10 SSC CGL Mocks – Just Rs. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 1:\u00a0<\/b>Find the value of $\\frac{(\\cot\\theta-cosec\\theta+1)(\\tan\\theta+\\sec\\theta+1)}{\\cos\\theta cosec\\theta}?$<\/p>\n a)\u00a0$2\\cos\\theta$<\/p>\n b)\u00a02<\/p>\n c)\u00a0$2cot\\theta$<\/p>\n d)\u00a0$2tan\\theta$<\/p>\n Question 2:\u00a0<\/b>What is the value of $sec12^{0}sin12^{0}tan38^{0}tan78^{0}tan52^{0}$?<\/p>\n a)\u00a01<\/p>\n b)\u00a03<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$\\frac{3}{2}$<\/p>\n Question 3:\u00a0<\/b>What is the value of $\\frac{\\cos x+cos y}{sinx + siny}$ ?<\/p>\n a)\u00a0$\\tan\\frac{x+y}{2}$<\/p>\n b)\u00a0$\\tan\\frac{x-y}{2}$<\/p>\n c)\u00a0$\\cot\\frac{x-y}{2}$<\/p>\n d)\u00a0$\\cot\\frac{x+y}{2}$<\/p>\n Question 4:\u00a0<\/b>If sin $x=\\frac{1}{2}$ and cos $y=\\frac{1}{2}$, what is the value of sin(x+y)?<\/p>\n a)\u00a0$\\frac{2}{3}$<\/p>\n b)\u00a0$\\frac{4}{9}$<\/p>\n c)\u00a0$\\frac{5}{9}$<\/p>\n d)\u00a01<\/p>\n Question 5:\u00a0<\/b>What is the value of $\\frac{(\\tan^{2}x-\\sin^{2}x)}{\\sec^{2}x}$ ?<\/p>\n a)\u00a0$\\sin^{4}x$<\/p>\n b)\u00a0$\\cos^{2}x$<\/p>\n c)\u00a0$\\sin^{2}x$<\/p>\n d)\u00a0$cos^{4}x$<\/p>\n SSC CGL Previous Papers Download PDF<\/a><\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Question 6:\u00a0<\/b>In the given figure, what is the value of $\\angle1+\\angle2+\\angle3+\\angle4+\\angle5+\\angle6+\\angle7+\\angle8+\\angle9+\\angle10$?<\/p>\n <\/p>\n a)\u00a0600<\/p>\n b)\u00a0720<\/p>\n c)\u00a0900<\/p>\n d)\u00a01080<\/p>\n Question 7:\u00a0<\/b>AB is a tangent to a circle with centre O. If the radius at the circle is 7 cm and the length of AB is 24 cm, the what is the length (in cm.) of OA ?<\/p>\n a)\u00a025<\/p>\n b)\u00a026<\/p>\n c)\u00a028<\/p>\n d)\u00a0 Question 8:\u00a0<\/b>In $\\triangle$PQR, S and T are the mid points of sides PQ and PR respectively. If $\\angle$QPR = $45^{0}$ and\u00a0$\\angle$PRQ = $55^{0}$, then what is the value (in degrees) of\u00a0$\\angle$QST ?<\/p>\n a)\u00a080<\/p>\n b)\u00a085<\/p>\n c)\u00a090<\/p>\n d)\u00a0100<\/p>\n Question 9:\u00a0<\/b>A,B and C are the three points on a circle such that $\\angle$ABC = $35^{0}$ and $\\angle$BAC =\u00a0$85^{0}$. What is the angle (in degrees) subtended by arc AB at the center of the circle?<\/p>\n a)\u00a060<\/p>\n b)\u00a090<\/p>\n c)\u00a0135<\/p>\n d)\u00a0120<\/p>\n Question 10:\u00a0<\/b>A circle passing through points Q and R of triangle PQR, cuts the sides PQ and PR at points X and Y respectively. If PQ = PR, then what is the value (in degrees) of $\\angle$PRQ + $\\angle$QXY ?<\/p>\n a)\u00a0120<\/p>\n b)\u00a0150<\/p>\n c)\u00a0240<\/p>\n d)\u00a0180<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 11:\u00a0<\/b>If $x^{2}-9x-1=0$, then what is the value of $\u00a0(x^{2}-\\frac{1}{x^{2}}+5x-\\frac{5}{x})$ ?<\/p>\n a)\u00a0115<\/p>\n b)\u00a0128<\/p>\n c)\u00a0124<\/p>\n d)\u00a0133<\/p>\n Question 12:\u00a0<\/b>If\u00a0 $(x-\\frac{1}{x})=3$, then what is the value of $(x^{3}-\\frac{1}{x^{3}})$ ?<\/p>\n a)\u00a036<\/p>\n b)\u00a021<\/p>\n c)\u00a09<\/p>\n d)\u00a027<\/p>\n Question 13:\u00a0<\/b>If $(3x^{2}-9x+3)=0$, then what is the value of $(x+\\frac{1}{x})^{3}$ ?<\/p>\n a)\u00a09<\/p>\n b)\u00a0729<\/p>\n c)\u00a081<\/p>\n d)\u00a027<\/p>\n Question 14:\u00a0<\/b>If $a^{2}+b^{2}+c^{2}+\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}=6$, then what is the value of $(a^{2}+b^{2}+c^{2})$ ?<\/p>\n a)\u00a03<\/p>\n b)\u00a06<\/p>\n c)\u00a0-3<\/p>\n d)\u00a02<\/p>\n Question 15:\u00a0<\/b>If $x=17-4\\sqrt{18}$, then find the value of $(\\sqrt{x}+\\frac{1}{\\sqrt{x}})$\u00a0 ?<\/p>\n a)\u00a0$7\\sqrt{2}$<\/p>\n b)\u00a09<\/p>\n c)\u00a022<\/p>\n d)\u00a06<\/p>\n Question 16:\u00a0<\/b>If the ratio of volume of two cubes is 11 : 13, then what is the ratio of the sides of the two cubes ?<\/p>\n a)\u00a011 : 13<\/p>\n b)\u00a0121 : 169<\/p>\n c)\u00a0$(11)^{\\frac{1}{2}}:(13)^{\\frac{1}{2}}$<\/p>\n d)\u00a0$(11)^{\\frac{1}{3}}:(13)^{\\frac{1}{3}}$<\/p>\n Question 17:\u00a0<\/b>If the diameter of a sphere is 14 cm., then what is the curved surface area (in $cm.^{2}$) of the sphere?<\/p>\n a)\u00a0616<\/p>\n b)\u00a01232<\/p>\n c)\u00a02464<\/p>\n d)\u00a0576<\/p>\n Question 18:\u00a0<\/b>The difference between circumference and the radius of a circle is 111 cm. What is the area (in $cm^{2}$) of the circle?<\/p>\n a)\u00a0469<\/p>\n b)\u00a01386<\/p>\n c)\u00a0912<\/p>\n d)\u00a01086<\/p>\n Question 19:\u00a0<\/b>Three circles of radius 9 cm are kept touching each other. The string is tightly tied around the three circles. What is the length (in cm.) of the string ?<\/p>\n a)\u00a0$48+18\\pi$<\/p>\n b)\u00a0$48+24\\pi$<\/p>\n c)\u00a0$54+18\\pi$<\/p>\n d)\u00a0$54+24\\pi$<\/p>\n Question 20:\u00a0<\/b>The perimetre of a rhombus in 20 cm and one of the diagonal is 8 cm. What is the area (in $cm^{2}$) of the rhombus?<\/p>\n a)\u00a012<\/p>\n b)\u00a024<\/p>\n c)\u00a048<\/p>\n d)\u00a096<\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression :\u00a0$\\frac{(\\cot\\theta-cosec\\theta+1)(\\tan\\theta+\\sec\\theta+1)}{\\cos\\theta cosec\\theta}$<\/p>\n = $\\frac{sin\\theta}{cos\\theta}\\times[(\\frac{cos\\theta}{sin\\theta}-\\frac{1}{sin\\theta}+1)\\times(\\frac{sin\\theta}{cos\\theta}+\\frac{1}{cos\\theta}+1)]$<\/p>\n =\u00a0$\\frac{sin\\theta}{cos\\theta}\\times[(\\frac{cos\\theta+sin\\theta-1}{sin\\theta})\\times(\\frac{cos\\theta+sin\\theta+1}{cos\\theta})]$<\/p>\n Using, $(x-y)(x+y)=x^2-y^2$, where $x=cos\\theta+sin\\theta$ and $y=1$<\/p>\n = $\\frac{1}{cos^2\\theta}\\times[(cos\\theta+sin\\theta)^2-(1)^2]$<\/p>\n = $\\frac{1}{cos^2\\theta}\\times[cos^2\\theta+sin^2\\theta+2cos\\theta.sin\\theta-1]$<\/p>\n = $\\frac{1}{cos^2\\theta}\\times[1+2cos\\theta.sin\\theta-1]$<\/p>\n = $\\frac{1}{cos^2\\theta}\\times(2cos\\theta.sin\\theta)$<\/p>\n = $\\frac{2sin\\theta}{cos\\theta}=2tan\\theta$<\/p>\n => Ans – (D)<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression\u00a0:\u00a0$sec12^{0}sin12^{0}tan38^{0}tan78^{0}tan52^{0}$<\/p>\n = $\\frac{1}{cos(12^\\circ)}.sin(12^\\circ).tan(38^\\circ).tan(78^\\circ).tan(52^\\circ)$<\/p>\n = $[tan(12^\\circ).tan(78^\\circ)].[tan(38^\\circ).tan(52^\\circ)]$<\/p>\n Using, $tan(90^\\circ-\\theta)=cot(\\theta)$<\/p>\n = $[tan(12^\\circ).cot(12^\\circ)].[tan(38^\\circ).cot(38^\\circ)]$<\/p>\n Also, $tan(\\theta)cot(\\theta)=1$<\/p>\n = $1\\times1=1$<\/p>\n => Ans – (A)<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{\\cos x+cos y}{sinx + siny}$<\/p>\n = $\\frac{cos(\\frac{x+y}{2})cos(\\frac{x-y}{2})}{sin(\\frac{x+y}{2})cos(\\frac{x-y}{2})}$<\/p>\n =\u00a0$\\frac{cos(\\frac{x+y}{2})}{sin(\\frac{x+y}{2})}$<\/p>\n =\u00a0$cot(\\frac{x+y}{2})$<\/p>\n => Ans – (D)<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0: $sin x=\\frac{1}{2}$ and $cos y=\\frac{1}{2}$<\/p>\n => $sin(x)=sin(30^\\circ)$<\/p>\n => $x=30^\\circ$<\/p>\n Similarly, => $cos(y)=cos(60^\\circ)$<\/p>\n => $y=60^\\circ$<\/p>\n $\\therefore$ $sin(x+y)$<\/p>\n = $sin(30^\\circ+60^\\circ)=sin(90^\\circ)=1$<\/p>\n => Ans – (D)<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{(\\tan^{2}x-\\sin^{2}x)}{\\sec^{2}x}$<\/p>\n =\u00a0$\\frac{(\\frac{sin^{2}x}{cos^2x}-sin^{2}x)}{sec^{2}x}$<\/p>\n = $\\frac{sin^2x}{sec^2x}(\\frac{1}{cos^2x}-1)$<\/p>\n = $sin^2x cos^2x(\\frac{1-cos^2x}{cos^2x})$<\/p>\n = $sin^2x\\times(sin^2x)=sin^4x$<\/p>\n => Ans – (A)<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0: OB is radius of circle = 7 cm and tangent AB = 24 cm<\/p>\n To find\u00a0: OA = ?<\/p>\n Solution\u00a0: The radius of a circle intersects the tangent at right angle, => $\\angle OBA = 90^\\circ$<\/p>\n Thus in $\\triangle$ OAB,<\/p>\n => $(OA)^2=(OB)^2+(AB)^2$<\/p>\n => $(OA)^2=(7)^2+(24)^2$<\/p>\n => $(OA)^2=49+576=625$<\/p>\n => $OA=\\sqrt{625}=25$ cm<\/p>\n => Ans – (A)<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0:\u00a0$\\angle$QPR = $45^\\circ$ and\u00a0$\\angle$PRQ = $55^\\circ$<\/p>\n To find\u00a0:\u00a0$\\angle$QST = ?<\/p>\n Solution : In triangle, PQR<\/p>\n => $\\angle P+\\angle Q+\\angle R=180^\\circ$<\/p>\n => $45^\\circ+55^\\circ+\\angle Q=180^\\circ$<\/p>\n => $\\angle Q=180^\\circ-100^\\circ=80^\\circ$<\/p>\n Now, since ST divides PQ and PR equally, thus ST is parallel to QR.<\/p>\n $\\therefore$ Angles on the same side of transversal are supplementary, => $\\angle PQR+\\angle QST=180^\\circ$<\/p>\n => $\\angle QST=180^\\circ-80^\\circ=100^\\circ$<\/p>\n => Ans – (D)<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0:\u00a0$\\angle$ABC = $35^\\circ$ and $\\angle$BAC =\u00a0$85^\\circ$<\/p>\n To find\u00a0:\u00a0$\\angle$ AOB = ?<\/p>\n Solution\u00a0: In triangle, ABC<\/p>\n => $\\angle A+\\angle B+\\angle C=180^\\circ$<\/p>\n => $85^\\circ+35^\\circ+\\angle C=180^\\circ$<\/p>\n => $\\angle C=180^\\circ-120^\\circ=60^\\circ$<\/p>\n Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.<\/p>\n => $\\angle AOB=2\\times\\angle ACB$<\/p>\n = $2\\times60^\\circ=120^\\circ$<\/p>\n => Ans – (D)<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0: PQR is an isosceles triangle, PQ = PR<\/p>\n To find\u00a0:\u00a0$\\angle$PRQ + $\\angle$QXY = ?<\/p>\n Solution\u00a0: Since, $\\triangle$ PQR is isosceles, we have $\\angle Q=\\angle R$<\/p>\n Now, XY is parallel to QR, and sum of angles on the same side of transversal is supplementary, => $\\angle PQR+\\angle QXY=180^\\circ$<\/p>\n =>\u00a0$\\angle$PRQ + $\\angle$QXY = $180^\\circ$<\/p>\n II method\u00a0: XYRQ is a cyclic quadrilateral and opposite angles in a cyclic quadrilateral are supplementary.<\/p>\n => Ans – (D)<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0:\u00a0$x^{2}-9x-1=0$<\/p>\n => $x^2-1=9x$<\/p>\n Dividing both sides by $’x’$,<\/p>\n => $x-\\frac{1}{x}=9$ ————–(i)<\/p>\n Squaring both sides, we get\u00a0:<\/p>\n => $(x-\\frac{1}{x})^2=(9)^2$<\/p>\n => $x^2-\\frac{1}{x^2}-2(x)(\\frac{1}{x})=81$<\/p>\n => $x^2-\\frac{1}{x^2}=81+2=83$ ———–(ii)<\/p>\n $\\therefore$\u00a0$\u00a0(x^{2}-\\frac{1}{x^{2}}+5x-\\frac{5}{x})$<\/p>\n =\u00a0$\u00a0(x^{2}-\\frac{1}{x^{2}})+5(x-\\frac{1}{x})$<\/p>\n Substituting values from equation (i) and (ii),<\/p>\n =\u00a0$83+5(9)=128$<\/p>\n => Ans – (B)<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0:\u00a0$(x-\\frac{1}{x})=3$ ———-(i)<\/p>\n Cubing both sides, we get\u00a0:<\/p>\n =>\u00a0$(x-\\frac{1}{x})^3=(3)^3$<\/p>\n => $x^3-\\frac{1}{x^3}-3(x)(\\frac{1}{x})(x-\\frac{1}{x})=27$<\/p>\n => $x^3-\\frac{1}{x^3}-3(1)(3)=27$<\/p>\n =>\u00a0$(x^{3}-\\frac{1}{x^{3}})=27+9=36$<\/p>\n => Ans – (A)<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0:\u00a0$(3x^{2}-9x+3)=0$<\/p>\n =>\u00a0$(3x^{2}+3)=9x$<\/p>\n Dividing both sides by $’3x’$, we get\u00a0:<\/p>\n => $x+\\frac{1}{x}=3$ ————-(i)<\/p>\n Cubing both sides,<\/p>\n => $(x+\\frac{1}{x})^3=(3)^3$<\/p>\n => $(x+\\frac{1}{x})^3=27$<\/p>\n => Ans – (D)<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0:\u00a0$a^{2}+b^{2}+c^{2}+\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}=6$<\/p>\n => $(a^2+\\frac{1}{a^2}-2)+(b^2+\\frac{1}{b^2}-2)+(c^2+\\frac{1}{c^2}-2)=0$<\/p>\n => $(a-\\frac{1}{a})^2+(b-\\frac{1}{b})^2+(c-\\frac{1}{c})^2=0$<\/p>\n $\\because$ Sum of three positive terms is zero, hence each term is equal to 0.<\/p>\n => $(a-\\frac{1}{a})=$\u00a0$(b-\\frac{1}{b})=$\u00a0$(c-\\frac{1}{c})=0$<\/p>\n => $\\frac{a^2-1}{a}=0$<\/p>\n => $a^2=1$<\/p>\n Similarly,\u00a0$b^2=c^2=1$<\/p>\n $\\therefore$\u00a0$(a^{2}+b^{2}+c^{2})=1+1+1=3$<\/p>\n => Ans – (A)<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0: $x=17-4\\sqrt{18}$<\/p>\n =>\u00a0$x=17-2\\sqrt{72}$<\/p>\n => $x=(\\sqrt9)^2+(\\sqrt8)^2+2(\\sqrt9)(\\sqrt8)$<\/p>\n Using, $a^2+b^2+2ab=(a+b)^2$<\/p>\n => $x=(\\sqrt9+\\sqrt8)^2$<\/p>\n => $\\sqrt{x}=3+2\\sqrt2$ ————-(i)<\/p>\n Also, $\\frac{1}{\\sqrt{x}}=\\frac{1}{3+2\\sqrt2}$<\/p>\n Rationalizing the denominator, we get :<\/p>\n =>\u00a0$\\frac{1}{\\sqrt{x}}=\\frac{1}{3+2\\sqrt2}\\times(\\frac{3-2\\sqrt2}{3-2\\sqrt2})$<\/p>\n =>\u00a0$\\frac{1}{\\sqrt{x}}=\\frac{3-2\\sqrt2}{9-8}$<\/p>\n =>\u00a0$\\frac{1}{\\sqrt{x}}=3-2\\sqrt2$ ———(ii)<\/p>\n Adding equation (i) and (ii),<\/p>\n $\\therefore$ $(\\sqrt{x}+\\frac{1}{\\sqrt{x}})=(3+2\\sqrt2)+(3-2\\sqrt2)=6$<\/p>\n => Ans – (D)<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let side of the two cubes be $a$ and $b$ units respectively<\/p>\n Ratio of volumes = $\\frac{a^3}{b^3}=\\frac{11}{13}$<\/p>\n => $\\frac{a}{b}=(\\sqrt[3]{\\frac{11}{13}})$<\/p>\n => $\\frac{a}{b}=(11)^{\\frac{1}{3}}:(13)^{\\frac{1}{3}}$<\/p>\n => Ans – (D)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Radius of sphere = 7 cm<\/p>\n Curved surface area = $4\\pi r^2$<\/p>\n = $4\\times\\frac{22}{7}\\times(7)^2=616$ $cm^2$<\/p>\n => Ans – (A)<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let radius of circle = $r$ cm<\/p>\n => $2\\pi r-r=111$<\/p>\n => $r(2\\times\\frac{22}{7}-1)=111$<\/p>\n => $r\\times\\frac{44-7}{7}=111$<\/p>\n => $r=111\\times\\frac{7}{37}=21$ cm<\/p>\n $\\therefore$ Area of circle = $\\pi r^2$<\/p>\n = $\\frac{22}{7}\\times(21)^2=1386$ $cm^2$<\/p>\n => Ans – (B)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0: ABCD is the rhombus whose diagonals bisect at O and the\u00a0diagonals of a rhombus bisect each other at right angle. BD = 8 cm<\/p>\n => OB = 4 cm<\/p>\n Perimeter of rhombus = 20 cm<\/p>\n => BC = $\\frac{20}{4}=5$ cm<\/p>\n Thus, in $\\triangle$ BOC,<\/p>\n => $(OC)^2=(BC)^2-(OB)^2$<\/p>\n => $(OC)^2 = (5)^2-(4)^2$<\/p>\n => $(OC)^2=25-16=9$<\/p>\n => $OC=\\sqrt{9}=3$ cm<\/p>\n Thus, AC = 6 cm and BD = 8 cm<\/p>\n $\\therefore$ Area of rhombus = $\\frac{1}{2}\\times d_1\\times d_2$<\/p>\n = $\\frac{1}{2}\\times6\\times8=24$ $cm^2$<\/p>\n => Ans – (B)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/140.png\")
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