Expected Quant Questions for SSC CHSL download PDF based on previous year question paper of SSC exams. 20 Very important Expected Quant questions for SSC CHSL Exam.<\/p>\n
Download Expected Quant Questions For SSC CHSL PDF<\/a><\/p>\n\n Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n More SSC CHSL Important Questions and Answers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>Find the value of $\\frac{(243)^\\frac{n}{5}\\times3^{2n+1}}{9^{n}\\times3^{n-1}}$<\/p>\n a)\u00a03<\/p>\n b)\u00a09<\/p>\n c)\u00a027<\/p>\n d)\u00a04<\/p>\n Question 2:\u00a0<\/b>If for a non-zero x, 3$x^{2}+5x + 3 = 0$, then the value of $\\ x^{3}+\\frac{1}{x^{3}}$ is:<\/p>\n a)\u00a0$\\frac{10}{27}$<\/p>\n b)\u00a0$-(\\frac{10}{27})$<\/p>\n c)\u00a0$\\frac{2}{3}$<\/p>\n d)\u00a0$-(\\frac{2}{3})$<\/p>\n Question 3:\u00a0<\/b>The ratio of the volume of a cube to that of a sphere which will fit inside the cube is<\/p>\n a)\u00a04 : $\\pi$<\/p>\n b)\u00a04 : $3\\pi$<\/p>\n c)\u00a06: $\\pi$<\/p>\n d)\u00a02 :\u00a0$\\pi$<\/p>\n Question 4:\u00a0<\/b>A string of length 24 cm is bent first into a square and then into a right-angled triangle by keeping one side of the square fixed as its base. Then the area of triangle equals to:<\/p>\n a)\u00a024 cm$^{2}$<\/p>\n b)\u00a060 cm$^{2}$<\/p>\n c)\u00a040 cm$^{2}$<\/p>\n d)\u00a028 cm$^{2}$<\/p>\n Question 5:\u00a0<\/b>The red blood cells in a blood sample grows by 10% per hour in first two hours, decreases by 10% in next one hour, remains constant in next one hour and again increases by 5% per hour in next two hours. If the original count of the red blood cells in the sample is 40000, find the approximate red blood cell count at the end of 6 hours.<\/p>\n a)\u00a040000<\/p>\n b)\u00a045025<\/p>\n c)\u00a048025<\/p>\n d)\u00a050025<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>The compound interest on a sum of Rs. 5000 at 8% per annum for 9 months when interest is compound quarterly is:<\/p>\n a)\u00a0Rs. 300<\/p>\n b)\u00a0Rs. 300.12<\/p>\n c)\u00a0Rs. 306.04<\/p>\n d)\u00a0Rs. 308<\/p>\n Instructions<\/b><\/p>\n Question 7:\u00a0<\/b>If an arc subtends an angle of 45 degrees at the center of the circle then what is the ratio of length of the arc to the diameter of the circle ?<\/p>\n a)\u00a011:26<\/p>\n b)\u00a011:24<\/p>\n c)\u00a011:28<\/p>\n d)\u00a011:20<\/p>\n Question 8:\u00a0<\/b>Perimeter of a rectangle is 24 cm and area of the square having diagonal equal to the breadth of the rectangle is 32 sq cm. What is the area of the rectangle ?<\/p>\n a)\u00a032 sq cm<\/p>\n b)\u00a064 sq cm<\/p>\n c)\u00a016 sq cm<\/p>\n d)\u00a024 sq cm<\/p>\n Question 9:\u00a0<\/b>In a triangle ABC,Angle bisector of A intersects the side BC at E and if AB=6 cm and AC =14 cm, EC=7 cm then the value of BC is<\/p>\n a)\u00a011<\/p>\n b)\u00a013<\/p>\n c)\u00a014<\/p>\n d)\u00a010<\/p>\n Question 10:\u00a0<\/b>Two sides of the triangle are 8 cm and 10 cm and the angle included between them is 60 degrees then what is the value of third side ?<\/p>\n a)\u00a0$\\sqrt{84}$<\/p>\n b)\u00a0$\\sqrt{82}$<\/p>\n c)\u00a0$\\sqrt{80}$<\/p>\n d)\u00a0$\\sqrt{86}$<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Instructions<\/b><\/p>\n Question 11:\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 Question 12:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{43}{2}$<\/p>\n b)\u00a0$\\frac{982}{45}$<\/p>\n c)\u00a0$\\frac{598}{28}$<\/p>\n d)\u00a0$\\frac{1000}{47}$<\/p>\n Question 13:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{37}{2}$<\/p>\n b)\u00a0$\\frac{427}{24}$<\/p>\n c)\u00a0$\\frac{207}{11}$<\/p>\n d)\u00a0$\\frac{117}{6}$<\/p>\n Question 14:\u00a0<\/b>Which of the following is least ? a)\u00a0$\\frac{11}{5}$<\/p>\n b)\u00a0$\\frac{10}{4}$<\/p>\n c)\u00a0$\\frac{23}{11}$<\/p>\n d)\u00a0$\\frac{31}{14}$<\/p>\n Question 15:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{58}{8}$<\/p>\n b)\u00a0$\\frac{96}{13}$<\/p>\n c)\u00a0$\\frac{38}{5}$<\/p>\n d)\u00a0$\\frac{89}{14}$<\/p>\n Question 16:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{123}{11}$<\/p>\n b)\u00a0$\\frac{80}{7}$<\/p>\n c)\u00a0$\\frac{57}{5}$<\/p>\n d)\u00a0$\\frac{213}{19}$<\/p>\n Question 17:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{73}{13}$<\/p>\n b)\u00a0$\\frac{87}{16}$<\/p>\n c)\u00a0$\\frac{51}{9}$<\/p>\n d)\u00a0$\\frac{120}{23}$<\/p>\n Question 18:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{57}{36}$<\/p>\n b)\u00a0$\\frac{31}{20}$<\/p>\n c)\u00a0$\\frac{42}{23}$<\/p>\n d)\u00a0$\\frac{82}{51}$<\/p>\n Question 19:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{17}{21}$<\/p>\n b)\u00a0$\\frac{23}{27}$<\/p>\n c)\u00a0$\\frac{19}{23}$<\/p>\n d)\u00a0$\\frac{21}{25}$<\/p>\n Question 20:\u00a0<\/b>Which of the following is greatest ? a)\u00a0$\\frac{7}{13}$<\/p>\n b)\u00a0$\\frac{8}{15}$<\/p>\n c)\u00a0$\\frac{5}{9}$<\/p>\n d)\u00a0$\\frac{12}{21}$<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{(243)^\\frac{n}{5}3^{2n+1}}{9^{n}3^{n-1}}$<\/p>\n =\u00a0$\\frac{(3^5)^\\frac{n}{5}\\times3^{2n}\\times3}{3^{2n}\\times3^{n}\\times3^{-1}}$<\/p>\n = $\\frac{3^{n+2n}\\times3}{3^{n+2n}\\times3^{-1}}$<\/p>\n = $\\frac{3^{3n}}{3^{3n}}\\times3\\times\\frac{1}{\\frac{1}{3}}$<\/p>\n = $3\\times3=9$<\/p>\n => Ans – (B)<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0: $3x^2+5x+3=0$<\/p>\n Divide by $x$, => $3x+\\frac{3}{x}=-5$<\/p>\n => $x+\\frac{1}{x}=\\frac{-5}{3}$ ————-(i)<\/p>\n Cubing both sides, we get\u00a0:<\/p>\n => $(x+\\frac{1}{x})^3=(\\frac{-5}{3})^3$<\/p>\n => $x^3+\\frac{1}{x^3}+3(x)(\\frac{1}{x})(x+\\frac{1}{x})=\\frac{-125}{27}$<\/p>\n => $x^3+\\frac{1}{x^3}+3(1)(x+\\frac{1}{x})=\\frac{-125}{27}$<\/p>\n => $x^3+\\frac{1}{x^3}+3(\\frac{-5}{3})=\\frac{-125}{27}$<\/p>\n => $x^3+\\frac{1}{x^3}=\\frac{-125}{27}+5$<\/p>\n => $x^3+\\frac{1}{x^3}=\\frac{-125+135}{27}$<\/p>\n => $x^3+\\frac{1}{x^3}=\\frac{10}{27}$<\/p>\n => Ans – (A)<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let edge of cube be $2a$ cm and thus diameter of sphere = $2a$ cm<\/p>\n => Radius of sphere = $\\frac{2a}{2}=a$ cm<\/p>\n Volume of cube = $(2a)^3=8a^3$ $cm^3$ ———–(i)<\/p>\n Volume of sphere = $\\frac{4}{3}\\pi r^3$<\/p>\n = $\\frac{4}{3} \\pi\u00a0\\times(a)^3=\\frac{4a^3\\pi}{3}$ $cm^3$ ———–(ii)<\/p>\n Dividing equation (i) by (ii), we get\u00a0:<\/p>\n => Required ratio = $\\frac{8a^3}{\\frac{4a^3\\pi}{3}}$<\/p>\n = $\\frac{8\\times3}{4\\pi}=\\frac{6}{\\pi}$<\/p>\n $\\therefore$ Ratio of the volume of a cube to that of a sphere which will fit inside the cube\u00a0= $6:\\pi$<\/p>\n => Ans – (C)<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n SSC CHSL Important Q&A PDF<\/a><\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n String of length 24 cm is bent into square, => Perimeter of square = 24 cm<\/p>\n Let side of square = $a$ cm<\/p>\n => $4a=24$<\/p>\n => $a=\\frac{24}{4}=6$ cm<\/p>\n Let the other side of triangle be $b$ and hypotenuse be $c$ cm<\/p>\n => Perimeter of triangle = $a+b+c=24$<\/p>\n => $b+c=24-6=18$<\/p>\n => $c=18-b$ ————(i)<\/p>\n Also, using Pythagoras Theorem<\/p>\n => $6^2+b^2=c^2$<\/p>\n => $c^2-b^2=36$ ———–(ii)<\/p>\n Solving equations (i) and (ii), we get\u00a0: $b=8$ cm and $c=10$ cm<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} ab$<\/p>\n = $\\frac{1}{2}\\times6\\times8=24$ $cm^2$<\/p>\n => Ans – (A)<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Original count = 40,000<\/p>\n In the next 2 hours, it increases by 10%<\/p>\n => Blood cell count after 2<\/strong> hours = $40,000(1+\\frac{10}{100})^2=40,000(\\frac{11}{10})^2$<\/p>\n = $40,000\\times\\frac{121}{100}=48,400$<\/p>\n It decreases by 10% in next hour<\/p>\n => Blood cell count after 3<\/strong> hours = $48,400(1-\\frac{10}{100})^1$<\/p>\n = $48,400\\times\\frac{9}{10}=43,560$<\/p>\n It remains constant in the next hour, => Blood cell count after 4<\/strong> hours = $43,560$<\/p>\n In the next 2 hours, it increases by 5%<\/p>\n => Blood cell count after 6<\/strong> hours = $43,560(1+\\frac{5}{100})^2=43,560(\\frac{21}{20})^2$<\/p>\n = $43,560\\times\\frac{441}{400}=48,024.9\\approx48,025$<\/p>\n => Ans – (C)<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Principal sum = Rs. 5000<\/p>\n Rate of interest = 8% and time period = $\\frac{9}{12}=\\frac{3}{4}$ years<\/p>\n Compound interest\u00a0when interest is compound quarterly\u00a0= $P[(1+\\frac{R}{400})^{4T}-1]$<\/p>\n = $5000[(1+\\frac{8}{400})^{\\frac{3}{4}\\times4}-1]$<\/p>\n = $5000[(1+\\frac{1}{50})^3-1]$<\/p>\n = $5000[(\\frac{51}{50})^3-1]$<\/p>\n = $5000\\times(\\frac{132651-125000}{125000})$<\/p>\n = $\\frac{7651}{25}=Rs.$ $306.04$<\/p>\n => Ans – (C)<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Length of the arc if it subtends x degrees at the center is $\\frac{x}{360}\\times2\\times\\pi\\times r$ 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given 2(l+b)=24 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In a triangle internal angular bisector divides the opposite side in the ratio of other sides. 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n From cosine rule we have $a^{2}$=$b^{2}+c^{2}-2bc\\cos A$ 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{17}{13}$=1.30 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{43}{2}$=21.5 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\frac{37}{2}$=18.5 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{11}{5}$=2.10 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{58}{8}$=7.25 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{123}{11}$=11.18 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{73}{13}$=5.61 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{42}{23}$=1.82 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{17}{21}$=0.809 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\frac{7}{13}$=0.538
\n$\\frac{17}{13},\\frac{18}{15},\\frac{15}{9},\\frac{12}{7}$<\/p>\n
\n$\\frac{982}{45},\\frac{43}{2},\\frac{598}{28},\\frac{1000}{47}$<\/p>\n
\n$\\frac{207}{11},\\frac{37}{2},\\frac{117}{6},\\frac{427}{24}$<\/p>\n
\n$\\frac{11}{5},\\frac{10}{4},\\frac{23}{11},\\frac{31}{14}$<\/p>\n
\n$\\frac{58}{8},\\frac{96}{13},\\frac{38}{5},\\frac{89}{14}$<\/p>\n
\n$\\frac{123}{11},\\frac{80}{7},\\frac{57}{5},\\frac{213}{19}$<\/p>\n
\n$\\frac{73}{13},\\frac{87}{16},\\frac{51}{9},\\frac{120}{23}$<\/p>\n
\n$\\frac{42}{23},\\frac{31}{20},\\frac{57}{36},\\frac{82}{51}$<\/p>\n
\n$\\frac{17}{21},\\frac{23}{27},\\frac{19}{23},\\frac{21}{25}$<\/p>\n
\n$\\frac{7}{13},\\frac{8}{15},\\frac{5}{9},\\frac{12}{21}$<\/p>\n
\nHere x=45 degrees
\nRequired ratio= $\\frac{\\frac{45}{360}\\times2\\times\\pi\\times r}{2r}$
\n=$\\pi\/16$
\n=11:28<\/p>\n
\nl+b=12
\nGiven b=diagonal of square
\n$d^{2}\/2$=32
\n$d^{2}$=64
\nd=8 cm
\nb=8 cm
\nl+8=12
\nl=4 cm
\nArea of the rectangle=l*b
\n=4*8
\n=32 sq cm<\/p>\n
\nTherefore AB\/AC=BE\/EC
\n6\/14=BE\/7
\nBE=3 cm
\nBC=BE+EC
\n=7+3
\n=10 cm<\/p>\n
\n$a^{2}$=64+100-(2*8*10*1\/2)
\n$a^{2}$=84
\na=$\\sqrt{84}$<\/p>\n
\n$\\frac{18}{15}$=1.2
\n$\\frac{15}{9}$=1.66
\n$\\frac{12}{7}$=1.71<\/p>\n
\n$\\frac{982}{45}$=21.82
\n$\\frac{598}{28}$=21.35
\n$\\frac{1000}{47}$=21.27<\/p>\n
\n$\\frac{427}{24}$=17.8
\n$\\frac{207}{11}$=18.8
\n$\\frac{117}{6}$=19.5<\/p>\n
\n$\\frac{23}{11}$=2.09
\n$\\frac{10}{4}$=2.5
\n$\\frac{31}{14}$=2.2<\/p>\n
\n$\\frac{96}{13}$=7.38
\n$\\frac{38}{5}$=7.6
\n$\\frac{89}{14}$=6.35<\/p>\n
\n$\\frac{80}{7}$=11.42
\n$\\frac{57}{5}$=11.40
\n$\\frac{213}{19}$=11.21<\/p>\n
\n$\\frac{87}{16}$=5.43
\n$\\frac{51}{9}$=5.66
\n$\\frac{120}{23}$=5.21<\/p>\n
\n$\\frac{31}{20}$=1.55
\n$\\frac{57}{36}$=1.58
\n$\\frac{82}{51}$=1.60<\/p>\n
\n$\\frac{23}{27}$=0.851
\n$\\frac{19}{23}$=0.826
\n$\\frac{21}{25}$=0.84<\/p>\n
\n$\\frac{8}{15}$=0.533
\n$\\frac{5}{9}$=0.555
\n$\\frac{12}{21}$=0.57<\/p>\n\n