20 RRB NTPC Mocks – Just Rs. 149<\/a><\/p>\n RRB NTPC Previous Papers [Download PDF]<\/a><\/p>\n Question 1:\u00a0<\/b>In an exam, Amit is supposed to calculate the product of 34.1 * 4.67. He instead calculates the product of 0.0341 * 46.7 as 1.59. What is the actual answer in the exam?<\/p>\n a)\u00a01.59<\/p>\n b)\u00a015.90<\/p>\n c)\u00a00.159<\/p>\n d)\u00a0159.0<\/p>\n Question 2:\u00a0<\/b>If $x^{2}-5x-24=0$, then find the value of x<\/p>\n a)\u00a0+8,-3<\/p>\n b)\u00a0+8,+3<\/p>\n c)\u00a0-8,-3<\/p>\n d)\u00a0-8,+3<\/p>\n Question 3:\u00a0<\/b>If A = 13, B = 24 and C = 19, which of the following is true? I. A+B-C = C a)\u00a0II and IV only<\/p>\n b)\u00a0I and III only<\/p>\n c)\u00a0I, II and IV only<\/p>\n d)\u00a0II, III and IV only<\/p>\n Question 4:\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 FREE RRB NTPC YOUTUBE VIDEOS<\/a><\/p>\n RRB NTPC Free Mock Test<\/a><\/p>\n Question 5:\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 6:\u00a0<\/b>The simplest form of $\\frac{391}{667}$ is<\/p>\n a)\u00a0$\\frac{23}{31}$<\/p>\n b)\u00a0$\\frac{19}{23}$<\/p>\n c)\u00a0$\\frac{15}{19}$<\/p>\n d)\u00a0$\\frac{17}{29}$<\/p>\n Question 7:\u00a0<\/b>When polynomial $x^{4}-3x^{2}+2x+5$ is divided by (x-1), the remainder is<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n Question 8:\u00a0<\/b>What will be the value of – a)\u00a0470.5<\/p>\n b)\u00a0480.5<\/p>\n c)\u00a0490.5<\/p>\n d)\u00a0481.5<\/p>\n Question 9:\u00a0<\/b>29% of 6500 is what percentage of 2500?<\/p>\n a)\u00a075.4%<\/p>\n b)\u00a076.5%<\/p>\n c)\u00a080.4%<\/p>\n d)\u00a072.5%<\/p>\n Question 10:\u00a0<\/b>If the number 24a53b1 is divisible by 9, then what can be the value of a+b?<\/p>\n a)\u00a03<\/p>\n b)\u00a012<\/p>\n c)\u00a021<\/p>\n d)\u00a0Cannot be determined<\/p>\n RRB NTPC Previous Papers Download PDF<\/a><\/p>\n RRB JE PREVIOUS PAPERS DOWNLOAD<\/a><\/p>\n Question 11:\u00a0<\/b>4.23232323… = ?<\/p>\n a)\u00a0422\/99<\/p>\n b)\u00a0440\/9<\/p>\n c)\u00a0419\/99<\/p>\n d)\u00a0415\/99<\/p>\n Question 12:\u00a0<\/b>$\\frac{32}{3}+\\frac{46}{9}-\\frac{14}{3}$ = ?<\/p>\n a)\u00a0100\/9<\/p>\n b)\u00a010<\/p>\n c)\u00a011<\/p>\n d)\u00a0103\/9<\/p>\n Question 13:\u00a0<\/b>Find the smallest positive integer that should be multiplied to 2352 to make it a perfect square.<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a07<\/p>\n d)\u00a011<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 14:\u00a0<\/b>What is the value of 21*2.9 + 72?<\/p>\n a)\u00a0122.9<\/p>\n b)\u00a0132.9<\/p>\n c)\u00a0133.8<\/p>\n d)\u00a0143.8<\/p>\n Instructions<\/b><\/p>\n What approximate value should come in place of question-mark (?) in the following questions ? (You are expected to calculate the exact value)<\/p>\n Question 15:\u00a0<\/b>$561204\\times58 = ? \\times 55555$<\/p>\n a)\u00a0606<\/p>\n b)\u00a0646<\/p>\n c)\u00a0586<\/p>\n d)\u00a0716<\/p>\n Question 16:\u00a0<\/b>$(9321+5406+1001)\\div(498+929+660)$ = ?<\/p>\n a)\u00a013.5<\/p>\n b)\u00a04.5<\/p>\n c)\u00a016.5<\/p>\n d)\u00a07.5<\/p>\n Question 17:\u00a0<\/b>What is the HCF of the fractions: \u00bd, 5\/7, 8\/11, \u00be ?<\/p>\n a)\u00a02\/121<\/p>\n b)\u00a01\/154<\/p>\n c)\u00a01\/121<\/p>\n d)\u00a01\/308<\/p>\n Question 18:\u00a0<\/b>Find the value of the following expression:126\\div 9 + 4*3-17<\/p>\n a)\u00a05<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n Question 19:\u00a0<\/b>What will be the value of – a)\u00a05500<\/p>\n b)\u00a05515<\/p>\n c)\u00a05510<\/p>\n d)\u00a05502<\/p>\n RRB NTPC Free Mock Test<\/a><\/p>\n RRB JE Free Mock Test<\/a><\/p>\n Question 20:\u00a0<\/b>Given that log2 = 0.3 approx, one billion would be approximately<\/p>\n a)\u00a0$2^{9}$<\/p>\n b)\u00a0$2^{10}$<\/p>\n c)\u00a0$2^{20}$<\/p>\n d)\u00a0$2^{30}$<\/p>\n Question 21:\u00a0<\/b>75% of a number from which 10 has been deducted is 180. The number is<\/p>\n a)\u00a0200<\/p>\n b)\u00a0240<\/p>\n c)\u00a0330<\/p>\n d)\u00a0250<\/p>\n Question 22:\u00a0<\/b>Which of the following is the arithmetic mean of $ 2X^{2}-10, 30-X^{2} and -X^{2}+6X+10 $ ?<\/p>\n a)\u00a03X+15<\/p>\n b)\u00a06X+10<\/p>\n c)\u00a02X+30<\/p>\n d)\u00a02X+10<\/p>\n Question 23:\u00a0<\/b>Ram was asked to calculate the sum of the first 40 numbers. He reported the sum as 799. He then realized that he missed adding a number during the summation. The number is<\/p>\n a)\u00a019<\/p>\n b)\u00a020<\/p>\n c)\u00a021<\/p>\n d)\u00a022<\/p>\n Question 24:\u00a0<\/b>$100 \\times 10 – 100 + 2000 \\div 100 = ?$<\/p>\n a)\u00a029<\/p>\n b)\u00a0920<\/p>\n c)\u00a0980<\/p>\n d)\u00a01000<\/p>\n Daily Free RRB Online Tests for RRB Exams<\/a><\/p>\n 100+ Free Online G.K Tests<\/a><\/p>\n Question 25:\u00a0<\/b>If \u2018P\u2019 means \u2018+\u2019, \u2018Q\u2019 means \u2018-\u2019, \u2018R\u2019 means \u2018\u00f7\u2019 and \u2018S\u2019 means \u2018x\u2019, then which of the following equation is correct?<\/p>\n a)\u00a014 R 7 S 6 P 4 Q 3 = 11<\/p>\n b)\u00a03 S 6 P 2 Q 3 R 6 = 35\/2<\/p>\n c)\u00a011 R 12 S 48 P 10 Q 6 = 48<\/p>\n d)\u00a09 S 8 P 6 R 4 S 8 = 80<\/p>\n Download General Science Notes PDF<\/a><\/p>\n Question 26:\u00a0<\/b>Find the least number among $\\frac{5}{9},\\sqrt\\frac{9}{49},0.43$ and $(0.7)^{2}.$<\/p>\n a)\u00a0$\\frac{9}{5}$<\/p>\n b)\u00a0$(0.7)^{2}$<\/p>\n c)\u00a0$\\sqrt\\frac{9}{49}$<\/p>\n d)\u00a00.43<\/p>\n Question 27:\u00a0<\/b>What is the value of $(\\frac{\\sqrt{2}}{3}-Cosec 60^\\circ)?$<\/p>\n a)\u00a0$\\frac{(\\sqrt{6}-6)}{3\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{(2-2\\sqrt{3})}{\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{(1-\\sqrt{6})}{\\sqrt{2}}$<\/p>\n d)\u00a0$\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n Question 28:\u00a0<\/b>Find the value of $\\sqrt{(2x-5)^{2}}+2\\sqrt{(x-1)^{2}}$, if 1<x<2.<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Question 29:\u00a0<\/b>What is the value of $(cosec 60^\\circ – \\frac{1}{2})$<\/p>\n a)\u00a0$\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{(2\\sqrt{3}-1)}{\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{(3\\sqrt{3}-1)}{3}$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n Question 30:\u00a0<\/b>Which of the following relation is CORRECT? a)\u00a0Only I<\/p>\n b)\u00a0Only II<\/p>\n c)\u00a0Neither I nor II<\/p>\n d)\u00a0Both I and II<\/p>\n Free RRB Daily Practice Set<\/a><\/p>\n Download Current Affairs Questions & Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 0.0341 * 46.7 = 1.59 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $x^{2}-5x-24=0$<\/p>\n So, $x^{2}-8x + 3x -24=0$<\/p>\n ie x(x- 8) + 3(x-8) = 0<\/p>\n => (x+3)(x-8)=0<\/p>\n or x = 8 or -3<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A+B-C = 13+24-19 = 18 which is not equal to 19. Hence, statement I. is wrong. Therefore, statement II and IV are true and the correct answer is option (a)<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The value of $\\sqrt{2025}$ = 45 5)\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 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\frac{391}{667}$ = $\\frac{17 \\times 23}{29 \\times 23}$ = $\\frac{17}{29}$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $f(x)=x^{4}-3x^{3}+2x +5$<\/p>\n to find the remainder put x=1 in $f(x)$<\/p>\n $f(1)$=1-3+2+5=5<\/p>\n \u22345 is the remainder<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 45% of 550 can be calculated as (40+5)% of 550 i.e. 220+27.5 = 247.5 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 29% of 6500 can be calculated as (30-1) % of 6500 i.e. 1950-65 = 1885 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 24a53b1 is divisible by 9 if (2+4+a+5+3+b+1) is divisible by 9. 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n x = 4.232323…. 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\frac{32}{3}+\\frac{46}{9}-\\frac{14}{3} = 6+ \\frac{46}{9}$ = $\\frac{100}{9}$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let\u2019s factorize 2352 into its prime factors. 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 21*2.9 = 60.9 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $561204\\times58 = ? \\times 55555$<\/p>\n implies ? = $561204\\times58 \\div 55555$ = 586<\/p>\n RRB General Science Notes PDF<\/a><\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 9321+5406+1001= 15728 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The HCF of fractions is equal to HCF of numerators\/LCM of denominators. 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 126\\div 9 + 4*3-17 = 14+4*3-17 = 14+12-17 = 26-17 = 9<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 59% of 6190 can be calculated as (50+9)% of 6190 i.e. equal to 3652.1 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 1 billion = $10^9$<\/p>\n log (1 billion) = 9<\/p>\n log 2 = 0.3<\/p>\n $log_2$1billion = 30<\/p>\n => 1 billion = $2^{30}$<\/p>\n 21)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the number be x. Hence, 75%*(x-10)=180. Hence, x=250.<\/p>\n 22)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Mean = sum\/3<\/p>\n = $ (2X^{2}-10 + 30-X^{2} -X^{2}+6X+10) $\/3<\/p>\n = 2x + 10<\/p>\n 23)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Sum of the numbers from 1 to 40 = 40*41\/2 = 20*41 = 820 24)\u00a0Answer\u00a0(B)<\/strong><\/p>\n By BODMAS, Division and Multiplication will be done before addition and subtraction.<\/p>\n Hence, 100 x 10 – 100 + 2000 \/ 100 = 1000 – 100 + 20 = 920<\/p>\n 25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : \u2018P\u2019 means \u2018+\u2019, \u2018Q\u2019 means \u2018-\u2019, \u2018R\u2019 means \u2018\u00f7\u2019 and \u2018S\u2019 means \u2018x\u2019<\/p>\n (A) : 14 R 7 S 6 P 4 Q 3 = 11<\/p>\n L.H.S. = $14\\div7\\times6+4-3$<\/p>\n = $12+1=13\\neq$ R.H.S.<\/p>\n (B) : 3 S 6 P 2 Q 3 R 6 = 35\/2<\/p>\n L.H.S. = $3\\times6+2-3\\div6$<\/p>\n = $18+2-0.5=19.5=\\frac{39}{2}\\neq$ R.H.S.<\/p>\n (C) : 11 R 12 S 48 P 10 Q 6 = 48<\/p>\n L.H.S. = $11\\div12\\times48+10-6$<\/p>\n = $44+4=48=$ R.H.S.<\/p>\n => Ans – (C)<\/p>\n 26)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Terms : $\\frac{5}{9}\\approx0.56$<\/p>\n $\\sqrt{\\frac{9}{49}}$ $=\\frac{3}{7}\\approx0.42$<\/p>\n $0.43$<\/p>\n $(0.7)^2=0.49$<\/p>\n Thus, the least number is = $\\sqrt\\frac{9}{49}$<\/p>\n => Ans – (C)<\/p>\n 27)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $(\\frac{\\sqrt{2}}{3}-Cosec 60^\\circ)$<\/p>\n = $\\frac{\\sqrt2}{3}-\\frac{2}{\\sqrt3}$<\/p>\n = $\\frac{(\\sqrt{6}-6)}{3\\sqrt{3}}$<\/p>\n => Ans – (A)<\/p>\n 28)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 29)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $(cosec 60^\\circ – \\frac{1}{2})$<\/p>\n = $\\frac{2}{\\sqrt3}-\\frac{1}{2}$<\/p>\n = $\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n => Ans – (A)<\/p>\n 30)\u00a0Answer\u00a0(A)<\/strong><\/p>\n I : $(\\sqrt{15}+\\sqrt{7})<(2\\sqrt{22})$<\/p>\n Squaring both sides, we get :<\/p>\n L.H.S. = $(\\sqrt{15}+\\sqrt{7})^2=15+7+2\\sqrt{105}=(22+2\\sqrt{105})\\approx(22+2\\times10)=42$<\/p>\n R.H.S. = $(2\\sqrt{22})^2=88$<\/p>\n Thus, L.H.S. < R.H.S., which is correct.<\/p>\n II : $(\\sqrt{17}+\\sqrt{5})<(\\sqrt{20}+\\sqrt{2})$<\/p>\n Squaring both sides, we get :<\/p>\n L.H.S. = $(\\sqrt{17}+\\sqrt{5})^2=17+5+2\\sqrt{85}=(22+2\\sqrt{85})$<\/p>\n R.H.S. = $(\\sqrt{20}+\\sqrt{2})^2=20+2+2\\sqrt{40}=(22+2\\sqrt{40})$<\/p>\n $\\because$ $\\sqrt{85}>\\sqrt{40}$, then L.H.S. > R.H.S.<\/p>\n Thus, only I is correct.<\/p>\n => Ans – (A)<\/p>\n![\"RRB](\"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/RRB-NTPC-Mock-Test.png\")
<\/a>
\nII. B+C-A = 30
\nIII. A+B+C = 2B
\nIV. A+C-B = 8<\/p>\n<\/a><\/p>\n
\n45% of 550 + 39% of 600<\/p>\n<\/a><\/p>\n
\n59% of 6190 + 37% of 5000 =<\/p>\n<\/a><\/p>\n
\nI. $(\\sqrt{15}+\\sqrt{7})<(2\\sqrt{22})$
\nII. $(\\sqrt{17}+\\sqrt{5})<(\\sqrt{20}+\\sqrt{2})$<\/p>\n<\/a><\/p>\n
\nSo, 34.1 * 46.7 = 1.59 * 1000
\n34.1 * 4.67 = 1.59 * 100
\nHence the answer is 159<\/p>\n
\nB+C-A = 24+19-13 = 30. Hence, statement II. is true.
\nA+B+C = 13+24+19 = 56 which is not equal to 2*24=48. Hence, statement III. is wrong.
\nA+C-B = 13+19-24 = 8. Hence, statement IV. is true.<\/p>\n
\nThe value of $\\sqrt{1152}$ = 34
\nSo, the value closest to $\\sqrt{2028}$ – $\\sqrt{1152}$ = 11<\/p>\n
\nAnd 39% of 600 can be calculated as (40-1)% of 600 i.e. 240-6 = 234
\nHence, their summation will be = 481.5<\/p>\n
\nHence, 1885 will be $\\frac{1885}{2500} \\times 100$ % of 2500
\ni.e. 75.4%<\/p>\n
\n(15+a+b) should be divisible by 9.
\nThe maximum value of each of \u201ca\u201d and \u201cb\u201d should be 9.
\na+b should not be more than 18.
\n15+a+b should not be more than 18+15 = 33.
\n15+a+b = 18 or 27
\na+b = 18-15 or 27-15 = 3 or 12
\na+b should be either 3 or 12.<\/p>\n
\n100x = 423.23232323….
\n100x-x = 423.2323…-4.2323… = 419
\nx = 419\/99<\/p>\n
\n2352 = 16 * 3 * 49 = $2^4 * 3 * 7^2 $.
\nTo be a perfect square, the number should have prime factors with even indices. Hence, for 3 to have an even index, the number should be multiplied with 3.<\/p>\n
\nThe expression boils down to 60.9 + 72 = 132.9<\/p>\n<\/a><\/p>\n
\n498+929+660= 2087
\n15728$\\div$2087= 7.536
\nSo option D is the right answer.<\/p>\n
\nHCF of numerators = HCF of 1, 5, 8 and 3 = 1
\nLCM of numerators = LCM of 2, 7, 11 and 4 = 7*11*4 = 308
\nSo, the HCF of the fractions is 1\/308<\/p>\n
\nAnd 37% of 5000 can be calculated as (30+7)% of 5000 i.e. equal to 1850
\nHence, their summation will be = 1850+3652.1 = 5502.1<\/p>\n
\nSo, difference = 820 – 799 = 21
\nThe number that Ram forgot to add was 21.<\/p>\n