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Questions Of Quantitative Aptitude Tests For Competitive Examinations

Most common questions and resources for preparation of competitive examinations

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Questions Of Quantitative Aptitude Tests For Competitive Examinations

  1. 1. Questions of Quantitative aptitude tests for competitive examinations by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – CSE & PGPSE STUDENTS (CSE & PGPSE are free online programmes open for all, free for all) mobile : 91+9414430763
  2. 2. My words..... My purpose here is to give a few questions, which are often asked in aptitude tests and competitive examinations. Please prepare well for your examinations. Please pass this presentation to all those who might need it. Let us spread knowledge as widely as possible. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we stand....
  3. 3. 13,16,49,169, 256, .. what will come next, if it will be a repeatition out of this series ? 1+3 = 4 and square of 4 = 16 now 1+6 = 7 and square of 7 = 49 now 4+9 = 13 square of 13 = 169 1+6+9 = 16, square of 16 is 256, 2+5+6 = 13, square of 13 is 169 so the next digit is 169. answer
  4. 4. 1,1,4,8,9,27 what will be the next 2 number? 1^2 =1 1^3 = 1 2^2 = 4 2^3 = 8 3^2 = 9 3^3 =27 4^2 = 16 , 4^3 = 64 so next numbers are 16, 64 answer.
  5. 5. What will be the angle in watch at 2.15 pm? Minute hand is at 15, hour hand is at (10*60/55 ) = 10.91 thus there is a gap of 15 – 10.91 = 4.09 the angle is : (4.09/60) *360 =24.54 degree answer
  6. 6. Rain = Drizzle is equal to 1. Boil = Sizzle, 2. Smile = laugh, 3. match = tournament, 4. flow = trickle Rain is bigger form of Drizzle. Thus we are going from bigger to smaller. Similar analogy can be observed in Flow =Trickle
  7. 7. In a family each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many daughters and sons does the family have? Options : A 4,3 B. 2,1 C. 3,2 D 5,3 first option gives the answer. For this type of questions, solve through options. In the case of 1 st option, we can see that each daughter has 3 brothers and 3 sisters but each son has 2 brothers and 4 sisters. So this is the answer
  8. 8. If 72 * 23 = 45, 13*12 =12, 23*14 = 25, 64*12 =?? Add 7 and 2 and also 2 and 3 now multiply these : we get : 9*5 = 45 similarly add 1+3 and 1+2 now multiply these we get : 4*3 = 12 similarly, add 2+3 and 1+4, we get : 5*5 = 25, similarly , 6+4 =10 and 1+2 = 3 multiply these : 10*3 = 30 answer
  9. 9. A train passes through a tunnel, whose length is 500 meters, in 1 minute moving at 72 km per hour, the length of the train is ? Solution : - Step 1. convert speed of train in meters per second : 72 * 5/18 = 20 m. p.s. Step 2 : total distance travelled in 1 minute = 20*60 = 1200 step 3. deduct the length of tunnel from this distance : 1200-500 = 700 meters. So answer = 700 meters.
  10. 10. The least number which is a perfect square and is divisible by each of numbers 16,20,24 is ? Look at the factors of these numbers : 16=2^4 20=2^2*5 24=2^3*3 thus taking them all : 2^4*5^2*3^2 = =16*25*9= 3600 answer note : here you have to take all the prime factors in power of multiple of 2.
  11. 11. The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ? Step 1. take LCM of 5,6,4,3 =product of all prime factors 5*2*3*2 = 60 step 2 : divide 2497 by 60 = remainder is 37 step 3. find difference of 60 and 37 = 23 step 4 : add 23 to 2497 and now it is divisible by 60, so it is divisible by each of the four numbers. Answer=23
  12. 12. .The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and pencils is ? Solution : - find HCF of 1001 and 910 divide 1001 by 910, remainder is 91, 910 is divisible by 91 so HCF is 91 the number of students is 91 answer
  13. 13. LCM of two numbers is 48..The numbers are in ratio 2:3. The sum of numbers is The two numbers are 2X and 3X 2*3*X = 48 or X = 8 the two numbers are : 16 and 24 the sum of the numbers = 40 answer
  14. 14. LCM and HCFof two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is ? The numbers are 1X and 4X and their HCF is 21 so the numbers are 21 and 84 respectively. Answer =84
  15. 15. A positive number which when added to 1000 gives a sum , which is greater than when it is multiplied by 1000.The positive integer is? It will be 1. answer
  16. 16. The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by 9 / sum of digits? Let us take 1,2,3 for each digit of the two digits, we have 3 options, so total we have 3*3 =9 options. 9/3 = 3 times – so each digit is repeated 3 times. Thus as a rule it will be 11 * 3 * (1+2+3) = 198 here. Thus divided by 9 gives 22 and divided by 6 gives 33. answer
  17. 17. How many terms are there in 2,4,6,8..........,1024? It is a question on AP (Arithmetic progression). Basic formula is : Tn = a + (n-1)*d a = first term, n= number of terms, d = difference between two terms 1024 = 2 + (n-1)2 2n= 1024 or n =512 answer
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  21. 21. If A+B+C =0, then A^3+B^3+c^3+ABC= ? We know that if A+B+C =0, then : A^3+B^3+C^3=3ABC thus 3ABC+ABC = 4ABC answer
  22. 22. The difference between compound interest and simple interest for 3 years @5% p.a. Can be found out by multiplying the principal by ? Let us assume the principal is 100. simple interest = 15, so the amount will be 115. compound interest = 100 * 1.05*1.05*1.05 = =115.76 Thus the difference between these is : .76 we can find the difference by multiplying the principal with (.76/100) answer
  23. 23. Simple interest on a sum of money is 1/9 th of sum and the number of years and the rate % per annum are equal. What is rate? Formula :( P*R*T)/100 = interest p=principal, r=rate, t=time, here r=t let us assume principal = 9, interest = 1 9 * R^2 / 100 =1 R^2= 100/9 R=10/3 = 3.33 answer
  24. 24. A person invested Rs. 5000 at some rate (simple interest) and 4000 at 1% higher rate of interest. If the interest in both the cases is equal after 4 years. What is the rate of interest in former case? (5000*R*4)/100=(4000*(r+1)*4)/100 5r=4r+4 r=4 answer
  25. 25. On a certain map of India, actual distance is 1450 Km. Between two cities. It is shown as 5 CM on map. What scale is being used? 5Cm represents 1450 Km. Thus 1Cm will represent : 290 Km. Or 1: 29 *10^6
  26. 26. The average price of 10 books is increased by 17 Rupees when one of them whose value is Rs.400 is replaced by a new book. What is the price of new book? Solution : - Total increase in price = 10*17 = 170 price of old item = 400 price of new item = 400+170 = 570 answer
  27. 27. The average marks of girls in a class is 62.5. The average marks of 4 girls among them is 60.The average marks of remaining girls is 63,then what is the number of girls in the class? Solution ...- 62.5X= 240+(X-4)63 62.5X = 240 +63X-252 -.5X = -12 X=24 answer
  28. 28. The average of 2,7,6 and x is 5 and the average of and the average of 18,1,6,x and y is 10 . what is the value of y? Solution : = 2+7+6+X=20, so X = 5 18+1+6+5+Y=50 so Y =20 answer
  29. 29. A team of 8 persons joins in a shooting competition. The best marksman scored 85 points.If he had scored 92 points ,the average score for the team would have been 84.The team scored was? Solution : 84*8= 672 but only 85 was scored instead of 92, so 672-7=665 (team score) average of team:665/8=83.125
  30. 30. 16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of groups B? Solution : 16*76 =1216 – (10*75) = 466 avg. For 6 = 466/6 =77.66 answer shortcut : 10/6 * (76-75) = 1.66, add it to 76= 77.66 answer.
  31. 31. The average temperature of the town in the first four days of a month was 58 degrees. The average for the second ,third,fourth and fifth days was 60 degree .If the temperature of the first and fifth days were in the ratio 7:8 then what is the temperature on the fifth day? (T+W+T+F)-(M+T+W+T) = (60*4)-(58*4)= 8 Difference is just due to 1 st and 5 th day. F-M or 8X-7X = 8 or X =8 temperature on 1 st day : 8*7 = 56 temperature on 5 th day = 8*8 = 64 answer
  32. 32. The average age of a class is 15 years .The average age of boys in the class is 16 years while that of the girls is 14.5 years .What is the ratio of boys to girls in the class. For boys : 16-15 = 1 for girls : 14.5 – 15 = .5 (ignore + - sign) now while calculating final ratio, put these in reverse order : boys : girls = .5:1 (here we have put .5 of girls first and then we have put 1 of boys) thus their ratio is 1:2 answer
  33. 33. A library has an average of 510 visitors on Sundays and 240 on other days .The average number of visitors per day in a month 30 days beginning with a Sunday is? Solution : - We have 5 Sundays : so 5*510=2550 for other days : 25*240=6000 total : 8550, avg=8550/30=285 answer
  34. 34. The difference between C.I and S.I. on a certain sum at 10% per annum for 2 yrs is Rs.631.find the sum Solution : let us assume the sum to be 100. on compound interest it will be : 100*1.1*1.1 = 121 on simple iterest it will be 100+(10*2) = 120 thus difference is 1. sum is : 631*100/1 = 63100 answer
  35. 35. A sum of money doubles itself at C.I in 15yrs.In how many yrs will it become 8 times? Solution : - 1 becomes 2 in 15 years 2 becomes 4 in 30 years 4 becomes 8 in 45 years. Thus it will take 45 years.
  36. 36. Solve the following puzzle : There are six cities A,B,C,D,E and F A is not a hill station. B and E are not historical places. D is not an industrial city. A and D are not historical cities. A and B are not alike. Assuming that there are 2 cities of each type : hill station, historical city and industrial town, Which two cities seem to be industrial centres?
  37. 37. Solution
  38. 38. solution... Thus only A and E are industrial towns.
  39. 39. Solve the following puzzle Five friends A,B,C,D,E travelled to five different cities of Chennai,Calcutta,Delhi,Bangalore and Hyderabad by five differnet modes of transport of Bus,Train,Aeroplane,Car and Boat from Mumbai. The person who travelled to Delhi did not travel by Boat. C went to Banalore by Car and B went to Calcutta by Aeroplane. D travelled by Boat whereas E travelled by Train. Mumbai is not connected by Bus to Delhi and Chennai.Person travelling to Delhi went by which mode?
  40. 40. SOLUTION...
  42. 42. SOLVE THIS PUZZLE Four youngmen A,B,C,D are friendly with four girls E,M,S and L.E and S are friends.C's girlfriend does not like E and S . M does not care for C.B's girlfriend is friendly with E.E does not like A. Who is A's girlfriend?
  43. 43. SOLUTION ...
  45. 45. SOLVE THIS There are five friends-A,S ,R ,Z and Y .A is shorter than S but taller than Y.R is the tallest. Z is little shorter than S and little taller than A. Who is the shortest? ANSWER : Y<A<Z<S<R thus Y is the shortest. Answer
  46. 46. Solve the following puzzle Among five boys V is taller than M,but not as tall as R,J is taller than D but shorter than M. Who is the tallest in their group? Answer : D<J<M<V<R answer : R is the tallest.
  47. 47. Solve this puzzle S is as much older than K as he is younger than P . N is as old as K .Which of the following is wrong? a)K is younger than P . b)N is younger than P . c)S is older than N . d)P is not the oldest. e)K is younger than S. ANSWER : N=K < S < P SO d IS WRONG
  48. 48. SOLVE THIS PUZZLE From amongst five doctors A,B,C,D and E;four engineers G,H,K and L; six teachers M,N,O,P,Q and R some teams are to be selected.A,B,G,H,O,P and Q are females and the rest are males. Where ever there is a male doctor ,no female teacher. Where ever there is a male engineer, no female doctor. There shall not be more than two male teachers in any team. If the team consists of 3 doctors,2 male engineers and 2 teachers, the members of the team could be: a)A B C K L M R b)B C D K L N R c)C D E K L M N d)C D E K L P R
  49. 49. SOLUTION It is given in question, that we have to take only male engineers. Thus we will be able to take only male doctors. If we take only male doctors, we shall be able to take only male teachers. Thus C is the only option possible (all males). Answer
  50. 50. Solve this puzzle There is a group of six persons A,B,C,D,E and F from a family. They are programmer,manager,lawyer,jeweler,doctor and engineer. The doctor is the grandfather of F who is a programmer. The manager D is married to A. C,the jeweler is married to the lawyer. B is the mother of F and E. There are two married couples in the family. What is the profession of E?
  51. 51. solution...
  52. 52. SOLUTION It is clear that the two couples are CB and AD. Thus the profession of A is doctor. Thus only profession left after all these is engineer. Thus profession of E is engineer. Answer
  53. 53. Solve this puzzle In a family of six persons,there are three generations.Each person has seperate profession and also they like different colours.There are two couples in the family.R is a CA and his wife neither is a doctor nor likes green colour. Engineer likes red colour and his wife is a teacher.M is mother-in-law of S and she likes orange colour. V is grandfather of T ,who is principal likes black colour.z is granddaughter of M and she likes blue colour. Z's mother likes white colour. Who is an Engineer?
  54. 54. Solution ....
  55. 55. Solution Thus from the table, it is clear that V is engineer
  56. 56. 14 students in a class failed in both QT and Accounts. 20 failed in QT, but 32 failed in Accounts. How many students were there in total ? 20+32 – 14 = 38 students. Answer
  57. 57. Solve the following : All tables are chairs All chairs are almirahs Some almirahs are pens All pens are bikes. Which conclusion can be drawn : 1. all almirahs are tables 2. some bikes are not pens 3.. Some tables are pens 4. some almirahs are bikes..
  58. 58. solution.... You cant be sure that 1 st or 2 nd or 3 rd conclusion is right. There is a possibility that this conclusion is not valid. Thus only 4 th conclusion is a valid conclusion – it is bound to be there. Answer
  59. 59. Solve the following question Statements : 1. some tables are fans 2. some fans are balls 3. Some balls are bats. Which conclusion can be drawn : 1. No table is ball 2. some bats are fans 3. some tables are not bats 4. some bats are tables. SOLUTION : NO CONCLUSION CAN BE DRAWN.
  60. 60. V is 2 years older to T. T is 4 years younger to H. H is of the same age as N. N is 1 year younger to M. What is the age of M, if their total age is 100 years? T seems to be youngest, so let us start with T. V=T+2, H = T+4, N = T+4, M=T+5 Now let us add their age : T+T+2+T+4+T+4+T+5 = 100 5T+15=100, or 5T=85 or T=17 the age of M = 17+5 = 22 answer
  61. 61. Statement v/s argument... Statement : Banks should be privatised. Argument 1 : Yes, they should be privatised, because other countries have also privatised banks. Argument 2: No – they should not be privatised, because government has lots of deposits in Banks. ANSWER : NO ARGUMENT IS STRONG ENOUGH.
  62. 62. Statement v/s argument Statement : Banks should be privatised. Argument 1 : Yes, they should be privatised, because only private sector can be efficient Argument 2: No – they should not be privatised, because they generate employment ANSWER : NO ARGUMENT IS STRONG ENOUGH. IN THE FIRST ARGUMENT THE WORD ONLY CREATES PROBLEMS. IN 2ND ARGUMENT, IT IS NOT CLEAR, WHY EMPLOYMENT IS A PRIORITY FOR BANKS.
  63. 63.
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  68. 68. There are five friends Lalu, Kalu, Monu, Anuj and raju. Lalu ia shorter than Kalu but taller than raju. Monu is tallest. Anuj is a little shorter than Kalu and a little taller than Lalu. Who is the shortest? R< L <A< K <M answer : Raju
  69. 69. (1)There is a group of five girls. (2)Lata is second in height but younger than Rupa. (3)Pooja is taller than Mem but younger in age. (4)Rupa and Mem are of the same age but Rupa is tallest between them. (5)Niru is taller than Puja and elder to Rupa. Arrange them in age and height AGE : N R M P/ L HEIGHT R L N P M
  70. 70. X knows more than A. Y knows as much as B. Z knows less than C. A nows more than Y. The best knowledge person amongst all is: X>A > Y=B C>Z we are not able to know who knows more Z or A. Thus answer cant be drawn.
  71. 71. Solve the following : Five children were administrated psychological tests to know their intellectual levels. In the report, psychologists pointed out that the child A is less intelligent than the child B. The child C is less intelligent than the child D. The child B is less intelligent than the child C and child A is more intelligent than child E. Which child is most intelligent? D>C>B>A>E thus D is the most intelligent
  72. 72. Solve the following : In an examination, Raj got more marks than Goti but not as many as Neha.Neha got more marks than Jitu and Kamal. Jitu got less marks than Goti but his marks are not the lowest in the group. Who is the second in the descending order of marks? N>R>G>J>K answer = Raj
  73. 73. Solve the following : (A)Gop is shorter than Nanu but taller than Kalu. (B)Lalu is shorter than Kalu. (C)Jitu is taller than Lalu. (D)Nanu is taller than Jitu . Who among them is the tallest? N>G>K / J>L so Nanu is the tallest.
  74. 74. Solve it ?? There are seven students P, Q, R, S, T, U and V (1) they take a series of tests (2)No two students get similar marks. (3)V always scores more than P. (4)P always scores more than Q. (5)Each time either R scores the highest and t gets the least or alternatively S scores the highest and U or Q scores the least. If S is ranked sixth and Q is ranked fifth,solve the questions on next slide ??
  75. 75. Which of these is possible ? a) V is ranked first or fourth (b) R is ranked second or third (c)Pis ranked second or fifth (d) U is ranked third or fourth (e) T is ranked fourth or fifth. Possible solutions : 1 2 3 4 5 6 7 R V P Q S T thus only U can be at 3 rd or 4 th rank. No other options is possible
  76. 76. Who is the oldest ? (1)A, B, C, D and E are five friends. (2)B is elder to E, but not as tall as C. (3)C is younger to A, and is taller to D and E. (4)A is taller to D, But younger to E. (5)D is elder to A but is shorter in the group. B / D>E / D > A>C the oldest can be either B or D.
  77. 77. QUESTIONS... A * B means A and B are the same age. A – B means B is younger than A. A + B means A is younger than B. SOLVE IT : A * b – c means (a) c is youngest (b) c is the oldest (c) b is younger than c (d)None of these ANSWER : c is youngest.
  78. 78. Solve the following Six persons A,B,C,D,E and F are sitting in two rows, three in each. E is not at the end of any row Dis second to the left of F. C the neighbour of E, is sitting diagonally opposite to D. B is the neighbour of F. Who are sitting diagonally opposite? A E C D B F ans : AF and DC are sitting diagonally opposite
  79. 79. Solve it ..... Eight books are kept one over the other counting from the top the second, fifth and sixth books are on plays. Two books on plays are between two books on composition. One book of plays is between two books on poetry. While the book at the top of the book of literature is a book of composition .Which book is fourth from the top? 1 2 3 4 5 6 7 8 P p P C p p C L Thus 4 th book is COMPOSITION.
  80. 80. SOLVE IT ... Four boys are sitting on a bench to be photographed. A is to the left of B. D is to the right of B. C is between B and D. Who would be second from the left in the photograph? A B C D ANS : C In the photo, A will be extreme right and D will be extreme left.
  81. 81. Solve it .... There are five different houses, A to E in a row. A is to the right of B and E is to the left of C and right of A. B is to the right of D. Which of the houses is in the middle? ANSWER : D B A E C THUS A IS IN THE MIDDLE.
  82. 82. SOLVE IT ... In a march past , seven persons are standing in a row. Q is standing left to R but right to P. O is standing right to N and left to P. Similarly , S is standing right to R and left to T. Find out who is standing in middle? ANS : N O P Q R S T the person in MIDDLE Is Q
  83. 83. Solve it ... Five children are sitting in a row. S is sitting next to P but not T who is sitting next to R who is sitting on extreme left and T is not sitting next to K. Who are sitting adjacent to S? Answer : RTPSK answer : P and K are adjacent to S.
  84. 84. Solve it Five Boys are sitting in a row, A is not adjacent to B or C.D is not adjacent to B. A is adjacent to E. E is at the middle in the row. Then, D is adjacent to whom ? DAECB or BCEAD ANSWER : A
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  87. 87. Up to the time the last vote was recorded, it was difficult to decide whether victory lay with the ruling party or the opposition Instead of lay, use lied remember the formats of lie and lay Lay : lay, laid laid lie : lie lay lain lay = to put lie = to rest here we will use lie as an infinitive not as verb
  88. 88. We should always side with those who is true and unselfish and work for others We should always side with those who are true and unselfish and work for others
  89. 89. Before giving the mixture to the child shake it thouroughly. Here it has been used for mixture, but it is coming just after child, thus it seems that child has to be shaked. So correct it : Before giving to the child the mixture, shake it thouroughly.
  90. 90. The short story should not exceed more than two hundred words. Dont use more than exceeds itself means more than, so there is no need of more than or instead of exceed use be of
  91. 91. He looks at everything from their best side. Instead of their, use its everything is singular so use its instead of their.
  92. 92. Page after Page of Gita were read and it gave great consolation to his mind. Instead of were, use was. Page is singular so use was instead of were.
  93. 93. His manners indicate that he has no other intention than to steal his money. Instead of than, use but
  94. 94. I will now deal with him in a manner different from the one I have adopted so far. Instead of I have adopted so far, use : I have been adopting so far (to show continuation).
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