{"id":30615,"date":"2019-06-19T19:25:48","date_gmt":"2019-06-19T13:55:48","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=30615"},"modified":"2019-06-19T19:41:08","modified_gmt":"2019-06-19T14:11:08","slug":"sbi-clerk-quant-previous-questions-with-video-explanations-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/sbi-clerk-quant-previous-questions-with-video-explanations-pdf\/","title":{"rendered":"SBI Clerk Quant Previous Year 2018 Questions (With Video Explanations PDF)"},"content":{"rendered":"<h1>SBI Clerk\u00a0Quant\u00a0Previous year 2018 asked Questions With Video Explanations PDF<\/h1>\n<p>For Previous year Quant questions of SBI clerk 2018 prelims exam download PDF. Go through the video of Quant questions explanations.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/downloads\/5012\" target=\"_blank\" class=\"btn btn-danger  download\">Download Expected Reasoning Questions for SBI Clerk PDF<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/pass\" target=\"_blank\" class=\"btn btn-info \">790+ Mocks &#8211; Just Rs. 194. Use coupon: SBIDREAM70<\/a><\/p>\n<p><a href=\"https:\/\/cracku.in\/blog\/sbi-clerk-important-questions-answers-pdf\/\" target=\"_blank\" rel=\"noopener\">SBI Clerk Topic-Wise Important Questions<\/a> (Download PDF)<br \/>\n<iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/W3hW_FNOEdQ\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\n<a href=\"https:\/\/youtu.be\/W3hW_FNOEdQ\" target=\"_blank\" rel=\"noopener\"><strong>Watch the video here<\/strong><\/a> (If not loaded below)<\/p>\n<p><b>Instructions<\/b><\/p>\n<p>What should come in place of the question mark (?) in the following number series?<\/p>\n<p><b>Question 1:\u00a0<\/b>17, 19, 23, 29, 37, ?<\/p>\n<p>a)\u00a046<\/p>\n<p>b)\u00a049<\/p>\n<p>c)\u00a047<\/p>\n<p>d)\u00a048<\/p>\n<p>e)\u00a045<\/p>\n<p><b>Question 2:\u00a0<\/b>900, 899, 891, 864, 800, ?<\/p>\n<p>a)\u00a0695<\/p>\n<p>b)\u00a0685<\/p>\n<p>c)\u00a0665<\/p>\n<p>d)\u00a0675<\/p>\n<p>e)\u00a0655<\/p>\n<p><b>Question 3:\u00a0<\/b>4, 32, 224, 1344, 6720, ?<\/p>\n<p>a)\u00a026885<\/p>\n<p>b)\u00a026880<\/p>\n<p>c)\u00a026882<\/p>\n<p>d)\u00a026888<\/p>\n<p>e)\u00a026883<\/p>\n<p><b>Question 4:\u00a0<\/b>56, 54, 58, 50, 66, ?<\/p>\n<p>a)\u00a034<\/p>\n<p>b)\u00a098<\/p>\n<p>c)\u00a038<\/p>\n<p>d)\u00a094<\/p>\n<p>e)\u00a044<\/p>\n<p><b>Question 5:\u00a0<\/b>655, 637, 622, 610, 601, ?<\/p>\n<p>a)\u00a0598<\/p>\n<p>b)\u00a0593<\/p>\n<p>c)\u00a0595<\/p>\n<p>d)\u00a0597<\/p>\n<p>e)\u00a0594<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/sbi-clerk-previous-papers\" target=\"_blank\" class=\"btn btn-info \">Download SBI Clerk Previous Papers PDF<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/sbi-clerk-mock-test\" target=\"_blank\" class=\"btn btn-danger \">Take a free mock test for SBI Clerk<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/blog\/sbi-clerk-reasoning-expected-questions-with-video-explanations-pdf\/\" target=\"_blank\" class=\"btn btn-danger \">Download SBI Clerk Quant Expected Questions\u00a0 PDF<\/a><\/p>\n<p><b>Instructions<\/b><\/p>\n<p>approximate value will sonar place of the question mark (1 in ilk. following questions ? (you are not required to calculate the exact value.).<\/p>\n<p><b>Question 6:\u00a0<\/b>10303.88 $\\div$ 55.94 + 62.95 = ?<\/p>\n<p>a)\u00a0247<\/p>\n<p>b)\u00a0250<\/p>\n<p>c)\u00a0260<\/p>\n<p>d)\u00a0220<\/p>\n<p>e)\u00a0None of these<\/p>\n<p><b>Question 7:\u00a0<\/b>$\\sqrt{\\frac{12321}{36.07}} = ?$<\/p>\n<p>a)\u00a020<\/p>\n<p>b)\u00a018.5<\/p>\n<p>c)\u00a017<\/p>\n<p>d)\u00a017.5<\/p>\n<p>e)\u00a0None of these<\/p>\n<p><b>Question 8:\u00a0<\/b>19.03 $\\times$ 16.98 $\\times$ 13.01 = ?<\/p>\n<p>a)\u00a04000<\/p>\n<p>b)\u00a04100<\/p>\n<p>c)\u00a04200<\/p>\n<p>d)\u00a04250<\/p>\n<p>e)\u00a0None of these<\/p>\n<p><b>Question 9:\u00a0<\/b>117% of 459.88 &#8211; 162% of 143.02 = ?<\/p>\n<p>a)\u00a0290<\/p>\n<p>b)\u00a0280<\/p>\n<p>c)\u00a0300<\/p>\n<p>d)\u00a0306<\/p>\n<p>e)\u00a0None of these<\/p>\n<p><b>Question 10:\u00a0<\/b>3\/5 $\\times$ 4\/9 $\\times$ 5894.92=?<\/p>\n<p>a)\u00a01527<\/p>\n<p>b)\u00a01572<\/p>\n<p>c)\u00a01752<\/p>\n<p>d)\u00a01725<\/p>\n<p>e)\u00a0None of these<\/p>\n<p><b>Instructions<\/b><\/p>\n<p>In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. Give answer :<br \/>\na:If x &lt; y<br \/>\nb: If x &gt; y<br \/>\nc: If x \u2264 y<br \/>\nd: If x \u2265 y<br \/>\ne: If relationship between x and y cannot be determined ,<\/p>\n<p><b>Question 11:\u00a0<\/b>I. $15x^{2} + 26x + 8 = 0$<br \/>\nII. $25y^{2} + 15y + 2 = 0$<\/p>\n<p>a)\u00a0If x &lt; y<\/p>\n<p>b)\u00a0If x &gt; y<\/p>\n<p>c)\u00a0If x \u2264 y<\/p>\n<p>d)\u00a0If x \u2265 y<\/p>\n<p>e)\u00a0If relationship between x and y cannot be determined<\/p>\n<p><b>Question 12:\u00a0<\/b>I. $6x^{2} &#8211; 19x + 15 = 0$<br \/>\nII. $5y^{2} &#8211; 22y + 24 = 0$<\/p>\n<p>a)\u00a0If x &lt; y<\/p>\n<p>b)\u00a0If x &gt; y<\/p>\n<p>c)\u00a0If x \u2264 y<\/p>\n<p>d)\u00a0If x \u2265 y<\/p>\n<p>e)\u00a0If relationship between x and y cannot be determined<\/p>\n<p><b>Question 13:\u00a0<\/b>I. $4x^{2} &#8211; 12x + 5 = 0$<br \/>\nII. $4y^{2} &#8211; 8y + 3 = 0$<\/p>\n<p>a)\u00a0If x &lt; y<\/p>\n<p>b)\u00a0If x &gt; y<\/p>\n<p>c)\u00a0If x \u2264 y<\/p>\n<p>d)\u00a0If x \u2265 y<\/p>\n<p>e)\u00a0If relationship between x and y cannot be determined<\/p>\n<p><b>Question 14:\u00a0<\/b>I. $10x^{2} + 21x + 8 = 0$<br \/>\nII. $5y^{2} + 19y + 18 = 0$<\/p>\n<p>a)\u00a0If x &lt; y<\/p>\n<p>b)\u00a0If x &gt; y<\/p>\n<p>c)\u00a0If x \u2264 y<\/p>\n<p>d)\u00a0If x \u2265 y<\/p>\n<p>e)\u00a0If relationship between x and y cannot be determined<\/p>\n<p><b>Question 15:\u00a0<\/b>I. $6x^{2} &#8211; 5x + 1 = 0$<br \/>\nII. $12y^{2} &#8211; 23y + 10 = 0$<\/p>\n<p>a)\u00a0If x &lt; y<\/p>\n<p>b)\u00a0If x &gt; y<\/p>\n<p>c)\u00a0If x \u2264 y<\/p>\n<p>d)\u00a0If x \u2265 y<\/p>\n<p>e)\u00a0If relationship between x and y cannot be determined<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/banking-study-material\" target=\"_blank\" class=\"btn btn-danger \">Banking Study Material &#8211; 15000 Questions<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/banking-online-test\" target=\"_blank\" class=\"btn btn-info \">Daily Free SBI Online Tests<\/a><\/p>\n<p><b>Instructions<\/b><\/p>\n<p>In the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option.<\/p>\n<p>a: If quantity I \u2265 quantity II<br \/>\nb: If quantity I &gt; quantity II<br \/>\nc: If quantity I &lt; quantity II<br \/>\nd: If quantity I = quantity II or the relationship cannot be established from the information that is given<br \/>\ne: If quantity quantity II<\/p>\n<p><b>Question 16:\u00a0<\/b>Arun and Bhadra are brothers. In how many years from now will Bhadra\u2019s age be 50 years ? \u2022<br \/>\nI. The ratio of the current ages of Arun and Bhadra is 5 : 7 respectively.<br \/>\nII. Bhadra was born 10 years before Arun.<br \/>\nIII. 5 years hence, Arun\u2019s age would be three-fourth of Bhadra\u2019s age at that time.<\/p>\n<p>a)\u00a0Any two of the three<\/p>\n<p>b)\u00a0Only II and either I or III<\/p>\n<p>c)\u00a0All I, II and III<\/p>\n<p>d)\u00a0Only II and III<\/p>\n<p>e)\u00a0Only I and III<\/p>\n<p><b>Question 17:\u00a0<\/b>A right-angled triangle is inscribed in a given circle. What is the area of the given circle (in cm2) ?<br \/>\nI. The base and height of the triangle (in cm) are both the roots of the equation $x^{2}-23x+120=0$<br \/>\nII. The sum of the base and height of the triangle is 23 cm.<br \/>\nIII. The height of the right-angled triangle is greater than the base of the same.<\/p>\n<p>a)\u00a0III and either omly I or only II<\/p>\n<p>b)\u00a0ALL I,II and III<\/p>\n<p>c)\u00a0Only II and III<\/p>\n<p>d)\u00a0Only I<\/p>\n<p>e)\u00a0Either I or II<\/p>\n<p><span style=\"text-decoration: underline;\"><strong>Answers &amp; Solutions:<\/strong><\/span><\/p>\n<p><strong>1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>Let the missing number be x<\/p>\n<p>Here the pattern is<\/p>\n<p>19-17 = 2<\/p>\n<p>23-19 = 4<\/p>\n<p>29-23 = 6<\/p>\n<p>37-29 = 8<\/p>\n<p>As we can see that the difference is geeting increased by 2 every time so ,<\/p>\n<p>x-37 =10,it implies x= 47<\/p>\n<p><strong>2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>Let the missing number be x<\/p>\n<p>900-899 = (1)^2<\/p>\n<p>899-891 = 8 = (2)^3<\/p>\n<p>891-864 = 27 = (3)^3<\/p>\n<p>864-800 = 64 = (4)^3<\/p>\n<p>as we can see the pattern between differences of consecutive numbers is of type (n)^3<\/p>\n<p>so x-800 = (5)^3,this implies that x= 675<\/p>\n<p><strong>3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>Let the missing number be x<\/p>\n<p>32<span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"mo\">\u00f7<\/span><\/span><\/span><\/span>4 = 8<\/p>\n<p>224<span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"mo\">\u00f7<\/span><\/span><\/span><\/span>32 = 7<\/p>\n<p>1344<span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"mo\">\u00f7<\/span><\/span><\/span><\/span>224 = 6<\/p>\n<p>6720<span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"mo\">\u00f7<\/span><\/span><\/span><\/span>1344 = 5<\/p>\n<p>As we can see that the division of consecutive numbers is forming a pattern of decreasing numbers by 1, so<\/p>\n<p>x<span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"mo\">\u00f7<\/span><\/span><\/span><\/span>6720 = 4, it implies x is 26880<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/pay\/64XyT\" target=\"_blank\" class=\"btn btn-danger \">25 SBI Clerk Prelims Mocks &#8211; Rs. 149<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/sbi-clerk-mock-test\" target=\"_blank\" class=\"btn btn-info \">Take a free mock test for SBI Clerk<\/a><\/p>\n<p><strong>4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>Let the missing number be x<\/p>\n<p>Here , a pattern is getting formed between consecutive numbers.<\/p>\n<p><span class=\"mn\" style=\"background-color: initial;\">54<\/span><span class=\"mo\" style=\"background-color: initial;\">\u2212<\/span><span class=\"mn\" style=\"background-color: initial;\">56 <\/span><span class=\"mo\" style=\"background-color: initial;\">=$(-2)^1$<\/span><\/p>\n<p><span class=\"mn\">58<\/span><span class=\"mo\">\u2212<\/span><span class=\"mn\">54<\/span><span class=\"mo\">= $(-2)^2$<\/span><\/p>\n<p><span class=\"mn\">50<\/span><span class=\"mo\">\u2212<\/span><span class=\"mn\">58<\/span><span class=\"mo\">= $(-2)^3$<\/span><\/p>\n<p><span class=\"mn\">66<\/span><span class=\"mo\">\u2212<\/span><span class=\"mn\">50<\/span><span class=\"mo\">= $(-2)^4$<\/span><\/p>\n<p><span class=\"mo\">As we can see that the difference is of the form $(-2)^{n}$<\/span><\/p>\n<p><span class=\"mi\">s<\/span><span class=\"mi\">o <\/span><span class=\"mi\">f<\/span><span class=\"mi\">o<\/span><span class=\"mi\">r <\/span><span class=\"mi\">m<\/span><span class=\"mi\">i<\/span><span class=\"mi\">s<\/span><span class=\"mi\">s<\/span><span class=\"mi\">i<\/span><span class=\"mi\">n<\/span><span class=\"mi\">g <\/span><span class=\"mi\">n<\/span><span class=\"mi\">u<\/span><span class=\"mi\">m<\/span><span class=\"mi\">b<\/span><span class=\"mi\">e<\/span><span class=\"mi\">r <\/span><span class=\"mi\">x<\/span><span class=\"mo\">,<\/span><\/p>\n<p><span class=\"mo\">x\u221266= $(-2)^{5}$<br \/>\n<\/span><\/p>\n<p><span class=\"msubsup\"><span class=\"mn\">This implies that x is 34<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p><strong>5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>Let the missing number be x.<\/p>\n<p>The difference between every two consecutive numbers is forming a pattern.<\/p>\n<p>655-637 = 18<\/p>\n<p>637-622 = 15<\/p>\n<p>622-610 = 12<\/p>\n<p>610-601 = 9<\/p>\n<p>As the difference is getting reduced by 3 every time . So 601-x = 6 which implies that the missing number ,x =595<\/p>\n<p>&nbsp;<\/p>\n<p><strong>6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>10303.88$\\div$55.94 + 62.95.<br \/>\n~10304$\\div$56+63.<br \/>\n=184+63.<br \/>\n=247.<br \/>\nHence, Option A is correct.<\/p>\n<p><strong>7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>$\\sqrt{\\frac{12321}{36.07}}$.<br \/>\n~$\\sqrt{\\frac{12321}{36}}$.<br \/>\nWe know that,<br \/>\n$111^{2}=12321, 6^{2}=36$.<br \/>\n~$\\sqrt{\\frac{12321}{36}}=\\frac{111}{6}$.<br \/>\n=$18.5$.<br \/>\nHence, Option B is correct.<\/p>\n<p><strong>8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>19.03 $\\times$ 16.98 $\\times$ 13.01.<br \/>\n~19 $\\times$ 17 $\\times$ 13.<br \/>\n=4199.<br \/>\n~4200.<br \/>\nHence, Option C is correct.<\/p>\n<p><strong>9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>117% of 459.88 &#8211; 162% of 143.02.<br \/>\n~117% of 460-162% of 143.<br \/>\n=$\\frac{117\\times460}{100}-\\frac{162\\times143}{100}$.<br \/>\n=$\\frac{53820}{100}-\\frac{23166}{100}$.<br \/>\n=$538.20-231.66$.<br \/>\n~$306$.<br \/>\nHence, Option D is correct.<\/p>\n<p><strong>10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>3\/5 $\\times$ 4\/9 $\\times$ 5894.92.<br \/>\n~3\/5 $\\times$ 4\/9 $\\times$ 5895.<br \/>\n=$3\\times4\\times131$.<br \/>\n=$1572$.<br \/>\nHence, Option B is correct.<\/p>\n<p><strong>11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>I. $15x^{2} + 26x + 8 = 0$<\/p>\n<p>=&gt; $15x^2 + 6x + 20x + 8 = 0$<\/p>\n<p>=&gt; $(3x + 4) (5x + 2) = 0$<\/p>\n<p>=&gt; $x = \\frac{-4}{3} , \\frac{-2}{5}$<\/p>\n<p>II. $25y^{2} + 15y + 2 = 0$<\/p>\n<p>=&gt; $25y^2 + 5y + 10y + 2 = 0$<\/p>\n<p>=&gt; $(5y + 2) (5y + 1) = 0$<\/p>\n<p>=&gt; $y = \\frac{-2}{5} , \\frac{-1}{5}$<\/p>\n<p>$\\therefore x \\leq y$<\/p>\n<p><strong>12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>I. $6x^{2} &#8211; 19x + 15 = 0$<\/p>\n<p>=&gt; $6x^2 &#8211; 9x &#8211; 10x + 15 = 0$<\/p>\n<p>=&gt; $(3x &#8211; 5) (2x &#8211; 3) = 0$<\/p>\n<p>=&gt; $x = \\frac{5}{3} , \\frac{3}{2}$<\/p>\n<p>II. $5y^{2} &#8211; 22y + 24 = 0$<\/p>\n<p>=&gt; $5y^2 &#8211; 10y &#8211; 12y + 24 = 0$<\/p>\n<p>=&gt; $(5y &#8211; 12) (y + 2) = 0$<\/p>\n<p>=&gt; $y = \\frac{12}{5} , 2$<\/p>\n<p>$\\therefore y &gt; x$<\/p>\n<p><strong>13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n<p>I. $4x^{2} &#8211; 12x + 5 = 0$<\/p>\n<p>=&gt; $4x^2 &#8211; 2x &#8211; 10x + 5 = 0$<\/p>\n<p>=&gt; $(2x &#8211; 5) (2x &#8211; 1) = 0$<\/p>\n<p>=&gt; $x = \\frac{5}{2} , \\frac{1}{2}$<\/p>\n<p>II. $4y^{2} &#8211; 8y + 3 = 0$<\/p>\n<p>=&gt; $4y^2 &#8211; 2y &#8211; 6y + 3 = 0$<\/p>\n<p>=&gt; $(2y &#8211; 3) (2y &#8211; 1) = 0$<\/p>\n<p>=&gt; $y = \\frac{3}{2} , \\frac{1}{2}$<\/p>\n<p>$\\therefore$ No relation can be established.<\/p>\n<p><strong>14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>I. $10x^{2} + 21x + 8 = 0$<\/p>\n<p>=&gt; $10x^2 + 5x + 16x + 8 = 0$<\/p>\n<p>=&gt; $(5x + 8) (2x + 1) = 0$<\/p>\n<p>=&gt; $x = \\frac{-8}{5} , \\frac{-1}{2}$<\/p>\n<p>II. $5y^{2} + 19y + 18 = 0$<\/p>\n<p>=&gt; $5y^2 + 10y + 9y + 18 = 0$<\/p>\n<p>=&gt; $(5y + 9) (y + 2) = 0$<\/p>\n<p>=&gt; $y = \\frac{-9}{5} , -2$<\/p>\n<p>$\\therefore x &gt; y$<\/p>\n<p><strong>15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>I. $6x^{2} &#8211; 5x + 1 = 0$<\/p>\n<p>=&gt; $6x^2 &#8211; 2x &#8211; 3x + 1 = 0$<\/p>\n<p>=&gt; $2x (3x &#8211; 1) &#8211; 1(3x &#8211; 1) = 0$<\/p>\n<p>=&gt; $(2x &#8211; 1) (3x &#8211; 1) = 0$<\/p>\n<p>=&gt; $x = \\frac{1}{2} , \\frac{1}{3}$<\/p>\n<p>II. $12y^{2} &#8211; 23y + 10 = 0$<\/p>\n<p>=&gt; $12y^2 &#8211; 8y &#8211; 15y + 10 = 0$<\/p>\n<p>=&gt; $4y (3y &#8211; 2) &#8211; 3 (3y &#8211; 2) = 0$<\/p>\n<p>=&gt; $(4y &#8211; 3) (3y &#8211; 2) = 0$<\/p>\n<p>=&gt; $y = \\frac{3}{4} , \\frac{2}{3}$<\/p>\n<p>$\\therefore y &gt; x$<\/p>\n<p><strong>16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>I &amp; II : Let Arun&#8217;s age = $x$ years<\/p>\n<p>=&gt; Bhadra&#8217;s age = $x + 10$ years<\/p>\n<p>$\\therefore \\frac{x}{x + 10} = \\frac{5}{7}$<\/p>\n<p>=&gt; $7x = 5x + 50$<\/p>\n<p>=&gt; $7x &#8211; 5x = 2x = 50$<\/p>\n<p>=&gt; $x = \\frac{50}{2} = 25$<\/p>\n<p>=&gt; Bhadra&#8217;s age = $25 + 10 = 35$ years<\/p>\n<p>Thus, I &amp; II are sufficient.<\/p>\n<hr \/>\n<p>II &amp; III : Let Arun&#8217;s age = $x$ years<\/p>\n<p>=&gt; Bhadra&#8217;s age = $x + 10$ years<\/p>\n<p>$\\therefore \\frac{x + 5}{x + 15} = \\frac{3}{4}$<\/p>\n<p>=&gt; $4x + 20 = 3x + 45$<\/p>\n<p>=&gt; $4x &#8211; 3x = 45 &#8211; 20$<\/p>\n<p>=&gt; $x = 25$<\/p>\n<p>=&gt; Bhadra&#8217;s age = $25 + 10 = 35$ years<\/p>\n<p>Thus, II &amp; III are sufficient.<\/p>\n<hr \/>\n<p>I &amp; III : \u00a0Let Arun&#8217;s age = $5x$ years<\/p>\n<p>=&gt; Bhadra&#8217;s age = $7x$ years<\/p>\n<p>$\\therefore \\frac{5x + 5}{7x + 5} = \\frac{3}{4}$<\/p>\n<p>=&gt; $20x + 20 = 21x + 15$<\/p>\n<p>=&gt; $21x &#8211; 20x = 20 &#8211; 15$<\/p>\n<p>=&gt; $x = 5$<\/p>\n<p>=&gt; Bhadra&#8217;s age = $7 \\times 5 = 35$ years<\/p>\n<p>Thus, I &amp; III are sufficient.<\/p>\n<p><strong>$\\therefore$ Any two of the three statements are sufficient.<\/strong><\/p>\n<p><strong>17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>I : $x^2 &#8211; 23x + 120 = 0$<\/p>\n<p>=&gt; $x^2 &#8211; 8x &#8211; 15x + 120 = 0$<\/p>\n<p>=&gt; $x (x &#8211; 8) &#8211; 15 (x &#8211; 8) = 0$<\/p>\n<p>=&gt; $(x &#8211; 8) (x &#8211; 15) = 0$<\/p>\n<p>=&gt; $x = 8 , 15$<\/p>\n<p>Thus, base = 8 cm and height = 15 cm (or vice versa)<\/p>\n<p>=&gt; Hypotenuse of right angled triangle = $\\sqrt{(8)^2 + (15)^2}$<\/p>\n<p>= $\\sqrt{64 + 225} = \\sqrt{289} = 17 cm$<\/p>\n<p>Since, triangle is inscribed in circle, =&gt; Radius of circle = half of hypotenuse<\/p>\n<p>=&gt; $r = \\frac{17}{2} = 8.5$ cm<\/p>\n<p>$\\therefore$ Area of circle = $\\pi r^2$<\/p>\n<p>= $\\frac{22}{7} \\times 8.5 \\times 8.5 \\approx 227 cm^2$<\/p>\n<p><strong>Thus, I alone is sufficient.<\/strong><\/p>\n<p>Clearly, we cannot find base and height from statements II or III. Thus, they are insufficient.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/sbi-clerk-mock-test\" target=\"_blank\" class=\"btn btn-primary \">SBI Clerk Free Mock Test<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.cracku.app&amp;hl=en_IN\" target=\"_blank\" class=\"btn btn-danger \">SBI Free Preparation App<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>SBI Clerk\u00a0Quant\u00a0Previous year 2018 asked Questions With Video Explanations PDF For Previous year Quant questions of SBI clerk 2018 prelims exam download PDF. Go through the video of Quant questions explanations. SBI Clerk Topic-Wise Important Questions (Download PDF) Watch the video here (If not loaded below) Instructions What should come in place of the question [&hellip;]<\/p>\n","protected":false},"author":43,"featured_media":30636,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[232],"tags":[49],"class_list":{"0":"post-30615","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-sbiclerk","8":"tag-sbi-clerk"},"better_featured_image":{"id":30636,"alt_text":"SBI Quant Previous Year Questions","caption":"SBI Quant Previous Year Questions","description":"SBI Quant Previous Year 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