Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>If $\\sec \\theta +\\tan \\theta=2$, then the value of $\\sin \\theta$ is:<\/p>\n a)\u00a0$\\frac{4}{5}$<\/p>\n b)\u00a0$\\frac{\\sqrt{3}}{5}$<\/p>\n c)\u00a0$\\frac{2}{5}$<\/p>\n d)\u00a0$\\frac{3}{5}$<\/p>\n Question 2:\u00a0<\/b>If $\\alpha + \\theta$ =\u00a0 $\\frac{7\\pi}{12}$ and $\\tan \\theta = \\sqrt{3}$, then the value of $\\tan \\alpha$ is:<\/p>\n a)\u00a0$\\sqrt{3}$<\/p>\n b)\u00a01<\/p>\n c)\u00a00<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n Question 3:\u00a0<\/b>A person travelled from his home to his office at 48 km\/hr and returned from his office to his home at 32 km\/hr. Find his average speed.<\/p>\n a)\u00a040 km\/hr<\/p>\n b)\u00a037.5 km\/hr<\/p>\n c)\u00a036 km\/hr<\/p>\n d)\u00a038.4 km\/hr<\/p>\n Question 4:\u00a0<\/b>A train travelled from A to B at a speed of 60 km\/hr and returned from B to A at a speed of 40 km\/hr. Find the average speed of the train.<\/p>\n a)\u00a050 km\/hr<\/p>\n b)\u00a048 km\/hr<\/p>\n c)\u00a042 km\/hr<\/p>\n d)\u00a056 km\/hr<\/p>\n Question 5:\u00a0<\/b>A trader has 800 kg of sugar of which a part of it has been sold at 4% profit and the rest has been sold at 12% profit. On the whole, the profit is 8%. Find the quantity of sugar which he had sold it at 12% profit.<\/p>\n a)\u00a0360 kg<\/p>\n b)\u00a0500 kg<\/p>\n c)\u00a0280 kg<\/p>\n d)\u00a0400 kg<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>A shopkeeper has 600 kg of rice of which a part of it has been sold at 6% profit and the rest has been sold at 4% loss. On the whole, he had no profit and no loss. Find the quantity which he had sold at loss.<\/p>\n a)\u00a0240 kg<\/p>\n b)\u00a0360 kg<\/p>\n c)\u00a0420 kg<\/p>\n d)\u00a0480 kg<\/p>\n Question 7:\u00a0<\/b>If \u2018a\u2019 is the root of the equation $x^{2}-5x+5=0$ and \u2018b\u2019 is the root of the equation $y^{2}-5y+5=0$, the what is the value of a+b ?<\/p>\n a)\u00a010<\/p>\n b)\u00a020<\/p>\n c)\u00a00<\/p>\n d)\u00a05<\/p>\n Question 8:\u00a0<\/b>Equation $x^{2}-bx+a=0$ has both of its roots as two consecutive numbers which are also prime numbers, then the value of $a^{2}-b^{2}$ ?<\/p>\n a)\u00a011<\/p>\n b)\u00a09<\/p>\n c)\u00a0-11<\/p>\n d)\u00a0-9<\/p>\n Question 9:\u00a0<\/b>Third term of an AP is 8 and the 7th term is 20. what is the ratio of sum of first 10 terms to the sum of first 20 terms ?<\/p>\n a)\u00a0155:620<\/p>\n b)\u00a031:121<\/p>\n c)\u00a031:122<\/p>\n d)\u00a0155:625<\/p>\n Question 10:\u00a0<\/b>If the sum of 31st term and 7th term of an arithmetic progression is 76 then find out the sum of first 37 terms of the A.P.?<\/p>\n a)\u00a02812<\/p>\n b)\u00a01330<\/p>\n c)\u00a01406<\/p>\n d)\u00a01735<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Instructions<\/b><\/p>\n Question 11:\u00a0<\/b>Marked price of an article is Rs 1000 and three successive discounts of 20% each are given. What is the profit\/loss if cost price=500 ?<\/p>\n a)\u00a0Rs 14<\/p>\n b)\u00a0Rs 12<\/p>\n c)\u00a0Rs 13<\/p>\n d)\u00a0Rs 11<\/p>\n Question 12:\u00a0<\/b>Marked price of an article is Rs 480 and cost price of article is marked up by 20%. If the profit percent is 20% then what is the discount offered ?<\/p>\n a)\u00a010<\/p>\n b)\u00a020<\/p>\n c)\u00a030<\/p>\n d)\u00a00<\/p>\n Instructions<\/b><\/p>\n Question 13:\u00a0<\/b>Which of the following number is divisible by both 3 and 9 ?<\/p>\n a)\u00a0349827<\/p>\n b)\u00a0739473<\/p>\n c)\u00a0650874<\/p>\n d)\u00a0835749<\/p>\n Question 14:\u00a0<\/b>If 83683h120 is divisible by 11 then what is the smallest value of h ?<\/p>\n a)\u00a02<\/p>\n b)\u00a05<\/p>\n c)\u00a04<\/p>\n d)\u00a06<\/p>\n Instructions<\/b><\/p>\n Question 15:\u00a0<\/b>Which of the following is least ? a)\u00a0$\\frac{17}{13}$<\/p>\n b)\u00a0$\\frac{18}{15}$<\/p>\n c)\u00a0$\\frac{15}{9}$<\/p>\n d)\u00a0$\\frac{12}{7}$<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\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 17:\u00a0<\/b>Find the number of factors of 13x if\u00a0 (7-$\\frac{4x}{3}$)($\\frac{3}{2}$)=$\\frac{x}{6}$.<\/p>\n a)\u00a012<\/p>\n b)\u00a04<\/p>\n c)\u00a08<\/p>\n d)\u00a06<\/p>\n Instructions<\/b><\/p>\n Read the following pie chart and answer the questions given below.<\/p>\n Question 18:\u00a0<\/b>Population of City A is how much percent more than that of City E?<\/p>\n a)\u00a025%<\/p>\n b)\u00a033.33%<\/p>\n c)\u00a050%<\/p>\n d)\u00a016.67%<\/p>\n Question 19:\u00a0<\/b>If the population of City C is 5 lakh less than that of City E, then what the population of City A?<\/p>\n a)\u00a02 lakh<\/p>\n b)\u00a020 lakh<\/p>\n c)\u00a015 lakh<\/p>\n d)\u00a010 lakh<\/p>\n Question 20:\u00a0<\/b>If the population of City B is 54 lakh, then what is the population of City E?<\/p>\n a)\u00a040 lakh<\/p>\n b)\u00a048 lakh<\/p>\n c)\u00a045 lakh<\/p>\n d)\u00a036 lakh<\/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(D)<\/strong><\/p>\n Given\u00a0:\u00a0$\\sec \\theta +\\tan \\theta=2$ ———–(i)<\/p>\n Also, $sec^2\\theta-tan^2\\theta=1$<\/p>\n => $(sec\\theta-tan\\theta)(sec\\theta+tan\\theta)=1$<\/p>\n =>\u00a0$\\sec \\theta -\\tan \\theta=\\frac{1}{2}$ ———-(ii)<\/p>\n Adding equations (i) and (ii), => $2sec\\theta=2+\\frac{1}{2}=\\frac{5}{2}$<\/p>\n => $sec\\theta=\\frac{5}{4}$<\/p>\n =>\u00a0$cos\\theta=\\frac{4}{5}$<\/p>\n $\\therefore$ $sin\\theta=\\sqrt{1-cos^2\\theta}$<\/p>\n = $\\sqrt{1-(\\frac{4}{5})^2}=\\sqrt{1-\\frac{16}{25}}$<\/p>\n = $\\sqrt{\\frac{9}{25}}=\\frac{3}{5}$<\/p>\n => Ans – (D)<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0:\u00a0$\\tan \\theta = \\sqrt{3}$<\/p>\n =>\u00a0$\\tan \\theta = tan(\\frac{\\pi}{3})$<\/p>\n => $\\theta=\\frac{\\pi}{3}$<\/p>\n Also,\u00a0$\\alpha + \\theta$ =\u00a0 $\\frac{7\\pi}{12}$<\/p>\n => $\\alpha=\\frac{7\\pi}{12}-\\frac{\\pi}{3}$<\/p>\n => $\\alpha=\\frac{7\\pi-4\\pi}{12}=\\frac{\\pi}{4}$<\/p>\n $\\therefore$ $tan\\alpha=tan(\\frac{\\pi}{4})=1$<\/p>\n => Ans – (B)<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the distance between his home and his office be 96 km. (LCM of 48 and 32) 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the distance between A and B be 120 km each (LCM of 60 and 40). 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Quantity sold at 4% profit : Quantity sold at 12% profit = 4 : 4 = 1 : 1 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Quantity at 6% profit : Quantity at 4% loss = 4 : 6 = 2 : 3<\/p>\n Hence, The quantity of rice sold at loss = $\\dfrac{3}{5} \\times 600 = 360 kg$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Substituting \u2018a\u2019 and \u2018b\u2019 in the equation we get $a^{2}-5a+5=0$ 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The only prime numbers that are consecutive are 2,3. 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given a+2d=8 and a+6d=20 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us assume that first term and common difference of A.P. are \u2018a\u2019 and \u2018d\u2019 respectively. GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n MP=1000 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n MP=480 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Sum of digits should be divisible by 9 is the required criteria and so 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Divisibility test of 11 says that if the difference of sum of odd and even terms is divisible by 11 or 0 then it is divisible by 11 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{17}{13}$=1.30 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n For the given two equations 3x+5y=20 and 24x+40y=k 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n solving the equation for x, $\\frac{21}{2}$=$\\frac{13x}{6}$<\/p>\n $x$=$\\frac{63}{13}$<\/p>\n $13x$=63 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total population of all the cities together be x 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total population of all the cities together be x 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the total population of all the cities together be x
\n$\\frac{17}{13},\\frac{18}{15},\\frac{15}{9},\\frac{12}{7}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_14_Q0gr7bz.png\")
<\/figure>\n
\nTotal distance = 96+96 = 192 km
\nTime required to travel 96 km at 48 km\/hr = 2 hours
\nTime required to travel 96 km at 32 km\/hr = 3 hours
\nTotal time = 2+3 = 5 hours
\nAverage speed $= \\dfrac{\\text{Total Distance}}{\\text{Total Time}} = \\dfrac{192}{5} = 38.4 km\/hr$<\/p>\n
\nTime taken by the train to travel 120 km at 60 km\/hr = 2 hours
\nTime taken by the train to travel 120 km at 40 km\/hr = 3 hours
\nTotal distance = 120+120 = 240 km
\nTotal time = 2+3 = 5 hours
\nAverage speed $= \\dfrac{\\text{Total Distance}}{\\text{Total Time}} = \\dfrac{240}{5} = 48 km\/hr$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209959.png\")
<\/figure>\n
\nHence, the quantity of sugar he sold at 12% profit = $\\dfrac{1}{2} \\times 800 = 400 kg$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/209958.png\")
<\/figure>\n
\n$b^{2}-5b+5=0$ so let us assume a equation $z^{2}-5z+5=0$
\nFor this equation both a and b are the roots since they satisfy the equation.
\nTherefore sum of the roots =5.<\/p>\n
\nSubstituting the values of $x=2 and x=3 $
\nWe get $3^{2}-3b+a=0$ and
\n$2^{2}-2b+a=0$
\nSolving for a and b we get a=6 and b=5.
\nSo $6^{2}-5^{2}=11$<\/p>\n
\nWe have d=3,a=2
\nsum of n terms of an AP=$\\frac{n}{2}(2a+(n-1)d)$
\nFor n=10,S=155
\nFor n=20,S=610
\nTherefore required ratio =31:122<\/p>\n
\n$T_{7} + T_{31} = (a+6d) + (a+30d) = 76$
\n$2a + 36d = 76$ … (1)
\nSum of first n terms of an A.P. = $\\frac{n\\times [2a+(n-1)d]}{2}$
\nTherefore sum of first 37 terms of the A.P. = $\\frac{37\\times [2a+(37-1)d]}{2}$
\n$\\Rightarrow$ $\\frac{37\\times [2a+36d]}{2}$
\n$\\Rightarrow$ $\\frac{37\\times 76}{2}$
\n$\\Rightarrow$ $1406$ (Answer :C)<\/p>\n
\nThree successive discounts of 20% each and so SP=1000*(80\/100)(80\/100)(80\/100)
\nSP=512
\nProfit=512-500
\n=12<\/p>\n
\n1.2*CP=480
\nCP=400
\nSP=MP-d
\nSP=480-d
\n$\\frac{480-d-400}{400}\\times100$=20
\n80-d=80
\nd=0<\/p>\n
\n349827=3+4+9+8+2+7=33
\n739473=7+3+9+4+7+3=33
\n650874=6+5+0+8+7+4=30
\n835749=8+3+5+7+4+9=36
\nHence d is the correct answer<\/p>\n
\nSum of odd terms=8+6+3+1+0=18
\nSum of even terms=3+8+h+2=13+h
\ndifference=18-13-h
\n=5-h
\nTherefore for h=5 we have difference =0<\/p>\n
\n$\\frac{18}{15}$=1.2
\n$\\frac{15}{9}$=1.66
\n$\\frac{12}{7}$=1.71<\/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
\nWe get $\\frac{21}{2}$=$2x+\\frac{x}{6}$<\/p>\n
\nFactorising 63 we get 63=$3^{2}\\times 7$
\nNumber of factors=(2+1)(1+1)
\n=6<\/p>\n
\nPopulation of City A = 20% of x = 0.2x
\nPopulation of City E = 15% of x = 0.15x
\nRequired percentage = $\\dfrac{0.20x-0.15x}{0.15x} \\times 100 = \\dfrac{0.5x}{0.15x} \\times 100 = 33.33$%<\/p>\n
\nPopulation of City C = 10% of x
\nPopulation of City E = 15% of x
\nGiven, 15% of x – 10% of x = 5 lakh
\n\u21d2 5% of x =5 lakh
\n\u21d2 x = 100 lakh or 1 crore
\nPopulation of City A = 20% of 1 crore = 20 lakh<\/p>\n
\nPopulation of City B = 18% of x
\nGiven, 18% of x = 54 lakh
\n\u21d2 $x = 5400000 \\times \\dfrac{100}{18} = 300$ lakh or 3 crore
\nHence, Population of E = 15% of 300 lakh = 45 lakh<\/p>\n\n