Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>In \u0394ABC measure of angle B is 90\u00b0. If cosecA = 13\/12, and AB = 10cm, then what is the length (in cm) of side AC?<\/p>\n a)\u00a024<\/p>\n b)\u00a012<\/p>\n c)\u00a014<\/p>\n d)\u00a026<\/p>\n Question 2:\u00a0<\/b>Two whole numbers are such that the cube of first number exceeds the cube of second by 61 and the ratio of the numbers is 5:4. What is the value of larger number?<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a05<\/p>\n d)\u00a06<\/p>\n Question 3:\u00a0<\/b>Chose the CORRECT option for the figure shoown below.<\/p>\n a)\u00a0AD = AF<\/p>\n b)\u00a0BD = BE<\/p>\n c)\u00a0CE = FC<\/p>\n d)\u00a0All options are correct.<\/p>\n Question 4:\u00a0<\/b>Calculate the length of the tangent (in cm) which is drawn from a point at a distance of 13 cm from the centre and the largest chord of that circle is 10 cm.<\/p>\n a)\u00a010<\/p>\n b)\u00a012<\/p>\n c)\u00a015<\/p>\n d)\u00a016<\/p>\n Question 5:\u00a0<\/b>The area and the length of one of the diagonals of a rhombus is 54 cm2 and 9 cm respectively. Find the length of its other diagonal (in cm).<\/p>\n a)\u00a06<\/p>\n b)\u00a024<\/p>\n c)\u00a012<\/p>\n d)\u00a018<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>Calculate the area (in $cm^{2}$ ) of a circle of radius 10.5 cm.<\/p>\n a)\u00a0693<\/p>\n b)\u00a0157.5<\/p>\n c)\u00a0315<\/p>\n d)\u00a0346.5<\/p>\n Question 7:\u00a0<\/b>Find the total surface area (in $cm^{2}$ ) of a hemisphere of diameter 42 cm.<\/p>\n a)\u00a04158<\/p>\n b)\u00a05782<\/p>\n c)\u00a06321<\/p>\n d)\u00a07782<\/p>\n Question 8:\u00a0<\/b>In \u0394XYZ measure of angle Y is 90\u00b0 . If sinX = 4\/5, and XY = 6cm, then what is the length (in cm) of side YZ?<\/p>\n a)\u00a010<\/p>\n b)\u00a05<\/p>\n c)\u00a08<\/p>\n d)\u00a04<\/p>\n Question 9:\u00a0<\/b>Find the square root of 1485961.<\/p>\n a)\u00a01213<\/p>\n b)\u00a01219<\/p>\n c)\u00a01229<\/p>\n d)\u00a01239<\/p>\n Question 10:\u00a0<\/b>The triangle in which the orthocenter, circumcentre, Incenter and the centroid coincide each other at a point is known as _____________.<\/p>\n a)\u00a0isosceles triangle<\/p>\n b)\u00a0equilateral triangle<\/p>\n c)\u00a0right angled triangle<\/p>\n d)\u00a0obtuse angled triangle<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>If the radius of a circle is decreased to 25% of its original value, calculate the percentage decrease in the area of the circle.<\/p>\n a)\u00a025%<\/p>\n b)\u00a043.75%<\/p>\n c)\u00a050%<\/p>\n d)\u00a093.75%<\/p>\n Question 12:\u00a0<\/b>The perimeter and the breadth of a rectangle are 60 cm and 14 cm respectively. Find its area (in cm\u00b2).<\/p>\n a)\u00a0112<\/p>\n b)\u00a0448<\/p>\n c)\u00a0224<\/p>\n d)\u00a0336<\/p>\n Question 13:\u00a0<\/b>If the perimeter of a semi-circle is 108 cm, then find its radius (in cm).<\/p>\n a)\u00a042<\/p>\n b)\u00a028<\/p>\n c)\u00a056<\/p>\n d)\u00a021<\/p>\n Question 14:\u00a0<\/b>Find the volume (in $cm^{3}$) of a cube of side 7.5 cm.<\/p>\n a)\u00a0421.875<\/p>\n b)\u00a077.145<\/p>\n c)\u00a039.245<\/p>\n d)\u00a024.435<\/p>\n Question 15:\u00a0<\/b>The height of the equilateral triangle is 9 cm. What is the radius (in cm) of the circle circumscribing the three vertices?<\/p>\n a)\u00a03<\/p>\n b)\u00a06<\/p>\n c)\u00a09<\/p>\n d)\u00a012<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>What is that least number that must be added to the product 684 \u00d7 686 to make it a perfect square?<\/p>\n a)\u00a0685<\/p>\n b)\u00a01<\/p>\n c)\u00a0684<\/p>\n d)\u00a0686<\/p>\n Question 17:\u00a0<\/b>The length of the diagonal and the breadth of a rectangle are 25 cm and 7 cm respectively. Find its perimeter (in cm).<\/p>\n a)\u00a0124<\/p>\n b)\u00a041<\/p>\n c)\u00a082<\/p>\n d)\u00a062<\/p>\n Question 18:\u00a0<\/b>If a regular polygon has 6 sides then the measure of its interior angle is greater than the measure of its exterior angle by how many degrees?<\/p>\n a)\u00a0$90^\\circ$<\/p>\n b)\u00a0$100^\\circ$<\/p>\n c)\u00a0$60^\\circ$<\/p>\n d)\u00a0$108^\\circ$<\/p>\n Question 19:\u00a0<\/b>The curved surface area and the slant height of a right circular cone are 99 cm2 and 9 cm respectively. Find its diameter (in cm).<\/p>\n a)\u00a03.5<\/p>\n b)\u00a07<\/p>\n c)\u00a014<\/p>\n d)\u00a010.5<\/p>\n Question 20:\u00a0<\/b>Determine the largest 4 digit number which is a perfect square.<\/p>\n a)\u00a09999<\/p>\n b)\u00a09702<\/p>\n c)\u00a09604<\/p>\n d)\u00a09801<\/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: AB = 10 cm and\u00a0cosec A = 13\/12<\/p>\n => $cosecA=\\frac{AC}{BC}=\\frac{13}{12}$<\/p>\n Let $AC=13x$ and $BC=12x$<\/p>\n In right triangle ABC, => $(AC)^2=(AB)^2+(BC)^2$<\/p>\n => $(13x)^2=(10)^2+(12x)^2$<\/p>\n => $169x^2-144x^2=100$<\/p>\n => $25x^2=100$<\/p>\n => $x^2=\\frac{100}{25}=4$<\/p>\n => $x=\\sqrt4=2$<\/p>\n $\\therefore$ $AC=13\\times2=26$ cm<\/p>\n => Ans – (D)<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the numbers be $5x$ and $4x$<\/p>\n According to ques,<\/p>\n => $(5x)^3-(4x)^3=61$<\/p>\n => $125x^3-64x^3=61x^3=61$<\/p>\n => $x^3=\\frac{61}{61}=1$<\/p>\n => $x=\\sqrt[3]{1}=1$<\/p>\n $\\therefore$ Larger number = $5\\times1=5$<\/p>\n => Ans – (C)<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n => AD = AF , BD = BE , CE = CF<\/p>\n Thus, all options are correct.<\/p>\n => Ans – (D)<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0: Largest chord in a circle is the diameter, => radius OA = $\\frac{10}{2}=5$ cm and OB = $13$ cm<\/p>\n To find\u00a0: AB = ?<\/p>\n Solution\u00a0: In right $\\triangle$ OAB,<\/p>\n => $(AB)^2=(OB)^2-(OA)^2$<\/p>\n => $(AB)^2=(13)^2-(5)^2$<\/p>\n => $(AB)^2=169-25=144$<\/p>\n => $AB=\\sqrt{144}=12$ cm<\/p>\n => Ans – (B)<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let length of one diagonal = $d$ cm and length of other diagonal = 9 cm<\/p>\n => Area of rhombus = $\\frac{1}{2}\\times d_1d_2$<\/p>\n => $\\frac{1}{2}\\times9\\times d=54$<\/p>\n => $d=\\frac{54\\times2}{9}$<\/p>\n => $d=12$ cm<\/p>\n => Ans – (C)<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Radius of circle = $r=10.5$ cm<\/p>\n => Area = $\\pi r^2$<\/p>\n = $\\frac{22}{7}\\times10.5\\times10.5$<\/p>\n = $22\\times1.5\\times10.5=346.5$ $cm^2$<\/p>\n => Ans – (D)<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Radius of hemisphere,\u00a0$r=\\frac{42}{2}=21$ cm<\/p>\n Total surface area = $3\\pi r^2$<\/p>\n = $3\\times\\frac{22}{7}\\times (21)^2$<\/p>\n = $66\\times3\\times21=4158$ $cm^2$<\/p>\n => Ans – (A)<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0: XY = 6 cm and\u00a0sinX = 4\/5 ———(i)<\/p>\n To find\u00a0: YZ = ?<\/p>\n Solution\u00a0: $cosX=\\sqrt{1-sin^2X}$<\/p>\n => $cosX=\\sqrt{1-\\frac{16}{25}}=\\sqrt{\\frac{9}{25}}=\\frac{3}{5}$ ———-(ii)<\/p>\n Dividing equation (i) by (ii), => $tanX=\\frac{4}{3}$<\/p>\n In right $\\triangle$ XYZ,<\/p>\n => $tan(X)=\\frac{YZ}{XY}$<\/p>\n => $\\frac{4}{3}=\\frac{YZ}{6}$<\/p>\n => $YZ=\\frac{4}{3}\\times6=8$ cm<\/p>\n => Ans – (C)<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Factorization of $1485961 = 23^2\\times53^2$<\/p>\n => Square root = $\\sqrt{1485961}=\\sqrt{23^2\\times53^2}$<\/p>\n = $23\\times53=1219$<\/p>\n => Ans – (B)<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Only in an\u00a0equilateral triangle<\/span><\/em>,\u00a0the orthocenter, circumcentre, Incenter and the centroid coincide each other at a point.<\/p>\n => Ans – (B)<\/p>\n GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the radius of the circle initially = $r=20$ cm<\/p>\n => Area = $A=\\pi r^2=\\pi(20)^2=400\\pi$<\/p>\n New radius = $r’=20-(\\frac{25}{100}\\times20)=15$ cm<\/p>\n => New Area = $A’=\\pi(15)^2=225\\pi$<\/p>\n $\\therefore$ Percent decrease in area = $\\frac{(400\\pi-225\\pi)}{400\\pi}\\times100$<\/p>\n = $\\frac{175}{4}=43.75\\%$<\/p>\n => Ans – (B)<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let $l$ be the length of rectangle and breadth,\u00a0$b=14$ cm<\/p>\n => Perimeter = $2(l+b)=60$<\/p>\n => $l+14=\\frac{60}{2}=30$<\/p>\n => $l=30-14=16$ cm<\/p>\n $\\therefore$ Area = $16\\times14=224$ $cm^2$<\/p>\n => Ans – (C)<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let radius of semi circle = $r$ cm<\/p>\n => Perimeter of semi circle = $\\pi r+2r=108$<\/p>\n => $r(\\frac{22}{7}+2)=108$<\/p>\n => $r(\\frac{22+14}{7})=108$<\/p>\n => $r=108\\times\\frac{7}{36}$<\/p>\n => $r=3\\times7=21$ cm<\/p>\n => Ans – (D)<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Side of cube =\u00a0$a=7.5$ cm<\/p>\n => Volume of cube = $a^3$<\/p>\n = $(7.5)^3=421.875$ $cm^3$<\/p>\n => Ans – (A)<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n ABC is an equilateral triangle with height = median = 9 cm<\/p>\n Now, O is the circumcentre of the triangle and it divides the median in the ratio 2:1<\/p>\n => $AO=\\frac{2}{(2+1)}\\times9$<\/p>\n => $AO=\\frac{2}{3}\\times3$<\/p>\n => $AO=2\\times3=6$ cm<\/p>\n => Ans – (B)<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Number = $684\\times686$<\/p>\n = $684\\times(684+2)$<\/p>\n = $(684)^2+2(684)$<\/p>\n Now, to make it a perfect square, =\u00a0$(684)^2+2(684)(1)+(1)^2$<\/p>\n = $(684+1)^2=(685)^2$<\/p>\n $\\therefore$ 1<\/strong> is added to make it a perfect square.<\/p>\n => Ans – (B)<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let length of rectangle = $l$ cm and breadth = $7$ cm<\/p>\n Also, diagonal = $d=25$ cm<\/p>\n => $l^2=d^2-b^2$<\/p>\n => $l=\\sqrt{(25)^2-(7)^2}$<\/p>\n => $l=\\sqrt{625-49}=\\sqrt{576}$<\/p>\n => $l=24$ cm<\/p>\n $\\therefore$ Perimeter = $2(l+b)$<\/p>\n = $2\\times(24+7)=62$ cm<\/p>\n => Ans – (D)<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Sum of all interior angles of regular polygon with $n$ sides = $(n-2)\\times180^\\circ$<\/p>\n => Sum of interior angles of a regular hexagon = $(6-2)\\times180^\\circ=720^\\circ$<\/p>\n Also, sum of exterior angle of an polygon = $360^\\circ$<\/p>\n $\\therefore$ Required difference = $\\frac{(720-360)}{6}$<\/p>\n = $\\frac{360^\\circ}{6}=60^\\circ$<\/p>\n => Ans – (C)<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let radius of cone = $r$ cm and slant height of cone = $l=9$ cm<\/p>\n Curved surface area of cone = $\\pi rl$<\/p>\n => $\\frac{22}{7}\\times r\\times9=99$<\/p>\n => $r=\\frac{99}{9}\\times\\frac{7}{22}$<\/p>\n => $r=\\frac{7}{2}=3.5$ cm<\/p>\n $\\therefore$ Diameter of cone = $2\\times3.5=7$ cm<\/p>\n => Ans – (B)<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 9999 is the largest 4 digit number and $100^2= 10000$<\/p>\n This means that the closest square root of the largest perfect square is most likely 99. So $99^2 =9801$ is the largest perfect square of four digits.<\/p>\n => Ans – (D)<\/p>\n\n![](\"https:\/\/cracku.in\/media\/uploads\/p_1pHVlWH.png\")
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If two tangents are drawn to a circle from one external point, then their tangent\u00a0segments (lines joining the external point and the\u00a0points of tangency) are equal.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/2138_OTOgSRv.PNG\")
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