Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>What is the value of a)\u00a0$\\frac{107}{30}$<\/p>\n b)\u00a0$\\frac{103}{25}$<\/p>\n c)\u00a0$\\frac{109}{17}$<\/p>\n d)\u00a0$\\frac{101}{6}$<\/p>\n Question 2:\u00a0<\/b>What is the value of $\\frac{2 \\div 3 \\times (1 + 3) + 5 – 6}{2\u00a0 of\u00a0 3 \\div 5 \\times 4 + 3 – 2}$?<\/p>\n a)\u00a0$\\frac{36}{89}$<\/p>\n b)\u00a0$\\frac{31}{73}$<\/p>\n c)\u00a0$\\frac{25}{87}$<\/p>\n d)\u00a0$\\frac{27}{92}$<\/p>\n Question 3:\u00a0<\/b>What is the mode of given data? a)\u00a09<\/p>\n b)\u00a04<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n Question 4:\u00a0<\/b>What is the mode of the given data? a)\u00a03<\/p>\n b)\u00a02<\/p>\n c)\u00a05<\/p>\n d)\u00a04<\/p>\n Question 5:\u00a0<\/b>What is the value of a)\u00a0$\\frac{31}{2}$<\/p>\n b)\u00a0$18$<\/p>\n c)\u00a0$\\frac{35}{2}$<\/p>\n d)\u00a0$16$<\/p>\n Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>What is the value of $\\frac{39 \\div 26 + 22 \\div 11 \\times 2 + 4 \\times 3}{2\u00a0 of\u00a0 5 – 3(7 + 10 \\div 2 – 3 \\times 3)}$?<\/p>\n a)\u00a0$\\frac{39}{2}$<\/p>\n b)\u00a0$\\frac{49}{2}$<\/p>\n c)\u00a0$\\frac{61}{2}$<\/p>\n d)\u00a0$\\frac{35}{2}$<\/p>\n Question 7:\u00a0<\/b>What is the value of $(24 + 16 \\times 5 – 8\u00a0 of\u00a0 4) \\div 84 \\times 48 \\div 24 \\times 6 + 4 + 3 ?$<\/p>\n a)\u00a0$\\frac{139}{3}$<\/p>\n b)\u00a0$\\frac{121}{7}$<\/p>\n c)\u00a0$\\frac{56}{3}$<\/p>\n d)\u00a0$\\frac{156}{5}$<\/p>\n Question 8:\u00a0<\/b>If $X : Y : Z = 1 : 2 : 3$ and, $X^2 + Y^2 + Z^2 = 224$, then what is the value of $X + Y + Z$?<\/p>\n a)\u00a024<\/p>\n b)\u00a048<\/p>\n c)\u00a036<\/p>\n d)\u00a032<\/p>\n Question 9:\u00a0<\/b>What is the value of $(3 \\times 4\u00a0 of\u00a0 12 \\div 2) \\div 9 \\times 4 + 4 \\div 8 + 3 \\times 2 ?$<\/p>\n a)\u00a0$\\frac{37}{2}$<\/p>\n b)\u00a0$\\frac{77}{2}$<\/p>\n c)\u00a0$\\frac{89}{3}$<\/p>\n d)\u00a0$\\frac{94}{3}$<\/p>\n Question 10:\u00a0<\/b>If $A = 8 \\div 4 \\times (3 – 1) + 6 \\times 3 \\div 2\u00a0 of\u00a0 3\u00a0 and\u00a0 B = 4 \\div 8 \\times 2 + 7 \\times 3$, then what is the value of $A + B$?<\/p>\n a)\u00a033<\/p>\n b)\u00a029<\/p>\n c)\u00a031<\/p>\n d)\u00a035<\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 11:\u00a0<\/b>What is the least number of four digits which is exactly divisible by 2, 4, 6 and 8?<\/p>\n a)\u00a01016<\/p>\n b)\u00a01024<\/p>\n c)\u00a01008<\/p>\n d)\u00a01096<\/p>\n Question 12:\u00a0<\/b>What is the value of $\\frac{3}{4} \\div \\left(\\frac{1}{2} + \\frac{1}{16}\\right) + \\frac{2}{3} of \\frac{4}{9} \\div \\left(\\frac{1}{3} – \\frac{11}{81}\\right) + \\frac{1}{4} \\times \\frac{2}{3}$?<\/p>\n a)\u00a0 b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a04<\/p>\n Question 13:\u00a0<\/b>What is the difference of mean and median of the given data : 4, 13, 8, 15, 9, 21, 18, 23, 35, 1?<\/p>\n a)\u00a00.7<\/p>\n b)\u00a01.7<\/p>\n c)\u00a01.2<\/p>\n d)\u00a02.1<\/p>\n Question 14:\u00a0<\/b>60% of a number is 168, then what is the number?<\/p>\n a)\u00a0280<\/p>\n b)\u00a0320<\/p>\n c)\u00a0240<\/p>\n d)\u00a0200<\/p>\n Question 15:\u00a0<\/b>What is the value of: $5\u00a0\u00a0of\u00a0 5\u00a0 of\u00a0 5 \\div 5 + 5 – 6 \\div 3 \\times 4 + 2 + (3 \\div 6 \\times 2)?$<\/p>\n a)\u00a021<\/p>\n b)\u00a025<\/p>\n c)\u00a028<\/p>\n d)\u00a019<\/p>\n Question 16:\u00a0<\/b>The mode of 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10 is:<\/p>\n a)\u00a05<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n Question 17:\u00a0<\/b>The mode of the following data is 36. What is the value of x ? a)\u00a011<\/p>\n b)\u00a015<\/p>\n c)\u00a013<\/p>\n d)\u00a012<\/p>\n Question 18:\u00a0<\/b>When 6892, 7105 and 7531 are divided by the greatest number x, then the remainder in each case is y. What is the value of (x – y)?<\/p>\n a)\u00a0123<\/p>\n b)\u00a0137<\/p>\n c)\u00a0147<\/p>\n d)\u00a0113<\/p>\n Question 19:\u00a0<\/b>The sum of the perfect square between 120 and 300 is:<\/p>\n a)\u00a01400<\/p>\n b)\u00a01024<\/p>\n c)\u00a01296<\/p>\n d)\u00a01204<\/p>\n Question 20:\u00a0<\/b>The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 is:<\/p>\n a)\u00a0990<\/p>\n b)\u00a0900<\/p>\n c)\u00a0909<\/p>\n d)\u00a0999<\/p>\n Free SSC Preparation Videos<\/a><\/p>\n Download SSC CHSL\u00a0 Previous Papers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\frac{3}{4}\u00a0 of\u00a0 \\left(\\frac{1}{3} \\div \\frac{1}{2}\\right) + \\left(2 – \\frac{2}{5}\\right) \\times \\frac{3}{2} + \\frac{2}{3}$<\/p>\n $\\frac{3}{4}\\times(\\frac{1}{3}\\times\\frac{2}{1}) + (\\frac{10 – 2}{5})\\times\\frac{3}{2} + \\frac{2}{3}$<\/p>\n $\\frac{3}{4}\\times\\frac{2}{3} + \\frac{8}{5}\\times\\frac{3}{2} + \\frac{2}{3}$<\/p>\n $\\frac{6}{12} + \\frac{24}{10} + \\frac{2}{3}$<\/p>\n $\\frac{30 + 144 + 40}{60}$<\/p>\n $\\frac{214}{60}$ = $\\frac{107}{30}$<\/p>\n Therefore option A is the answer.<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{2 \\div 3 \\times (1 + 3) + 5 – 6}{2\u00a0 of\u00a0 3 \\div 5 \\times 4 + 3 – 2}$<\/p>\n $\\frac{\\frac{2}{3}\\times4 + 5 – 6}{2\\times\\frac{3}{5}\\times 4 + 3 – 2}$<\/p>\n $\\frac{\\frac{8}{3} + 5 – 6}{\\frac{24}{5} + 3 – 2}$<\/p>\n $\\frac{\\frac{8 + 15 – 18}{3}}{\\frac{24 + 15 – 10}{5}}$<\/p>\n $\\frac{8 + 15 – 18}{3}\\times\\frac{5}{24 + 15 – 10}$<\/p>\n $\\frac{5}{3}\\times\\frac{5}{29}$<\/p>\n $\\frac{25}{87}$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Number 3 is repeated more number of times when compared to other numbers.Therefore 3 is the answer.<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The mode of a data set is the number that occurs most frequent in the set<\/p>\n To find the mode :<\/p>\n Step 1: arrange numbers in ascending order<\/p>\n 2, 2 , 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6<\/p>\n Step 2 : count how many times each number occurs<\/p>\n 2 three times<\/p>\n 3 five times<\/p>\n 4 four times<\/p>\n 5 two times<\/p>\n 6 one time<\/p>\n Step 3 : The number that occurs the most is the mode<\/p>\n 3 is the mode<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $36 \\div 8 \\times 4 + 2 \\div 4 – 1 + 5 of 3 \\div (4 \\times 2 – 3) – 3 $<\/p>\n $ = 36 \\div 8 \\times 4 + 2 \\div 4 – 1 + 5 of 3 \\div 5 – 3 $<\/p>\n $ =36 \\div 8 \\times 4 + 2 \\div 4 – 1 + 15 \\div 5 – 3 $<\/p>\n $ =\\frac{9}{2} \\times 4 + \\frac{1}{2} – 1 + 3\u00a0– 3 $<\/p>\n $ =18 +\u00a0\\frac{1}{2} – 1 + 3 – 3 $<\/p>\n $ =18 + \\frac{1}{2} + 3 – 4 $<\/p>\n $ = \\frac{35}{2} $<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\frac{39 \\div 26 + 22 \\div 11 \\times 2 + 4 \\times 3}{2\u00a0 of\u00a0 5 – 3(7 + 10 \\div 2 – 3 \\times 3)}$<\/p>\n $\\frac{\\frac{39}{26} + \\frac{22}{11} \\times 2 + 4 \\times 3}{2 \\times 5 – 3(7 + \\frac{10}{2} – 3\\times 3)}$<\/p>\n $\\frac{\\frac{3}{2} + 4 + 12}{10 – 3(7 + 5 -9)}$<\/p>\n $\\frac{\\frac{3 + 8 + 24}{2}}{10 – 3(3)}$<\/p>\n $\\frac{\\frac{35}{2}}{10 – 9}$<\/p>\n $\\frac{35}{2}$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $(24 + 16 \\times 5 – 8\u00a0 of\u00a0 4) \\div 84 \\times 48 \\div 24 \\times 6 + 4 + 3 $<\/p>\n $\\frac{(24 + 16\\times5 – 8\\times4)}{84}\\times\\frac{48}{24}\\times6 + 4 + 3 $<\/p>\n $\\frac{24 + 80 + 32}{7} + 4 + 3$<\/p>\n $\\frac{72 + 28 + 21}{7}$<\/p>\n $\\frac{121}{7}$<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $X : Y : Z = 1 : 2 : 3$<\/p>\n let X = a, Y= 2a , Z= 3a<\/p>\n now\u00a0$X^2 + Y^2 + Z^2 = 224$ =\u00a0$\\left(a\\right)^{2} +\\left(2a\\right)^{2} +\\left(3a\\right)^{2} = 224$<\/p>\n $a^2 + 4a^2 + 9a^2 = 224$<\/p>\n $ 14a^2 = 224$\u00a0, $ a^2 = \\frac{224}{14}$\u00a0,\u00a0$ a^2 = 16$<\/p>\n a=4<\/p>\n $X + Y + Z$ = $a+2a+3a= 6a $ = $6\\times 4$ = 24<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n using the BODMAS rule { priority brackets > of > division > multiplication > addition > subtraction}<\/p>\n solving the bracket first (1st priority brackets)<\/p>\n $(3 \\times 4\u00a0 of\u00a0 12 \\div 2)$ , now since ‘of’ is the priority hence it should be solved first<\/p>\n simplifying\u00a0 it we get<\/p>\n $(3 \\times 4\\times12 \\div 2)$ (here 4 of 12 is $4\\times12$ ) =$(3 \\times 4\\times 6)$<\/p>\n substituting in original\u00a0 question we get<\/p>\n $(3 \\times 4\\times 6) \\div 9 \\times 4 + 4 \\div 8 + 3 \\times 2 $<\/p>\n simpliying it further we get<\/p>\n $\\frac{(3 \\times 4\\times 6)}{9} \\times 4 +\\frac{4}{8}+ 3 \\times 2 $<\/p>\n = 32+$\\frac{1}{2}$+ 6 = $\\frac{77}{2}$<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Applying the BODMAS { priority brackets > of > division > multiplication > addition > subtraction }<\/p>\n To solve A , first solve the subtraction in the brackets i.e (3-1) = 2<\/p>\n simplifying A, we get<\/p>\n $A = 8 \\div 4 \\times 2 + 6 \\times 3 \\div 2\u00a0 of\u00a0 3\u00a0$ = $\\frac{8}{4}\\times 2 + \\frac{6 \\times 3}{6}$ ( here 2 of 3 is $2\\times 3$ = 6)<\/p>\n A= 7<\/p>\n similarly applying BODMAS we solve for B<\/p>\n $B = 4 \\div 8 \\times 2 + 7 \\times 3$ =\u00a0$B = \\frac{4}{8} \\times 2 + 7 \\times 3$ = 22<\/p>\n B = 22<\/p>\n A+B = 7+22 = 29<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n For a number to be divisible 2,4,6,8 should be multiple of 2 and 3,as numbers 2,4,8 have common factor 2 and number 6 is a multiple of 2 and 3.<\/p>\n So, from the options given we get 1008 as a multiple of 2 and 3 both.<\/p>\n Hence option C is a correct choice<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\frac{3}{4} \\div \\left(\\frac{1}{2} + \\frac{1}{16}\\right) + \\frac{2}{3} of \\frac{4}{9} \\div \\left(\\frac{1}{3} – \\frac{11}{81}\\right) + \\frac{1}{4} \\times \\frac{2}{3}$<\/p>\n $ \\Rightarrow$ $\\frac{3}{4} \\div \\frac{9}{16} + \\frac{8}{27} \\div \\frac{16}{81} + \\frac{1}{6}$<\/p>\n $ \\Rightarrow$ $\\frac{3}{4}\\times\\frac{16}{9}+\\frac{8}{27}\\times\\frac{81}{16}+\\frac{1}{6}$<\/p>\n $ \\Rightarrow$ $\\frac{4}{3}+\\frac{3}{2}+\\frac{1}{6}$<\/p>\n $ \\Rightarrow$ 3.<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Mean:<\/p>\n No. of samples (n) = 10<\/p>\n Mean = $\\frac{\\sum x}{n} = \\frac{4+13+8+15+9+21+18+23+35+1}{10}= \\frac{147}{10} = 14.7$<\/p>\n Median:<\/p>\n Arranging the data in ascending order, we get:<\/p>\n 1, 4, 8, 9, 13, 15, 18, 21, 23, 35<\/p>\n n = 10 (even)<\/p>\n Therefore, median is the average of 5th and 6th term.<\/p>\n Median = $\u00a0\u00a0\\frac{13+15}{2} \u00a0= 14$<\/p>\n Mean – Median = 14.7 – 14 = 0.7<\/p>\n Therefore, Option A is correct.<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 60% of the number is 168.<\/p>\n Let’s assume the number is ‘y’.<\/p>\n 60% of y =\u00a0168<\/p>\n 0.6y = 168<\/p>\n y =\u00a0280<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $5\\times5\\times\\frac{5}{5}+5-\\frac{6}{3}\\times4+2+\\frac{3}{6}\\times2$<\/p>\n $25+5-8+2+1$<\/p>\n 25<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Mode : The value that appears most often in a set of given data values.<\/p>\n Given Data :\u00a02, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10<\/p>\n So, Mode of the given data is 5.<\/p>\n Hence,\u00a0Option\u00a0<\/strong>A\u00a0<\/strong>is correct.<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n As per given data,<\/p>\n Class interval of 30-40 has highest frequency, thatswhy it is modal class<\/p>\n As we know,<\/p>\n M =\u00a0$l+\\left\\{\\frac{\\left(f_1-f_0\\right)}{2f_1-f_0-f_2}\\right\\}\\times\\ h$<\/p>\n where, h= size of the class interval,<\/p>\n l = lower limit of the modal class,<\/p>\n $f_{1=}$ frequency of the modal class,<\/p>\n $f_{_0=}$ frequency of the class preceding the modal class<\/p>\n $f_{_2=}$ frequency of the class succeeding the modal class<\/p>\n putting the values from the given data\u00a0:<\/p>\n $36=30+\\frac{\\left(16-10\\right)}{2\\times\\ 16-10-x}\\times\\ 10$<\/p>\n $36-30=\\frac{6}{22-x}\\times\\ \\ 10$<\/p>\n $22-x=10$<\/p>\n $x=12$<\/p>\n Hence, Option D<\/strong> is correct.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We have to find HCF of given numbers : 6892, 7105, 7531<\/p>\n 7105 – 6892 = 213<\/p>\n 7531 – 7105 = 426<\/p>\n 426 – 213 = 213<\/p>\n So, Either the difference or the factor of difference is the\u00a0HCF of those given number.<\/p>\n Here , 213 is the HCF.<\/p>\n When 6892, 7105, 7531 is divided by 213 we get 76 as an remainder<\/p>\n So, x = 213 and y = 76<\/p>\n According to Question :<\/p>\n x – y = 213 – 76 = 137<\/p>\n Hence,\u00a0Option<\/strong> B<\/strong> is correct.\u00a0\u00a0<\/strong><\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Sum of the squares of n consecutive numbers =<\/p>\n The sum of the perfect square between 120 and 300 =\u00a0$11^2+12^2+13^2+14^2+15^2+16^2+17^2$<\/p>\n $=\\frac{17\\left(17+1\\right)\\left(2\\left(17+1\\right)\\right)}{6}-\\frac{10\\left(10+1\\right)\\left(2\\left(10\\right)+1\\right)}{6}$<\/p>\n $=\\frac{17\\left(18\\right)\\left(35\\right)}{6}-\\frac{10\\left(11\\right)\\left(21\\right)}{6}$<\/p>\n $=51\\times35-11\\times35$<\/p>\n $=35\\left(51-11\\right)$<\/p>\n $=35\\left(40\\right)$<\/p>\n $=1400$<\/p>\n Hence, the correct answer is Option A<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The greatest four digit number that begins with 3 and ends with 5 = 3995<\/p>\n The least four digit number that begins with 3 and ends with 5 = 3005<\/p>\n $\\therefore\\ $The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 = 3995 – 3005 = 990<\/p>\n
\n$\\frac{3}{4}\u00a0 of\u00a0 \\left(\\frac{1}{3} \\div \\frac{1}{2}\\right) + \\left(2 – \\frac{2}{5}\\right) \\times \\frac{3}{2} + \\frac{2}{3}$?<\/p>\n
\n4, 3, 7, 13, 16, 23, 3, 4, 7, 4, 3, 3, 9, 6, 9, 6<\/p>\n
\n4, 3, 4, 3, 2, 2, 2, 5, 5, 3, 4, 6, 4, 3, 3<\/p>\n
\n$36 \\div 8 \\times 4 + 2 \\div 4 – 1 + 5\u00a0 of\u00a0 3 \\div (4 \\times 2 – 3) – 3$?<\/p>\n<\/del>3<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_5_rSrqvwb.png\")
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