Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is<\/p>\n a)\u00a0$16\\pi$<\/p>\n b)\u00a0$16(\\pi – 1)$<\/p>\n c)\u00a0$32(\\pi – 1)$<\/p>\n d)\u00a0$32\\pi$<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> BC is the diameter of circle C2 so we can say that\u00a0$\\angle BAC=90^{\\circ\\ }$ as angle in the semi circle is\u00a0$90^{\\circ\\ }$<\/p>\n Therefore overlapping area =\u00a0$\\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1<\/p>\n AB= AC = 8 cm and as\u00a0$\\angle BAC=90^{\\circ\\ }$, so we can conclude that BC=\u00a0$8\\sqrt{2}$ cm<\/p>\n Radius of C2 = Half of length of\u00a0BC =\u00a0$4\\sqrt{2}$ cm<\/p>\n Area of C2 =\u00a0$\\pi\\left(4\\sqrt{2}\\right)^2=32\\pi$\u00a0$cm^2$<\/p>\n A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm<\/p>\n Area of the minor sector made be BC in C1 =\u00a0$\\frac{1}{4}$(Area of circle C1) – Area of triangle ABC =\u00a0$\\frac{1}{4}\\pi\\left(8\\right)^2-\\left(\\frac{1}{2}\\times8\\times8\\right)=16\\pi-32$\u00a0$cm^2$<\/p>\n Therefore,<\/p>\n Overlapping area between the two circles= $\\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1<\/p>\n =\u00a0$\\frac{1}{2}\\left(32\\pi\\right)\\ +\\left(16\\pi-32\\right)=32\\left(\\pi-1\\right)$\u00a0$cm^2$<\/p>\n Question 2:\u00a0<\/b>The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a05<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Sum of the three sides of a quadrilateral is greater than the fourth side.<\/p>\n Therefore, let the fourth side be<\/p>\n 1+2+4>d or d<7<\/p>\n 1+2+d>4 or d>1<\/p>\n Possible values of d are 2, 3, 4, 5 and 6.<\/p>\n Question 3:\u00a0<\/b>Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is<\/p>\n 3)\u00a0Answer:\u00a09<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Area of ABD : Area of BDC = 1:1<\/p>\n Therefore, area of ABD = 54<\/p>\n Area of ADE : Area of EDB = 1:1<\/p>\n Therefore, area of ADE = 27<\/p>\n O is the centroid and it divides the medians in the ratio of 2:1<\/p>\n Area of BEO : Area of EOD = 2:1<\/p>\n Area of EOD = 9<\/p>\n Question 4:\u00a0<\/b>The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is<\/p>\n a)\u00a0$\\sqrt{8}$<\/p>\n b)\u00a0$\\sqrt{6}$<\/p>\n c)\u00a0$\\sqrt{7}$<\/p>\n d)\u00a0$\\sqrt{5}$<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Area of triangle ABD is twice the area of triangle ACD<\/p>\n $\\angle\\ ADB=\\theta\\ $<\/p>\n $\\frac{1}{2}\\left(AD\\right)\\left(BD\\right)\\sin\\theta\\ =2\\left(\\frac{1}{2}\\left(AD\\left(CD\\right)\\sin\\left(180-\\theta\\ \\right)\\right)\\right)$<\/p>\n $BD\\ =2CD$<\/p>\n Therefore, BD = 2 and CD = 1<\/p>\n $\\angle\\ ABC=\\angle\\ ACB=60^{\\circ\\ }$<\/p>\n Applying cosine rule in triangle ADC, we get<\/p>\n $\\cos\\angle\\ ACD=\\ \\frac{\\ AC^2+CD^2-AD^2}{2\\left(AC\\right)\\left(CD\\right)}$<\/p>\n $\\frac{1}{2}=\\ \\frac{\\ 9+1-AD^2}{6}$<\/p>\n $AD^2=7$<\/p>\n $AD=\\sqrt{\\ 7}$<\/p>\n The answer is option C.<\/p>\n Question 5:\u00a0<\/b>Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals<\/p>\n 5)\u00a0Answer:\u00a010<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of sides of polygons A and B be n and 2n, respectively.<\/p>\n $\\ \\dfrac{\\ \\ \\dfrac{\\ \\left(n-2\\right)180}{n}}{\\ \\dfrac{\\ \\left(2n-2\\right)180}{2n}}=\\dfrac{3}{4}$<\/p>\n $\\ \\dfrac{\\ \\ \\ n-2}{\\ \\ n-1}=\\dfrac{3}{4}$<\/p>\n $\\ \\ \\ \\ 4n-8=3n-3$<\/p>\n $n=5$<\/p>\n The number of sides of polygon B is 2*5, i.e. 10.<\/p>\n Question 6:\u00a0<\/b>In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\\angle BAC = 45^{\\circ}$ and $\\angle ABC=\\theta\\ $, then $\\frac{AD}{BE}$ equals<\/p>\n a)\u00a0$\\sqrt{2} \\cos \\theta$<\/p>\n b)\u00a0$\\frac{(\\sin \\theta + \\cos \\theta)}{\\sqrt{2}}$<\/p>\n c)\u00a01<\/p>\n d)\u00a0$\\sqrt{2} \\sin \\theta$<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> It is given, Angle BAE = 45 degrees<\/p>\n This implies AE = BE<\/p>\n Let AE = BE = x<\/p>\n In right-angled triangle ABD, it is given $\\angle ABC=\\theta\\ $<\/p>\n $\\sin\\theta=\\frac{AD}{AB}\\ $<\/p>\n $\\sin\\theta=\\frac{AD}{x\\sqrt{\\ 2}}\\ $<\/p>\n $\\sqrt{\\ 2}\\sin\\theta=\\frac{AD}{BE}\\ $<\/p>\n The answer is option D.<\/p>\n Question 7:\u00a0<\/b>Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (\u22122, 8), respectively. Then, the coordinates of the vertex D are<\/p>\n a)\u00a0(0, 11)<\/p>\n b)\u00a0(4, 5)<\/p>\n c)\u00a0(\u22123, 4)<\/p>\n d)\u00a0(\u22124, 5)<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.<\/p>\n Midpoint of AC =\u00a0$\\left(\\ \\frac{\\ 1-2}{2},\\ \\frac{\\ 1+8}{2}\\right)$<\/p>\n Let the coordinates of vertex D be (x,y)<\/p>\n $\\left(\\ \\frac{\\ x+3}{2},\\ \\frac{\\ y+4}{2}\\right)=\\left(\\ \\frac{\\ 1-2}{2},\\ \\frac{\\ 1+8}{2}\\right)$<\/p>\n x = -4 and y = 5<\/p>\n The answer is option D.<\/p>\n Question 8:\u00a0<\/b>All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is<\/p>\n a)\u00a0$\\frac{P^2}{16} – R^2$<\/p>\n b)\u00a0$\\frac{P^2}{8} – 2R^2$<\/p>\n c)\u00a0$\\frac{P^2}{2} – 2PR$<\/p>\n d)\u00a0$\\frac{P^2}{8} – \\frac{R^2}{2}$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> $A=lb$<\/p>\n $l^2+b^2=4r^2$<\/p>\n $P\\ =2\\left(l+b\\right)$<\/p>\n $\\frac{P}{2}=l+b$<\/p>\n Squaring on both the sides, we get<\/p>\n $\\frac{P^2}{4}=l^2+b^2+2lb$<\/p>\n $\\frac{P^2}{4}=4r^2+2lb$<\/p>\n $\\frac{P^2}{8}-2r^2=lb$<\/p>\n The answer is option B.<\/p>\n Question 9:\u00a0<\/b>A trapezium $ABCD$ has side $AD$ parallel to $BC, \\angle BAD = 90^\\circ, BC = 3$ cm and $AD= 8$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is<\/p>\n 9)\u00a0Answer:\u00a066<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> CD =\u00a0$\\sqrt{\\ y^2+25}$<\/p>\n $11+y+\\sqrt{y^2+25}=36$<\/p>\n $\\sqrt{y^2+25}=25-y$<\/p>\n $y^2+25=25^2+y^2-50y$<\/p>\n 2y = 24<\/p>\n y = 12<\/p>\n Area of trapezium =\u00a0$3y+\\frac{5y}{2}=\\frac{11y}{2}=\\frac{11}{2}\\left(12\\right)=66$<\/p>\n Question 10:\u00a0<\/b>If a 30 meter ladder is placed against a wall such that it just reaches the top of the wall, if the horizontal distance between the wall and the base of the ladder is 1\/3rd of the length of ladder, then the height of wall is :<\/p>\n a)\u00a025 meter<\/p>\n b)\u00a0$20\\sqrt{2}$ meter<\/p>\n c)\u00a0$20\\sqrt{3}$ meter<\/p>\n d)\u00a020 meter<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> AC is ladder, C is base of ladder, AB is wall<\/p>\n Given AC = 30 m and BC =\u00a0$\\frac{1}{3}AC$ =\u00a0$\\frac{1}{3}\\left(30\\right)$ = 10 m<\/p>\n Applying pythagoras theorem,<\/p>\n $AB^2+\\ BC^2=\\ AC^2$<\/p>\n $AB^2$ =\u00a0$30^2-10^2$ = 800<\/p>\n AB =\u00a0$20\\sqrt{\\ 2}$<\/p>\n Answer is option B.<\/p>\n Question 11:\u00a0<\/b>A copper wire having length of 243m and diameter 4 mm was melted to form a sphere. Find the diameter of the sphere :<\/p>\n a)\u00a017 m<\/p>\n b)\u00a018 cm<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a020 cm<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Volume of copper wire melted will be equal to volume of sphere.<\/p>\n Volume of copper wire = Area\u00a0$\\times\\ $ length =\u00a0$\\pi\\ \\left(\\frac{2}{10}\\right)^2\\times\\ 24300\\ cm^3$<\/p>\n Let the radius of sphere be r<\/p>\n $\\frac{4}{3}\\pi\\ r^{3\\ \\ }=\\ \\pi\\ \\left(\\frac{2}{10}\\right)^2\\times\\ 24300\\ cm^3$<\/p>\n $r^3$ = 729<\/p>\n r = 9 cm<\/p>\n Diameter of sphere = 2(9) = 18 cm<\/p>\n Answer is option B.<\/p>\n Question 12:\u00a0<\/b>If a cuboidal box has height, length and width as 20 cm, 15 cm and 10 cm respectively.\u00a0Then its total surface area is__________.<\/p>\n a)\u00a01300$cm^{2}$<\/p>\n b)\u00a01400$cm^{2}$<\/p>\n c)\u00a01200$cm^{2}$<\/p>\n d)\u00a01100$cm^{2}$<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total surface area = 2(lb\u00a0+ bh + hl) = 2(150 + 200 + 300) = 1300\u00a0$cm^2$<\/p>\n Answer is option A.<\/p>\n Question 13:\u00a0<\/b>Angle bisectors of a parallelogram form a _____.<\/p>\n a)\u00a0parallelogram<\/p>\n b)\u00a0square<\/p>\n c)\u00a0rhombus<\/p>\n d)\u00a0rectangle<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Angular bisectors from A and B intersect at P, B and D intersect at Q, C and D intersect at R, and A and C intersect at S. Answer is option D.<\/p>\n Question 14:\u00a0<\/b>A tree bends due to heavy storm and now its peak touches the ground making an\u00a0angle of $30^{\\circ}$ with it. If the bent part is 20 metre long, find the original height of the tree in metres.<\/p>\n a)\u00a010<\/p>\n b)\u00a020\/\u221a3<\/p>\n c)\u00a040<\/p>\n d)\u00a030<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> BC is ground and it is given that\u00a0$\\angle\\ ACB\\ =30^{\\circ\\ }$ Answer is option D.<\/p>\n Question 15:\u00a0<\/b>$ABC$ is a triangle, $\\angle B=60^\\circ, \\angle C=45^\\circ$, BC is produced to \/ extended till D so that $\\angle ADB=30^\\circ$, then given $\\sqrt{3}=\\frac{19}{11}$ and $BC \\times CD=3^{4}\\times 2^{3}\\times\\frac{11}{19}$, what will be the square of the altitude from A to BC?<\/p>\n a)\u00a0144<\/p>\n b)\u00a0324<\/p>\n c)\u00a0484<\/p>\n d)\u00a01254<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> AE is altitude from A on BC.\u00a0$\\angle\\ AEC\\ =\\ 90^{\\circ\\ }$<\/p>\n It is given,<\/p>\n $\\angle\\ ABC\\ =\\ 60^{\\circ\\ },\\ \\angle\\ ACB\\ =\\ 45^{\\circ\\ }$<\/p>\n In triangle ABE,<\/p>\n AE =\u00a0$\\sqrt{\\ 3}x$<\/p>\n As $\\angle\\ EAC\\ =\\angle\\ ECA$, EC =\u00a0$\\sqrt{\\ 3}x$<\/p>\n In triangle AED,\u00a0$\\angle\\ EAD\\ =\\ 60^{\\circ\\ }$<\/p>\n Therefore, CD =\u00a0$3x-\\sqrt{\\ 3}x$<\/p>\n It is given,<\/p>\n $BC\\ \\cdot\\ CD\\ =\\ \\frac{3^4.2^3.11}{19}$<\/p>\n $x\\left(\\sqrt{\\ 3}+1\\right)\\sqrt{\\ 3}\\left(\\sqrt{\\ 3}-1\\right)x\\ =\\ \\frac{3^4.2^3.11}{19}$<\/p>\n $x\\ =\\ \\frac{3^2.2.11}{19}$<\/p>\n AE = 9*2 = 18<\/p>\n $AE^2$ = 324<\/p>\n Answer is option B.<\/p>\n Question 16:\u00a0<\/b>ABCD is a rectangle in the clockwise direction. The coordinates of A are (1,3) and the coordinates of C are (5,1), the coordinates of vertices B and D satisfy the line $y=2x+c$, then what will be the coordinates of the mid-point of BC.<\/p>\n a)\u00a0$(\\frac{5}{2},\\frac{7}{2})$<\/p>\n b)\u00a0$(\\frac{9}{2},\\frac{5}{2})$<\/p>\n c)\u00a0$(\\frac{9}{5},\\frac{7}{2})$<\/p>\n d)\u00a0$(3,2)$<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> Midpoint of AC passes through the line y = 2x + c<\/p>\n Midpoint of AC = (3,2)<\/p>\n 2 = 6 + c<\/p>\n c = -4<\/p>\n Line passing through B is y = 2x – 4<\/p>\n (slope of AB)(slope of BC) = -1<\/p>\n $\\left(\\ \\frac{\\ y-1}{x-5}\\right)\\left(\\ \\frac{\\ y-3}{x-1}\\right)=-1$<\/p>\n Substituting y = 2x – 4 and solving, we get<\/p>\n B(2,0) or B(4,4)<\/p>\n B cannot lie on X-axis. Therefore, co-ordinates of B = (4,4)<\/p>\n Midpoint of BC = $\\left(\\frac{9}{2},\\frac{5}{2}\\right)$<\/p>\n The answer is option B.<\/p>\n Question 17:\u00a0<\/b>In a figure, $\\triangle ABC$ is a right angled triangle at $C$, semicircles are drawn on $AC$, $BC$ and $AB$’s diameter. Find the area of the shaded region. Note: Figure not as per scale<\/p>\n a)\u00a0336<\/p>\n b)\u00a0676<\/p>\n c)\u00a0196<\/p>\n d)\u00a0556<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Area of shaded region = area of semicircle on BC + area of semicircle of AC – area of unshaded semicircle regions The answer is option A.<\/p>\n Question 18:\u00a0<\/b>In the figure (not drawn to scale) given below, if $AD=CD=BC$ and angle $BCE = 81^\\circ$, how much is the value of angle $DBC$? a)\u00a0$27^\\circ$<\/p>\n b)\u00a0$33^\\circ$<\/p>\n c)\u00a0$54^\\circ$<\/p>\n d)\u00a0$66^\\circ$<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Question 19:\u00a0<\/b>A triangular park named ABC, is required to be protected by green fencing. The length of the side BC is 293. If the length of side AB is a perfect square, the length of the side AC is a power of two (2), and the length of side AC is twice the length of side AB. Determine how much fencing is required to cover the triangular park.<\/p>\n a)\u00a01079<\/p>\n b)\u00a01024<\/p>\n c)\u00a01096<\/p>\n d)\u00a01061<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n ABC is a triangle and BC =\u00a0 293 Question 20:\u00a0<\/b>The angle of elevation of a glider from the deck of ship 25 meters above the sea surface is $30^\\circ$ nd the angle of depression of the reflection of glider in the sea is $75^\\circ$. Find the approximate height of the glider from the sea surface. $\\sqrt{3}=\\frac{19}{11}$<\/p>\n a)\u00a046<\/p>\n b)\u00a034<\/p>\n c)\u00a029<\/p>\n d)\u00a052<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> It is given, angle ABC = 30 degrees and angle FBC = 75 degrees<\/p>\n $\\tan\\ 30^{\\circ\\ }=\\frac{x}{BC}$<\/p>\n $BC\\ =\\ x\\sqrt{\\ 3}$<\/p>\n $\\tan75^{\\circ\\ }=\\ \\frac{\\ 50+x}{x\\sqrt{\\ 3}}$<\/p>\n $2+\\sqrt{\\ 3}\\ =\\ \\frac{\\ 50+x}{x\\sqrt{\\ 3}}$<\/p>\n Solving we get x = 9.1<\/p>\n Height of glinder from sea surface = 25+x\u00a0$\\approx\\ $ 34<\/p>\n Answer is option B.<\/p>\n
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\nIn a parallelogram ABCD, AC is parallel to BD.
\n$\\angle\\ CAB\\ +\\angle\\ ACD\\ =\\ 180^{\\circ\\ }$
\n$\\angle\\ CAS\\ +\\angle\\ ACS\\ =\\ 90^{\\circ\\ }$
\nIn triangle CAS,
\n$\\angle\\ ASC\\ =\\ 180^{\\circ\\ }-90^{\\circ\\ }=90^{\\circ\\ }$
\nAs all the angles of PQRS are $90^{\\circ\\ }$,\u00a0PQRS is a rectangle.<\/p>\n
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\nActual length of the tree = AB + AC
\n$\\sin30^{\\circ\\ }=\\frac{AB}{AC}$
\nAB = 20\/2 = 10 cm
\nLength of the tree = AB + AC = 10 + 20 = 30 cm<\/p>\n
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\nArea of unshaded semicircle regions = area of semicircle on AB – area of triangle ABC
\nGiven,
\nAB = 50, BC = 14
\nIn triangle ABC, $AC^2+BC^2 = AB^2$
\nSolving we get,
\nAC = 48
\nArea of unshaded semicircle regions =\u00a0$\\ \\frac{\\pi\\ \\left(25\\right)^2}{2}\\ -\\frac{1}{2}\\times\\ 48\\times\\ 14=\\frac{625\\pi\\ }{2}-336$
\nArea of shaded region =\u00a0$\\frac{576\\pi\\ }{2}\\ +\\frac{49\\pi}{2}\\ -\\frac{625\\pi}{2}\\ +336$ = 336<\/p>\n
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\nAD = DC, this implies angle ACD = angle DAC = x
\nDC = BC, this implies angle CDB = angle CBD = 2x
\n$\\angle\\ DCB$ = 180 – x – 81 = 99 – x
\nIn triangle BCD, sum of the angles is 180 degrees
\n99 – x + 2x + 2x = 180
\n3x = 81
\nx = 27
\n$\\angle\\ DBC$ = 2x = 54 degrees
\nThe answer is option C.<\/p>\n
\nLet the length of AB = $x^2$
\nThe length of AC = $2^y$
\nAC = 2AB
\n$2^y = 2.x^2$
\n$2^{y-1} = x^2$
\n$2^{y-1}$ is a perfect square when y is odd.
\nSum of any two sides of a triangle should be larger than the third side. Therefore,
\n$2^y + x^2$ > 293, 293 + $2^y$ > $x^2$ and\u00a0293 + $x^2$ > $2^y$
\ny > 8
\nIf y = 9, AC = 512, x = 16; satisfies above equations.
\nIf y = 11, AC = 2048, x = 32; doesn’t satisfy above equations.
\nTherefore, AB = 256 and AC = 512
\nFencing required = 256 + 512 + 293 = 1061
\nAnswer is option D.<\/p>\n
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