QUANTITATIVE APTITUDE

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

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

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