Download Top-20 IBPS RRB Clerk Approximation Questions PDF. Approximation questions based on asked questions in previous year exam papers very important for the IBPS RRB Assistant exam<\/p>\n
Download Approximation Questions For IBPS RRB Clerk<\/a><\/p>\n 35 IBPS RRB Clerk Mocks @ Rs. 149<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB Clerk<\/a><\/p>\n Download IBPS RRB Clerk Previous Papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following questions? a)\u00a025<\/p>\n b)\u00a035<\/p>\n c)\u00a040<\/p>\n d)\u00a010<\/p>\n Question 2:\u00a0<\/b>What approximate value should comein place of the question mark (?) in the following equation? a)\u00a0$\\frac{1}{6}$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0$\\frac{1}{9}$<\/p>\n d)\u00a0$\\frac{1}{3}$<\/p>\n Question 3:\u00a0<\/b>What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) a)\u00a0$6$<\/p>\n b)\u00a0$7$<\/p>\n c)\u00a0$8$<\/p>\n d)\u00a0$9$<\/p>\n e)\u00a0$10$<\/p>\n Question 4:\u00a0<\/b>What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) a)\u00a0$6$<\/p>\n b)\u00a0$7$<\/p>\n c)\u00a0$8$<\/p>\n d)\u00a0$9$<\/p>\n e)\u00a0$10$<\/p>\n Question 5:\u00a0<\/b>What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) a)\u00a0$64$<\/p>\n b)\u00a0$72$<\/p>\n c)\u00a0$80$<\/p>\n d)\u00a0$81$<\/p>\n e)\u00a0$63$<\/p>\n Question 6:\u00a0<\/b>What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) a)\u00a0$6$<\/p>\n b)\u00a0$7$<\/p>\n c)\u00a0$8$<\/p>\n d)\u00a0$9$<\/p>\n e)\u00a0$10$<\/p>\n Question 7:\u00a0<\/b>What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) a)\u00a0$24$<\/p>\n b)\u00a0$27$<\/p>\n c)\u00a0$30$<\/p>\n d)\u00a0$32$<\/p>\n e)\u00a0$36$<\/p>\n Question 8:\u00a0<\/b>What approximate value should come in place of (?) in the following equation ? a)\u00a0395<\/p>\n b)\u00a0295<\/p>\n c)\u00a0300<\/p>\n d)\u00a0315<\/p>\n e)\u00a0310<\/p>\n Question 9:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question ? a)\u00a088<\/p>\n b)\u00a046<\/p>\n c)\u00a0112<\/p>\n d)\u00a056<\/p>\n e)\u00a027<\/p>\n Question 10:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question? a)\u00a084<\/p>\n b)\u00a048<\/p>\n c)\u00a0118<\/p>\n d)\u00a058<\/p>\n e)\u00a024<\/p>\n Free Mock Test for IBPS RRB Clerk<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 11:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following questions? a)\u00a049<\/p>\n b)\u00a058<\/p>\n c)\u00a022<\/p>\n d)\u00a066<\/p>\n e)\u00a034<\/p>\n Question 12:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question? a)\u00a03795<\/p>\n b)\u00a03800<\/p>\n c)\u00a03810<\/p>\n d)\u00a03875<\/p>\n e)\u00a03995<\/p>\n Question 13:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value) a)\u00a03800<\/p>\n b)\u00a04500<\/p>\n c)\u00a02600<\/p>\n d)\u00a03000<\/p>\n e)\u00a04080<\/p>\n Question 14:\u00a0<\/b>What is the approximate value of $\\frac{(\\sqrt{334} + \\sqrt{223}) * (\\sqrt{334} – \\sqrt{223})}{(\\sqrt{111} + \\sqrt{11}) * (\\sqrt{111} – \\sqrt{11})}$?<\/p>\n a)\u00a01.01<\/p>\n b)\u00a01.10<\/p>\n c)\u00a01.11<\/p>\n d)\u00a01.21<\/p>\n Question 15:\u00a0<\/b>What is the approximate value of 29\/182?<\/p>\n a)\u00a0.14<\/p>\n b)\u00a0.13<\/p>\n c)\u00a0.17<\/p>\n d)\u00a0.16<\/p>\n e)\u00a0.18<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n By applying BODMAS we have 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n ($\\dfrac{1}{9}\\div\\dfrac{1}{9}$\u00a0 of\u00a0 $\\dfrac{1}{3}$) of $\\dfrac{1}{6} = (\\dfrac{1}{9} \\div \\dfrac{1}{27}$)\u00a0 of\u00a0 $\\dfrac{1}{6}$<\/p>\n $= \\dfrac{1}{9} \\times 27$\u00a0 of\u00a0 $\\dfrac{1}{6}$<\/p>\n $= 3 \\times \\dfrac{1}{6} = \\dfrac{1}{2}$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The given expression can be approximated to $\\frac{30}{5} – \\frac{8}{4} + \\frac{32}{8}$ 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The given expression can be approximated to $\\frac{108}{3*3} – \\frac{45}{9}$ 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The given expression can be approximated to $\\frac{72}{9} * \\frac{63}{7}$. 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The given expression can be simplified as $50 – \\frac{36}{42} *49$ 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The given expression can be approximated to $\\frac{\\sqrt{81 * 441}}{7}$ 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a09876 $\\div$ \u00a024.96 + 215.005 – ? = 309.99<\/p>\n => $\\frac{9875}{25} + 215 – ? = 310$<\/p>\n => $395 + 215 – 310 = ?$<\/p>\n => $? = 610 – 310 = 300$<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Expression\u00a0:\u00a0$45.709 \\div 11.342 \\times 6.667 = ?$<\/p>\n = $\\frac{44}{11} \\times 7$<\/p>\n = $4 \\times 7 = 28$<\/p>\n $\\approx$ 27<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Expression\u00a0: $54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n = $\\frac{55}{10} \\times 4.5$<\/p>\n = $24.75 \\approx 24$<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Expression\u00a0:\u00a0$59.786 \\div 14.444 \\times 8.321 = ?$<\/p>\n = $\\frac{60}{15} \\times 8$<\/p>\n = $4 \\times 8$<\/p>\n = $32 \\approx 34$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a0$4275 \\div 496 \\times (21)^{2} = ?$<\/p>\n = $\\frac{4275 \\times 441}{495}$<\/p>\n = 3808.63 $\\approx$ 3810<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Since an approximate answer is sufficient, you can rewrite the question as $10000 \\div 50 \\times 5^{2} – 1130$<\/p>\n Following BDMAS rule the result is 3870.<\/p>\n The nearest value is 3800.<\/p>\n Hence Option A is the correct answer.<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The formula we use to simplify the expression is (a+b)(a-b) = $a^2- b^2$ 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n For this fraction, the numerator and denominator do not have any common factors and thus it is irreducible. Thus, it is better to first approximate the denominator to a number which can divide the numerator easily.<\/p>\n 29\/182 $\\approx$ 30\/180 = 0.16<\/p>\n
\n$21 + 68 \\div 17 = ? $<\/p>\n
\n($\\frac{1}{9}\\div\\frac{1}{9}$ of $\\frac{1}{3}$) of $\\frac{1}{6}=?$<\/p>\n
\n$\\frac{29.96}{5.08} – \\frac{7.99}{4.01} + \\frac{31.97}{7.94}$<\/p>\n
\n$\\frac{107.94}{2.99*3.02} – \\frac{44.92}{8.97}$<\/p>\n
\n$\\frac{71.94}{8.99} * \\frac{62.91}{7.02}$<\/p>\n
\n$50.011 – \\frac{35.996}{42.007} *48.998$<\/p>\n
\n$\\frac{\\sqrt{80.98 * 440}}{6.99}$<\/p>\n
\n9876 $\\div$ \u00a024.96 + 215.005 – ? = 309.99<\/p>\n
\n$45.709 \\div 11.342 \\times 6.667 = ?$<\/p>\n
\n$54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n
\n$59.786 \\div 14.444 \\times 8.321 = ?$<\/p>\n
\n$4275 \\div 496 \\times (21)^{2} = ?$<\/p>\n
\n$9980 \\div 49 \\times (4.9)^{2} – 1130 = ?$<\/p>\n
\n=$21+(68\\div17)$
\n=21+4
\n=25<\/p>\n
\n= $6 – 2 + 4$
\n=$8$.
\nTherefore, option C is the right answer.<\/p>\n
\n=$\\frac{108}{9} – \\frac{45}{9}$
\n=$\\frac{63}{9}$
\n=$7$.
\nTherefore, option B is the right answer.<\/p>\n
\n=$8*9$
\n=$72$
\nTherefore, option B is the right answer.<\/p>\n
\n=$50 – \\frac{6}{7} *49$
\n=$50 – 42$
\n= $8$.
\nTherefore, option C is the right answer.<\/p>\n
\n=$\\frac{9*21}{7}$
\n=$9*3$
\n=$27$.
\nTherefore, option B is the right answer.<\/p>\n
\nSo, $\\frac{(\\sqrt{334} + \\sqrt{223}) * (\\sqrt{334} – \\sqrt{223})}{(\\sqrt{111} + \\sqrt{11}) * (\\sqrt{111} – \\sqrt{11})}$ = $\\frac{334 – 223}{111-11}$ = 1.11<\/p>\n