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First Job. Dream Job. Freshersworld.com

Aptitude Questions1.One of the following is my secret word:AIM DUE MOD OAT TIE.With the list in front of you, if I were to tell you any one of my secret word, then you would be able to tell me the number of vowels in my secret word.Which is my secret word? Ans.TIE

2.In the following figure:A B C D E F G H I Each of the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 is: a)Represented by a different letter in the figure above. b)Positioned in the figure above so that each of A + B + C,C + D +E,E + F + G, and G + H + I is equal to 13. Which digit does E represent? Ans.E is 4 3.One of Mr. Horton,his wife,their son,and Mr. Horton's mother is a doctor and another is a lawyer. a)If the doctor is younger than the lawyer, then the doctor and the lawyer are not blood relatives. b)If the doctor is a woman, then the doctor and the lawyer are blood relatives. c)If the lawyer is a man, then the doctor is a man. Whose occupation you know? Ans.Mr. Horton:he is the doctor. 4.Here is a picture of two cubes:

a)The two cubes are exactly alike. b)The hidden faces indicated by the dots have the same alphabet on them. Which alphabet-q, r, w, or k is on the faces indicated by the dots? Ans.q

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5.In the following figure: A B C G D E F

Each of the seven digits from 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is: a)Represented by a different letter in the figure above. b)Positioned in the figure above so that A*B*C,B*G*E, and D*E*F are equal. Which digit does G represent? Ans.G represents the digit 2. 6.Mr. and Mrs. Aye and Mr. and Mrs. Bee competed in a chess tournament.Of the three games played: a)In only the first game werethe two players married to each other. b)The men won two games and the women won one game. c)The Ayes won more games than the Bees. d)Anyone who lost game did not play the subsequent game. Who did not lose a game? Ans.Mrs.Bee did not lose a game. 7.Three piles of chips--pile I consists one chip, pile II consists of chips, and pile III consists of three chips--are to be used in game played by Anita and Brinda.The game requires: a)That each player in turn take only one chip or all chips from just one pile. b)That the player who has to take the last chip loses. c)That Anita now have her turn. From which pile should Anita draw in order to win? Ans.Pile II 8.Of Abdul, Binoy, and Chandini: a)Each member belongs to the Tee family whose members always tell the truth or to the El family whose members always lie. b)Abdul says ''Either I belong or Binoy belongs to a different family from the other two." Whose family do you name of? Ans.Binoy's family--El. 9.In a class composed of x girls and y boys what part of the class is composed of girls A.y/(x + y) B.x/xy C.x/(x + y) D.y/xy Ans.CFreshersworld.com Resource Center

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10.What is the maximum number of half-pint bottles of cream that can be filled with a 4-gallon can of cream(2 pt.=1 qt. and 4 qt.=1 gal) A.16 B.24 C.30 D.64 Ans.D 11.If the operation,^ is defined by the equation x ^ y = 2x + y,what is the value of a in 2 ^ a = a ^ 3 A.0 B.1 C.-1 D.4 Ans.B 12.A coffee shop blends 2 kinds of coffee,putting in 2 parts of a 33p. a gm. grade to 1 part of a 24p. a gm.If the mixture is changed to 1 part of the 33p. a gm. to 2 parts of the less expensive grade,how much will the shop save in blending 100 gms. A.Rs.90 B.Rs.1.00 C.Rs.3.00 D.Rs.8.00 Ans.C 13.There are 200 questions on a 3 hr examination.Among these questions are 50 mathematics problems.It is suggested that twice as much time be spent on each maths problem as for each other question.How many minutes should be spent on mathematics problems A.36 B.72 C.60 D.100 Ans.B


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14.In a group of 15,7 have studied Latin, 8 have studied Greek, and 3 have not studied either.How many of these studied both Latin and Greek A.0 B.3 C.4 D.5 Ans.B 15.If 13 = 13w/(1-w) ,then (2w)2 = A.1/4 B.1/2 C.1 D.2 Ans.C 16. If a and b are positive integers and (a-b)/3.5 = 4/7, then (A) b < a (B) b > a (C) b = a (D) b >= a Ans. A 17. In june a baseball team that played 60 games had won 30% of its game played. After a phenomenal winning streak this team raised its average to 50% .How many games must the team have won in a row to attain this average? A. 12 B. 20 C. 24 D. 30 Ans. C 18. M men agree to purchase a gift for Rs. D. If three men drop out how much more will each have to contribute towards the purchase of the gift/ A. D/(M-3) B. MD/3


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First Job. Dream Job. Freshersworld.comC. M/(D-3) D. 3D/(M2-3M) Ans. D 19. A company contracts to paint 3 houses. Mr.Brown can paint a house in 6 days while Mr.Black would take 8 days and Mr.Blue 12 days. After 8 days Mr.Brown goes on vacation and Mr. Black begins to work for a period of 6 days. How many days will it take Mr.Blue to complete the contract? A. 7 B. 8 C. 11 D. 12 Ans.C

20. 2 hours after a freight train leaves Delhi a passenger train leaves the same station travelling in the same direction at an average speed of 16 km/hr. After travelling 4 hrs the passenger train overtakes the freight train. The average speed of the freight train was? A. 30 B. 40 C.58 D. 60 Ans. B

21. If 9x-3y=12 and 3x-5y=7 then 6x-2y = ? A.-5 B. 4 C. 2 D. 8 Ans. D 22. There are 5 red shoes, 4 green shoes. If one draw randomly a shoe what is the probability of getting a red shoe Ans 5c1/ 9c1 23. What is the selling price of a car? If the cost of the car is Rs.60 and a profit of 10% over selling price is earned Ans: Rs 66/Freshersworld.com Resource Center

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24. 1/3 of girls , 1/2 of boys go to canteen .What factor and total number of classmates go to canteen. Ans: Cannot be determined. 25. The price of a product is reduced by 30% . By what percentage should it be increased to make it 100% Ans: 42.857% 26. There is a square of side 6cm . A circle is inscribed inside the square. Find the ratio of the area of circle to square. Ans. 11/14 27. There are two candles of equal lengths and of different thickness. The thicker one lasts of six hours. The thinner 2 hours less than the thicker one. Ramesh lights the two candles at the same time. When he went to bed he saw the thicker one is twice the length of the thinner one. How long ago did Ramesh light the two candles . Ans: 3 hours. 28. If M/N = 6/5,then 3M+2N = ? 29. If p/q = 5/4 , then 2p+q= ? 30. If PQRST is a parallelogram what it the ratio of triangle PQS & parallelogram PQRST . Ans: 1:2 31. The cost of an item is Rs 12.60. If the profit is 10% over selling price what is the selling price ? Ans: Rs 13.86/32. There are 6 red shoes & 4 green shoes . If two of red shoes are drawn what is the probability of getting red shoes Ans: 6c2/10c2 33. To 15 lts of water containing 20% alcohol, we add 5 lts of pure water. What is % alcohol.


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First Job. Dream Job. Freshersworld.comAns : 15% 34. A worker is paid Rs.20/- for a full days work. He works 1,1/3,2/3,1/8.3/4 days in a week. What is the total amount paid for that worker ? Ans : 57.50 35. If the value of x lies between 0 & 1 which of the following is the largest? (a) x (b) x2 (c) -x (d) 1/x Ans : (d) 36. If the total distance of a journey is 120 km .If one goes by 60 kmph and comes back at 40kmph what is the average speed during the journey? Ans: 48kmph 37. A school has 30% students from Maharashtra .Out of these 20% are Bombey students. Find the total percentage of Bombay? Ans: 6% 38. An equilateral triangle of sides 3 inch each is given. How many equilateral triangles of side 1 inch can be formed from it? Ans: 9 39. If A/B = 3/5,then 15A = ? Ans : 9B 40. Each side of a rectangle is increased by 100% .By what percentage does the area increase? Ans : 300% 41. Perimeter of the back wheel = 9 feet, front wheel = 7 feet on a certain distance, the front wheel gets 10 revolutions more than the back wheel .What is the distance? Ans : 315 feet. 42. Perimeter of front wheel =30, back wheel = 20. If front wheel revolves 240 times. How many revolutions will the back wheel take? Ans: 360 timesFreshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.com43. 20% of a 6 litre solution and 60% of 4 litre solution are mixed. What percentage of the mixture of solution Ans: 36% 44City A's population is 68000, decreasing at a rate of 80 people per year. City B having population 42000 is increasing at a rate of 120 people per year. In how many years both the cities will have same population? Ans: 130 years 45Two cars are 15 kms apart. One is turning at a speed of 50kmph and the other at 40kmph . How much time will it take for the two cars to meet? Ans: 3/2 hours 46A person wants to buy 3 paise and 5 paise stamps costing exactly one rupee. If he buys which of the following number of stamps he won't able to buy 3 paise stamps. Ans: 9 47There are 12 boys and 15 girls, How many different dancing groups can be formed with 2 boys and 3 girls. 48Which of the following fractions is less than 1/3 (a) 22/62 (b) 15/46 (c) 2/3 (d) 1 Ans: (b) 49There are two circles, one circle is inscribed and another circle is circumscribed over a square. What is the ratio of area of inner to outer circle? Ans: 1 : 2 50Three types of tea the a,b,c costs Rs. 95/kg,100/kg and70/kg respectively. How many kgs of each should be blended to produce 100 kg of mixture worth Rs.90/kg, given that the quntities of band c are equal a)70,15,15 b)50,25,25 c)60,20,20 d)40,30,30 Ans. (b)Freshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.com51. in a class, except 18 all are above 50 years. 15 are below 50 years of age. How many people are there (a) 30 (b) 33 (c) 36 (d) none of these. Ans. (d) 52. If a boat is moving in upstream with velocity of 14 km/hr and goes downstream with a velocity of 40 km/hr, then what is the speed of the stream ? (a) 13 km/hr (b) 26 km/hr (c) 34 km/hr (d) none of these Ans. A 53. Find the value of ( 0.75 * 0.75 * 0.75 - 0.001 ) / ( 0.75 * 0.75 - 0.075 + 0.01) (a) 0.845 (b) 1.908 (c) 2.312 (d) 0.001 Ans. A 54. A can have a piece of work done in 8 days, B can work three times faster than the A, C can work five times faster than A. How many days will they take to do the work together ? (a) 3 days (b) 8/9 days (c) 4 days (d) can't say Ans. B 55. A car travels a certain distance taking 7 hrs in forward journey, during the return journey increased speed 12km/hr takes the times 5 hrs.What is the distance travelled (a) 210 kms (b) 30 kms (c) 20 kms (c) none of these Ans. BFreshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.com56. Instead of multiplying a number by 7, the number is divided by 7. What is the percentage of error obtained ? 57. Find (7x + 4y ) / (x-2y) if x/2y = 3/2 ? (a) 6 (b) 8 (c) 7 (d) data insufficient Ans. C 58. A man buys 12 lts of liquid which contains 20% of the liquid and the rest is water. He then mixes it with 10 lts of another mixture with 30% of liquid.What is the % of water in the new mixture? 59. If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain? 60. Pipe A can fill a tank in 30 mins and Pipe B can fill it in 28 mins.If 3/4th of the tank is filled by Pipe B alone and both are opened, how much time is required by both the pipes to fill the tank completely ? 61. If on an item a company gives 25% discount, they earn 25% profit. If they now give 10% discount then what is the profit percentage. (a) 40% (b) 55% (c) 35% (d) 30% Ans. D 62. A certain number of men can finish a piece of work in 10 days. If however there were 10 men less it will take 10 days more for the work to be finished. How many men were there originally? (a) 110 men (b) 130 men (c) 100 men (d) none of these Ans. A 63. In simple interest what sum amounts of Rs.1120/- in 4 years and Rs.1200/in 5 years ? (a) Rs. 500 (b) Rs. 600 (c) Rs. 800 (d) Rs. 900Freshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.comAns. C 64. If a sum of money compound annually amounts of thrice itself in 3 years. In how many years will it become 9 times itself. (a) 6 (b) 8 (c) 10 (d) 12 Ans A 65. Two trains move in the same direction at 50 kmph and 32 kmph respectively. A man in the slower train observes the 15 seconds elapse before the faster train completely passes by him. What is the length of faster train ? (a) 100m (b) 75m (c) 120m (d) 50m Ans B 66. How many mashes are there in 1 squrare meter of wire gauge if each mesh is 8mm long and 5mm wide ? (a) 2500 (b) 25000 (c) 250 (d) 250000 Ans B 67. x% of y is y% of ? (a) x/y (b) 2y (c) x (d) can't be determined Ans. C 68. The price of sugar increases by 20%, by what % should a housewife reduce the consumption of sugar so that expenditure on sugar can be same as before ? (a) 15% (b) 16.66% (c) 12% (d) 9% Ans BFreshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.com69. A man spends half of his salary on household expenses, 1/4th for rent, 1/5th for travel expenses, the man deposits the rest in a bank. If his monthly deposits in the bank amount 50, what is his monthly salary ? (a) Rs.500 (b) Rs.1500 (c) Rs.1000 (d) Rs. 900 Ans C 70. The population of a city increases @ 4% p.a. There is an additional annual increase of 4% of the population due to the influx of job seekers, find the % increase in population after 2 years ? 71. The ratio of the number of boys and girls in a school is 3:2 Out of these 10% the boys and 25% of girls are scholarship holders. % of students who are not scholarship holders.? 72. 15 men take 21 days of 8 hrs. each to do a piece of work. How many days of 6 hrs. each would it take for 21 women if 3 women do as much work as 2 men? (a) 30 (b) 20 (c) 19 (d) 29 Ans. A 73. A cylinder is 6 cms in diameter and 6 cms in height. If spheres of the same size are made from the material obtained, what is the diameter of each sphere? (a) 5 cms (b) 2 cms (c) 3 cms (d) 4 cms Ans C 74. A rectangular plank (2)1/2 meters wide can be placed so that it is on either side of the diagonal of a square shown below.(Figure is not available)What is the area of the plank? Ans :7*(2)1/2 75. The difference b/w the compound interest payble half yearly and the simple interest on a certain sum lent out at 10% p.a for 1 year is Rs 25. What is the sum? (a) Rs. 15000 (b) Rs. 12000 (c) Rs. 10000 (d) none of these Ans CFreshersworld.com Resource Center

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First Job. Dream Job. Freshersworld.com76. What is the smallest number by which 2880 must be divided in order to make it into a perfect square ? (a) 3 (b) 4 (c) 5 (d) 6 Ans. C 77. A father is 30 years older than his son however he will be only thrice as old as the son after 5 years what is father's present age ? (a) 40 yrs (b) 30 yrs (c) 50 yrs (d) none of these Ans. A 78. An article sold at a profit of 20% if both the cost price and selling price would be Rs.20/- the profit would be 10% more. What is the cost price of that article? 29. If an item costs Rs.3 in '99 and Rs.203 in '00.What is the % increase in price? (a) 200/3 % (b) 200/6 % (c) 100% (d) none of these Ans. A 80. 5 men or 8 women do equal amount of work in a day. a job requires 3 men and 5 women to finish the job in 10 days how many woman are required to finish the job in 14 days. a) 10 b) 7 c) 6 d) 12 Ans 7 81. A simple interest amount of rs 5000 for six month is rs 200. what is the anual rate of interest? a) 10% b) 6%


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First Job. Dream Job. Freshersworld.comc) 8% d) 9% Ans 8% 82. In objective test a correct ans score 4 marks and on a wrong ans 2 marks are ---. a student score 480 marks from 150 question. how many ans were correct? a) 120 b) 130 c) 110 d) 150 Ans130. 83. An artical sold at amount of 50% the net sale price is rs 425 .what is the list price of the artical? a) 500 b) 488 c) 480 d) 510 Ans 500 84. A man leaves office daily at 7pm A driver with car comes from his home to pick him from office and bring back home One day he gets free at 5:30 and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car He reaches home 20 minutes earlier than usual. In how much time does the man reach home usually?? Ans. 1hr 20min 85. A works thrice as much as B. If A takes 60 days less than B to do a work then find the number of days it would take to complete the work if both work together? Ans. 22days 86. How many 1's are there in the binary form of 8*1024 + 3*64 + 3 Ans. 4 87. In a digital circuit which was to implement (A B) + (A)XOR(B), the designer implements (A B) (A)XOR(B) What is the probability of error in it ?


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First Job. Dream Job. Freshersworld.com88. A boy has Rs 2. He wins or loses Re 1 at a time If he wins he gets Re 1 and if he loses the game he loses Re 1. He can loose only 5 times. He is out of the game if he earns Rs 5. Find the number of ways in which this is possible? Ans. 16 89. If there are 1024*1280 pixels on a screen and each pixel can have around 16 million colors Find the memory required for this? Ans. 4MB 90. On a particular day A and B decide that they would either speak the truth or will lie. C asks A whether he is speaking truth or lying? He answers and B listens to what he said. C then asks B what A has said B says "A says that he is a liar" What is B speaking ? (a) Truth (b) Lie (c) Truth when A lies (d) Cannot be determined Ans. (b) 91. What is the angle between the two hands of a clock when time is 8:30 Ans. 75(approx) 92. A student is ranked 13th from right and 8th from left. How many students are there in totality ? 93. A man walks east and turns right and then from there to his left and then 45degrees to his right.In which direction did he go Ans. North west 94. A student gets 70% in one subject, 80% in the other. To get an overall of 75% how much should get in third subject. 95. A man shows his friend a woman sitting in a park and says that she the daughter of my grandmother's only son. What is the relation between the two Ans. Daughter


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First Job. Dream Job. Freshersworld.com96. How many squares with sides 1/2 inch long are needed to cover a rectangle that is 4 ft long and 6 ft wide (a) 24 (b) 96 (c) 3456 (d) 13824 (e) 14266 97. If a=2/3b , b=2/3c, and c=2/3d what part of d is b/ (a) 8/27 (b) 4/9 (c) 2/3 (d) 75% (e) 4/3 Ans. (b) 2598Successive discounts of 20% and 15% are equal to a single discount of (a) 30% (b) 32% (c) 34% (d) 35% (e) 36 Ans. (b) 99. The petrol tank of an automobile can hold g liters.If a liters was removed when the tank was full, what part of the full tank was removed? (a)g-a (b)g/a (c) a/g (d) (g-a)/a (e) (g-a)/g Ans. (c) 100. If x/y=4 and y is not '0' what % of x is 2x-y (a)150% (b)175% (c)200% (d)250% Ans. (b)


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NumbersIntroduction: Natural Numbers: All positive integers are natural numbers. Ex 1,2,3,4,8,...... There are infinite natural numbers and number 1 is the least natural number. Based on divisibility there would be two types of natural numbers. They are Prime and composite. Prime Numbers: A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity. Note:-Unity i e,1 is not a prime number. Properties Of Prime Numbers: ->The lowest prime number is 2. ->2 is also the only even prime number. ->The lowest odd prime number is 3. ->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime. ->The remainder of division of the square of a prime number p>=5 divide by 24 is 1. ->For prime numbers p>3, p-1 is divided by 24. ->If a and b are any 2 odd primes then a-b is composite. Also a+b is composite. ->The remainder of the division of the square of a prime number p>=5 divided by 12 is 1. Process to Check A Number s Prime or not: Take the square root of the number. Round of the square root to the next highest integer call this number as Z. Check for divisibility of the number N by all prime numbers below Z. If there is no numbers below the value of Z which divides N then the number will be prime. Example 239 is prime or 239 lies between 15 or Prime numbers less than 239 is not divisible by is a prime number. not? 16.Hence take the value of Z=16. 16 are 2,3,5,7,11 and 13. any of these. Hence we can conclude that 239

Composite Numbers: The numbers which are not prime are known as composite numbers. Co-Primes: Two numbers a an b are said to be co-primes,if their H.C.F is 1. Example (2,3),(4,5),(7,9),(8,11)..... Place value or Local value of a digit in a Number: place value: Example 689745132 Place value of 2 is (2*1)=2 Place value of 3 is (3*10)=30 and so on. Face value:-It is the value of the digit itself at whatever place it may be. Example 689745132 Face value of 2 is 2. Face value of 3 is 3 and so on.

TopTests of Divisibility: Divisibility by 2:-A number is divisible by 2,if its unit's digit is any of 0,2,4,6,8. Example 84932 is divisible by 2,while 65935 is not. Divisibility by 3:-A number is divisible by 3,if the sum of its digits is divisible by 3. Example 1.592482 is divisible by 3,since sum of its digits 5+9+2+4+8+2=30 which is divisible by 3. Example 2.864329 is not divisible by 3,since sum of its digits 8+6+4+3+2+9=32 which is not divisible by 3. Divisibility by 4:-A number is divisible by 4,if the number formed by last two digits is divisible by 4. Example 1.892648 is divisible by 4,since the number formed by the last two digits is 48 divisible by 4. Example 2.But 749282 is not divisible by 4,since the number formed by the last two digits is 82 is not divisible by 4. Divisibility by 5:-A number divisible by 5,if its unit's digit is either 0 or 5. Example 20820,50345 Divisibility by 6:-If the number is divisible by both 2 and 3. example 35256 is clearly divisible by 2

sum of digits =3+5+2+5+21,which is divisible by 3 Thus the given number is divisible by 6. Divisibility by 8:-A number is divisible by 8 if the last 3 digits of the number are divisible by 8. Divisibility by 11:-If the difference of the sum of the digits in the odd places and the sum of the digitsin the even places is zero or divisible by 11. Example 4832718 (8+7+3+4) - (1+2+8)=11 which is divisible by 11. Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12. Divisibility by 7,11,13:-The difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7,11,13. BASIC FORMULAE: ->(a+b)=a+b+2ab ->(a-b)=a+b-2ab ->(a+b)-(a-b)=4ab ->(a+b)+(a-b)=2(a+b) ->a-b=(a+b)(a-b) ->(a+b+c)=a+b+c+2(ab+bc+ca) ->a+b=(a+b)(a+b-ab) ->a-b=(a-b)(a+b+ab) ->a+b+c-3abc=(a+b+c)(a+b+c-ab-b c-ca) ->If a+b+c=0 then a+b+c=3a b c DIVISION ALGORITHM If we divide a number by another number ,then Dividend = (Divisor * quotient) + Remainder

TopMULTIPLICATION BY SHORT CUT METHODS 1.Multiplication by distributive law: a)a*(b+c)=a*b+a*c b)a*(b-c)=a*b-a*c Example a)567958*99999=567958*(100000-1) 567958*100000-567958*1 56795800000-567958 56795232042 b)978*184+978*816=978*(184+816) 978*1000=978000 2.Multiplication of a number by 5n:-Put n zeros to the right of the multiplicand and divide the number so formed by 2n

Example 975436*625=975436*54=9754360000/16=609647500. PROGRESSION: A succession of numbers formed and arranged in a definite order according to certain definite rule is called a progression. 1.Arithmetic Progression:-If each term of a progression differs from its preceding term by a constant. This constant difference is called the common difference of the A.P. The n th term of this A.P is Tn=a(n-1)+d. The sum of n terms of A.P Sn=n/2[2a+(n-1)d]. xImportant Results: a.1+2+3+4+5......................=n(n+1)/2. b.12+22+32+42+52......................=n(n+1)(2n+1)/6. c.13+23+33+43+53......................=n2(n+1)2/4 2.Geometric Progression:-A progression of numbers in which every term bears a constant ratio with ts preceding term. i.e a,a r,a r2,a r3............... In G.P Tn=a r n-1 Sum of n terms Sn=a(1-r n)/1-r Problems 1.Simplify a.8888+888+88+8 b.11992-7823-456 Solution: a.8888 888 88 8 9872 b.11992-7823-456=11992-(7823+456) =11992-8279=3713 2.What could be the maximum value of Q in the following equation? 5PQ+3R7+2Q8=1114 Solution: 5 P Q 3 R 7 2 Q 8 11 1 4 2+P+Q+R=11 Maximum value of Q =11-2=9 (P=0,R=0) 3.Simplify: a.5793405*9999 b.839478*625 Solution: a. 5793405*9999=5793405*(10000-1)

57934050000-5793405=57928256595 b. 839478*625=839478*54=8394780000/16=524673750. 4.Evaluate 313*313+287*287 Solution: a+b=1/2((a+b)+(a-b)) 1/2(313+287) +(313-287)=1/2(600 +26 ) (360000+676)=180338

Top5.Which of the following is a prime number? a.241 b.337 c.391 Solution: a.241 16>241.Hence take the value of Z=16. Prime numbers less than 16 are 2,3,5,7,11 and 13. 241 is not divisible by any of these. Hence we can conclude that 241 is a prime number. b. 337 19>337.Hence take the value of Z=19. Prime numbers less than 16 are 2,3,5,7,11,13 and 17. 337 is not divisible by any of these. Hence we can conclude that 337 is a prime number. c. 391 20>391.Hence take the value of Z=20. Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19. 391 is divisible by 17. Hence we can conclude that 391 is not a prime number. 6.Find the unit's digit n the product 24 7153 * 34172? Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172 Now 7 4 gives unit digit 1 7 152 gives unit digit 1 7 153 gives 1*7=7.Also 172 gives 1 Hence unit's digit in the product =7*1=7. 7.Find the total number of prime factors in 411 *7 5 *112 ? Solution: 411 7 5 112= (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors=22+5+2=29 8.Which of the following numbers s divisible by 3? a.541326 b.5967013 Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3. b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3. 9.What least value must be assigned to * so that th number 197*5462 is divisible by 9? Solution: Let the missing digit be x Sum of digits = (1+9+7+x+5+4+6+2)=34+x For 34+x to be divisible by 9 , x must be replaced by 2

The digit in place of x must be 2. 10.What least number must be added to 3000 to obtain a number exactly divisible by 19? Solution:On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19-17=2. 11.Find the smallest number of 6 digits which is exactly divisible by 111? Solution:Smallest number of 6 digits is 100000 On dividing 10000 by 111 we get 100 as remainder Number to be added =111-100=11. Hence,required number =10011. 12.On dividing 15968 by a certain number the quotient is 89 and the remainder is 37.Find the divisor? Solution:Divisor = (Dividend-Remainder)/Quotient =(15968-37) / 89 =179. 13.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder? Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9. The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.

Top14.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed? Solution:Let the number be x. 3 x 5 y - 1 8 z y=5z+4 = 5*15+4 = 79 x=3y+1 = 3*79+1=238 Now 8 238 5 29 - 6 3 5 - 4 1 - 2 Respective remainders are 6,4,2. - 4 1 - 7 z=8*1+7=15

15.Find the remainder when 231 is divided by 5? Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as 4*4*4 gives unit digit 4 unit digit of 231 is 8. Now 8 when divided by 5 gives 3 as remainder. 231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7? Solution:The required numbers are 14,21,28,...........,84 This is an A.P with a=14,d=7. Let it contain n terms then T =84=a+(n-1)d =14+(n-1)7 =7+7n 7n=77 =>n=11. 17.Find the sum of all odd numbers up to 100? Solution:The given numbers are 1,3,5.........99. This is an A.P with a=1,d=2. Let it contain n terms 1+(n-1)2=99 =>n=50 Then required sum =n/2(first term +last term) =50/2(1+99)=2500. 18.How many terms are there in 2,4,6,8..........,1024? Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2 Let the number of terms be n then 2*2 n-1=1024 2n-1 =512=29 n-1=9 n=10. 19.2+22+23+24+25..........+28=? Solution:Given series is a G.P with a=2,r=2 and n=8. Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2. =2*255=510. 20.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? a.1 b.3 c.5 d.7 Solution:1000+N>1000N clearly N=1. 21.The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to? a.18 b.22 c.36 d. none Solution:Let the one digit numbers x,y,z Sum of all possible two digit numbers= =(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z) Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22. 22.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is?

a.7

b.29

c.41

d67.

Solution:x+(x+36)+y=100 2x+y=64 Therefore y must be even prime which is 2 2x+2=64=>x=31. Third prime number =x+36=31+36=67. 23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is? a.1220 b.1250 c.22030 d.220030. Solution:Number=(555+445)*(555-445)*2+30 =(555+445)*2*110+30 =220000+30=220030. 24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15. The smaller number is? a.240 b.270 c.295 d.360 Solution:Let the smaller number be x, then larger number =1365+x Therefore 1365+x=6x+15 5x=1350 => x=270 Required number is 270. 25.In doing a division of a question with zero remainder,a candidate took 12 as divisor instead of 21.The quotient obtained by him was 35. The correct quotient is? a.0 b.12 c.13 d.20 Solution:Dividend=12*35=420. Now dividend =420 and divisor =21. Therefore correct quotient =420/21=20.

H.C.F and L.C.MFacts And Formulae: Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) : The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods : i.Factorization method: Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives HCF. Example : Solution: Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72 The prime numbers given common numbers are 2,5,7

Therefore HCF

is

22 * 5 *72 .

ii.Division Method : Divide the larger number by smaller one. Now divide the divisor by remainder. Repeat the process of dividing preceding number last obtained till zero is obtained as number. The last divisor is HCF. Example: Solution: 1134) 1215(1 1134 ---------81)1134(14 81 ----------324 324 ----------0 ----------HCF of this two numbers is 81. 81)513(6 486 -------27)81(3 81 ----0 --HCF of 81 and 513 is 27. Find HCF of 513, 1134, 1215

Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM. There are two methods for this: i.Factorization method : Resolve each one into product of prime factors. Then LCM is product of highest powers of all factors. ii.Common Problems: 1.The HCF of 2 numbers is 77.find other. Sol: Other number = 11 * is 11 and LCM is 693/77=99. divisible by 12,15,18,27 693.If one of numbers division method.

Top

2.Find largest number of 4 digits Sol: The largest number is 9999. LCM of 12,15,18,27 is 540.

on dividing 9999 by 540 we get 279 as remainder. Therefore number =9999 279 =9720. 3.Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34. Sol: 2014=6 25-19=6 35-29=6 40-34=6 Therefore number =LCM of (20,25,35,40) - 6=1394 can be expressed as prime as : 2 252 2 126 3 63 3 21 7 prime factor is 2 *2 * 3 * 3 *7 5.1095/1168 when expressed in simple form is 1095)1168(1 1095 -----73)1095(15 73 --------365 365 --------0 ---------So, HCF is 73 Therefore 1095/1168 = 1095/73/1168/73= 15/16 6.GCD of 1.08,0.36,0.9 is Sol: HCF of 108,36,90 72 ---18)36(2 36 ---0 ---HCF is 18. HCF of 18 and 108 is 18 18)108(6 108 ------0 -------Therefore HCF =0.18 36)90(2 4.252

7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers. Sol: Let the numbers be x. Three numbers are x,2x,3x Therefore HCF is 2x)3x(1 2x ----x)2x(2 2x -------0 ------------HCF is x so, x is 12 Therefore numbers are 12,24,36.

Top8.The Sol: sum of two numbers is 216 and HCF is 27. Let numbers are 27a + 27 b =216 a + b =216/27=8 Co-primes of 8 are (1,7) and (3,5) numbers=(27 * 1 ), (27 * 7) =27,89 9.LCM of two numbers is of numbers is Sol: Let the number be x. Numbers are 2x,3x LCM of 2x,3x is 6x Therefore 6x=48 x=8. Numbers are 16 and 24 Sum=16 +24=40. 10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is Sol: Let the numbers be x,4x Then x * 4x = 84 * 21 x2 =84 * 21 /4 x = 21 Largest number is 4 * 21. 11.HCF of two numbers is 23,and other factors of LCM Largest number is Sol: 23 * 14 is Largest number. 12.The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets are 13,14. 48..The numbers are in ratio 2:3. The sum

same number of pens and pencils is ? Sol: HCF of 1001 and 910 910)1001(1 910 -----------91)910(10 910 -------0 --------Therefore HCF=91 13.The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ? Sol: LCM of 5,6,4,3 is 60. On dividing 2497 by 60 we get 37 as remainder. Therefore number to added is 60 37 =23. Answer is 23. 14.The least number which is a perfect square and is each of numbers 16,20,24 is ? Sol: LCM of 16,20,24 is 240. 2 * 2*2*2*3*5=240 To make it a perfect square multiply by 3 * 5 Therefore 240 * 3 * 5=3600 Answer is 3600. divisible by

Decimal Fractions1.Decimal fractions: Fractionin 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 = 1/4 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 porducts= 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

Top8.4.036 divided by 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 -------- (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

Simplifications

Introduction: 'BODMAS' rule: This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of a given expression. Here B stands for Bracket, O for Of, D for Division, M for Multiplication, A for Addition and S for Subtraction. First of all the brackets must be removed, strictly in the order () , {} , []. After removing the brackets, we want use the following operations: 1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction Modulus of a real number: Modulus of a real number is a defined as |a| = a, if a>0 or -a, if a < 0; Problems: 1.(5004 /139) 6= ? Sol: Expression = 5004/ 139 6 = 36 6 = 30; 2.What mathematical operations should come at the place of ? in the equation : (2 ? 6 12 / 4 + 2 = 11) ? Sol: 2 ? 6 = 11 + 12 / 4 2 = 11 + 3 2 = 12 2 * 6 = 12 3.( 8 / 88) * 8888088 = ? Sol : (1/11) * 8888088 = 808008 4.How many 1/8's are there in 371/2 ? Sol: (371/2) /(1/8)= (75/2) /(1/8) = 300 5.Find the values of 1/2*3 +1/3*4 +1/4*5+ .................+1/9*10 ? Sol: 1/2*3 +1/3*4+1/4*5+ ..................+1/9*10 = [ -1/3] +[ 1/3 ] + [- 1/5] +...............+[1/9-1/10] = [ 1/10] = 4/15 = 2/5

Top6.The value of 999 of 995/999* 999 is: Sol: [1000- 4/1000]*999 = 999000-4

= 998996 7.Along a yard 225m long, 26 trees are planted at equal distance, one tree being at each end of the yard. what is the distance between two consecutive trees ? Sol: 26 trees have 25 gaps between them. Hence , required distance = 225/ 25 m= 9m 8.In a garden , there are 10 rows and 12 columns of mango trees. the distance between the two trees is 2 m and a distance of one meter is left from all sides of the boundary of the length of the garden is : Sol: Each row contains 12 plants. leaving 2 corner plants, 10 plants in between have 10 * 2 meters and 1 meter on each side is left. length = (20 + 2) m = 22m 9.Eight people are planning to share equally the cost of a rental car, if one person with draws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by? Sol: Original share of one person = 1/8 new share of one person = 1/7 increase = 1/7 1/8 = 1/56 required fractions = (1/56)/(1/8) = 1/7 10.A piece of cloth cost Rs 35. if the length of the piece would have been 4m longer and each meter cost Re 1 less , the cost would have remained unchanged. how long is the piece? Sol: Left the length of the piece be x m. then, cost of 1m of piece = Rs [35 / x] 35/ x 35 /x+4 = 1 x + 4 x = x(x+ 4)/35 x2 + 4x 140 = 0 x= 10 11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five as much as each nephew. how much does each daughter receive ? Sol: Let the share of each nephew be Rs x. then, share of each daughter Rs 4x. share of each son = 5x Rs so, 5 *5x+ 4 * 4x + 2x =8600 2x + 16x + 25x= 8600 43x = 8600 x = 200 share of each daughter = 4 * 200 = Rs 800

12.A man spends 2/5 of his salary on house rent, 3/10 of his salary on food, and 1/8 of his salary on conveyance. if he has Rs 1400 left with him, find his expenditure on food and conveyance? Sol: Part of the salary left = 1-[2/5 +3/10+1/9] = 1- 33/40 =7/40 Let the monthly salary be rs x then, 7/40 of x = 1400 x= [1400*40]/7 x= 8000 Expenditure on food = 3/10*8000 =Rs 2400 Expenditure on conveyance= 1/8*8000 =Rs 1000

Square and Cube RootsFormula: 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 division with 100. here consider the number 50 which is nearer to 37 when compared with the number 50 we get the now subtract the number 13 from the 37, we get 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 1t as 13 then place 69 beside the 13 we get 1369. 3 7

gets when the number given. difference of 13. 24.

2 we get

1 3 ----------------2 4 24/2=12 square 13* 13=169

as

50

=

100/2

1 2 + 1 6 9 ----------------------1 3 6 9 -------------------------

TopExample: 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 solutin: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: (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: (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

Top7.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: (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.

Averages

Formula: 1.Average=Sum of quantities/Number of quantities. 2.Suppose a man covers a certain distance at x kmph and an equal distance at y kmph ,then the average speed during the whole journey is (2xy/x+y) kmph. Examples: 1.Find the average of all these numbers.142,147,153,165,157. Solution: 142 147 153 165 157 Here consider the least number i.e, 142 comparing with others, 142 147 153 165 157 +5 +11 +23 +15 Now add 5+11+23+15 = 52/5 = 10.8 Now add 10.8 to 142 we get 152.8 (Average of all these numbers). Answer is 152.8 2.Find the average of all these numbers.4,10,16,22,28 Solution: 4,10,16,22,28 As the difference of number is 6 Then the average of these numbers is central one i.e, 16. Answer is 16. 3.Find the average of all these numbers.4,10,16,22,28,34. Solution: Here also difference is 6. Then middle numbers 16,22 take average of these two numbers 16+22/2=19 Therefore the average of these numbers is 19. Answer is 19. 4.The average marks of a marks of a student in 4 Examination is 40.If he got 80 marks in 5th Exam then what is his new average. Solution: 4*40+80=240 Then average means 240/5=48. Answer is 48. 5.In a group the average income of 6 men is 500 and that of 5 women is 280, then what is average income of the group. Solution: 6*500+5*280=4400 then average is 4400/11=400. Another Method: here consider for 6 men 6 men each 500. so 5th women is 280.

then 500-280=220. then 220*6/11=120. therefore 120+280=400. Answer is 400. 6.The average weight of a class of 30 students is 40 kgs if the teacher weight is included then average increases by 2 kgs then find the weight of the teacher? Solution: 30 students average weight is 40 kgs. So,when teacher weight is added it increases by 2 kgs so total 31 persons ,therefore 31*2=62. Now add the average weight of all student to it we get teachers weight i.e, 62+40=102 kgs. Answer is 102 kgs. 7.The average age of Mr and Mrs Sharma 4 years ago is 28 years . If the present average age of Mr and Mrs Sharma and their son is 22 years. What is the age of their son. Solution: 4 years ago their average age is 28 years. So their present average age is 32 years. 32 years for Mr and Mrs Sharma then 32*2=64 years. Then present age including their son is 22 years. So 22*3 =66 years. Therefore son age will be 66-64 = 2 years. Answer is 2 years. 8.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: 10 books Average increases by 17 Rupees so 10*17= 170. so the new book cost is more and by adding its cost average increase,therefore the cost of new book is 400+170=570Rs. Answer is 570 Rs. 9.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: Total number of girls be x+4. Average marks of 4 girls is 60. therefore 62.5-60=2.5 then 4*2.5 =10. the average of remaining girls is 63 here 0.5 difference therefore 0.5*x=10(since we got from 4 girls) (this is taken becoz both should be equal) x=10/0.5 x=20. This clear says that remaining are 20 girls therefore total is x+4=20+4=24 girls

Answer is 24 girls.

Top10.Find the average of first 50 natural numbers. Solution: Sum of the Natural Numbers is n(n+1)/2 therefore for 50 Natural numbers 50*51/2=775. the average is 775/50=15.5 Answer is 15.5 . 11.The average of the first nine prime number is? Solution: Prime numbers are 2,3,5,7,11,13,17,19,23 therefore 2+3+5+7+11+13+17+19+23=100 then the average 100/9= 11 1/9. Answer is 11 1/9. 12.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/4=5 =>15+x=20 =>x=5. 18+1+6+x+y/5=10 =>25+5+y=50 =>y=20. 13.The average of a non-zero number and its square is 5 times the number.The number is Solution: The number be x then x+x2/2=5x =>x2-9x=0 =>x(x-9)=0 therefore x=0 or x=9. The number is 9. 14.Nine persons went to a hotel for taking their meals . Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 then the average expenditure of all the nine. What was the total money spent by them? Solution: The average expenditure be x. then 8*12+(x+8)=9x =>96+x+8=9x. =>8x=104 =>x=13 Total money spent =9x=>9*13=117 Answer is Rs.117 15.The average weight of A.B.C is 45 Kgs.If the average weight of A and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?

Solution: The weight of A,B,Care 45*3=135 Kgs. The weight of A,B are 40*2=80 Kgs. The weight of B,C are 43*2=86 Kgs. To get the Weight of B. (A+B)+(B+C)-(A+B+C)=80+86-135 B=31 kgs. Answer is 31 Kgs. 16.The sum of three consecutive odd number is 48 more than the average of these number .What is the first of these numbers? Solution: let the three consecutive odd numbers are x, x+2, x+4. By adding them we get x+x+2+x+4=3x+6. Then 3x+6-(3x+6)/3=38(given) =>2(3x+6)=38*3. =>6x+12=114 =>6x=102 =>x=17. Answer is 17. 17.A family consists of grandparents,parents and three grandchildren. The average age of the grandparents is 67 years,that of parents is 35 years and that of the grand children is 6 years . What is the average age of the family? Solution: grandparents age is 67*2=134 parents age is 35*2=70 grandchildren age is 6*3=18 therefore age of family is 134+70+18=222 average is 222/7=31 5/7 years. Answer is 31 5/7 years.

Top18.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: Here specified that month starts with Sunday so, in a month there are 5 Sundays. Therefore remaining days will be 25 days. 510*5+240*25=2550+6000 =8550 visitors. The average visitors are 8550/30=285. Answer is 285. 19.The average age of a class of 39 students is 15 years . If the age of the teacher be included ,then average increases by 3 months. Find the age of the teacher. Solution: Total age for 39 persons is 39*15=585 years. Now 40 persons is 40* 61/4=610 years (since 15 years 3 months=15 3/12=61/4) Age of the teacher =610-585 years

=>25 years. Answer is 25 years. 20.The average weight of a 10 oarsmen in a boat is increases by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is replaced by new man. Find the weight of the new man. Solution: Weight of 10 oars men is increases by 1.8 Kgs so, 10*1.8=18 Kgs therefore 53+18=71 Kgs will be the weight of the man. Answer is 71 Kgs. 21.A bats man makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find the average after 17th inning. Solution: Average after 17 th inning =x then for 16th inning is x-3. Therefore 16(x-3)+87 =17x =>x=87-48 =>x=39. Answer is 39. 22.The average age of a class is 15.8 years .The average age of boys in the class is 16.4 years while that of the girls is 15.4 years .What is the ratio of boys to girls in the class. Solution: Ratio be k:1 then k*16.4 + 1*15.4 = (k+1)*15.8 =>(16.4-15.8)k=15.8-15.4 =>k=0.4/0.6 =>k=2/3 therefore 2/3:1=>2:3 Answer is 2:3 23.In a cricket eleven ,the average of eleven players is 28 years .Out of these ,the average ages of three groups of players each are 25 years,28 years, and 30 years respectively. If in these groups ,the captain and the youngest player are not included and the captain is eleven years older than the youngest players , what is the age of the captain? Solution: let the age of youngest player be x then ,age of the captain =(x+11) therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28 =>75+84+90+2x+11=308 =>2x=48 =>x=24. Therefore age of the captain =(x+11)= 24+11= 35 years. Answer is 35 years.

Top24.The average age of the boys in the class is the number of girls in the class .If the ratio boys and girls in the class of 36 be 5:1, what the total of the age (in years) of the boys in twice of is the class?

Solution: Number of boys=36*5/6=30 Number of girls =6 Average age of boys =2*6=12 years Total age of the boys=30*12=360 years Answer is 360 years. 25.Five years ago, the average age of P and Q was 15 years ,average age of P,Q, and R today is 20 years,how old will R be after 10 years? Solution: Age of P and Q are 15*2=30 years Present age of P and Q is 30+5*2=40 years. Age of P Q and R is 20*3= 60 years. R ,present age is 60-40=20 years After 10 years =20+10=30 years. Answer is 30 years. 26.The average weight of 3 men A,B and C is 84 Kgs. Another man D joins the group and the average now becomes 80 Kgs.If another man E whose weight is 3 Kgs more than that of D ,replaces A then the average weight B,C,D and E becomes 79 Kgs. The weight of A is. Solution:Total weight of A, B and C is 84 * 3 =252 Kgs. Total weight of A,B,C and Dis 80*4=320 Kgs Therefore D=320-252=68 Kgs. E weight (68+3)=71 kgs Total weight of B,C,D and E = 79*4=316 Kgs (A+B+C+D)-(B+C+D+E)=320-316 =4Kgs A-E=4Kgs A-71=4 kgs A=75 Kgs Answer is 75 kgs 27.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: Here consider the total score be x. therefore x+92-85/8=84 =>x+7=672 =>x=665. Answer is 665 28.A man whose bowling average is 12.4,takes 5 wickets for 26 runs and there by decrease his average by 0.4. The number of wickets,taken by him before his last match is: Solution: Number of wickets taken before last match be x. therefore 12.4x26/x+5=12(since average decrease by 0.4 therefore 12.4-0.4=12) =>12.4x+2612x+60 =>0.4x=34 =>x=340/4 =>x=85.

Answer is 85. 29.The mean temperature of Monday to Wednesday was 37 degrees and of Tuesday to Thursday was 34 degrees .If the temperature on Thursday was 4/5th that of Monday. The temperature on Thursday was: Solution: The total temperature recorded on Monday,Wednesday was 37*3=111. The total temperature recorded on Tuesday, Wednesday,Thursday was 34*3=102. and also given that Th=4/5M =>M=5/4Th (M+T+W)-(T+W+Th)=111-102=9 M-Th=9 5/4Th-Th=9 Th(1/4)=9 =>Th=36 degrees. 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: Here given average of group A and whole groups . So,(76*16)-(75*10)/6 =>1216-750/6 =>466/6=233/3=77 2/3 Answer is 77 2/3. 31.Of the three numbers the first is twice the second and the second is twice the third .The average of the reciprocal of the numbers is 7/72,the number are. Solution:Let the third number be x Let the second number be 2x. Let the first number be 4x. Therefore average of the reciprocal means 1/x+1/2x+1/4x=(7/72*3) 7/4x=7/24 =>4x=24 x=6. Therefore First number is 4*6=24. Second number is 2*6=12 Third number is 1*6=6 Answer is 24,12,6. 32.The average of 5 numbers is 7.When 3 new numbers are added the average of the eight numbers is 8.5. The average of the three new number is: Solution: Sum of three new numbers=(8*8.5-5*7)=33 Their average =33/3=11. Answer is 11.

33.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? Solution : Sum of temperature on 1st 2nd 3rd and 4th days =58*4=232 degrees. Sum of temperature on 2nd 3rd 4th and 5th days =60*4=240 degrees Therefore 5th day temperature is 240-232=8 degrees. The ratio given for 1st and 5th days be 7x and 8x degrees then 8x-7x=8 =>x=8. therefore temperature on the 5th day =8x=8*8=64 degrees.

Problems on NumbersSimple problems: 1.What least number must be added to 3000 to obtain a number exactly divisible by 19? Solution: On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19-17=2. 2.Find the unit's digit n the product 2467 153 * 34172? Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172 Now 7 4 gives unit digit 1 7 152 gives unit digit 1 7 153 gives 1*7=7.Also 172 gives 1 Hence unit's digit in the product =7*1=7. 3.Find the total number of prime factors in 411 *7 5 *112 ? Solution: 411 7 5 112= (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors=22+5+2=29 4.The least umber of five digits which is exactly divisible by 12,15 and 18 is? a.10010 b.10015 c.10020 d.10080 Solution: Least number of five digits is 10000 L.C.Mof 12,15,18 s 180.

On dividing 10000 by 180,the remainder is 100. Therefore required number=10000+(180-100) =10080. Ans (d). 5.The least number which is perfect square and is divisible by each of the numbers 16,20 and 24 is? a.1600 b.3600 c.6400 d.14400 Solution: The least number divisible by 16,20,24 = L.C.M of 16,20,24=240 =2*2*2*2*3*5 To make it a perfect square it must be multiplied by 3*5. Therefore required number =240*3*5=3600. Ans (b). 6.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? a.1 b.3 c.5 d.7 Solution: 1000+N>1000N clearly N=1. 7.How many numbers between 11 and 90 are divisible by 7? Solution: The required numbers are 14,21,28,...........,84. This is an A.P with a=14,d=7. Let it contain n terms then T =84=a+(n-1)d =14+(n-1)7 =7+7n 7n=77 =>n=11. 8.Find the sum of all odd numbers up to 100? Solution: The given numbers are 1,3,5.........99. This is an A.P with a=1,d=2. Let it contain n terms 1+(n-1)2=99 =>n=50 Then required sum =n/2(first term +last term) =50/2(1+99)=2500. 9.How many terms are there in 2,4,6,8..........,1024? Solution: Clearly 2,4,6........1024 form a G.P with a=2,r=2 Let the number of terms be n then 2*2 n-1=1024 2n-1 =512=29 n-1=9 n=10. 10.2+22+23+24+25..........+28=?

Solution: Given series is a G.P with a=2,r=2 and n=8. Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2. =2*255=510. 11.Find the number of zeros in 27!? Solution: Short cut method : number of zeros in 27!=27/5 + 27/25 =5+1=6zeros.

TopMedium Problems: 12.The difference between two numbers 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15.The smaller number is? a.240 b.270 c.295 d.360 Solution: Let the smaller number be x, then larger number =1365+x Therefore 1365+x=6x+15 5x=1350 => x=270 Required number is 270. 13.Find the remainder when 231 is divided by 5? Solution: 210 =1024. unit digit of 210 * 210 * 210 is 4 as 4*4*4 gives unit digit 4 unit digit of 231 is 8. Now 8 when divided by 5 gives 3 as remainder. 231 when divided by 5 gives 3 as remainder. 14.The largest four digit number which when divided by 4,7 or 13 leaves a remainder of 3 in each case is? a.8739 b.9831 c.9834 d.9893. Solution: solution: Greatest number of four digits is 9999 L.C.M of 4,7, and 13=364. On dividing 9999 by 364 remainder obtained is 171. Therefore greatest number of four digits divisible by 4,7,13 =9999-171=9828. Hence required number=9828+3=9831. Ans (b). 15.What least value must be assigned to * so that th number 197*5462 is divisible by 9? Solution: Let the missing digit be x Sum of digits = (1+9+7+x+5+4+6+2)=34+x For 34+x to be divisible by 9 , x must be replaced by 2 The digit in place of x must be 2.

16.Find the smallest number of 6 digits which is exactly divisible by 111? Solution: Smallest number of 6 digits is 100000 On dividing 10000 by 111 we get 100 as remainder Number to be added =111-100=11. Hence,required number =10011. 17.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder? Solution: Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9. The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder. 18.In doing a division of a question with zero remainder,a candidate took 12 as divisor instead of 21.The quotient obtained by him was 35. The correct quotient is? a.0 b.12 c.13 d.20 Solution: Dividend=12*35=420. Now dividend =420 and divisor =21. Therefore correct quotient =420/21=20. 19.If a number is multiplied by 22 and the same number is added to it then we get a number that is half the square of that number. Find the number. a.45 b.46 c.47 d. none Solution: Let the required number be x. Given that x*22+x = 1/2 x2 23x = 1/2 x2 x = 2*23=46 Ans (b) 20.Find the number of zeros in the factorial of the number 18? Solution: 18! contains 15 and 5,which combined with one even number gives zeros. Also 10 is also contained in 18! which will give additional zero .Hence 18! contains 3 zeros and the last digit will always be zero. 21.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is? a.7 b.29 c.41 d67. Solution: x+(x+36)+y=100 2x+y=64 Therefore y must be even prime which is 2 2x+2=64=>x=31.

Third prime number =x+36=31+36=67. 22.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder . The number is? a.1220 b.1250 c.22030 d.220030. Solution: Number=(555+445)*(555-445)*2+30 =(555+445)*2*110+30 =220000+30=220030. 23.The difference of 1025-7 and 1024+x is divisible by 3 for x=? a.3 b.2 c.4 d.6 Solution: The difference of 1025-7 and 1024+x is =(1025-7)-(1024-x) =1025-7-1024-x =10.1024-7 -1024-x =1024(10-1)-(7-x) =1024*9-(7+x) The above expression is divisible by 3 so we have to replace x with 2. Ans (b).

TopComplex Problems: 24.Six bells commence tolling together and toll at intervals of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many times do they toll together? Solution: To find the time that the bells will toll together we have to take L.C.M of 2,4,6,8,10,12 is 120. So,the bells will toll together after every 120 seconds i e, 2 minutes In 30 minutes they will toll together [30/2 +1]=16 times 25.The sum of two numbers is 15 and their geometric mean is 20% lower than their arithmetic mean. Find the numbers? a.11,4 b.12,3 c.13,2 d.10,5 Solution: Sum of the two numbers is a+b=15. their A.M = a+b / 2 and G.M = (ab)1/2 Given G.M = 20% lower than A.M =80/100 A.M (ab)1/2=4/5 a+b/2 = 2*15/5= 6 (ab)1/2=6 ab=36 =>b=36/a a+b=15 a+36/a=15 a2+36=15a a2-15a+36=0 a2-3a-12a+36=0 a(a-3)-12(a-3)=0

a=12 or 3. If a=3 and a+b=15 then b=12. If a=12 and a+b=15 then b=3. Ans (b). 26.When we multiply a certain two digit number by the sum of its digits 405 is achieved. If we multiply the number written in reverse order of the same digits by the sum of the digits,we get 486.Find the number? a.81 b.45 c.36 d. none Solution: Let the number be x y. When we multiply the number by the sum of its digit 405 is achieved. (10x+y)(x+y)=405....................1 If we multiply the number written in reverse order by its sum of digits we get 486. (10y+x)(x+y)=486......................2 dividing 1 and 2 (10x+y)(x+y)/(10y+x)(x+y) = 405/486. 10x+y / 10y+x = 5/6. 60x+6y = 50y+5x 55x=44y 5x = 4y. From the above condition we conclude that the above condition is satisfied by the second option i e b. 45. Ans (b). 27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10? a.(x-2),(x-2)(x-3)(x-5) b.(x-2),(x-2)(x-3) c.(x-3),(x-2)(x-3)(x-5) d. none Solution: The given polynomials are x2-5x+6=0................1 x2-7x+10=0...............2 we have to find the factors of the polynomials x2-5x+6 and x2-7x+10 x2-2x-3x+6 x2-5x-2x+10 x(x-2)-3(x-2) x(x-5)-2(x-5) (x-3)(x-2) (x-2)(x-5) From the above factors of the polynomials we can easily find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5). Ans (c) 28.The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to? a.18 b.22 c.36 d. none Solution: Let the one digit numbers x,y,z Sum of all possible two digit numbers =(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)

Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22. 29.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed? Solution: Let the number be x. 5 8 z=8*1+7=15 y=5z+4 = 5*15+4 = 79 x=3y+1 = 3*79+1=238 Now 8 3 y z 1 x 1 4 7

29 5 1 Respective remainders are 6,4,2.

5 3

238

-

6 4 2

30.The arithmetic mean of two numbers is smaller by 24 than the larger of the two numbers and the GM of the same numbers exceeds by 12 the smaller of the numbers. Find the numbers? a.6,54 b.8,56 c.12,60 d.7,55 Solution: Let the numbers be a,b where a is smaller and b is larger number. The AM of two numbers is smaller by 24 than the larger of the two numbers. AM=b-24 AM of two numbers is a+b/2. a+b/2 = b-24 a+b = 2b-48 a = b-48...................1 The GM of the two numbers exceeds by 12 the smaller of the numbers GM = a+12 GM of two numbers is (ab)1/2 (ab) 1/2= a+12 ab = a2+144+24a from 1 b=a+48 a(a+48)= a2+144+24a a2+48a = a2+144+24a 24a=144=>a=6 Therefore b=a+48=54. Ans (a). 31.The sum of squares of the digits constituting a positive two digit number is 13,If we subtract 9 from that number we shall get a number written by the same digits in the reverse order. Find the number? a.12 b.32 c.42 d.52.

Solution: Let the number be x y. the sum of the squares of the digits of the number is 13 x2+y2=13 If we subtract 9 from the number we get the number in reverse order x y-9=y x. 10x+y-9=10y+x. 9x-9y=9 x-y=1 (x-y)2 =x2+y2-2x y 1 =13-2x y 2x y = 12 x y = 6 =>y=6/x x-y=1 x-6/x=1 x2-6=x x2-x-6=0 x+2x-3x-6=0 x(x+2)-3(x+2)=0 x=3,-2. If x=3 and x-y=1 then y=2. If x=-2 and x-y=1 then y=-3. Therefore the number is 32. Ans (b).

Top32.If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number? a.18 b.39 c.49 d.28 Solution: Let the digit number be x y Given that if we add square of the digit in the of a number to the product of the digits we get x2+x y=52. x(x+y)=52....................1 Given that if we add the square of the digit in e to the product is 117. y2+x y= 117 y(x+y)=117.........................2 dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9 x/y=4/9 from the options we conclude that the two digit because the condition is satisfied by the third Ans (c)

tens place 52. the unit's plac

number is 49 option.

33.The denominators of an irreducible fraction is greater than the numerator by 2.If we reduce the numerator of the reciprocal fraction by 3 and subtract the given fraction from the resulting one,we get 1/15.Find the given fraction? Solution: Let the given fraction be x / (x+2) because given that

denominator of the fraction is greater than the numerator by 2 1 [(x 1/(x+2))/3] = 1/15. 1 (x2+2x-1) /3(x+2) = 1/15 (3x+6-x2-2x+1)/3(x+2) = 1/15 (7-x2+2x)/(x+2) = 1/5 -5x2+5x+35 = x+2 5x2-4x-33 = 0 5x2-15x+11x-33 = 0 5x(x-3)+11(x-3) = 0 (5x+11)(x-3) = 0 Therefore x=-11/5 or 3 Therefore the fraction is x/(x+2) = 3/5. 34.Three numbers are such that the second is as much lesser than the third as the first is lesser than the second. If the product of the two smaller numbers is 85 and the product of two larger numbers is 115. Find the middle number? Solution: Let the three numbers be x,y,z Given that z y = y x 2y = x+z.....................1 Given that the product of two smaller numbers is 85 x y = 85................2 Given that the product of two larger numbers is 115 y z = 115...............3 Dividing 2 and 3 x y /y z = 85/115 x / z = 17 / 23 From 1 2y = x+z 2y = 85/y + 115/y 2y2 = 200 y2 = 100 y = 10 35.If we divide a two digit number by the sum of its digits we get 4 as a quotient and 3 as a remainder. Now if we divide that two digit number by the product of its digits we get 3 as a quotient and 5 as a remainder . Find the two digit number? Solution: Let the two digit number is x y. Given that x y / (x+y) quotient=4 and remainder = 3 we can write the number as x y = 4(x+y) +3...........1 Given that x y /(x*y) quotient = 3 and remainder = 5 we can write the number as x y = 3 x*y +5...............2 By trail and error method For example take x=1,y=2 1............12=4(2+3)+3 =4*3+3 ! =15 let us take x=2 y=3

1..............23=4(2+3)+3 =20+3 =23 2.............23=3*2*3+5 =18+5 =23 the above two equations are satisfied by x=2 and y=3 Therefore the required number is 23. 36.First we increased the denominator of a positive fraction by 3 and then it by 5.The sum of the resulting fractions proves to be equal to 2/3. Find the denominator of the fraction if its numerator is 2. Solution: Let us assume the fraction is x/y First we increasing the denominator by 3 we get x/(y-3) Then decrease it by 5 we get the fraction as x/(y-5) Given that the sum of the resulting fraction is 2/3 x/(y+3) + x/(y-5) = 2/3 Given numerator equal to 2 2*[ 1/y+3 + 1/y-5] =2/3 (y-5+y+3) / (y-3)(y+5) =1/3 6y 6 = y2-5y+3y-15 y2-8y-9 = 0 y2-9y+y-9 = 0 y(y-9)+1(y-9) = 0 Therefore y =-1 or 9. 37.If we divide a two digit number by a number consisting of the same digits written in the reverse order,we get 4 as quotient and 15 as a remainder. If we subtract 1 from the given number we get the sum of the squares of the digits constituting that number. Find the number? a.71 b.83 c.99 d. none Solution: Let the number be x y. If we divide 10x+y by a number in reverse order i e,10y+x we get 4 as quotient and 15 as remainder. We can write as 10x+y = 4(10y+x)+15......................1 If we subtract 1 from the given number we get square of the digits 10x+y = x2+y2.....................................2 By using above two equations and trail and error method we get the required number. From the options also we can solve the problem. In this no option is satisfied so answer is d. Ans (d)

Problems on AgesSimple problems:

1.The present age of a father is 3 years more than three times the age of his son.Three years hence,fathers age will be 10 years more than twice the age of the son.Find the present age of the father. Solution: Let the present age be 'x' years. Then father's present age is 3x+3 years. Three years hence (3x+3)+3=2(x+3)+10 x=10 Hence father's present age = 3x+3 = 33 years. 2. One year ago the ratio of Ramu & Somu age was 6:7respectively. Four years hence their ratio would become 7:8. How old is Somu. Solution: Let us assume Ramu &Somu ages are x &y respectively. One year ago their ratio was 6:7 i.e x-1 / y-1 = 7x-6y=1 Four years hence their ratios,would become 7:8 i.e x-4 / y-4 = 7 / 8 8x-7y=-4 From the above two equations we get y= 36 years. i.e Somu present age is 36 years. 3. The total age of A &B is 12 years more than the total age of B&C. C is how many year younger than A. Solution: From the given data A+B = 12+(B+C) A+B-(B+C) = 12 A-C=12 years. C is 12 years younger than A

4. The ratio of the present age of P & Q is 6:7. If Q is 4 years old than P. what will be the ratio of the ages of P & Q after 4 years. Solution: The present age of P & Q is 6:7 i.e P / Q = 6 / 7 Q is 4 years old than P i.e Q = P+4. P/ P+4 = 6/7 7P-6P = 24, P = 24 , Q = P+4 =24+4 = 28 After 4 years the ratio of P &Q is P+4:Q+4 24+4 : 28+4 = 28:32 = 7:8

5. The ratio of the age of a man & his wife is 4:3.After 4 years this ratio will be 9:7. If the time of marriage the ratio was 5:3, then how many years ago were they married.

Solution:

The age of a man is 4x . The age of his wife is 3x. After 4 years their ratio's will be 9:7 i.e 4x+4 / 3x+4 = 9 / 7 28x-27x=36-28 x = 8. Age of a man is 4x = 4*8 = 32 years. Age of his wife is 3x = 3*8 = 24 years. Let us assume 'y' years ago they were married , the ratio was 5:3 ,i.e 32-y / 24-y = 5/ 3 y=12 years i.e 12 years ago they were married Top

6. Sneh's age is 1/6th of her father's age.Sneh's father's age will be twice the age of Vimal's age after 10 years. If Vimals eight birthday was celebrated two years before,then what is Sneh's present age. a) 6 2/3 years b) 24 years c) 30 years d) None of the above

Solution:

Assume Snehs age is 'x' years. Assume her fathers age is 'y' years. Snehs age is 1/6 of her fathers age i.e x = y /6. Fathers age will be twice of Vimal's age after 10 years. i.e y+10 = 2( V+10)( where 'V' is the Vimal's age) Vimal's eight birthday was celebrated two years before, Then the Vimal's present age is 10 years. Y+10 = 2(10+10) Y=30 years. Sneh's present age x = y/6 x = 30/6 = 5 years. Sneh's present age is 5 years.

7.The sum of the ages of the 5 children's born at the intervals of 3 years each is 50 years what is the age of the youngest child. a) 4 years Solution: b) 8 years c) 10 years d)None of the above

Let the age of the children's be x ,x+3, x+6, x+9, x+12. x+(x+3)+(x+6)+(x+9)+(x+12) = 50 5x+30 = 50 5x = 20 x=4. Age of the youngest child is x = 4 years.

8. If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18,then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan

whose age is 5 years,then a) 48 years Solution: b)60 years

what is Gagan's present age. c)84 years d)65 years

Let us assume Anup age (x-6) x-6 x=60

Gagan present age is 'x' years. = 5-2 = 3 years. / 18 = 3 = 54 years

9.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my father was 26 years of age when i was born. If my sister was 4 years of age when my brother was born,then what was the age my father and mother respectively when my brother was born. a) 32 yrs, 23yrs b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs

Solution: My brother was born 3 years before I was born & 4 years after my sister was born. Father's age when brother was born = 28+4 = 32 years. Mother's age when brother was born = 26-3 = 23 years.

SURDS AND INDICESSimple problems: 1. Laws of Indices: (i) (ii) (iii) (iv) (v) (vi) am * an am / an (am)n (ab)n (a/b)n a0 = = = = = = a(m+n) a(m-n) a(m*n) an * bn an / bn 1

2.Surds :Let 'a' be a rational number & 'n' be a positive integer such that a1/n = nth root a is irrational.Then nth root a is called 'a' surd of 'n'. Problems:(1) (i) (27)2/3 = (33)2/3 = 32 = 9. (ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256. (iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16 (2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .

Solution: 2x/2+2x.2 = 1280 2x(1+22) = 2*1280 2x = 2560/5 2x = 512 => 2x = 29 x = 9 (3) Find the value of [5[81/3+271/3]3]1/4 Solution: [5[(23)1/3+(33)1/3]3]1/4 [5[2+3]3]1/4 [54]1/4 => 5. (4) If (1/5)3y= 0.008 then find the value of (0.25)y Solution: (1/5)3y = 0.008 (1/5)3y =[0.2]3 (1/5)3y =(1/5)3 3y= 3 => y=1. (0.25)y = (0.25)1 =>

0.25 = 25/100 32n+1 / / (32)n 9n

= 1/4

(5) Find the value of (243)n/5 * Solution: (35)n/5 * 32n +1 33n+1 / 33n-1 3 33n+1 * 3-3n+1

* 3 n-1 * 3n-1

=> 32

=>9.

(6) Find the value of Solution:

(21/4-1)( 23/4 +21/2+21/4+1)

Let us say 21/4 = x (x-1)(x3+x2+x+1) (x-1)(x2(x+1)+(x+1)) (x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)] (x2+1) (x2-1) => (x4-1) ((21/4))4 - 1) = > (2-1) = > 1. , z = xc then find the value

(7)

If x= ya , y = zb of abc. Solution:

z= xc z= (ya)c [ z= (y)ac z= (zb)ac z= zabc abc = 1

x= ya ] [y= zb]

(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2 Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2 [ (a-b)(a2+ab+b2) = a3-b3]

from the above formula => xa3-b3 xb3-c3 xc3-a3 => xa3-b3+b3-c3+c3-a3 => x0 = 1 (9) (1000)7 (a) 10 /1018 (b) = ? 100 (c ) 1000 = > (10)21 / (d) 10000 (10)18 1000

Solution: (1000)7 / 1018 (103)7 / (10)18 Ans :( c ) (10) The => (10)21-18

=> (10)3 =>

value

of

(8-25-8-26)

is (d) None

(a) 7* 8-25

(b) 7*8-26

(c ) 8* 8-26

Solution: ( 8-25 - 8-26 ) => 8-26 (8-1 ) => 7* 8-26 Ans: (b) Top (11) 1 / (1+ an-m ) +1/ (1+am-n) (a) 0 (b) 1/2 (c ) 1 1/ => => => = ? (d) an+m

Solution:

Ans: ( c)

(1+ an/am) + 1/ ( 1+ am/an) am / (am+ an ) + an /(am +an ) (am +an ) /(am + an) 1

(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=? (a) 0 (b) 1 ( c ) xa-b-c (d) None of the above

Solution:

1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) + 1/(1+xb/xc +xa/xc) => xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc) =>(xa +xb+xc) /(xa +xb+xc) =>1 Ans: (b) (13) If x=3+2 2 then the value of (x 1/ x) is [ =root] (a) 1 (b) 2 (c ) = 22 ( d) 33

Solution:

(x-1/x)2

x+ 1/x-2

=> => =>

Ans : (b) (14)

3+22 + (1/3+22 )-2 3+22 + 3-22 -2 6-2 = 4 (x-1/x)2 = 4 =>(x-1/x)2 = 22 (x-1/x) = 2.

(xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c (a) xabc (b) 1

= ?

( c) xab+bc+ca (d) xa+b+c [xc-a]c+a-b [xa-b]a+b-c

Solution: [xb-c]b+c-a

=>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c) =>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc) =>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc) => x0 =>1 Ans: (b) (15) If 3x-y (a) Solution: 0 = 27 and (b) 2 3x+y = 243 then x is equal to (d) 6

(c ) 4

3x-y = 27 => 3x-y = 33 x-y= 3 3x+y = 243 => 3x+y = 35 x+y = 5 From above two equations x = 4 , y=1

Ans: (c ) (16) If ax = by = cz and b2 = ac then yequals

(a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z Solution: Let us say ax = by = cz = k ax =k => [ax]1/x = k1/x => a = k1/x Simillarly b = k1/y c = k1/z b2 = ac [k1/y]2=k1/xk1/z =>k2/y = k1/x+1/z => 2/y = 1/x+1/y =>y= 2xz/x+z

Ans: (17)

(d) ax = b,by = c ,cz = a then the value of xyz is is

(a) 0

(b) 1

(c ) 1/abc

(d) abc

Solution:

ax = b (cz)x = b by)xz = b =>xyz =1

[cz = a] [by = c]

Ans: (b) (18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then the value of 'z' is (a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64

Solution:

Ans: ( c)

2x = 4y=8z 2x = 22y = 23z x= 2y = 3z Multiply above equation with 2 2x = 4y= 6z (1/2x+1/4y+1/6z) = 24/7 =>(1/6z+1/6z+1/6z) = 24/7 => 3 / 6z = 24/7 => z= 7/48

Problems on PercentagesSimple problems: 1 . Express the following a) 56% SOLUTION: b) 4% 56/100=14/25 as a fraction.

SOLUTION: 4/100=1/25

c) 0.6%

SOLUTION:

0.6/100=6/1000=3/500

d) 0.08% SOLUTION: 0.08/100=8/10000=1/1250

2. a)

Express the following as decimals 6% SOLUTION: 6% = 6/100=0.06

b) 0.04%

SOLUTION: 0.04% = 0.04/100=0.0004

3 . Express the following as rate percent. i).23/36 SOLUTION: ii).6 = (23/36*100) % = 63 8/9%

SOLUTION: 6 =27/4 (27/4 *100) % =675 % 4. Evaluate the following: 28% of 450 + 45% of 280 ? SOLUTION: =(28/100) *450 + (45/100) *280 = 28 * 45 / 5 = 252

5. 2 is what percent of 50? SOLUTION: Formula : (IS / OF ) *100 % = 2/50 *100 = 4%

6. is what percent of 1/3? SOLUTION: =( ) / (1/3) *100 % = 3/2 *100 % = 150 % is 40 Quintals?

7. What percent of 2 Metric tonnes

SOLUTION: 1 metric tonne =10 Quintals So required percentage=(40/(2*10))*100% = 200% 8. Find the missing figure . i) ? % of 25 = 2.125 SOLUTION : Let x% of 25 = 2.125.then (x/100) *25 =2.125 x = 2.125 * 4 = 8.5 9% of ? =6.3

ii)

SOLUTION:

Let 9 % of x = 6.3. Then 9/100 of x= 6.3 so x = 6.3 *100/7 = 70.

9. Which is the greatest in 16 2/3 %, 2/15,0.17? SOLUTION: 16 2/3 % = 50/3 % =50/3 * 1/100 =1/6 = 0.166 2 / 15 =0.133 So 0.17 is greatest number in the given series. 10.If the sales tax be reduced from 3 % to 3 1/3 % , then what difference does it make to a person who purchases an article with marked price of RS 8400? SOLUTION: Required difference = 3 % of 8400 3 1/3 % of 8400 =(7/2-10/3)% of 8400 =1/6 % of 8400 = 1/6* 1/100* 8400 = Rs 14. 11. A rejects 0.08% of the meters as defective .How many will he examine to reject 2? SOLUTION: Let the number of meters to be examined be x. Then 0.08% of x=2. 0.08/100*x= 2 x= 2 * 100/0.08 =2 * 100 * 100/8 = 2500 12.65 % of a number is 21 less than 4/5 of that number. What is the number? SOLUTION: Let the number be x. 4/5 x- (65% of x) = 21 4/5x 65/100 x=21 15x=2100 x=140

13. Difference of two numbers is 1660.If 7.5 % of one number is 12.5% of the other number. Find two numbers? SOLUTION: Let the two numbers be x and y. 7.5% of x=12.5% of y' So 75x=125 y 3x=5y x=5/3y.

Now x-y=1660 5/3y-y=1660 2/3y=1660 y=2490 So x= 2490+1660 =4150. So the numbers are 4150 , 1660. 14. In expressing a length 81.472 KM as nearly as possible with 3 significant digits ,Find the % error? SOLUTION: Error= 81.5-81.472=0.028 So the required percentage = 0.028/81.472*100% = 0.034% Top 15. In an election between two persons ,75% of the voters cast their votes out of which 2% are invalid. A got 9261 which 75% of the total valid votes. Find total number of votes? SOLUTION: Let x be the total votes. valid votes are 98% of 75% of x. So 75%(98%(75% of x))) = 9261 ==> 75/100 *98 /100 * 75 100 *x = 9261 x= 1029 * 4 *100 *4 / 9 = 16800 So total no of votes = 16800

16 . A's maths test had 75 problems i.e 10 arithmetic, 30 algebra and 35 geometry problems.Although he answered 70% of arithmetic ,40% of algebra and 60 % of geometry problems correctly he didn't pass the test because he got less than 60% of the problems right. How many more questions he would have needed to answer correctly to get a 60% passing grade. SOLUTION: 70% of 10 =70/100 * 10 =7 40% of 30 = 40 / 100 * 30 = 12 60 % of 35 = 60 / 100 *35 = 21 So correctly attempted questions = 7 + 12 + 21 =40. Questions to be answered correctly for 60% grade =60% of 75 = 60/100 *75 =45. So required questions=45-40 = 5

17 . If 50% of (x y) = 30% of (x + y) then what percent of x is y ?

SOLUTION: 50/100(x-y) =30/100(x+y) (x-y)= 3/10(x+y) 5x-5y=3x+3y x=4y So Required percentage =y/x*100 % =y/4y *100 % = 25%. 18 .If the price of tea is increased by 20% ,find how much percent must a householder reduce her consumption of tea so as not to increase the expenditure? SOLUTION: Reduction in consumption= R/(100+R) *100% =20/120 *100 = 16 2/3 % 19.The population of a town is 176400 . If it increases at the rate of 5% per annum ,what will be the population 2 years hence? What was it 2 years ago? SOLUTION: Population After 2 years = 176400[1+5/100]2 =176400 * 21/20 *21/20 =194481 Population 2 years ago = 176400/(1+5/100)2 = 176400 * 20/21 *20/ 21 =160000 20.1 liter of water is add to 5 liters of a 20 % solution of alcohol in water . Find the strength of alcohol in new solution? SOLUTION: Alcohol in 5 liters = 20% of 5 =1 liter Alcohol in 6 liters of new mixture = 1liter So % of alcohol is =1/6 *100=16 2/3% 21.If A earns 33 1/3 more than B .Then B earns less than A by what percent? SOLUTION: 33 1/3 =100 / Required Percentage = (100/3)/(100 + (100/3)) *100 % = 100/400 *100 = 25 % 22.A school has only three classes which contain 40,50,60 students respectively.The pass percent of these classes are 10, 20 and 10 respectively . Then

find the pass percent in the school. SOLUTION: Number of passed candidates = 10/100*40+20/100 *50+10/100 * 60 =4+10+6 =20 Total students in school = 40+50+60 =150 So required percentage = 20/150 *100 = 40 /3 =13 1/3 % 23. There are 600 boys in a hostel . Each plays either hockey or football or both .If 75% play hockey and 45 % play football ,Find how many play both? SOLUTION: n(A)=75/100 *600 =450 n(B) = 45/100 *600 = 270 n(A^B)=n(A) + n(B) n(AUB) =450 + 270 -600 =120 So 120 boys play both the games.

24.A bag contains 600 coins of 25p denomination and 1200 coins of 50p denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the percentage of money removed from the bag ? SOLUTION: Total money = (600 * 25/100 +1200 *50/100) =Rs 750 25p coins removed = 12/100 *600 =72 50p coins removed = 24/100 *1200 =288 So money removed =72 *1/4 +288 *1/2 = Rs 162 So required percentage=162/750 *100 =21 .6% 25. P is six times as large as Q.Find the percent that Q is less than P? SOLUTION: Given that P= 6Q So Q is less than P by 5Q. Required percentage= 5Q/P*100 % =5/6 * 100 % =83 1/3%

26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume? SOLUTION: Surface area = 4 *22/7 *r2 = 3/r(4/3 * 22/7 * r3) =3/r * VOLUME Where r = 10 cm So we have S= 3/10 V =3/10 *100 % of V = 30 % of V So surface area is 30 % of Volume.

27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5 kg more for Rs 100.What is the reduced price per kg. SOLUTION: Let the original price = Rs x/kg Reduced price =79/100x /kg ==> 100/(79x/100)-100/x =10.5 ==> 10000/79x-100/x=10.5 ==> 10000-7900=10.5 * 79 x ==> x= 2100/10.5 *79 So required price = Rs (79/100 *2100/10.5 *79) /kg = Rs 2 per kg. 28.The length of a rectangle is increased by 60 % . By what percent would the width have to be decreased to maintain the same area? SOLUTION: Let the length =l,Breadth= b. Let the required decrease in breadth then 160/100 l *(100-x)/100 b=lb 160(100-x)=100 *100 or 100-x =10000/160 =125/2 so x = 100-125/2 =75/2=37.5

be x %

Profit and LossImportant Facts: Cost Price: The price at which an article is purchased, is called its cost price,abbreviated as C.P. Selling Price: The price at which an article is sold, is called its selling price,abbreviated as S.P. Profit or Gain: If S.P. Is greater than C.P. The seller is said to have a profit or gain.

Loss:if S.P. Is less than C.P., the seller is said to have incurred a loss. Formulae 1.Gain=(S.P-C.P) 2.Loss=(C.P-S.P) 3.Loss or Gain is always reckoned on C.P. 4.Gain%=(gain*100)/C.P 5.Loss%=(loss*100)/C.P 6.S.P=[(100+gain%)/100]*C.P 7.S.P=[(100-loss%)/100]*C.P 8.C.P=(100*S.P)/(100+gain%) 9.C.P=(100*S.P)/(100-loss%) 10.If an article is sold at a gain of say,35%,then S.P=135% of

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