SSC Stenographer Constable Linear Equation Question and Answers download PDF based on previous year question paper of SSC Stenographer exam. 20 Very important Linear Equation questions for Stenographer Constable.<\/p>\n
LINEAR EQUATION QUESTIONS FOR SSC STENOGRAPHER PDF<\/a><\/p>\n 10 Stenographer Mock Tests – Just Rs. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Download All Important SSC Stenographer Questions PDF<\/a> (Topic-Wise)<\/p>\n SSC Stenographer Free Mock Test<\/a> (Latest Pattern)<\/p>\n SSC Stenographer Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Two linear equations 3x+5y=20, 24x+40y=K will have no solution for x and y when k = ?<\/p>\n a)\u00a0160<\/p>\n b)\u00a08<\/p>\n c)\u00a040<\/p>\n d)\u00a0both b and c<\/p>\n Question 2:\u00a0<\/b>x+3y=12 and 5x+15y=3k are two linear equations then for what value of k will the equations have infinite solutions ?<\/p>\n a)\u00a00<\/p>\n b)\u00a04<\/p>\n c)\u00a020<\/p>\n d)\u00a012<\/p>\n Question 3:\u00a0<\/b>if (2, 0) is a solution of the linear equation 2x+3y=K, then the value of K is<\/p>\n a)\u00a05<\/p>\n b)\u00a02<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n Question 4:\u00a0<\/b>What is the x – intercept of the linear equation 18x + 25y – 900 = 0?<\/p>\n a)\u00a018<\/p>\n b)\u00a025<\/p>\n c)\u00a050<\/p>\n d)\u00a0450<\/p>\n Question 5:\u00a0<\/b>How many solutions does a pair of linear equations will have, if the equations are a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0Infinite<\/p>\n Download SSC Stenographer Syllabus PDF<\/a><\/p>\n SSC STENO FREE MOCK TEST<\/a><\/p>\n Question 6:\u00a0<\/b>What is the y – intercept of the linear equation 59x + 14y – 112 = 0?<\/p>\n a)\u00a08<\/p>\n b)\u00a014<\/p>\n c)\u00a028<\/p>\n d)\u00a059<\/p>\n Question 7:\u00a0<\/b>If $x^{3} – y^{3} = 112$ and $x – y = 4$, then what is the value of $x^{2} + y^{2}$?<\/p>\n a)\u00a016<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a028<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 8:\u00a0<\/b>If $(1\/x) + (1\/y) + (1\/z) = 0$ and $x + y + z = 11$, then what is the value of $x^{3}+y^{3}+z^{3}-3xyz$ ?<\/p>\n a)\u00a01331<\/p>\n b)\u00a02662<\/p>\n c)\u00a03993<\/p>\n d)\u00a014641<\/p>\n Question 9:\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 Question 10:\u00a0<\/b>If$\\frac{1}{x+2}=\\frac{3}{y+3}=\\frac{1331}{z+1331}=\\frac{1}{3}$, then what is the value of $\\frac{x}{x+1}+\\frac{4}{y+2}+\\frac{z}{z+2662}?$<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a03\/2<\/p>\n d)\u00a03<\/p>\n Question 11:\u00a0<\/b>If x=a(b-c), y=b(c-a), z=c(a-b), then the value of $(\\frac{x}{a})^3\\ +\\ (\\frac{y}{b})^3\\ +\\ (\\frac{z}{c})^3\\ $is:<\/p>\n a)\u00a0$\\frac{xyz}{abc}$<\/p>\n b)\u00a00<\/p>\n c)\u00a0$\\frac{3xyz}{abc}$<\/p>\n d)\u00a0$\\frac{2xyz}{abc}$<\/p>\n Question 12:\u00a0<\/b>In an exam the sum of the scores of A and B is 120, that of B and C is 130 and that of C and A is 140. Then the score of C is<\/p>\n a)\u00a065<\/p>\n b)\u00a060<\/p>\n c)\u00a070<\/p>\n d)\u00a075<\/p>\n Question 13:\u00a0<\/b>If x (x+y+z) = 20, y (x+y+z) = 30, & z(x+y+z)=50, then the value of 2(x+y+z) is:<\/p>\n a)\u00a0-10<\/p>\n b)\u00a015<\/p>\n c)\u00a018<\/p>\n d)\u00a020<\/p>\n Question 14:\u00a0<\/b>If $Cos\\ \\theta + Sin\\ \\theta$ = m and $Sec\\ \\theta + Cosec\\ \\theta$= n then the value of n($m^{2}-1$) is equal to:<\/p>\n a)\u00a02n<\/p>\n b)\u00a04mn<\/p>\n c)\u00a0mn<\/p>\n d)\u00a02m<\/p>\n Question 15:\u00a0<\/b>If $x=a(sin\\ \\theta + cos\\ \\theta)$ and $y=(sin\\ \\theta – cos\\ \\theta)$, then the value of $\\frac{x^{2}}{a^{2}}+\\frac{v^{2}}{b^{2}}$ is:<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a02<\/p>\n d)\u00a01<\/p>\n SSC STENOGRAPHER PREVIOUS PAPERS<\/a><\/p>\n GK Questions and Answers PDF<\/a><\/p>\n Question 16:\u00a0<\/b>A man walks a certain distance in certain time, if he had gone 3 km an hour faster, he would have taken 1 hour less than the schedule time. If he had gone 2 km an hour slower, he would have been one hour longer on the road. The distance (in km) is<\/p>\n a)\u00a045<\/p>\n b)\u00a065<\/p>\n c)\u00a080<\/p>\n d)\u00a060<\/p>\n Question 17:\u00a0<\/b>If $xy=\\frac{a+2}{3}$ and $\\frac{x}{y}=\\frac{1}{3}$, then find the value of $\\frac{x^{2}+y^{2}}{x^{2-}y^{2}}$<\/p>\n a)\u00a01<\/p>\n b)\u00a04<\/p>\n c)\u00a03<\/p>\n d)\u00a0None of these<\/p>\n Question 18:\u00a0<\/b>What is the value of equation $a^3 + b^3 + c^3 – 3abc$ if $a^2 + b^2 + c^2 = ab + bc + ca + 4$ and $a + b + c = 4$<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a016<\/p>\n d)\u00a0256<\/p>\n Question 19:\u00a0<\/b>What is the value of $x^4 + y^4$ when the value of $x^3+y^3 = 8$ and $x + y = 2$?<\/p>\n a)\u00a02<\/p>\n b)\u00a08<\/p>\n c)\u00a016<\/p>\n d)\u00a032<\/p>\n Question 20:\u00a0<\/b>The pair of equations are 7x+8ky-16=0 and 14x+112y-21=0. Find the value of \u2018k\u2019 for which the system is inconsistent.<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a05<\/p>\n d)\u00a07<\/p>\n Stenographer Expected Cut off 2018-19<\/a><\/p>\n SSC Stenographer Salary after 7th Pay Commission<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n For the given two equations 3x+5y=20 and 24x+40y=k 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n x+3y=12$\\rightarrow$ 1 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given (2,0) is a solution of the equation 2x + 3y = k.<\/p>\n Substitute x = 2 and y = 0<\/p>\n k = 2(2) + 3(0) = 4<\/p>\n Hence, option D is the correct answer.<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Equation of line : $18x+25y-900=0$<\/p>\n To find x-intercept, we need to put $y=0$ (and vice-versa)<\/p>\n => $18x+25(0)=900$<\/p>\n => $x=\\frac{900}{18}=50$<\/p>\n => Ans – (C)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Equations : 7x+4y-16=0 and 14x+6y-32=0<\/p>\n Comparing the ratio of coefficients of both equations,<\/p>\n => $\\frac{7}{14}\\neq\\frac{4}{6}$<\/p>\n Thus, there is only 1 solution.<\/p>\n => Ans – (B)<\/p>\n DAILY FREE SSC PRACTICE SET<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Equation = $59x+14y-112=0$<\/p>\n => $14y=-59x+112$<\/p>\n => $y=\\frac{-59x+112}{14}$<\/p>\n => $y=\\frac{-59}{14}x+8$<\/p>\n Comparing above equation with $y=mx+c$, where $c$ is the y-intercept.<\/p>\n => y-intercept = 8<\/p>\n => Ans – (A)<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x^3-y^3=112$ ————–(i)<\/p>\n Also, $x-y=4$ ————-(ii)<\/p>\n Cubing both sides, we get :<\/p>\n => $(x-y)3=(4)^3$<\/p>\n => $(x^3-y^3)-3(x)(y)(x-y)=64$<\/p>\n Substituting values from equations (i) and (ii),<\/p>\n => $112-3xy(4)=64$<\/p>\n => $12xy=112-64=48$<\/p>\n => $xy=\\frac{48}{12}=4$ ———–(iii)<\/p>\n Now, squaring equation (ii), we get :<\/p>\n => $x^2+y^2-2xy=16$<\/p>\n => $x^2+y^2=16+8=24$<\/p>\n => Ans – (C)<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$<\/p>\n => $\\frac{yz+zx+xy}{xyz}=0$<\/p>\n => $xy+yz+zx=0$ ————-(i)<\/p>\n Also, $x+y+z=11$ ————(ii)<\/p>\n Squaring both sides, we get :<\/p>\n => $(x+y+z)^2=(11)^2$<\/p>\n => $(x^2+y^2+z^2)+2(xy+yz+zx)=121$<\/p>\n Substituting value from equation (i),<\/p>\n => $x^2+y^2+z^2=121$ ————(iii)<\/p>\n To find : $x^{3}+y^{3}+z^{3}-3xyz$<\/p>\n = $(x+y+z)[(x^2+y^2+z^2)-(xy+yz+zx)]$<\/p>\n Substituting values from equations (i), (ii) and (iii),<\/p>\n = $(11)(121-0)$<\/p>\n = $11\\times121=1331$<\/p>\n => Ans – (A)<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $P=\\frac{\\sqrt7-\\sqrt6}{\\sqrt7+\\sqrt6}$ —————(i)<\/p>\n => $\\frac{1}{P}=\\frac{\\sqrt7+\\sqrt6}{\\sqrt7-\\sqrt6}$ ———-(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 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $\\frac{1}{x+2}=\\frac{1}{3}$<\/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 FREE MOCK TEST FOR SSC STENOGRAPHER<\/a><\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x=a(b-c)$ , $y=b(c-a)$ , $z=c(a-b)$<\/p>\n => $\\frac{x}{a}=(b-c)$ ————-(i)<\/p>\n and $\\frac{y}{b}=(c-a)$ ————-(ii)<\/p>\n and $\\frac{z}{c}=(a-b)$ ————-(iii)<\/p>\n Adding equations (i), (ii) and (iii), we get :<\/p>\n => $\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}=(b-c)+(c-a)+(a-b)$<\/p>\n => $\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}=0$<\/p>\n Now, we know that if $(p+q+r)=0$, then $p^3+q^3+r^3=3pqr$<\/p>\n $\\therefore$ $(\\frac{x}{a})^3\\ +\\ (\\frac{y}{b})^3\\ +\\ (\\frac{z}{c})^3\\ $<\/p>\n = $3\\times(\\frac{x}{a})\\times(\\frac{y}{b})\\times(\\frac{z}{c})$<\/p>\n = $\\frac{3xyz}{abc}$<\/p>\n => Ans – (C)<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let scores of A, B and C are $a,b$ and $c$ respectively.<\/p>\n According to ques, => $a+b=120$ ————–(i)<\/p>\n $b+c=130$ ————–(ii)<\/p>\n $c+a=140$ ————–(iii)<\/p>\n Adding above equations, we get : $2(a+b+c)=(120+130+140)$<\/p>\n => $a+b+c=\\frac{390}{2}=195$<\/p>\n Substituting value from equation (i) in above equation,<\/p>\n => $120+c=195$<\/p>\n => $c=195-120=75$<\/p>\n => Ans – (D)<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $x(x+y+z)=20$<\/p>\n => $x^2+xy+xz=20$ —————–(i)<\/p>\n Similarly, => $y^2+xy+yz=30$ —————–(ii)<\/p>\n and $z^2+xz+yz=50$ —————–(iii)<\/p>\n Adding equations (i), (ii) and (iii), we get :<\/p>\n => $(x^2+y^2+z^2)+2(xy+yz+xz)=20+30+50$<\/p>\n => $(x+y+z)^2=100$<\/p>\n => $(x+y+z)=\\sqrt{100}=10$<\/p>\n $\\therefore$ $2(x+y+z)=2\\times10=20$<\/p>\n => Ans – (D)<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $cos\\ \\theta + sin\\ \\theta=m$ ————–(i)<\/p>\n Squaring both sides, we get :<\/p>\n => $(cos\\ \\theta + sin\\ \\theta)^2=(m)^2$<\/p>\n => $cos^2\\ \\theta+sin^2\\ \\theta+2sin\\ \\theta.cos\\ \\theta=m^2$<\/p>\n => $1+2sin\\ \\theta.cos\\ \\theta=m^2$<\/p>\n => $sin\\ \\theta.cos\\ \\theta=\\frac{m^2-1}{2}$ ————-(ii)<\/p>\n Also, it is given that : $sec\\ \\theta + cosec\\ \\theta=n$<\/p>\n => $\\frac{1}{cos\\ \\theta}+\\frac{1}{sin\\ \\theta}=n$<\/p>\n => $\\frac{sin\\ \\theta+cos\\ \\theta}{sin\\ \\theta.cos\\ \\theta}=n$<\/p>\n Using equations (i) and (ii), => $m=\\frac{m^2-1}{2}\\times n$<\/p>\n => $n(m^2-1)=2m$<\/p>\n => Ans – (D)<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x=a(sin\\ \\theta + cos\\ \\theta)$ and $y=(sin\\ \\theta – cos\\ \\theta)$<\/p>\n Squaring both sides, we get :<\/p>\n => $x^2=a^2(sin\\ \\theta+cos\\ \\theta)^2$<\/p>\n => $x^2=a^2(sin^2\\ \\theta+cos^2\\ \\theta+2sin\\ \\theta.cos\\ \\theta)$<\/p>\n => $x^2=a^2(1+2sin\\ \\theta.cos\\ \\theta)$<\/p>\n => $\\frac{x^2}{a^2}=1+2sin\\ \\theta.cos\\ \\theta$ —————-(i)<\/p>\n Similarly, $\\frac{y^2}{b^2}=1-2sin\\ \\theta.cos\\ \\theta$ —————-(i)<\/p>\n Adding both equations (i) and (ii),<\/p>\n => $\\frac{x^{2}}{a^{2}}+\\frac{v^{2}}{b^{2}}$ $=(1+2sin\\ \\theta.cos\\ \\theta)+(1-2sin\\ \\theta.cos\\ \\theta)$<\/p>\n = $1+1=2$<\/p>\n => Ans – (C)<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let ideal speed of man = $s$ km\/hr and ideal time = $t$ hours<\/p>\n => Distance = $d=st$ ————(i)<\/p>\n According to ques, => $d=(s+3)(t-1)$<\/p>\n => $d=st-s+3t-3$<\/p>\n Thus, using equation (i), => $3t-s=3$ ————(ii)<\/p>\n Also, $d=(s-2)(t+1)$<\/p>\n => $d=st+s-2t-2$<\/p>\n => $-2t+s=2$ ————(iii)<\/p>\n Adding equations (ii) and (iii), we get :<\/p>\n => $3t-2t=3+2$<\/p>\n => $t=5$<\/p>\n Substituting above value in equation (ii), => $s=3(5)-3=15-3=12$<\/p>\n $\\therefore$ Distance = $12\\times5=60$ km<\/p>\n => Ans – (D)<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $xy=\\frac{a+2}{3}$ and $\\frac{x}{y}=\\frac{1}{3}$<\/p>\n Multiplying both equations, we get : $x^2=\\frac{a+2}{9}$ ————-(i)<\/p>\n Dividing both equations, => $y^2=a+2$ ————-(ii)<\/p>\n Adding equations (i) and (ii),<\/p>\n => $x^2+y^2=(\\frac{a+2}{9})+(a+2)=\\frac{9a+18+a+2}{9}$<\/p>\n => $x^2+y^2=\\frac{10a+20}{9}$ ———–(iii)<\/p>\n Similarly, subtracting equation (ii) from (i), => $x^2-y^2=\\frac{-8a-16}{9}$ ————-(iv)<\/p>\n Dividing equation (iii) by (iv), we get :<\/p>\n => $\\frac{x^2+y^2}{x^2-y^2}=(\\frac{10a+20}{9})\\div(\\frac{-8a-16}{9})$<\/p>\n = $\\frac{10a+20}{-8a-16}=\\frac{10(a+2)}{-8(a+2)}$<\/p>\n = $\\frac{-5}{4}$<\/p>\n => Ans – (D)<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $a + b + c = 4$ ———–(i)<\/p>\n and $a^2 + b^2 + c^2 = ab + bc + ca + 4$<\/p>\n => $a^2 + b^2 + c^2 – ab – bc – ca = 4$ ————(ii)<\/p>\n To find : $a^3 + b^3 + c^3 – 3abc$<\/p>\n = $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$<\/p>\n Substituting values from equations (i) and (ii), we get :<\/p>\n = $4\\times4=16$<\/p>\n => Ans – (C)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x^3+y^3=8$ ———–(i)<\/p>\n and $x+y=2$ ————(ii)<\/p>\n Cubing both sides, we get :<\/p>\n => $(x+y)^3=(2)^3$<\/p>\n => $x^3+y^3+3xy(x+y)=8$<\/p>\n Substituting values from equations (i) and (ii),<\/p>\n => $8+3xy(2)=8$<\/p>\n => $6xy=8-8=0$<\/p>\n => $xy=0$ ———–(iii)<\/p>\n Now, squaring equation (ii), => $(x+y)^2=(2)^2$<\/p>\n => $x^2+y^2+2xy=4$<\/p>\n => $x^2+y^2=4$ $[\\because xy=0]$<\/p>\n Similarly, again squaring both sides, we get :<\/p>\n => $x^4+y^4+2x^2y^2=16$<\/p>\n => $x^4+y^4+2(xy)^2=16$<\/p>\n => $x^4+y^4=16$<\/p>\n => Ans – (C)<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Equations : 7x+8ky-16=0 and 14x+112y-21=0<\/p>\n For two equations to be inconsistent (no solution), the ratio of coefficients of both equations must be of the form = $\\frac{a_1}{b_1}=\\frac{a_2}{b_2}\\neq\\frac{c1}{c_2}$<\/p>\n => $\\frac{7}{14}=\\frac{8k}{112}\\neq\\frac{-16}{-21}$<\/p>\n => $\\frac{k}{14}=\\frac{1}{2}$<\/p>\n => $k=\\frac{14}{2}=7$<\/p>\n => Ans – (D)<\/p>\n
\n7x+4y-16=0 and 14x+6y-32=0?<\/p>\n
\n8(3x+5y)=k
\n3x+5y=($\\frac{k}{8}$)
\nThey have infinite solutions if k=160 and for all other values of k it will have no solution.<\/p>\n
\n5(x+3y)=3k
\nx+3y=$\\frac{3k}{5}\\rightarrow$ 2
\nThese both equations represent a single equation and have infinite solutions when $\\frac{3k}{5}$=12.
\n$\\therefore$ k=20.<\/p>\n