## LCM & HCF SET 3

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