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TIME DISTANCE SET 2

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Q.1 A person walking takes 26 steps to come down on a escalator & it takes 30 sec for walking. Same person while running takes 18 sec. & 34 steps. How many steps are there ?A) 46 stepsB) 47 stepsC) 48 stepsD) 49 stepsAns. ALet escalator moves at n steps/secTherefore,26y+30n = 18n + 34 y12n = 8n= 2/3Total no. of steps= 26+30*2/3 = 46 steps Q.2 A man walks from P to Q at the rate of 5 km an hour & returns from Q to P at the rate of 3 km an hour. What is the rate in km per hour for the whole journey ?A) 9/2B) 47/9C) 4D) 15/4Ans. DL.C.M. of 3 & 5 = 15 kmTime taken in going from P to Q = 15/5 = 3Time taken in coming from Q to P = 15/3 = 5Therefore, total time in going & coming back = 5 + 3 = 8 Thus, average speed = 30 / 8 =15/ 4 km/hr Q.3 A ship 40 km from shore springs a leak which admits 3 ¾ quintal of water in 12 min. 60 quintal would suffice to sink the ship. But its pumps can throw out 12 quintals of water in an hour. Find the average rate of sailing so that she may reach the shore just it begins to sink ?A) 4.5 km/hrB) 6 km/hrC) 5 km/hrD) 4 km/hrAns. AIn 12 min. leak admits = 15/4 quintalsIn one hour leak admits = 15/4 * 60/12=75/4 quintalsIn one hour pumps throw out = 12 quintalsWater left in the ship in one hour = 75/4 – 12=27/4 quintals27/4 quintals of water is left in the ship in 1 hour60 quintals of water is left in the ship in = 1*60*4 / 27 = 80/9Now in 80/9hrs ship runs = 40 km1 hr the ship runs = 40 * 9 / 80=4.5 km Q.4 Amit starts from a point A & walks to another point B, & then returns from B to A by his car & thus takes a total time of 6 hrs & 45 min. If he had driven both ways in his car, he would have taken 2hrs less. How long would it take for him to walk both the ways ?A) 8hr 15minB) 8h 45minC) 7h 45minD) 8h 30minAns. BLet w be the time taken in one way by walking & c be the time taken in one way by carIst caseW+c=6h 45minOr 2w +2c=13h 30minIInd case2c=4h 45minFrom these two equations,we get2w +2c=13h 30minOr 2w+4h 45min=13h 30minOr 2w=13h 30min – 4h 45minOr 2w=8h 45minHence, if he walks both the ways time = 8h 45min Q.5 Arya runs at a speed of 40 metres/min. Rahul follows him after an interval of 5 min. & runs at a speed of 50metres/min. Rahul’s dog runs at a speed of 60 metres/min & starts along with the Rahul. The dog reaches Aryans & then comes back to the Rahul, & continues to do so till Rahul reaches Aryan .What is the total distance covered by the dog ?A) 980 metersB) 750 metersC) 600 metersD) 1200 metersAns. DLet they meet at a distance of y from start after time t of Rahul start40*(5-t) = 50t200 + 40t = 50tT=200/10 = 20min.Y=50*t= 50*20 = 1000 metersNow see from the options except 1200 meters are smaller then 1000m & dog in any case has to move more than 1000m

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TIME DISTANCE SET 1

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Q.1 Satpal walks 2/3 km from a place P, towards east and then from there 1 5/7 km towards west. Where will he be now from P? A) Distance of 1 1/21 km towards west of P B) Distance of 1 7/21 km towards west of P C) Distance of 1 1/21 km towards east of P D) Distance of 1 7/21 km towards east of P Ans. A 2/3 + (-1 5/7) = 2/3 + (-12)/7= 2 * 7 / 3 * 7 + (-12) * 3 / 7 * 314-36 / 21 = 1 1/21 Since it is negative, it means Satpal is at a distance 1 1/21km towards west of P. Q.2 If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then what is the ratio of the distances travelled by them in one hour? A) 8 : 5 B) 5 : 8 C) 4 : 5 D) 1 : 2 Ans. B Required ratio= Distance travelled by bus in 1 hr/ Distance travelled by train in 1 hr= 160/4 / 320/5 =40/64 =5/8= 5:8 Q.3 A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 m ahead of the policeman, then the time required for the policeman to catch the thief will be A) 2 min B) 3 min C) 4 min D) 6 minAns. B60*100/2000=3 min. Q.4 The tank-full petrol in Arun’s motor-cycle lasts for 10 days. If he starts using 25% more everyday, how many days will the tank-full petrol last? A) 5 B) 6 C) 7 D) 8Ans. D125:100 //10:xOr x= 8 Q.5 A person can walk a certain distance and drive back in six hours. He can also walk both ways in 10 hours. How much time will he take to drive both ways? A) Two hours B) Two and a half hours C) Five and a half hours D) Four hours Ans. A

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LCM & HCF SET 3

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Q.1 A person has to completely put each of the three liquids 403 litres of petrol, 465litres of diesel & 496 litres of mobil oil in bottles of equal sizes without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible no. of bottles required ? A) 44 B) 46 C) 34 D) None of the above Ans. A H.C.F. of 403,465,496 is = 31 Therefore largest capacity of bottles must be 31 ltr Now, No. of bottles required to put 403 ltr of petrol i.e. 403/31 = 13 In the same way, No. of bottles required to put 465 ltr of diesel =465/31=15 No. of bottles required to put 496 litre of mobil oil 496/31 = 16 Therefore, least no. of bottle required of 31 litre size = 13+15+16 =44 Q.2 Four metal rods of lengths 78cm, 104cm, 117cm & 169cm are to be cut into the parts of equal length. Each part must be as long as possible. What is the maximum no. of pieces can be cut ? A) 36 B) 43 C) 480 D) 27 Ans. A 78 = 13*2*3 104 = 13*2*2*2 177 = 13*3*3 169 = 13*13 Since H.C.F of each piece 78,104,117,169 = 13 Therefore, number of pieces = 6+8+9+13 = 36 Q.3 Two numbers X & Y are 20 % & 28 % less than a third number Z. By what % is the number Y less that the number X ? A) 9 % B) 8 % C) 12 % D) 10 % Ans. D Let Z= 100 X= 80 Y=72 Y is less by X =8 Therefore, Y is less than X in % = 8 / 80 * 100 =10 % Q.4 A student has made as many identical bunches of flowers as he possibly could using a total of 100 carnations,150 Tulips & 200 lilies . How many flowers did he use in each bunch ? A) 9 B) 5 C) 20 D) 10 Ans. A HCF of 100,150 and 200 is 50. So she can make 50 bunches. Number of carnations in each bunch = 100/50=2 Number of tulips in each bunch =150/50=3 Number of lilies in each bunch =100/50=4 So number of flowers in each bunch = 2+3+4=9 flowers Q.5 The great Indian mathematician Aryabhatt formulated this problem in the twelfth century for his teenaged prime number aged daughter Amravati. He also authored the eponymous Amravati, a compendium of mathematical puzzles, in which the number of problems that use this formula is the sum of two prime numbers. The product of the two prime numbers is smaller than the total number of problems in the Amravati . Now, if the difference of any two numbers is 4 and their product is 18, what is the sum of their squares? A) 34 B) 40 C) 42 D) 52 Ans. D x-y = 4 x*y = 18 x^2 + y^2 = (x-y)^2 + 2*x*y = 16+36 = 52 Q.6 How many numbers between 50000 and 60000 can be formed using the digits 2 to 7 when any digit can occur any number of times? A) 1296 B) 625 C) 7776 D) 6667 Ans. A 1 6 6 6 6 the 1st place should be 5(only 1) remaining places in 6 ways each total no of numbers=6^4=1296 Q.7 The product of two numbers is 2028 and their H.C.F is 13. The number of such pair(s) is : A) 1 B) 2 C) 3 D) 4 Ans. B Let the numbers be 13 a and 13 b. Then 13a x 13 b = 2028 or ab=12 Now, co-prime with product 12 are (1, 12) and (3, 4) So, the required numbers are (13 x 1), (13 x 12 ) and (13 x 3), (13 x 4). Therefore there are two such pairs i.e. (13,156) and (39,52). Q.8 The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is: A) 64 B) 124 C) 156 D) 260 Ans. A N = H.C.F of (1356 – 12), (1868 – 12), (2764 – 12) H.C.F of 1344, 1856 and 2752 = 64

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LCM & HCF SET 2

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Q.1 A contract on construction job specifies a penalty for delay in completion of the work beyond a certain date is as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day etc., the penalty for each succeeding day being Rs. 50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days? A) Rs. 4950 B) Rs. 4250 C) Rs. 3600 D) Rs. 650 Ans. B Rs. 4250 (200 + 250 + 300 + …10 terms This is an Arithmetic Progression first term (a) = 200 difference (d) = 50 number of terms (n) = 10 sum = n/2 [2a+(n-1)d] 10/2*[2*200 + (9*50)]=4250 Q.2 How many nos. from 0 to 999 are not divisible by either 5 to 7 ? A) 341 B) 313 C) 686 D) 786 Ans. C S= (5+10+15+….+995)+(7+14+….+994)- (35+70+…+980) = 199+142-28 = 313 Therefore, no. not divisible by 5 or 7 = 999-313 = 686 Q.3 While adding first few continuous natural numbers, a candidate missed one of the numbers & wrote the answer as 177. What was the number missed ? A) 12 B) 11 C) 13 D) 14 Ans. C Sum of first natural nos less than 177 is : 18 = 18/2(18+1) = 9*19 = 171 (Just Take natural no. & put in the formula,& find whether is it less than given no. or not ) Sum of first natural nos. greater than 177 is: 19 = 19/2(19+1) = 19*10 = 190 Therefore, missing term is 190 – 177 = 13 Q.4 Five persons fire a bullets at a target at an interval of 6,7,8,9,& 12 seconds respectively. The no. of times they would fire the bullets together at the target in an hour is A) 6 B) 7 C) 8 D) 9 Ans. B Explanation: L.C.M. of 6,7,8,9 & 12 = 504 Reqd. no. = 60*60/504 =7 times Q.5 Three persons start walking together and their steps measure 40 cm, 42 cm and 45cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? A) 25 m 20 cm B) 50 m 40 cm C) 75 m 60 cm D) 100 m 80 cm Ans. A 25 m 20 cm LCM of 40, 42 and 45 is 2520

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LCM & HCF SET 1

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Q.1 The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? A) 40 & 24 B) 42 & 24 C) 42 & 20 D) None of the above Ans. B Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6). When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5). According to the given condition. (10x + y) + (10y + x) = 66 i.e., 11(x + y) = 66 i.e., x + y = 6 (1) We are also given that the digits differ by 2, therefore, either x – y = 2 (2) or y – x = 2 (3) If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24. Q.2 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.Find the rational number ? A) 13/ 21 B) 13/20 C) 11/21 D) 11/ 20 Ans. A Acc. to the question, 2(x + 17) = 3(x + 7) ? 2x + 34 = 3x + 21 ? 34 – 21 = 3x – 2x ?13 = x Numerator of the rational number = x = 13 Denominator of the rational number = x + 8 = 13 + 8 = 21 Q.3 Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number? A) 2 B) 3 C) 4 D) 5 Ans. C Let the number be x. According to the given question, 8x – 20 = 3x Transposing 3x to L.H.S and -20 to R.H.S, we obtain 8x – 3x = 20 5x = 20 Dividing both sides by 5, we obtain x = 4 Hence, the number is 4. Q.4 What is the number of terms in the series 117,120,123,126…333 ? A) 79 B) 76 C) 72 D) 7 Ans. D a=117, d=3 nth term = 333 Since, tn=a+(n-1)d Or 333=117+(n-1)3 333=117+3n-3 3n=333-114= 219 Or n=219/3 = 73 Q.5 If the numbers from 501 to 700 are written what is the total number of times does the digit 6 appear ? A) 141 B) 140 C) 139 D) 138 Ans. B No. of 6 between 501 to 999 at tens place = 10 No. of 6 between 501 to 599 at unit place = 10 No. of 6 between 600 to 699 at hundred place = 100 No. of 6 between 600 to 699 at tens place = 10 No. of 6 between 600 to 699 at unit place = 10 Total no. of times 6 appears =10+10+100+10+10=140

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AVERAGE SET 4

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Q.1 Four years ago, the average age of four friends was 25 years. Now, with a fifth friend joining the group, their average is 24 years. What is the present age of the new friend? A) 20 yrs B) 21 yrs C) 36 yrs D) 24 yrs Ans. A Total age of all friends including new friend = 24 5 = 120 years Present age of the new friend = new average age – old average age 120 – 100 = 20 years. Q.2 Gavaskar’s average in his first 50 innings was 50. After the 51st innings, his average was 51. How many runs did he score in his 51st innings. (supposing that he lost his wicket in his 51st innings) A) 101 B) 102 C) 103 D) 104 Ans. A Explanation : Total score after 50 innings = 50*50 = 2500 Total score after 51 innings = 51*51 = 2601 So, runs made in the 51st innings = 2601-2500 = 101 If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings. Q.3 In a class the average age is 16yrs.If the teacher who is 40 yrs of age is also included ,the average becomes 17yrs,how many students were there? A) 21 B) 22 C) 23 D) 24 Ans. C 40 – 17 = 23 students Q.4 Average salary of 100 employees in an office is Rs.16000 /month. Management decided to raise salary of every employee by 5% but stopped a transport allowance of Rs. 800 /month which was paid earlier to every employee. What will be the average monthly salary ? A) Rs.16500 B) Rs. 16000 C) Rs. 16800 D) Can’t be determined Ans. B Increament = 5% of 16000 16000*5/100 = Rs. 800 Therefore incremented salary = 16000+800 = 16800 New salary after deducting transport allowance = 16800-800 =Rs. 16000 Q.5 Average marks obtained by the 120 candidates in a certain exam is 35. If the average marks of the passed candidates is 39 & that of the failed candidates is 15. What is the no. of candidates passed the exams ? A) 100 B) 200 C) 120 D) 220 Ans. A Shortcut: No. of passed candidates = Total Candidates (Total average – Failed average) / Passed average – Failed Average Therefore, 120*(35-15) / 39-15 =100

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AVERAGE SET 3

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Q.1 Average marks in a class of 30 students are found to be 45. On checking two mistakes were found. After correction if one student got 45 marks more & another student got 15 marks less, then the corrected average marks are : A) 47 B) 44 C) 46 D) 45 Ans. C Total marks of 300 students before correction = 30*45 = 1350 & total marks of 30 students after correction = 1350+45-15 = 1380 Therefore, corrected average marks 1380/30 = 46 Q.2 Average salary of entire staff in an office is Rs. 120 per month. The average salary of officers is Rs. 460 & that of non-officers is Rs. 110. If the number of officers is 15, then find the number of non-officers in the office ? A) 500 B) 550 C) 510 D) 445 Ans. C Let number of non-officers be x 110x + 460 * 15 = 120(15+x) 120 – 110x = 460*15 – 120*15 = 15(460-120) 10x = 15*340 X= 15*34 = 510 Q.3 A cricketer has completed 10 innings & his average is 21.5 runs. How many runs must he make in his next Innings so as to raise his average to 24 ? A) 25 B) 45 C) 30 D) 49 Ans. D 10 (24 – 21.5) + 24 = 25 + 24 = 49 Q.4 Rahina did a Mathematic test. In this test Each correct answer scored 8 marks & each incorrect answer reduced score by 4 marks. The test contained 30 Questions & after completing it ,Rahina had a score of Zero. How many Questions did she answer correctly ? A) 8 B) 10 C) 15 D) 12 Ans. B 8x – 4y = 0 X + y = 30 On solving, We get x = 10 Q.5 A tree is 8.4 m tall, another is half its height and the third one is its quarter. What is the average height of the trees? A) 4.55 m B) 4.6 m C) 4.9 m D) 4.8 m Ans. C Height of the first tree = 8.4 m tall Height of the second tree = ½ 8.4 = 4.2 m Height of the third tree = ¼ 8.4 = 2.1 m Total height = 8.4 + 4.2 + 2.1 = 14.7 Average height = = 4.9 m

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AVERAGE SET 2

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Q.1 Average weight of 8 persons is increased by the 2.5kg when one of the weights 56kg is replaced by a new man. Find the weight of a new man ? A) 76 kg B) 77 kg C) 78 kg D) 79 kg Ans. A Shortcut: Weight of new comer = Previous weight + No. of persons*Increase in average weight Therefore, 56+8*2.5 = 76 kg Q.2 Average age of 30 boys is equal to the 14yrs, when the age of the class teacher is included average becomes 15 years . Find the average of the class teacher ? A) 45 yrs B) 46 yrs C) 47 yrs D) 48 yrs Ans. A Shortcut: Age of new comer = Previous age + No. of persons*Increase in average age 14+31(15-14) =45 Ans. Q.3 The average age of a group of 15 persons is 29 years. Among them, the age of each of the two persons is 55 years. What is the average age of remaining 13 persons? A) 23 B) 28 C) 26 D) 25 Ans. D 15*29 = 435 2 person of 55kg leaving = 435-110 Therefore new average = 325/13 = 25 Q.4 A man spends his 2 months earning in 3 months & if he saves Rs. 1500 a year, then his monthly income is: A) Rs. 450 B) Rs. 375 C) Rs. 425 D) Rs. 400 Ans. B A man spends his 2 months earning in 3 months & if he saves Rs. 1500 a year, then his monthly income is: Man saves one month’s earning in 3 months Therefore he will save in one year salary of 4 months But his saving in 1 year = Rs. 1500 His 4 month’s earning = Rs.1500 Therefore, His 1 month’s earning = Rs. 375 Q.5 The average score of a cricketer in certain number of innings is 15. When he scored a duck in one innings, his average dropped to 13.5. How many innings has he played in total, including the latest one? A) 13 B) 10 C) 11 D) 9 Ans. B Let the number of innings = x Total runs scored by the player = 15x New number of innings = x + 1 Also new average = = 13.5 ? 15x = 13.5x + 13.5 ? 1.5x = 13.5 ? x = 9. Number of innings played including the latest one = 9 + 1 = 10

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AVERAGE SET 1

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Q.1 A student on her first 3 tests received an average score of N points. If she exceeds her previous average score by 20 points on her fourth test, then what is the average score for the first 4 tests? A) N + 20 B) N + 10 C) N + 4 D) N + 5 Ans. D Required Average= 4N+20/4 = 4N/4 + 20/ = N+5 Q.2 If the weight of a group of 20 boys was calculated to be 89.4 kg& it was later discovered that one weight was misread as 78 kg instead of the correct one of 87 kg. Then the correct average weight (in kg ) is: A) 89.25 B) 89.85 C) 88.95 D) 89.55 Ans. B Total weight of 20 boys which is wrong = 89.4 * 20 = 1788 kg Therefore, correct weight of 20 boys = 1788 – 78 + 87 = 1797 kg Therefore, correct average weight = 1797 / 20 = 89.85 kg Q.3 Average salary of 100 employees in an office is Rs.16000 /month. Management decided to raise salary of every employee by 5% but stopped a transport allowance of Rs. 800 /month which was paid earlier to every employee. What will be the average monthly salary ? A) Rs.16500 B) Rs. 16000 C) Rs. 16800 D) Can’t be determined Ans. B Increament = 5% of 16000 16000*5/100 = Rs. 800 Therefore incremented salary = 16000+800 = 16800 New salary after deducting transport allowance = 16800-800 =Rs. 16000 Q.4 Average marks obtained by the 120 candidates in a certain exam is 35. If the average marks of the passed candidates is 39 & that of the failed candidates is 15. What is the no. of candidates passed the exams ? A) 100 B) 200 C) 120 D) 220 Ans. A Shortcut: No. of passed candidates = Total Candidates (Total average – Failed average) / Passed average – Failed Average Therefore, 120*(35-15) / 39-15 =100 Q.5 A man spends on an average Rs.450 for the first 8 months & Rs. 475 for the next 4 months. If he saves Rs.500. What is the monthly salary? A) 300 B) 400 C) 500 D) 600 Ans. C 450*8 = 3600 475*4 = 1900 Therefore, total expenditure = Rs.500 But saving is : 500 Thus total salary is=6000 So, Monthly salary = 6000/2 = 500 Rs.

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DATA INTERPRETATION SET 6

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1. The number of males in University Q is what per cent of the total number of students (males & females together) in University S? 1) 68 2) 62 3) 66 4) 64 5) None of these Ans: (4) 2. If the total number of males in University T increases by 50%. what would be the total number of students (males & females together) in that University? 1)75,26,000 2)7,62,50,000 3)76,25,000 4)7,52,60,000 5) None of these Ans: (3) 3. What is the ratio of the number of females from University P and Q together to the number of males in the University Rand T together? 1) 27:32 2) 27:28 3) 25:28 4) 28: 27 5) None of these Ans: (2) 4. What is the average number of females in all the universities together? 1) 33,00,000 2) 3,50,000 3) 3,20,000 4) 32,00,000 5) None of these Ans: (1) 5. What is the total number of students (males & females together) in University P and R together? 1) 13,00,000 2) 13,50,000 3) 14,00,000 4) 1,45,00,000 5) None of these Ans: (5)

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