Download RRB JE Arithmetic Questions and Answers PDF. Top 25 RRB JE Arithmetic questions based on asked questions in previous exam papers very important for the Railway JE exam.<\/p>\n
Download RRB JE Arithmetic Questions PDF<\/a><\/p>\n 10 RRB JE Mocks – Just Rs. 117<\/a><\/p>\n RRB JE Previous Papers [Download PDF]<\/a><\/p>\n Question 1:\u00a0<\/b>If \u2018+\u2019 means \u2018-\u2019, \u2018-\u2019 means \u2018*\u2019, \u2018*\u2019 means \u2018\/\u2019 and \u2018\/\u2019 means \u2018+\u2019, then what is the value of the following expression? 13 – 13 \/ 3 – 63 * 9 + 2 -5<\/p>\n a)\u00a0170<\/p>\n b)\u00a0180<\/p>\n c)\u00a0190<\/p>\n d)\u00a0200<\/p>\n Question 2:\u00a0<\/b>$\\frac{3}{4}\\times \\frac{16}{27} \\div \\frac{2}{3} = ?$<\/p>\n a)\u00a02\/3<\/p>\n b)\u00a03\/2<\/p>\n c)\u00a09\/4<\/p>\n d)\u00a04\/9<\/p>\n FREE RRB JE YOUTUBE VIDEOS<\/a><\/p>\n RRB NTPC Free Mock Test<\/a><\/p>\n Question 3:\u00a0<\/b>$\\sqrt{0.6+\\sqrt{0.0123-0.0082-0.0025}}$ = ?<\/p>\n a)\u00a00.6<\/p>\n b)\u00a00.7<\/p>\n c)\u00a00.8<\/p>\n d)\u00a00.9<\/p>\n Question 4:\u00a0<\/b>$\\frac{6}{8}\\times \\frac{16}{18} \\times \\frac{3}{4} \\times X = \\frac{4}{7}$, find $1\/X$ ?<\/p>\n a)\u00a04\/7<\/p>\n b)\u00a08\/7<\/p>\n c)\u00a07\/8<\/p>\n d)\u00a07\/4<\/p>\n Question 5:\u00a0<\/b>Find $((4096)^{1\/3})^{1\/2}=?$<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n\n Question 6:\u00a0<\/b>Find LCM of 2\/3 , 3\/6, 4\/12 ?<\/p>\n a)\u00a04<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n RRB NTPC Previous Papers Download PDF<\/a><\/p>\n Question 7:\u00a0<\/b>Find the LCM of 7\/2, 9\/4 and 5\/3 ?<\/p>\n a)\u00a0252<\/p>\n b)\u00a021<\/p>\n c)\u00a0126<\/p>\n d)\u00a0315<\/p>\n Question 8:\u00a0<\/b>Which of the following is divisible by 3, 7, 9 and 11 ?<\/p>\n a)\u00a02645<\/p>\n b)\u00a04158<\/p>\n c)\u00a03791<\/p>\n d)\u00a01188<\/p>\n Question 9:\u00a0<\/b>Find the angle between the hours hand and the minutes hand at 3:18 pm ?<\/p>\n a)\u00a013\u00b0<\/p>\n b)\u00a012\u00b0<\/p>\n c)\u00a09\u00b0<\/p>\n d)\u00a010\u00b0<\/p>\n\n Question 10:\u00a0<\/b>4591.15 – 528.116 = x + 456.123. Find x ?<\/p>\n a)\u00a03660.911<\/p>\n b)\u00a03666.911<\/p>\n c)\u00a03006.911<\/p>\n d)\u00a03606.911<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 11:\u00a0<\/b>Find the product of LCM and HCF of 48 and 72 ?<\/p>\n a)\u00a0432<\/p>\n b)\u00a0864<\/p>\n c)\u00a01728<\/p>\n d)\u00a03456<\/p>\n Question 12:\u00a0<\/b>$\\frac{2}{3}\\times \\frac{27}{36}\\times \\frac{4}{7} (X) = 20$, find $X$ ?<\/p>\n a)\u00a0210<\/p>\n b)\u00a0140<\/p>\n c)\u00a070<\/p>\n d)\u00a035<\/p>\n Question 13:\u00a0<\/b>The difference of two positive numbers is 10. 2 and 48 are the HCF and LCM of those two numbers, then find the smaller number ?<\/p>\n a)\u00a012<\/p>\n b)\u00a016<\/p>\n c)\u00a06<\/p>\n d)\u00a08<\/p>\n Question 14:\u00a0<\/b>Sum of two numbers is 99 and difference is 63, then find the smaller number ?<\/p>\n a)\u00a018<\/p>\n b)\u00a036<\/p>\n c)\u00a081<\/p>\n d)\u00a054<\/p>\n RRB NTPC Free Mock Test<\/a><\/p>\n RRB JE Free Mock Test<\/a><\/p>\n Question 15:\u00a0<\/b>What is the value closest to $2.0101^2$?<\/p>\n a)\u00a04.01<\/p>\n b)\u00a04.02<\/p>\n c)\u00a04.03<\/p>\n d)\u00a04.04<\/p>\n Question 16:\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 17:\u00a0<\/b>What is the value closest to $\\sqrt{2028}$ – $\\sqrt{1152}$?<\/p>\n a)\u00a08<\/p>\n b)\u00a09<\/p>\n c)\u00a010<\/p>\n d)\u00a011<\/p>\n\n Question 18:\u00a0<\/b>What is the greatest common divisor of 576 and 960?<\/p>\n a)\u00a096<\/p>\n b)\u00a0240<\/p>\n c)\u00a0144<\/p>\n d)\u00a0192<\/p>\n Question 19:\u00a0<\/b>What is the value of $2^5$ + 16 * 4 + 4 \/2 – 1?<\/p>\n a)\u00a097<\/p>\n b)\u00a067<\/p>\n c)\u00a047<\/p>\n d)\u00a027<\/p>\n Daily Free Online Tests for RRB Exams<\/a><\/p>\n Question 20:\u00a0<\/b>Find the total number of factors of $18^{25} \\times 24^{32}$<\/p>\n a)\u00a09360<\/p>\n b)\u00a09600<\/p>\n c)\u00a010126<\/p>\n d)\u00a08480<\/p>\n Question 21:\u00a0<\/b>The mean of five consecutive odd numbers is 87. What is the sum of the smallest two numbers in the group?<\/p>\n a)\u00a0170<\/p>\n b)\u00a0172<\/p>\n c)\u00a0168<\/p>\n d)\u00a0164<\/p>\n Download RRB GK Material PDF<\/a><\/p>\n Question 22:\u00a0<\/b>Given that the number 1234P is divisible by 11 and P is a single natural number, what is the value of P?<\/p>\n a)\u00a09<\/p>\n b)\u00a05<\/p>\n c)\u00a00<\/p>\n d)\u00a02<\/p>\n Question 23:\u00a0<\/b>What is the square root of 156.25<\/p>\n a)\u00a013.5<\/p>\n b)\u00a015.25<\/p>\n c)\u00a010.5<\/p>\n d)\u00a012.5<\/p>\n Question 24:\u00a0<\/b>What is the value of $7^3 + 8^3 + 3 \\times 7 \\times 8 \\times 15 – 15^3$?<\/p>\n a)\u00a0900<\/p>\n b)\u00a0169<\/p>\n c)\u00a0225<\/p>\n d)\u00a00<\/p>\n Question 25:\u00a0<\/b>If $X$ is a real number, what is the minimum value of $X^2 – 6X + 20$ ?<\/p>\n a)\u00a09<\/p>\n b)\u00a011<\/p>\n c)\u00a020<\/p>\n d)\u00a04<\/p>\n General Science Notes for RRB Exams (PDF)<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let us replace the symbols with the actual operators.<\/p>\n 13 – 13 \/ 3 – 63 * 9 + 2 -5 = 13 * 13 + 3*63\/9 – 2*5<\/p>\n Applying BODMAS rule, we get, 169 + 3*7 – 10 = 180 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\frac{3}{4}\\times \\frac{16}{27} \\div \\frac{2}{3}$ = $\\frac{3}{4}\\times \\frac{16}{27} \\times \\frac{3}{2}$ = $\\frac{4}{9} \\times \\frac{3}{2} = \\frac{2}{3}$<\/p>\n So the answer is option A.<\/p>\n\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\sqrt{0.6+\\sqrt{0.0123-0.0082-0.0025}}$ = $\\sqrt{0.6+\\sqrt{0.0016}}$ = $\\sqrt{0.6+0.04}$ = $\\sqrt{0.64} = 0.8$<\/p>\n So the answer is option C.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{6}{8}\\times \\frac{16}{18} \\times \\frac{3}{4} \\times X = \\frac{4}{7}$<\/p>\n $\\frac{3}{4}\\times \\frac{8}{9} \\times \\frac{3}{4} \\times X = \\frac{4}{7}$<\/p>\n $\\frac{1}{2} \\times X = \\frac{4}{7}$<\/p>\n $X = 8\/7$<\/p>\n $1\/X = 7\/8$<\/p>\n So the answer is option B.<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $((4096)^{1\/3})^{1\/2} = 16^{1\/2} = 4$<\/p>\n So the answer is option C.<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 2\/3 , 3\/6, 4\/12 = 2\/3 , 1\/2 , 1\/3<\/p>\n LCM of fractions = (LCM of numerators)\/(HCF of denominators) = (LCM of 2, 1, 1)\/(HCF of 3,2,3) = 2\/1 = 2<\/p>\n So the answer is option B.<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n LCM of fractions = LCM of numerators\/HCF of denominators<\/p>\n LCM of 7\/2, 9\/4 and 5\/3 = (LCM of 7,9,5)\/(HCF of 2,4,3) = 315\/1 = 315<\/p>\n So the answer is option A.<\/p>\n\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 3*7*9*11 = 2079<\/p>\n 2079*2 = 4158 is divisible by 3, 7, 9, 11<\/p>\n So the answer is option B.<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Hours hand rotates 30\u00b0 in 1 hour($\\because 360\/12 = 30$\u00b0) and rotates 0.5\u00b0 in 1 minute($\\because 30\/60 = 1\/2 = 0.5$\u00b0)<\/p>\n Minute hand rotates 6\u00b0 in 1 minute($\\because 360\/60 = 6$\u00b0)<\/p>\n At 3:18,<\/p>\n Angle made by hours hand = 3(30\u00b0)+18(0.5\u00b0) = 90\u00b0+ 9\u00b0= 99\u00b0<\/p>\n Angle made by minutes hand = 18(6\u00b0) = 108\u00b0<\/p>\n Hence the angle b\/w these two hands = 108\u00b0-99\u00b0 = 9\u00b0<\/p>\n So the answer is option C.<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 4591.15 – 528.116 = x + 456.123<\/p>\n 4063.034 = x + 456.123<\/p>\n 4063.034 – 456.123 = x<\/p>\n 3606.911 = x<\/p>\n So the answer is option D.<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Product of LCM & HCF of two numbers = product of those two numbers = 48*72 = 3456<\/p>\n So the answer is option D.<\/p>\n\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{2}{3}\\times \\frac{27}{36}\\times \\frac{4}{7} (X) = 20$<\/p>\n $\\frac{2}{7}(X) = 20$<\/p>\n $X = 10(7) = 70$<\/p>\n So the answer is option C.<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let one number is X, then another number is 10+X<\/p>\n Product of two numbers = (LCM)*(HCF)<\/p>\n (X)(X+10) = (2)(48)<\/p>\n X^2+10X = 96<\/p>\n X^2+10X-96 = 0<\/p>\n (X+16)(X-6) = 0<\/p>\n X = -16 or 6<\/p>\n Since X is positive, X = 6 = smaller number<\/p>\n So the answer is option C.<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n x+y = 99 —-(1)<\/p>\n x-y = 63 —–(2)<\/p>\n (1)-(2) ==> x+y-x+y = 99-63 ==> 2y = 36 ==> y = 18<\/p>\n So the answer is option A.<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let\u2019s calculate the value of 2.01 * 2.01 = 4.0401 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The formula we use to simplify the expression is (a+b)(a-b) = $a^2- b^2$ 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The value of $\\sqrt{2025}$ = 45 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $576 = 24*24 = 3*2^3*3*2^3 = 3^2*2^6$ 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We have to use BEDMAS rule in this situation. => 32 + 16 * 4 + 4\/2 – 1 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $18^{25} \\times 24^{32} = 2^{25} \\times 3^{50} \\times 2^{96} \\times 3^{32}$ 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the five odd numbers be equal to a, a+2, a+4, a+6 and a+8. 22)\u00a0Answer\u00a0(D)<\/strong><\/p>\n For a number to be divisible by 11, the sum of the odd digits should be equal to the sum of the even digits. 23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $15625 = 5*3125 = 25*625 = 25^3 = 5^6$ 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We know that $A^3 + B^3 + C^3 – 3ABC = (A+B+C) \\times (A^2 + B^2 + C^2 – 3ABC)$ 25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $X^2 – 6X + 20 = X^2 – 6X + 9 + 11$
\nTherefore, option B is the right answer.<\/p>\n
\nSince 2.0101 is greater than 2.01, the value of its square will be higher as well. So, it will be greater than 4.04.<\/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
\nThe value of $\\sqrt{1152}$ = 34
\nSo, the value closest to $\\sqrt{2028}$ – $\\sqrt{1152}$ = 11<\/p>\n
\n$960 = 24*40 = 3*2^3*5*2^3 = 3*5*2^6$
\nTherefore, their GCD equals $3*2^6 = 3*64 =192$<\/p>\n\n
\nSo we resolve Brackets, Exponents, Division, Multiplication, Addition, Subtraction in that order.<\/p>\n
\n=> 32 + 16 * 4 + 2 – 1
\n=> 32 + 64 + 2 – 1 = 98 -1 = 97<\/p>\n
\nThis equals $2^{121} \\times 3^{82}$
\nHence, the number of factors equals $122 \\times 83 = 10126$<\/p>\n
\nTherefore, their sum equals 5a + 20 and their average equals a+4
\nTherefore, a+4 = 87 and a = 83.
\nHence, the sum of the smallest two numbers in the group equals a + (a+2) = 83 + 85 = 168<\/p>\n
\nTherefore, 1+3+P = 2+4
\nOr P = 2<\/p>\n
\nHence, $\\sqrt{15625} = 5^3 = 125$
\nTherefore, $\\sqrt{156.25} = 12.5$<\/p>\n\n
\nIn this case, $A+B+C = 7+8-15 = 0$
\nTherefore, the value of $7^3 + 8^3 + 3 \\times 7 \\times 8 \\times 15 – 15^3 = 0$<\/p>\n
\nTherefore, this equals $(X-3)^2 + 11$
\nNote that $(X-3)^2 \\geq 0$
\nHence, the minimum value of $X^2 – 6X + 20$ is $11$<\/p>\n