Mathematical Reasoning is an important topic in the Quant section of the SNAP Exam. You can also download this Free Mathematical Reasoning Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Mathematical Reasoning questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Mathematical Reasoning Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>In a question on division, the divisor is 6 times the quotient and 3 times the remainder. If the remainder is 40, then find the dividend.<\/p>\n a)\u00a02445<\/p>\n b)\u00a02440<\/p>\n c)\u00a02455<\/p>\n d)\u00a02450<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, remainder = 40<\/p>\n Divisor is 3 times the remainder<\/p>\n $\\Rightarrow$\u00a0 divisor = 3 x 40 = 120<\/p>\n Divisor is 6 times the quotient<\/p>\n $\\Rightarrow$\u00a0 120 = 6 x quotient<\/p>\n $\\Rightarrow$\u00a0 quotient = 20<\/p>\n Dividend = divisor x quotient + remainder<\/p>\n $\\Rightarrow$ Dividend = 120 x 20 + 40<\/p>\n $\\Rightarrow$ Dividend = 2400 + 40<\/p>\n $\\Rightarrow$ Dividend = 2440<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 2:\u00a0<\/b>In an examination, Anita scored 31% marks and failed by 16 marks. Sunita scored 40% marks and obtained 56 marks more than those required to pass. Find the minimum marks required to pass.<\/p>\n a)\u00a03116<\/p>\n b)\u00a0264<\/p>\n c)\u00a03944<\/p>\n d)\u00a07100<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Anita scored 31% marks and failed by 16 marks and Sunita scored 40% marks and obtained 56 marks more than those required to pass.<\/p>\n Difference of percentage = difference of marks<\/p>\n 40% – 31% = 16 + 56 (Anita is failed and Sunita is passed so marks will be added)<\/p>\n 9% = 72 marks<\/p>\n 56 marks = $\\frac{9}{72} \\times 56 = 7%$<\/p>\n Sunita scored 40% marks and obtained 56 marks more than those required to pass so,<\/p>\n Minimum passing marks = 40% – 7% = 33%<\/p>\n 7% = 56 marks<\/p>\n 33% =\u00a0$\\frac{56}{7} \\times 33 = 264$<\/p>\n Question 3:\u00a0<\/b>The value of $\u00a0\\frac{(0.67 \\times 0.67 \\times0.67)\u00a0 \\times (0.33 \\times 0.33 \\times 0.33)}{(0.67 \\times 0.67) \\div (0.67 \\times 0.33) \\div (0.33 \\times 0.33)} $<\/p>\n a)\u00a011<\/p>\n b)\u00a00.34<\/p>\n c)\u00a01.1<\/p>\n d)\u00a03.4<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Mathematical Operators provided in the question seems to be incorrect. Please review the question again and provide correct data.<\/p>\n Question 4:\u00a0<\/b>In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder’r’. If r = 46, the dividend will be<\/p>\n a)\u00a05042<\/p>\n b)\u00a05348<\/p>\n c)\u00a05336<\/p>\n d)\u00a04276<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that<\/p>\n Dividend = $ (Divisor \\times Quotient) +Remainder $ —- (1)<\/p>\n Given that Divisor = 10 times the Quotient<\/p>\n => Divisor= 10Q —- (2)<\/p>\n and Divisor = 5 times the remainder<\/p>\n => Divisor = 5R = 5(46) = 230<\/p>\n Substituting divisor value in (2), we get, Q= 23<\/p>\n Substituting all values in equation (1), we get<\/p>\n Dividend = $(230 \\times 23) + 46 $ = 5336<\/p>\n Question 5:\u00a0<\/b>The value of $\\frac{18.43 \\times 18.43 – 6.57 \\times 6.57}{11.86}$ is:<\/p>\n a)\u00a023.62<\/p>\n b)\u00a025<\/p>\n c)\u00a026<\/p>\n d)\u00a024.12<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{18.43 \\times 18.43 – 6.57 \\times 6.57}{11.86}$<\/p>\n $\\frac{18.43^{2}-6.57^{2}}{11.86}$<\/p>\n $\\frac{(18.43+6.57)(18.43 – 6.57)}{11.86}$<\/p>\n $\\frac{25\\times11.86}{11.86}=25$<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>The value of the expression $\\frac{1}{4} \\left\\{\\left(a + \\frac{1}{a}\\right)^2 – \\left(a – \\frac{1}{a}\\right)^2\\right\\}$ is:<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a0$\\frac{1}{4}$<\/p>\n c)\u00a01<\/p>\n d)\u00a04<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{1}{4} \\left\\{\\left(a + \\frac{1}{a}\\right)^2 – \\left(a – \\frac{1}{a}\\right)^2\\right\\}$<\/p>\n $\\frac{1}{4}\\left\\{a^2+\\frac{1}{a^2}+2-\\left(a^2+\\frac{1}{a^2}-2\\right)\\right\\}$<\/p>\n $\\frac{1}{4}\\left\\{a^2+\\frac{1}{a^2}+2-a^2-\\frac{1}{a^2}+2\\right\\}$<\/p>\n $\\frac{1}{4}\\times4=1$<\/p>\n Question 7:\u00a0<\/b>The simplified value of $\\frac{\\left(3\\frac{1}{5} + \\frac{3}{5}\\right) \\div \\frac{8}{5}}{1\\frac{1}{7} \\div \\left\\{\\frac{5}{7} + \\left(\\frac{1}{7} \\div \\frac{1}{3}\\right)\\right\\}}$ is:<\/p>\n a)\u00a0$\\frac{19}{7}$<\/p>\n b)\u00a0$\\frac{19}{16}$<\/p>\n c)\u00a0$\\frac{19}{8}$<\/p>\n d)\u00a0$\\frac{19}{64}$<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\left(3\\frac{1}{5} + \\frac{3}{5}\\right) \\div \\frac{8}{5}}{1\\frac{1}{7} \\div \\left\\{\\frac{5}{7} + \\left(\\frac{1}{7} \\div \\frac{1}{3}\\right)\\right\\}}= \\frac{\\left(\\frac{16}{5} + \\frac{3}{5}\\right) \\div \\frac{8}{5}}{\\frac{8}{7} \\div \\{\\frac{5}{7}+\\frac{3}{7}\\}}$<\/p>\n $=\\frac{\\frac{19}{5}\\div\\frac{8}{5}}{\\frac{8}{7}\\div\\frac{8}{7}}$<\/p>\n $=\\frac{19}{5}\\div\\frac{8}{5}$<\/p>\n $=\\frac{19}{8}$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>The value of $3 \\times 2 \\div 3\u00a0 of\u00a0 12 – 3 \\div 2 \\times (2 – 3) \\times 2 + 3 \\div 2\u00a0 \u00a0of\u00a0 3$ is:<\/p>\n a)\u00a0$2\\frac{1}{3}$<\/p>\n b)\u00a0$-3\\frac{2}{3}$<\/p>\n c)\u00a0$-2\\frac{1}{3}$<\/p>\n d)\u00a0$3\\frac{2}{3}$<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Apply bodmas rule to this problem to simplify<\/p>\n BODMAS RULE<\/strong>.\u00a0BODMAS<\/strong>\u00a0is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. In certain regions, PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition and Subtraction) is the synonym of\u00a0BODMAS<\/strong>. It explains the order of operations to solve an expression<\/p>\n First solve the bracket then orders then division then multiplication then addition then subtraction<\/p>\n 3\u00d72\u00f73of12-3\u00f72\u00d7(2-3)\u00d72+3\u00f72of3<\/p>\n 3\u00d7$\\frac{2}{36}$-$\\frac{3}{2}$\u00d7(-1)\u00d72+3\u00f72\u00d73<\/p>\n 3\u00d7$\\frac{1}{18}$-$\\frac{3}{2}$\u00d7(-1)\u00d72+$\\frac{3}{6}$<\/p>\n $\\frac{1}{6}$+$\\frac{6}{2}$+$\\frac{1}{2}$<\/p>\n Take LCM of 6,2,2 you will get 6<\/p>\n =\u00a0$\\frac{1+18+3}{6}$<\/p>\n =$\\frac{22}{6}$<\/p>\n =$\\frac{11}{3}$<\/p>\n Question 9:\u00a0<\/b>The value of $\\frac{3 \\div \\left\\{5 – 5 \\div (6 – 7) \\times 8 + 9\\right\\}}{4 + 4 \\times 4 \\div 4\u00a0 of\u00a0 4}$ is:<\/p>\n a)\u00a0$\\frac{1}{3}$<\/p>\n b)\u00a0$\\frac{1}{45}$<\/p>\n c)\u00a0$\\frac{1}{90}$<\/p>\n d)\u00a0$\\frac{1}{18}$<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Apply BODMAS rule to this type of problems<\/p>\n According\u00a0to Bodmas\u00a0rule, if an expression contains brackets ((), {}, [])\u00a0we\u00a0have\u00a0to\u00a0first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left\u00a0to\u00a0right.<\/p>\n Numerator = $3 \\div \\left\\{5-5\\div(6-7)\\times8+9\\right\\} = 3\\div \\left\\{5-5\\div(-1)\\times8+9\\right\\}$ Denominator =\u00a0$4+4\\times4\\div4 \\text{of} 4 = 4+4\\times4\\div16 = 4+1 = 5$<\/p>\n $\\dfrac{3 \\div \\left\\{5 – 5 \\div (6 – 7) \\times 8 + 9\\right\\}}{4 + 4 \\times 4 \\div 4 of 4} = \\dfrac{(\\dfrac{1}{18})}{5} = \\dfrac{1}{90}$<\/p>\n Question 10:\u00a0<\/b>The value of $ 6 – 6 \\div 6 \\times 6 + (6 \\div 6\u00a0 of\u00a0 6) \\times 6 – \\left(3\\frac{2}{3} \\div \\frac{11}{30}\u00a0 of\u00a0 \\frac{2}{3}\\right) \\div 5$ is:<\/p>\n a)\u00a00<\/p>\n b)\u00a0-1<\/p>\n c)\u00a0-2<\/p>\n d)\u00a02<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Apply bodmas rule to this problem to simplify<\/p>\n BODMAS RULE<\/strong>.\u00a0BODMAS<\/strong>\u00a0is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. In certain regions, PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition and Subtraction) is the synonym of\u00a0BODMAS<\/strong>. It explains the order of operations to solve an expression<\/p>\n $ 6 – 6 \\div 6 \\times 6 + (6 \\div 6\u00a0 of\u00a0 6) \\times 6 – \\left(3\\frac{2}{3} \\div \\frac{11}{30}\u00a0 of\u00a0 \\frac{2}{3}\\right) \\div 5$<\/p>\n First solve the bracket then orders then division then multiplication then addition then subtraction<\/p>\n =6-1\u00d76+(6\u00f76\u00d76)\u00d76-($\\frac{11}{3}$\u00f7$\\frac{11}{30}$\u00d7$\\frac{2}{3}$)\u00f75<\/p>\n =6-6+($\\frac{6}{36}$)\u00d76-($\\frac{11}{3}$\u00d7$\\frac{90}{22}$)\u00f75<\/p>\n =0+$\\frac{1}{6}$\u00d76-($\\frac{30}{2}$)\u00f75<\/p>\n =0+1-$\\frac{15}{5}$<\/p>\n =0+1-3<\/p>\n =-2<\/p>\n Question 11:\u00a0<\/b>The value of $3.8 – (4.2 \\div 0.7 \\times 3) + 5 \\times 2 \\div 0.5$ is:<\/p>\n a)\u00a015.6<\/p>\n b)\u00a05.8<\/p>\n c)\u00a021.8<\/p>\n d)\u00a018.4<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n 4.2\/0.7 =6 Question 12:\u00a0<\/b>The value of $4.5 – (3.2 \\div 0.8 \\times 5) + 3 \\times 4 \\div 6$ is:<\/p>\n a)\u00a0-13.5<\/p>\n b)\u00a04.2<\/p>\n c)\u00a0-8.5<\/p>\n d)\u00a05.7<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n we have to solve this question according to the vbodmas rule<\/p>\n $4.5 – (3.2 \\div 0.8 \\times 5) + 3 \\times 4 \\div 6$<\/p>\n =$4.5-(4\\times5)+2$<\/p>\n =4.5-20+2<\/p>\n =6.5-20<\/p>\n = -13.5<\/p>\n Question 13:\u00a0<\/b>The value of a)\u00a0$\\frac{20}{9}$<\/p>\n b)\u00a0$\\frac{4}{25}$<\/p>\n c)\u00a0$\\frac{18}{125}$<\/p>\n d)\u00a0$\\frac{40}{9}$<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{9}{15} of \\left(\\frac{2}{3} \\div \\frac{2}{3} of \\frac{3}{2}\\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div \\frac{3}{4} of \\frac{4} {3}\\right) of \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{2}{5} of \\frac{4}{5}\\right)$<\/p>\n We know that “of” can be replace by multiply,<\/p>\n $\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\div 1 \\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div 1 \\right) of \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{8}{25} \\right)$<\/p>\n Now, solving small brakets,<\/p>\n $\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\right) \\div \\left(\\frac{9}{16} \\right) of \\left(\\frac{10}{20} \\times \\frac{8}{25} \\right)$<\/p>\n $\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\right) \\div \\left(\\frac{9}{16} \\right) of \\left(\\frac{4}{25} \\right)$<\/p>\n $\\Rightarrow \\dfrac{\\frac{18}{45}}{ \\left(\\frac{9}{100} \\right) }$<\/p>\n $\\Rightarrow \\frac{18\\times 100}{45\\times 9}$<\/p>\n $\\Rightarrow \\frac{2\\times 20}{9}$<\/p>\n $\\Rightarrow \\frac{40}{9}$<\/p>\n Question 14:\u00a0<\/b>The value of $(5 + 3 \\div 5 \\times 5) \\div (3 \\div 3\u00a0 of\u00a0 6)\u00a0 of\u00a0 (4 \\times 4 \\div 4\u00a0 of\u00a0 4 + 4 \\div 4 \\times 4)$ is:<\/p>\n a)\u00a0$7\\frac{1}{3}$<\/p>\n b)\u00a0$8\\frac{1}{5}$<\/p>\n c)\u00a0$9\\frac{3}{5}$<\/p>\n d)\u00a0$6\\frac{2}{3}$<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n As per the question,<\/p>\n $(5 + 3 \\div 5 \\times 5) \\div (3 \\div 3 of 6) of (4 \\times 4 \\div 4 of 4 + 4 \\div 4 \\times 4)$<\/p>\n Now,<\/p>\n $\\Rightarrow (5 + 3 \\div 5 \\times 5) \\div (3 \\div 3 of 6) of (4 \\times 4 \\div (4 \\times 4)+ \\dfrac{4}{4} \\times 4))$<\/p>\n $\\Rightarrow (5 + 3 \\div 5 \\times 5) \\div (3 \\div 3 of 6) of (4 \\times \\dfrac{4}{16} + 4)$<\/p>\n $\\Rightarrow (5 + 3 \\div 5 \\times 5) \\div (3 \\div 3 of 6) of (5)$<\/p>\n $\\Rightarrow (5 + 3 \\div 5 \\times 5) \\div (3\\div 18) of (5)$<\/p>\n $\\Rightarrow (5 + 3 \\div 5 \\times 5) \\div (\\dfrac{1}{6}) of (5)$<\/p>\n $\\Rightarrow \\dfrac{(5 + 3 \\div 5 \\times 5)}{(\\dfrac{5}{6})}$<\/p>\n $\\Rightarrow \\dfrac{(5 + \\dfrac{3}{5} \\times 5)}{ (\\dfrac{5}{6})}$<\/p>\n $\\Rightarrow \\dfrac{(5 + 3)\\times 6}{5}$<\/p>\n $\\Rightarrow \\dfrac{48}{ 5}$<\/p>\n $\\Rightarrow 9\\dfrac{3}{5}$<\/p>\n Question 15:\u00a0<\/b>The value of $2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7}\u00a0 of\u00a0 2\\frac{1}{3}\\right) \\times \\left[\\left(2\\frac{6}{7}\u00a0 of\u00a0 4\\frac{1}{5} \\div \\frac{2}{3}\\right) \\times \\frac{5}{9}\\right]$ is:<\/p>\n a)\u00a0$\\frac{1}{4}$<\/p>\n b)\u00a04<\/p>\n c)\u00a0$\\frac{1}{23}$<\/p>\n d)\u00a05<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that,<\/p>\n $2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7} of 2\\frac{1}{3}\\right) \\times \\left[\\left(2\\frac{6}{7} of 4\\frac{1}{5} \\div \\frac{2}{3}\\right) \\times \\frac{5}{9}\\right]$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7} of 2\\frac{1}{3}\\right) \\times\u00a0\\left[\\left(\\frac{20}{7} of \\frac{21}{5} \\div \\frac{2}{3}\\right) \\times \\frac{5}{9}\\right]$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7} of 2\\frac{1}{3}\\right) \\times \\left[\\left(\\frac{20\\times 21}{7\\times 5} \\div \\frac{2}{3}\\right) \\times \\frac{5}{9}\\right]$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7} of 2\\frac{1}{3}\\right) \\times \\left[\\left(\\frac{20\\times 21\\times 3}{7\\times 5\\times 2}\\right) \\times \\frac{5}{9}\\right]$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(3\\frac{5}{6} \\div \\frac{2}{7} of 2\\frac{1}{3}\\right) \\times 10$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(\\frac{23}{6} \\div \\frac{2}{7} of \\frac{7}{3}\\right) \\times 10$<\/p>\n $\\Rightarrow 2\\frac{7}{8} \\div \\left(\\frac{23}{6} \\div \\frac{2}{3}\\right) \\times 10$<\/p>\n $\\Rightarrow \\frac{23}{8} \\div \\left(\\frac{23\\times 3}{6\\times 2}\\right) \\times 10$<\/p>\n $\\Rightarrow \\frac{23}{8} \\div \\left(\\frac{23}{4}\\right) \\times 10$<\/p>\n $\\Rightarrow \\frac{23\\times 4}{8\\times 23} \\times 10$<\/p>\n $\\Rightarrow 5$<\/p>\n Question 16:\u00a0<\/b>The value of $5 \\div 5\u00a0 of\u00a0 5 \\times 2 + 2 \\div 2\u00a0 of\u00a0 2 \\times 5 – (5 – 2) \\div 6 \\times 2$ is:<\/p>\n a)\u00a0$\\frac{9}{5}$<\/p>\n b)\u00a0$\\frac{23}{2}$<\/p>\n c)\u00a019<\/p>\n d)\u00a0$\\frac{19}{10}$<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $5 \\div 5 of 5 \\times 2 + 2 \\div 2 of 2 \\times 5 – (5 – 2) \\div 6 \\times 2$<\/p>\n $\\Rightarrow 5 \\div 25 \\times 2 + 2 \\div 4 \\times 5 – (5 – 2) \\div 6 \\times 2$<\/p>\n $\\Rightarrow 5 \\div 25 \\times 2 + 2 \\div 4 \\times 5 – (3) \\div 6 \\times 2$<\/p>\n $\\Rightarrow 5 \\div 25 \\times 2 +\\dfrac{2}{4} \\times 5 – (3) \\div 6 \\times 2$<\/p>\n $\\Rightarrow \\dfrac{5}{25} \\times 2 +\\dfrac{2}{4} \\times 5 – \\dfrac{3}{6} \\times 2$<\/p>\n $\\Rightarrow \\dfrac{1}{5} \\times 2 +\\dfrac{1}{2} \\times 5 – \\dfrac{1}{2} \\times 2$<\/p>\n $\\Rightarrow \\dfrac{1}{5} \\times 2 +\\dfrac{1}{2} \\times 5 – \\dfrac{1}{2} \\times 2$<\/p>\n $\\Rightarrow \\dfrac{2}{5} +\\dfrac{ 5 }{2} – \\dfrac{2}{2}$<\/p>\n $\\Rightarrow \\dfrac{19}{10}$<\/p>\n Question 17:\u00a0<\/b>The value of $2 \\times 3 \\div 2\u00a0 of\u00a0 3 \\times 2 \\div (4 + 4 \\times 4 \\div 4\u00a0 of\u00a0 4 – 4 \\div 4 \\times 4)$ is:<\/p>\n a)\u00a08<\/p>\n b)\u00a01<\/p>\n c)\u00a04<\/p>\n d)\u00a02<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Following BODMAS ,<\/p>\n $2 \\times 3 \\div 2 of 3 \\times 2 \\div (4 + 4 \\times 4 \\div 4 of 4 – 4 \\div 4 \\times 4) $<\/p>\n =\u00a0$2 \\times 3 \\div 6 \\times 2 \\div (4 + 4 \\times 4 \\div 16 – 4 \\div 4 \\times 4) $<\/p>\n = $2 \\times\u00a0 \\frac{1}{2} \\times 2 \\div (4 + 4 \\times \\frac{1}{4} – 4 \\div 4 \\times 4) $<\/p>\n = $2 \\div (4 + 1 – 1 \\times 4) $<\/p>\n =\u00a0$2 \\div (5 – 4) $<\/p>\n =2<\/p>\n So , the answer would be option d) 2.<\/p>\n Question 18:\u00a0<\/b>The value of $15.2 + 5.8 \\div 2.9 \\times 2 – 3.5 of 2 \\div 0.5$ is equal to:<\/p>\n a)\u00a04.8<\/p>\n b)\u00a03.2<\/p>\n c)\u00a05.2<\/p>\n d)\u00a05.4<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $15.2 + 5.8 \\div 2.9 \\times 2 – 3.5 of 2 \\div 0.5$<\/p>\n =\u00a0$15.2 + 2\u00a0 \\times 2 – 14$<\/p>\n =\u00a0$15.2 + 4 – 14$<\/p>\n =5.2<\/p>\n So , the answer would be option c)5.2<\/p>\n Question 19:\u00a0<\/b>The value of $62 – 5$ of $(18 – 14) + 5 \\times 7$ is equal to:<\/p>\n a)\u00a018<\/p>\n b)\u00a028<\/p>\n c)\u00a015<\/p>\n d)\u00a077<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0$62 – 5$ of $(18 – 14) + 5 \\times 7$<\/p>\n = $62-(5\\times4)+(5\\times7)$<\/p>\n = $62-20+35=77$<\/p>\n => Ans – (D)<\/p>\n Question 20:\u00a0<\/b>The value of $(6 \\times 15) \\div (2 \\times 3) –\u00a0 2^2 + 3$ is:<\/p>\n a)\u00a012<\/p>\n b)\u00a014<\/p>\n c)\u00a04<\/p>\n d)\u00a06<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression :\u00a0$(6 \\times 15) \\div (2 \\times 3) –\u00a0 2^2 + 3$<\/p>\n = $\\frac{90}{6}-4+3$<\/p>\n = $15-1=14$<\/p>\n => Ans – (B)<\/p>\n
\n$= 3\\div\\left\\{5+5\\times8+9\\right\\} = 3\\div\\left\\{5+40+9\\right\\}$
\n$=\\dfrac{3}{54} = \\dfrac{1}{18}$<\/p>\n
\nusing BODMAS we get
\n3.8-6*3+5*4
\n=3.8+2
\n=5.8<\/p>\n
\n$\\frac{9}{15}\u00a0 of\u00a0 \\left(\\frac{2}{3} \\div \\frac{2}{3}\u00a0 of\u00a0 \\frac{3}{2}\\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div \\frac{3}{4}\u00a0 of\u00a0 \\frac{4} {3}\\right)\u00a0 of\u00a0 \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{2}{5}\u00a0 of\u00a0 \\frac{4}{5}\\right)$ is:<\/p>\n