(1) A lighthouse has two lights one that f lashes every 2 minutes, and another that f lashes every 1

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1 ID : sg-6-lcm-and-hcf [1] Grade 6 LCM and HCF For more such worksheets visit Answer t he quest ions (1) A lighthouse has two lights one that f lashes every 2 minutes, and another that f lashes every 1 2 minutes. Suppose the lights f lash together at noon. What is the f irst time af ter 3 PM that 2 they will f lash together again? (2) Cayden saves $6.8 every day. Find the minimum number of days in which he will be able to save an exact amount of money. (3) Two tankers contain 768 litres and 240 litres of petrol respectively. What is the capacity of the largest measuring container which can measure the petrol of either tanker exactly? (4) The length, breadth and height of a room are 15 m 60 cm, 13 m 60 cm and 11 m 40 cm respectively. What is the length of the longest rod which can measure the dimensions of the room exactly? (5) A rectangular courtyard with length 4 m 97 cm and breadth 2 m 52 cm is to be paved with square stones of the same size. Find the least number of such stones required. Choose correct answer(s) f rom given choice (6) Giselle has two sheets of paper. One sheet is 374 cms wide and the other sheet is 110 cms wide. She wants to divide the sheets into strips of equal width that are as wide as possible without wasting any paper. How wide should she cut the strips? a. 25 cms b. 28 cms c. 22 cms d. 21 cms (7) If a number is divisible by 24 and 8, it will be necessarily divisible by a. 24 b. 26 c. 28 d. 30 Fill in the blanks (8) The two numbers nearest to which are exactly divisible by 7, 9, 3, 4 are and. (9) The least common multiple (LCM) of 32 and 24 is. (10) Find the Least Common Multiple (LCM) of f ollowing:

2 A) LCM of 12, 48, 12 is. B) LCM of 28, 100, 32, 25, 16 is. ID : sg-6-lcm-and-hcf [2] C) LCM of 18, 126, 48 is. D) LCM of 36, 180, 36, 20 is. E) LCM of 45, 210, 10, 18, 35 is. F) LCM of 18, 120, 6 is. (11) The greatest number which divides 958, 221 and 1122 leaving remainder 4, 5 and 6 respectively is. (12) 260 cola cans and 600 f ruit juice cans need to be stacked in a school canteen. If each stack is of the same height and is to contain cans of the same type, the greatest number of cans each stack can have is. (13) The largest number that divides 664 and 352 leaving remainder 1 is. (14) Find the LCM (least common multiple) of the f ollowing numbers A) 32, 336, 36, 91 = B) 140, 126, 70, 108, 133 = C) 18, 324, 27 = D) 104, 192, 16, 40, 144 = E) 26, 44, 16 = F) 20, 256, 48, 72, 12 = (15) Two brands of chocolates are available in packs of 40 and 15 respectively. If Victoria wants to buy equal number of chocolates of each brands, then the least number of boxes of each chocolate brand that she needs to buy are and respectively Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : sg-6-lcm-and-hcf [3] (1) 3:10 The f irst light f lashes every 2 minutes and the other light f lashes every minutes. Once the two lights f lash together, the amount of time af ter which the two lights will f lash together again is equal to the LCM of 2 minutes and minutes. Bef ore we calculate the LCM of 2 minutes and minutes, let us translate both time periods f rom minutes to seconds. Since 1 minute = 60 seconds, 2 minutes = 2 60 = 120 seconds, and minutes = = 150 seconds. Let us now calculate the LCM of 120 and 150. All prime f actors of 120: is a factor of is a factor of is a factor of is a factor of is a factor of 5 Thus, 120 = All prime f actors of 150: is a factor of is a factor of is a factor of is a factor of 5 Thus, 150 = Step 5

4 The LCM of 120 and 150 = = 600. ID : sg-6-lcm-and-hcf [4] Step seconds in minutes = minutes = 10 minutes. Step 7 Now, we know that the two lights f lash together every 10 minutes. We have been told that the two lights f lashes together at noon. This means, that times when they f lash together again are 12:10, 12:20, 12:30, 12:40,... 3:10,... Theref ore, the time af ter 3 PM that the two lights will f lash together again = 3:10 PM.

5 (3) 48 litres ID : sg-6-lcm-and-hcf [5] The container which can measure petrol of both tanks, should be such that its volume in litres should f ully divide 768 and 240. T heref ore, capacity of the largest measuring container which can measure the petrol of either tanker exactly is the HCF of 768 and 240. Let us f ind all prime f actors of 768: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of = Let us now f ind all prime f actors of 240: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of = The HCF of 768 and 240 is = = 48 Step 5 T heref ore, the largest measuring container which can measure the petrol of either tanker exactly will have a capacity of 48 liters.

6 (4) 20 cm ID : sg-6-lcm-and-hcf [6] According to the question, the length, breadth and height of the room are 15 m 60 cm, 13 m 60 cm and 11 m 40 cm respectively. Now you have to convert each dimension into same unit. Since you know that 1m = 100 cm, the length, breadth and height of the room in centimeters is 1560 cm, 1360 cm and 1140 cm respectively. The length of the rod which can be used to exactly measure all three dimensions of a room should be a f actor of all three dimensions, that is, a f actor common to all three dimensions. Since we have been asked to f ind the length of longest such rod, the length should be equal to HCF of all three dimensions of the room. Let us now f ind the HCF of 1560, 1360 and All prime f actors of 1560: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 13 Theref ore, 1560 = Step 5 All prime f actors of 1360: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 17 Theref ore, 1360 = Step 6 All prime f actors of 1140:

7 is a factor of 1140 ID : sg-6-lcm-and-hcf [7] is a factor of is a factor of is a factor of is a factor of 19 Theref ore, 1140 = Step 7 The HCF of 1560, 1360 and 1140 is = = 20. Step 8 Hence, the length of the longest rod which can measure the dimensions of the room exactly is 20 cm.

8 (5) 2556 ID : sg-6-lcm-and-hcf [8] According to the question, the length and breadth of the courtyard is 4 m 97 cm and 2 m 52 cm respectively. Let us f irst convert each dimension into centimeters. Since we know that 1m = 100 cm, the length and breadth of the courtyard in centimeters is 497 cm and 252 cm respectively. The pavement has to be done with square stones. To perf ectly f it the courtyard, the size of the square stones should be a f actor of both the length and breadth of the courtyard. Please ref er to this f igure to see why it is so. Since we need to minimise the number of stones needed, the size of the stones should be the largest possible. This means the size of the stones should be the highest common f actor (HCF) of 497 cm and 252 cm, the length and breadth of the courtyard in centimeters. The HCF of 497 and 252 is 7. Thus, the needed size of square stones = 7 cm The number of stones needed = Area of the couryard Area of a single stone = 7 7 = 2556 Step 5 The least number of such stones required are 2556.

9 (6) c. 22 cms ID : sg-6-lcm-and-hcf [9] Let the width of strip be x cms Since no cloth should be wasted, 374 cms should be divisible by x cms Similarly, 110 cms should be divisible by x cms Also x has to be as large as possible, theref ore x should be HCF of 374 and 110 x = HCF(374, 110) = 22 cms (7) a. 24 If a number is divisible by two dif f erent numbers, it is necessarily divisible by their L.C.M. L.C.M of 24 and 8 = 24. Thus, if a number is divisible by 24 and 8, it will be necessarily divisible by 24. (8) (9) (10) A) B) C) D) E) F)

10 (11) 18 ID : sg-6-lcm-and-hcf [10] The biggest number that divides 958, 221 and 1122 leaving remainder 4, 5 and 6 respectively is the same number that divides the numbers (958-4), (221-5) and (1122-6) leaving no remainder. In other words, we need to f ind the HCF of the f ollowing three numbers: 954 [Simplify 958-4],, 216 [Simplify 221-5],, 1116 [Simplify ]. All prime f actors of 954: is a factor of is a factor of is a factor of is a factor of 53 Thus, 954 = All prime f actors of 216: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 3 Thus, 216 = Step 5 All prime f actors of 1116: is a factor of is a factor of is a factor of is a factor of is a factor of 31

11 ID : sg-6-lcm-and-hcf [11] Thus, 1116 = Step 6 The HCF of 954, 216 and 1116 is = = 18. Step 7 Theref ore, the greatest number which divides 958, 221 and 1122 leaving remainder 4, 5 and 6 respectively is 18.

12 (12) 20 ID : sg-6-lcm-and-hcf [12] Since each stack needs to be of the same height and is to contain cans of the same type, the number of cans in each stack should be a f actor of both 260 and 600. Since the number of cans in each stack should additionally be greatest possible, it should be the Highest Common Factor(HCF) of 260 and 600. Lets f ind the HCF of 260 and 600. All prime f actors of 260: is a factor of is a factor of is a factor of is a factor of 13 Thus, 260 = All prime f actors of 600: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 5 Thus, 600 = Step 5 The HCF of 260 and 600 is = = 20 Step 6 The greatest number of cans each stack can have 20.

13 (13) 39 ID : sg-6-lcm-and-hcf [13] We have to f ind the largest number that divides 664 and 352 leaving remainder 1. In other words, we have to f ind the largest number that divides (664-1) and (352-1) leaving no remainder. Such number is the Highest Common Factor (HCF) of : 663 [i.e., 664-1] and 351 [i.e., 352-1]. Let us now f ind the HCF of 663 and 351. All prime f actors of 663: is a factor of is a factor of is a factor of 17 Theref ore, 663 = All prime f actors of 351: is a factor of is a factor of is a factor of is a factor of 13 Theref ore, 351 = Step 5 The HCF of 663 and 351 is = 3 13 = 39. Step 6 Thus, the largest number which divides 664 and 352 leaving remainder 1 is 39. (14) A) B) C) D)

14 E) F) ID : sg-6-lcm-and-hcf [14] (15) 3 8 Victoria wants to buy equal number of chocolates of two brands, but she needs to buy them in multiples of 40 and 15 respectively. Also, she wants to buy the least number of boxes. Thus, the number of chocolates of each brand she will need to buy = L.C.M of 40 and 15. L.C.M of 40 and 15 = 120. Number of boxes of the f irst brand Victoria will need to buy = Number of chocolates of the f irst brand he will need to buy Number of chocolates in one box of f irst brand = = 3 Number of boxes of the second brand Victoria will need to buy= Number of chocolates of the second brand he will need to buy Number of chocolates in one box of second brand = = 8 Thus, she will need to buy 3 boxes of f irst brand and 8 boxes of second brand.