p>\n $n^4 +1$ $ + 49n^2$ $ = 51n^2 \u00a0+ 49n^2$<\/p>\n $n^4 + 49n^2 + 1$\u00a0$ = 100n^2 $<\/p>\n $\\large\\frac{n^4 + 49n^2 + 1}{n^2}$ $= 100$<\/p>\n $\\large\\frac{n^{2}}{n^{4}+49n^2 +1}$\u00a0$=\\large\\frac{1}{100}$<\/p>\n <\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given,<\/p>\n $n-5=2\\sqrt{6}$<\/p>\n $n= 5+2\\sqrt{6}$<\/p>\n $\\large\\frac{1}{n}=\\frac{1}{5+2\\sqrt{6}}$<\/p>\n $\\large\\frac{1}{n}=\\frac{1}{5+2\\sqrt{6}}$ $\\times\\large\\frac{5-2\\sqrt{6}}{5-2\\sqrt{6}}$<\/p>\n $\\large\\frac{1}{n}$ $=\u00a05-2\\sqrt{6}$<\/p>\n $(n+\\large\\frac{1}{n}) $ $= 10$<\/p>\n $(n+\\large\\frac{1}{n})^{2} $ $= 10^2$<\/p>\n $n^{2}+\\large\\frac{1}{n^{2}}$ $ + 2 = 100$<\/p>\n $n^{2}+\\large\\frac{1}{n^{2}} $ $= 100-2 = 98$<\/p>\n $\\large\\frac{n^{4}+1}{n^{2}}$ = $98$<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $9^{0.20} \\times 3^{1.2} \\times 9^{0.4} \\times 3^{0.6} = 81^{n}$<\/p>\n $3^{0.40} \\times 3^{1.2} \\times 3^{0.8} \\times 3^{0.6} = 3^{4n}$<\/p>\n $3^{0.40+1.2+0.8+0.6} = 3^{4n}$<\/p>\n $3^{0.40+1.2+0.8+0.6} = 3^{4n}$<\/p>\n $3^{3} = 3^{4n}$<\/p>\n $3 = 4n$<\/p>\n $n=0.75$<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given,<\/p>\n $\\large\\frac{p}{q}+\\frac{q}{p}$ $=-1$<\/p>\n $\\large\\frac{p^{2}+q^{2}}{pq}$ $ = -1$<\/p>\n $p^{2}+q^{2} = -pq$<\/p>\n $p^{2}+q^{2} + pq = 0$<\/p>\n We know $p^{3}-q^{3}={(p-q)}{(p^{2}+q^{2}}{+pq)} $<\/p>\n As $p^{2}-q^{2} + pq = 0$, therefore $p^{3}-q^{3}={(p-q)}{(p^{2}+q^{2}}{+pq)}=0$<\/p>\n\n
\n$p + \\large\\frac{1}{p}$ $=$ $3$<\/p>\n