Railway Recruitment Board Sample Paper

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Article Index Railway Recruitment Board Sample Paper Railway Recruitment Board Sample Paper -1 Railway Recruitment Board Sample Paper -2 Railway Recruitment Board Sample Paper -3 Railway Recruitment Board Sample Paper -4 Railway Recruitment Board Sample Paper -5 All Pages 1. A shopkeeper purchases a packet of 50 pens for Rs. 500. He sells a part of the packet at a profit of 30%. On the remaining part, he incurs a loss of 10%. If his overall profit on the entire packet is 10%, what is the number of pens sold at profit? a. 25 b. 30 c. 20 d. 15 e.None of these 2. For a loan of Rs.32,400 that I gave to my friend Anil, he promised to pay me the amount in monthly instalments starting with a certain amount & increasing the instalments every month a fixed amount of Rs.100. In this way he could clear off the loan in 24 months. How much was the first instalment? a. 100 b. 200 c. 400 d. 500 e.None of these Directions (Qs 3 to 5): Study the following information carefully & answer accordingly: (i) In a school a total of 220 students are studying together in two sections A & B in the ratio of 5:6. Students are studying only English or only Sanskrit or both English & Sanskrit. (ii) The number of students studying only English from section A & of those studying both English & Sanskrit from section B are equal & each of them is 40% of the students who are studying only English from section B. (iii) The number of students studying only Sanskrit from section A is 30% of the number of students studying in section B & 60% of the students who are studying only English from section B. 3. How many students are studying both English & Sanskrit from section A? a. 48 b. 16 c. 40 d. 36 e.None of these 4. How many students are studying only Sanskrit from section B? a. 36 b. 10 c. 12 d. 24 e.None of these 5. Number of students studying only English from section B is what percent more than that of the students studying only English from section A? a. 150% b. 100% c. 75% d. 20% e.None of these Directions (Qs 6 & 7): Study the following information carefully & answer accordingly: A survey was conducted by an agency in 25000 houses. It was found that 48% used ‘Head & Shoulder’, 48% used ‘Clinic Plus’ & 53% used ‘Pentene Shampoo’. 12% used both ‘Head & Shoulder’ & ‘Clinic Plus’ only & 10% used both ‘Clinic Plus’ & ‘Pentene’ only. 6. How many people used both Head & Shoulders & Pentene only if 8% used all the three? a. 2750 b. 2500 c. 2000 d. Data inadequate e.None of these 7. How many people used only Pentene if 8% used all the three shampoos? a. 5000 b. 6000 c. 8000 d. 8500 Directions (Qs 8 to 25): Following Questions are accompanied by three statements. You have to determine which statement is/are necessary/sufficient to answer the question, 8. What is Sangita’s present age? (1) Five years ago, Sangita’s age was double that of her son’s age that time. (2) Present age of Sangita & her son are in the ratio 11:6 respectively. (3) Five years hence, the respective ratio of Sangita’s age & her son’s age will become 12:7 a. Only 1 & 3 b. Only 2 & 3 c. Only 1 & 2 d. 1 , 2 & 3 All e.None of these 9. Find out Rahul’s percentage of marks in the examination: (1) He appeared in all the five papers of the exam carrying full marks of 100 for each paper. (2) He gained 6% marks more than the marks needed for the first division. (3) He got 30 marks more than the marks needed for first division. a. 1 & 3 are sufficient b. 2 alone is sufficient c. Either only 2 or 1 & 3 together are sufficient d. All together are necessary e.None of these 10. What is the price of a chair? (1) The price of three tables is equal to the price of seven chairs. (2) The difference between the price of a table & that of a chair is Rs.900. (3) The price of one table is 400/3% more than the price of one chair. a. Only 1 & 2 together are sufficient b. Only 1 & 3 together are sufficient c. all together are necessary d. 2 & either 1 or 3 together are sufficient. e.None of these 11. What is the total number of boys in the college? (1) The number of boys in the college is 30% more than that of girls. (2) The ratio between them is 13:10 (3) Total number of students in the college is 3910. a. Any two of them are sufficient b. 1 & 2 together are sufficient c. 1 & 3 together are sufficient d. Either 1 & 3 together or 2 & 3 together are sufficient. 12. What is the staff strength of Company ‘X’? (1) Male & female employees are in the ratio of 2:3 respectively (2) Of the officer employees 80% are males (3) Total number of officers is 132 a. 1 & 3 only b. 2 & either 3 or 1 only c. All 1, 2 & 3 d. Question cannot be answered e.None of these 13. Two persons X & Y entered into a partnership. Find the share of X’s profit if, (1) The ratio of the capitals of X & Y is 3:5 (2) X invested his money for two months more than Y (3) Y’s share in the profit is Rs. 1000 less than the share of X a. 1 & 2 are sufficient b. All together are necessary c. All together are not necessary d. Even all taken together are not sufficient e.None of these 14. Two persons L & M entered into a partnership. Find the share of L in the total profit (1) The ratio of the capitals of L & M is 4:5 (2)L invested his money two months more than M (3) M’s share in total profit is Rs.800 more than the share of L a. Questions cannot be answered even after using all the information b. All statements are required to answer the question c. 1 & 3 together d. 1 & 2 together e.None of these 15. Find the % discount on the market price offered by a shopkeeper (1) If the shopkeeper does not give the discount, he will gain a profit of 50% (2) If the shopkeeper gives the discount, the profit will be 38% (3) The cost price of the article is Rs.400 a. Any two of them b. All statements are required c. 3 & either 1 or 2 d. Only 1 & 2 together e.None of these 16. A shopkeeper sold an article & got Rs108 as profit. Find the profit percentage (1) Selling price of the article is Rs.828. (2) He gave 10% discount on the marked price, which is Rs.920. (3) Cost price of the article is Rs.720. a. Any two of them b. Either 1 or 3 c. Any of them d. Only 2 e.None of these 17. What is the cost price of an article? (1) After allowing a discount of 10% on marked price, the shopkeeper charges Rs.810. (2) If the shopkeeper does not give the discount, the shopkeeper gets a profit of 50%. (3) If the shopkeeper gives only 5% discount on marked price, he will have 42.50% profit. a. Any two of them. b. 1 & 2 together. c. 1 & either 2 or 3 d. Only 2 & 3 e.None of these 18. How much did an article originally cost? (1) It changed hands three times & each seller gained 25% (2) The third person sells it for Rs.250 more than the second. (3) The second person sells it for Rs.200 more than the first. a. 1 & either 2 or 3 b. Any two of them c. Any of them d. All statements are required e.None of these 19. What was the profit earned on the cost price by Mahesh by selling an article? (1) He got 15% concession on buying that article. (2) He sold it for Rs.3060/- (3) He earned a profit of 2% on the labeled price. a. Only 1 & 2 together are required. b. Only either 1 or 3 & 2 together are required c. Even with all 1, 2 & 3 the answer cannot be arrived at d. All 1, 2 & 3 together are required e.None of these 20. How many articles were sold? (1) Total profit earned was Rs.1596/- (2) Cost price per article was Rs.632/- (3) Selling price per article was Rs.765/- a. 1 & 2 only b. All 1, 2 & 3 c. Any two of the three d. Question cannot be answered even after the information in all the three statements. e.None of these 21. Rinku borrowed an amount of Rs.5000 from Milan & Rahul. What is the rate of interest? (1) Rinku returned the amount of Rs.5400 after due date to Milan. (2) Rinku returned the amount of Rs.5900 after due date to Rahul. (3) Rinku returned the money to Milan by simple interest whereas to Rahul by compound interest. a. Only 1 & 2 together are sufficient b. Only 2 & 3 together are sufficient c. Either 1 & 2 together or 2 & 3 together are sufficient d. 1, 2 & 3 given together are not sufficient e.None of these 22. The compound interest on a sum is Rs.573. Find the rate of interest (1) The simple interest on the same amount at the same rate is Rs.540 (2) The sum is Rs.3000 (3) The compound interest is compounded half yearly a. 1 & 2 are sufficient b. 2 & 3 are sufficient c. Any two statements are sufficient d. All given together are not sufficient e.None of these 23. Between P & Q who will get the maximum interest? (1) P deposits a certain sum at 15% p.a. for one & half years, when the interest is compounded half yearly. (2) Q deposits a certain sum of money at 15/2% per annum for 4 years, when the interest is compounded yearly. (3) The ratio of sums of money of P & Q is 2:3 a. Only 1 & 2 together are sufficient. b. Any two of the three statements are sufficient. c. Either 1 & 2 together or 2 & 3 together are sufficient. d. All 1, 2 & 3 together are necessary e.None of these 24. The compound interest on a sum is Rs.625. Find the rate of interest. (1) The sum is Rs.3500 (2) The simple interest on the same amount at the same rate is Rs.575 (3) The interest is compounded half yearly a. Only 2 & 3 together are sufficient b. Any two of the three statements are sufficient c. All 1, 2 & 3 together are necessary d. Even all together are not sufficient e.None of these 25. What is the rate of interest? (1) An mount becomes double in 5 years at simple interest (2) The difference between compound interest & simple interest at the end of 2 years is Rs.200 (3) The sum is Rs.5000 a. Only 1 b. 1 & 2 together c. Either 1 alone or 2 & 3 together d. 3 & either 1 or 2 e.None of these ANSWERS with Hints :- 1. a 2. b For Q. 3 to 5 From (1) Students in section A =5 x 220=100 11 Students in section B = 120 From (3) Students studying only Sanskrit from section A =30% of 120=36 »60% of students studying only English in section B=36 »Students studying only English in section B = 36x100 =60 60 From (2) Number of students studying only English from section A = Number of students studying both Sanskrit & English from section B = 40% of 60=24 Hence Subject Section A Section B Only English 24 60 Only Sanskrit 36 36 Both subjects 40 24 100 120 3. c 4. a 5. a 6. a 7. b 8. a, b & c all are correct. (Let M be the age of Sangita & N be her son’s present age. a. (M - 5) = 2(N - 5) »M = 2N – 10 + 5 = 2N – 5 M+5 = 12 N+5 7 Solving we get M = 55, N = 30 b.M = 11 or 6M = 11M N 6 M+5 = 12 N+5 7 Solving we get M = 55, N = 30 c. (M – 5) = 2(N – 5) M = 11 N 6 Solving we get M = 55, N = 30) 9. c 10. d 11. d 12. d (No data is available regarding the non officer employees) 13. c 14. b 15. d 16. c ((1)» Selling price = Rs.828 Cost price = Rs. (828-108) = Rs.720 Profit percentage = 108 x 100 = 15% 720 (3)» Profit percentage = 108 x 100 = 15% 720 (2)» Selling price = 90 x 920 = Rs.828 100 Profit = Rs.108 Cost price = Rs. (828-108) = Rs.720 Profit percentage = 108 x 100 = 15%) 720 17. c 18. a 19. d 20. b 21. d(No data is given for the time for which the amount was borrowed) 22. d 23. d 24. d 25. c

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Article Index For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -1 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -2 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -3 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -4 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -5 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -6 For Bank PO, MBA , BANK CLERK , RRB TC/ASM Exam -7 All Pages 1.Decimal fractions: Fraction in which denominations are powers of 10 are decimal fractions. Example:1 /10 = 0.1, 1 / 100 = 0.01 2.Convertion of Decimal into fraction:- Example: 0.25 = 25/100 = ¼ 3.i) If numerator and denominator contain same number of decimal places, then we remove decimal sign. Thus, 1.84/2.99 =184/299 Problems 1.0.75 =75/100 =3/4 2.Find products= 6.3204*100 = 632.04 3.2.61*1.3=261*13=3393 some of decimal places 2 +1 =3 sol: 3.393 4.If 1/3.718 =0.2689,then find value of 1/0.0003718 ? Sol: 10000/3.718 =10000*1/3.718 =10000*0.2689 = 2689 5.Find fractions : i) 0.37 = 37/99 ii)3.142857 =3+0.142857 =3 +142857/999999 = 3 142857/ 999999 iii) 0.17=17-1/90 =16/90=8/45 iv)0.1254 =1254 -12/9900 =1242/9900=69/550 6.Fraction 101 27/100000 Sol: 101+27/100000 =101+0.00027 =101.00027 7.If 47.2506 =4A + 7/B +2C + 5/D + 6E then 40+7+0.2+0.05+0.0006 Sol: compairing terms 4A= 40 => A=10 7/B = 7 => B=1 2C= 0.2=> C=0.1 5/D= 0.05=>D=5/0.05 =>5*100/5 =100 6E= 0.0006=> E= 0.0001 5A + 3B+6C+ D+ 3E = 5*10+ 3*1+ 6*0.1 + 100+ 3*0.0001 =50+3+0.6+100+0.0003 =153.6003 8.4.036 dividedby 0.04 Sol: 4.036/0.04 =4036/4 =100.9 9.[ 0.05/0.25 + 0.25/ 0.05]3 Sol: =>[5/25 + 25/5] = [1/5+ 5]3 =26/53 =5.23 = 140.603 10.The least among the following :- a. 0.2 b.1/0.2 c. 0.2 d. 0.22 sol:10/2 =5 0.2222 0.04 0.04 < 0.2 < 0.22 --------<5 Since 0.04 is least (0.2)2 is least. 11.Let F= 0.84181 Sol: when F is written as a fraction in lowest terms, denominator exceeds numerator by 84181 -841 /99000 = 83340/99000 =463/550 Required distance = (550 – 463) = 87 12.2 .75 + 3.78 Sol: [-2+0.75]+[-3+0.78] =-5+[0.75+0.78] = -5+1.53 =-5+1+0.53 = -4+0.53 = 4.53 13.the sum of first 20 terms of series is 1/5*6 +1/6*7+1/7*8—– Sol: [1/5 -1/6]+[1/6-1/7]+[1/7-1/8]+———————— = [1/5-1/25] =4/25=0.16 14.13 +23+ ————+93 =2025 Sol: value of (0.11) 3+ (0.22) 3+———(0.99)3 => (0.11) [1+2+--------+9] =0.001331*2025 =2.695275 15.(0.96)3 – (0.1)3/ (0.96)2 +0.096 +(0.1)2 Sol: formula => a3 -b3/a2 +ab +b2 =a -b (0.96-0.1)=0.86 16.3.6*0.48*2.50 / 0.12*0.09*0.5 Sol: 36*48*250/12*9*5=800 17.find x/y = 0.04/1.5 = 4/150 =2/75 find y-x/y+x (1- x/y) / (1+ x/y) 1 - 2/75 /1 +2/75 =73/77 18.0.3467+0.1333 Sol: 3467 -34/9900 + 1333-13/9900 = 3433 +1320/9900 = 4753/9900 = 4801 -48/9900 =0.4301 SQUARE AND CUBE ROOTS Formula: The Product of two same numbers in easiest way as follow. Example:let us calculate the product of 96*96 Solution: Here every number must be compare with the 100. See here the given number 96 which is 4 difference with the 100. so subtract 4 from the 96 we get 92 ,then the square of the number 4 it is 16 place the 16 beside the 92 we get answer as 9216. 9 6 - 4 ————– 9 2 ————– 4*4=16 9 2 1 6  therefore square of the two numbers 96*96=9216. Example: Calculate product for 98*98 Solution: Here the number 98 is having 2 difference when compare to 100 subtract 2 from the number then we get 96 square the number 2 it is 4 now place beside the 96 as 9604 9 8 - 2 ————- 9 6 ————- 2*2=4 9 6 0 4. so, we get the product of 98*98=9604. Example: Calculate product for 88*88 Solution: Here the number 88 is having 12 difference when compare to 100 subtract 12 from the 88 then we get 76 the square of the number 12 is 144 (which is three digit number but should place only two digit beside the 76) therefore in such case add one to 6 then it becomes 77 now place 44 beside the number 77 we will get 7744. 88 -12 ———— 76 ———– 12*12=144 76 + 144 ——————– 7744 ——————– Example: Find the product of the numbers 46 *46? Solution:consider the number 50=100/2. Now again go comparision with the number which gets when division with 100.here consider the number 50 which is nearer to the number given. 46 when compared with the number 50 we get the difference of 4. Now subtract the number 4 from the 46, we get 42. As 50 got when 100 get divided by 2. so, divided the number by 2 after subtraction. 42/2=21 now square the the number 4 i.e, 4*4=16 just place the number 16 beside the number 21 we get 2116. 4 6 4 —————- 4 2 as 50 = 100/2 42/2=21 now place 4*4=16 beside 21 2 1 1 6 Example: Find the product of the numbers 37*37 Solution: consider the number 50=100/2 now again go comparision with the number which gets when division with 100. here consider the number 50 which is nearer to the number given. 37 when compared with the number 50 we get the difference of 13. now subtract the number 13 from the 37, we get 24. as 50 got when 100 get divided by 2. so, divided the number by 2 after subtraction. 24/2=12 now square the the number 13 i.e, 13*13=169 just place the number 169 beside the number 21 now as 169 is three digit number then add 1 to 2 we get 1t as 13 then place 69 beside the 13 we get 1369. 3 7 1 3 —————– 2 4 as 50 = 100/2 24/2=12 square 13* 13=169 1 2 + 1 6 9 ———————– 1 3 6 9 ————————- Example: Find the product of 106*106 Solution: now compare it with 100 , The given number is more then 100 then add the extra number to the given number. That is 106+6=112 then square the number 6 that is 6*6=36 just place beside the number 36 beside the 112,then we get 11236. 1 0 6 + 6 ——————— 1 1 2 ——————– now 6* 6=36 place this beside the number 112, we get 1 1 2 3 6 Square root: If x2=y ,we say that the square root of y is x and we write ,√y=x. Cube root: The cube root of a given number x is the number whose cube is x. we denote the cube root of x by x1/3 . Examples: 1.Evaluate 60841/2 by factorization method. Solution: Express the given number as the product of prime factors. Now, take the product of these prime factors choosing one out of every pair of the same primes. This product gives the square root of the given number. Thus resolving 6084 in the prime factors ,we get 6084 2 6024 2 3042 3 1521 3 507 13 169 13 6084=21/2 *31/2 *131/2 60841/2=2*3*13=78. Answer is 78. 2.what will come in place of question mark in each of the following questions? i)(32.4/?)1/2 = 2 ii)86.491/2 + (5+?1/2)2 =12.3 Solution: 1) (32.4/x)1/2=2 Squaring on both sides we get 32.4/x=4 =>4x=32.4 =>x=8.1 Answer is 8.1 ii)86.491/2 + √(5+x2)=12.3 solution:86.491/2 + (5+x1/2 )=12.3 9.3+ √(5+x1/2 )=12.3 => √(5+x1/2 ) =12.3-9.3 => √(5+x1/2 )=3 Squaring on both sides we get (5+x1/2 )=9 x1/2 =9-5 x1/2 =4 x=2. Answer is 2. 3.√ 0.00004761 equals: Solution: âˆš (4761/108) √4761/√ 108 . 69/10000 0.0069. Answer is 0.0069 4.If √18225=135,then the value of  √182.25 + √1.8225 + √ 0.018225 + √0.00018225. Solution: âˆš(18225/100) +√(18225/10000) + √(18225/1000000) +√(18225/100000000) =√(18225)/10 + (18225)1/2/100 + √(18225)/1000 + √(18225)/10000 =135/10 + 135/100 + 135/1000 + 135/10000 =13.5+1.35+0.135+0.0135=14.9985. Answer is 14.9985. 5.what should come in place of both the question marks in the equation (?/ 1281/2= (162)1/2/?) ? Solution: x/ 1281/2= (162)1/2/x =>x1/2= (128*162)1/2 => x1/2= (64*2*18*9)1/2 =>x2= (82*62*32) =>x2=8*6*3 =>x2=144 =>x=12. 6.If 0.13 / p1/2=13 then p equals Solution: 0.13/p2=13 =>p2=0.13/13 =1/100 p2=√(1/100) =>p=1/10 therefore p=0.1 Answer is 0.1 7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to: Solution:37+(0.0615+x)1/2=37.25(since 37*37=1369) =>(0.0615+x)1/2=0.25 Squaring on both sides (0.0615+x)=0.0625 x=0.001 x=10-3. Answer is 10-3. 8.If √(x-1)(y+2)=7 x& y being positive whole numbers then values of x& y are? Solution: âˆš(x-1)(y+2)=7 Squaring on both sides we get (x-1)(y+2)=72 x-1=7 and y+2=7 therefore x=8 , y=5. Answer x=8 ,y=5. 9.If 3*51/2+1251/2=17.88.then what will be the value of 801/2+6*51/2? Solution: 3*51/2+1251/2=17.88 3*51/2+(25*5)1/2=17.88 3*51/2+5*51/2=17.88 8*51/2=17.88 51/2=2.235 therefore 801/2+6 51/2=(16*51/2)+6*1/25 =4 51/2+6 51/2 =10*2.235 =22.35 Answer is 22.35 10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is: Solution: 4b=6c =>b=3/2*c 3a=4b =>a=4/3b =>a=4/3(3/2c)=2c therefore a+b+c=27*291/2 2c+3/2c+c=27*291/2 =>4c+3c+2c/2=27*291/2 =>9/2c=27*291/2 c=27*291/2*2/9 c=6*291/2 11.If 2*3=131/2 and 3*4=5 then value of 5*12 is Solution:Here a*b=(a2+b2)1/2 therefore 5*12=(52+122)1/2 =(25+144)1/2 =1691/2 =13 Answer is 13. 12.The smallest number added to 680621 to make the sum a perfect square is Solution: Find the square root number which is nearest to this number 8 680621 824 64 162 406 324 1644 8221 6576 1645 therefore 824 is the number ,to get the nearest square root number take (825*825)-680621 therefore 680625-680621=4 hence 4 is the number added to 680621 to make it perfect square. 13.The greatest four digit perfect square number is  Solution: The greatest four digit number is 9999. now find the square root of 9999. 9 9999 99 81 189 1819 1701 198 therefore 9999-198=9801 which is required number. Answer is 9801. 14.A man plants 15376 apples trees in his garden and arranges them so, that there are as many rows as there are apples trees in each row .The number of rows is. Solution: Here find the square root of 15376. 1 15376 124 1 22 53 44 244 976 976 0 therefore the number of rows are 124.  15.A group of students decided to collect as many paise from each member of the group as is the number of members. If the total collection amounts to Rs 59.29.The number of members in the group is: Solution: Here convert Money into paise. 59.29*100=5929 paise. To know the number of member ,calculate the square root of 5929.7 5929 77 49 147 1029 1029 0 Therefore number of members are 77. 16.A general wishes to draw up his 36581 soldiers in the form of a solid square ,after arranging them ,he found that some of them are left over .How many are left? Solution: Here he asked about the left man ,So find the square root of given number the remainder will be the left man 1 36581 191 1 29 265 261 381 481 381 100(since remaining) Therefore the left men are 100. 17.By what least number 4320 be multiplied to obtain number which is a perfect cube? Solution: find l.c.m for 4320. 2 4320 2 2160 2 1080 2 540 2 270 3 135 3 45 3 15 5 4320=25 * 33 * 5 =23 * 33 * 22 *5 so make it a perfect cube ,it should be multiplied by 2*5*5=50 Answer is 50. 18.3(4*12/125)1/2=? Solution: 3(512/125)1/2 3(8*8*8)1/2/(5*5*5) 3(83)1/2/(53) ((83)/(53))1/3 =>8/5 or 1 3/5.

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