Enroll to XAT 2024 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b><\/p>\n In a circular field, AOB and COD are two mutually perpendicular diameters having length of 4 meters. X is the mid – point of OA. Y is the point on the circumference such that \u2220YOD = 30\u00b0. Which of the following correctly gives the relation among the three alternate paths from X to Y?<\/p>\n a)\u00a0XOBY : XODY : XADY :: 5.15 : 4.50 : 5.06<\/p>\n b)\u00a0XADY : XODY : XOBY :: 6.25 : 5.34 : 4.24<\/p>\n c)\u00a0XODY : XOBY : XADY :: 4.04 : 5.35 : 5.25<\/p>\n d)\u00a0XADY : XOBY : XODY :: 5.19 : 5.09 : 4.04<\/p>\n e)\u00a0XOBY : XADY : XODY :: 5.06 : 5.15 : 4.50<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n XADY = XA + AD + DY = 2\/2 + (2 * 3.14 * 2)\/4 + (30\/360) * (2 * 3.14 * 2) = 5.19 Question 2:\u00a0<\/b>Let $S_{1}, S_{2},…$ be the squares such that for each n \u2265 1, the length of the diagonal of $S_{n}$ is equal to the length of the side of $S_{n+1}$. If the length of the side of $S_{3}$ is 4 cm, what is the length of the side of $S_{n}$ ?<\/p>\n a)\u00a0$2^[{\\frac{2n+1}{2}}]$<\/p>\n b)\u00a0$2.(n-1)$<\/p>\n c)\u00a0$2^{n-1}$<\/p>\n d)\u00a0$2^[{\\frac{n+1}{2}}]$<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Length of side of $S_{n + 1}$ = Length of diagonal of $S_n$<\/p>\n =>\u00a0Length of side of $S_{n + 1}$ =\u00a0$\\sqrt{2}$ (Length of side of $S_{n}$)<\/p>\n => $\\frac{\\textrm{Length of side of }S_{n + 1}}{\\textrm{Length of side of }S_n} = \\sqrt{2}$<\/p>\n => Sides of $S_1 , S_2 , S_3 , S_4,…….., S_n$ form a G.P. with common ratio, $r = \\sqrt{2}$<\/p>\n It is given that, $S_3 = ar^2 = 4$<\/p>\n => $a (\\sqrt{2})^2 = 4$<\/p>\n => $a = \\frac{4}{2} = 2$<\/p>\n $\\therefore$ $n^{th}$ term of G.P. = $a (r^{n – 1})$<\/p>\n = $2 (\\sqrt{2})^{n – 1}$<\/p>\n =$2^[{\\frac{n+1}{2}}]$<\/p>\n Question 3:\u00a0<\/b>In the figure below, AB = AC = CD. If ADB = 20\u00b0, what is the value of BAD? a)\u00a040\u00b0<\/p>\n b)\u00a060\u00b0<\/p>\n c)\u00a070\u00b0<\/p>\n d)\u00a0120\u00b0<\/p>\n e)\u00a0140\u00b0<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> AB = AC = CD, => $\\angle CAD = \\angle CDA = 20^{\\circ}$<\/p>\n and $\\angle ABC = \\angle ACB$<\/p>\n In $\\triangle$ ACD<\/p>\n => $\\angle ACD + \\angle CAD + \\angle CDA = 180^{\\circ}$<\/p>\n => $\\angle ACD = 180^{\\circ} – 20^{\\circ} – 20^{\\circ} = 140^{\\circ}$<\/p>\n => $\\angle ACB = 180^{\\circ} – 140^{\\circ} = 40^{\\circ} = \\angle ABC$<\/p>\n Similarly, In $\\triangle$ ABC<\/p>\n => $\\angle BAC = 180^{\\circ} – 40^{\\circ} – 40^{\\circ} = 100^{\\circ}$<\/p>\n $\\therefore \\angle BAD = 100^{\\circ} + 20^{\\circ} = 120^{\\circ}$<\/p>\n Question 4:\u00a0<\/b>The difference between the area of the circumscribed circle and the area of the inscribed circle of an equilateral triangle is 2156 sq. cm. What is the area of the equilateral triangle?<\/p>\n a)\u00a0686\u221a3<\/p>\n b)\u00a01000<\/p>\n c)\u00a0961\u221a2<\/p>\n d)\u00a0650\u221a3<\/p>\n e)\u00a0None of the above<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let radius of incircle = $r$, => Radius of circumcircle = $2r$<\/p>\n Difference in area = $\\pi [(2r)^2 – (r)^2] = 2156$<\/p>\n => $3 \\times \\frac{22}{7} \\times r^2 = 2156$<\/p>\n => $r^2 = \\frac{2156 \\times 7}{66}$<\/p>\n => $r = \\sqrt{\\frac{686}{3}}$<\/p>\n Now, height of equilateral triangle = $3 r = \\frac{\\sqrt{3}}{2} a$ \u00a0 \u00a0(where $a$ is side of triangle)<\/p>\n => $3 \\times \\sqrt{\\frac{686}{3}} = \\frac{\\sqrt{3}}{2} a$<\/p>\n => $a = 2 \\sqrt{686}$<\/p>\n $\\therefore$ Area of triangle = $\\frac{\\sqrt{3}}{4} a^2$<\/p>\n = $\\frac{\\sqrt{3}}{4} \\times 4 \\times 686 = 686 \\sqrt{3} cm^2$<\/p>\n Question 5:\u00a0<\/b>A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re – cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?<\/p>\n a)\u00a09%<\/p>\n b)\u00a016%<\/p>\n c)\u00a025%<\/p>\n d)\u00a050%<\/p>\n e)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Volume of Cylinder = $\\pi r^2 h = \\pi \\times 7^2 \\times 10 = 490\\pi$<\/p>\n Now, The solid metal cylinder is re-cast into two cones in the proportion 3 : 4 i.e. the volumes of cone 1 and cone 2 is<\/p>\n $210 \\pi$ and $280 \\pi$ respectively.<\/p>\n So, flat Surface area of cylinder before melting = $2 \\pi r^2 = 2 \\pi \\times 7^2 = 98 \\pi$<\/p>\n Volume of cone 1 = $\\frac{1}{3} \\pi r_1^2 h = 210 \\pi$<\/p>\n => $r_1^2 = \\frac{210 \\times 3}{10} = 63$<\/p>\n Volume of cone 2 = $\\frac{1}{3} \\pi r_2^2 h = 280 \\pi$<\/p>\n => $r_2^2 = \\frac{280 \\times 3}{10} = 84$<\/p>\n Flat surface area of cones = $\\pi r_1^2 + \\pi r_2^2$<\/p>\n = $\\pi (63 + 84) = 147 \\pi$<\/p>\n $\\therefore$ Percentage change in surface area = $\\frac{147 \\pi – 98 \\pi}{98 \\pi} \\times 100$<\/p>\n = $\\frac{1}{2} \\times 100 = 50 \\%$<\/p>\n Question 6:\u00a0<\/b>Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals.<\/p>\n a)\u00a015 and 30<\/p>\n b)\u00a015 and 40<\/p>\n c)\u00a017 and 30<\/p>\n d)\u00a017 and 40<\/p>\n e)\u00a0Multiple solutions are possible<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Using distance formula,<\/p>\n $CX = \\sqrt{(17.5 – 5.5)^2 + (23.5 – 7.5)^2} = \\sqrt{12^2 + 16^2}$<\/p>\n = $\\sqrt{144 + 256} = \\sqrt{400} = 20$<\/p>\n => $AC = 2 \\times CX = 40$<\/p>\n $BX = \\sqrt{(17.5 – 13.5)^2 + (23.5 – 16)^2} = \\sqrt{4^2 + 7.5^2}$<\/p>\n = $\\sqrt{16 + 56.25} = \\sqrt{72.25} = 8.5$<\/p>\n => $BD = 2 \\times BX = 17$<\/p>\n Question 7:\u00a0<\/b>In quadrilateral PQRS, PQ = 5 units, QR = 17 units, RS = 5 units, and PS = 9 units. The length of the diagonal QS can be:<\/p>\n a)\u00a0> 10 and < 12<\/p>\n b)\u00a0> 12 and < 14<\/p>\n c)\u00a0> 14 and < 16<\/p>\n d)\u00a0> 16 and < 18<\/p>\n e)\u00a0cannot be determined<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Geometry View Video Solution<\/a><\/p><\/p>\n Solution:<\/b> In a triangle , sum of two sides is greater than the third side and difference of two sides is less than third side.<\/p>\n In $\\triangle$ PQS<\/p>\n => $QS < 9 + 5$ => $QS < 14$ ——Eqn(I)<\/p>\n In $\\triangle$ QRS<\/p>\n => $QS > 17 – 5$ => $QS > 12$ ——Eqn(II)<\/p>\n From eqn(I) & (II)<\/p>\n => $12 < QS < 14$<\/p>\n Question 8:\u00a0<\/b>There are two circles $C_{1}$ and $C_{2}$ of radii 3 and 8 units respectively. The common internal tangent, T, touches the circles at points $P_{1}$ and $P_{2}$ respectively. The line joining the centers of the circles intersects T at X. The distance of X from the center of the smaller circle is 5 units. What is the length of the line segment $P_{1} P_{2}$ ?<\/p>\n a)\u00a0\u2264 13<\/p>\n b)\u00a0> 13 and \u2264 14<\/p>\n c)\u00a0> 14 and \u2264 15<\/p>\n d)\u00a0> 15 and \u2264 16<\/p>\n e)\u00a0> 16<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given\u00a0: $OP_1 = 3 , O’P_2 = 8 , OX = 5$ units<\/p>\n To find\u00a0: $P_1P_2 = ?$<\/p>\n Solution\u00a0: In $\\triangle OP_1X$<\/p>\n => $(P_1X)^2 = (OP_1)^2 – (OX)^2$<\/p>\n => $(P_1X)^2 = 5^2 – 3^2 = 25 – 9$<\/p>\n => $P_1X = \\sqrt{16} = 4$<\/p>\n In $\\triangle OP_1X$ and $\\triangle O’P_2X$<\/p>\n $\\angle OXP_1 = O’XP_2$ \u00a0 (Vertically opposite angles)<\/p>\n $\\angle OP_1X = O’P_2X = 90$<\/p>\n => $\\triangle OP_1X \\sim \\triangle O’P_2X$<\/p>\n => $\\frac{XP_1}{XP_2} = \\frac{OP_1}{O’P_2}$<\/p>\n => $XP_2 = 4 \\times \\frac{8}{3} = 10.66$<\/p>\n $\\therefore P_1P_2 = P_1X + XP_2 = 4 + 10.66 = 14.66$ units<\/p>\n Question 9:\u00a0<\/b>There are two squares S 1 and S 2 with areas 8 and 9 units, respectively. S 1 is inscribed within S 2 , with one corner of S 1 on each side of S 2 . The corners of the smaller square divides the sides of a)\u00a0\u2265 5 and < 8<\/p>\n b)\u00a0\u2265 8 and < 11<\/p>\n c)\u00a0\u2265 11 and < 14<\/p>\n d)\u00a0\u2265 14 and < 17<\/p>\n e)\u00a0> 17<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Area of $S_1 = 8$ sq. units<\/p>\n => Side of $S_1 = PS = \\sqrt{8} = 2 \\sqrt{2}$ units<\/p>\n Similarly, Side of $S_2 = CD = \\sqrt{9} = 3$ units<\/p>\n => $a + b = 3$<\/p>\n In $\\triangle$ PDS<\/p>\n => $b^2 + a^2 = 8$<\/p>\n => $b^2 + (3 – b)^2 = 8$<\/p>\n => $b^2 + 9 + b^2 – 6b = 8$<\/p>\n => $2b^2 – 6b + 1 = 0$<\/p>\n => $b = \\frac{6 \\pm \\sqrt{36 – 8}}{4} = \\frac{6 \\pm \\sqrt{28}}{4}$<\/p>\n => $b = \\frac{3 + \\sqrt{7}}{2}$\u00a0\u00a0\u00a0\u00a0 $(\\because b > a)$<\/p>\n => $a = 3 – \\frac{3 + \\sqrt{7}}{2} = \\frac{3 – \\sqrt{7}}{2}$<\/p>\n $\\therefore \\frac{b}{a} = \\frac{\\frac{3 + \\sqrt{7}}{2}}{\\frac{3 – \\sqrt{7}}{2}}$<\/p>\n = $\\frac{3 + \\sqrt{7}}{3 – \\sqrt{7}} \\approx 15.9$<\/p>\n Question 10:\u00a0<\/b>A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:<\/p>\n a)\u00a01000 times<\/p>\n b)\u00a0100 times<\/p>\n c)\u00a010 times<\/p>\n d)\u00a0No change<\/p>\n e)\u00a0None of the above<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Radius of larger sphere = $R = 10$ cm<\/p>\n Let radius of each of the smaller spheres = $r$ cm<\/p>\n => $\\frac{4}{3} \\pi R^3 = 1000 \\times \\frac{4}{3} \\pi r^3$<\/p>\n => $10^3 = 1000 r^3$<\/p>\n => $r = \\sqrt[3]{1} = 1$ cm<\/p>\n Initial surface area of sphere = $4 \\pi R^2 = 4 \\pi \\times 100 = 400 \\pi$<\/p>\n Final surface area of 1000 spheres = $1000 \\times 4 \\pi r^2 = 1000 \\times 4 \\pi = 4000 \\pi$<\/p>\n $\\therefore$ Increase in surface area = $4000 \\pi – 400 \\pi = 3600 \\pi$<\/p>\n => $\\frac{3600 \\pi}{400 \\pi} = 9$ times<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/1_cGPckxd.png\")
<\/figure>\n
\nXOBY = XO + OB + BY = 2\/2 + 2 + (60\/360) * (2 * 3.14 * 2) = 5.09
\nXODY = XO + OD + DY = 2\/2 + 2 + (30\/360) * (2 * 3.14 * 2) =\u00a0 4.04
\nHence, option D is the correct answer.<\/p>\n
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\nthe bigger square into two segments, one of length \u2018a\u2019 and the other of length \u2018b\u2019, where, b > a. A possible value of \u2018b\/a\u2019, is:<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/5877.PNG\")
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