(7) The lowest natural number which when divided by 16, 24, 20 leaves the remainder of 4 in each case is a. 247 b. 244 c. 243 d.

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1 ID : ae-6-lcm-and-hcf [1] Grade 6 LCM and HCF For more such worksheets visit Answer t he quest ions (1) Farah and Hadil are f riends and cricket coach too. Farah goes to Lotus Valley school every 2 days and Hadil goes to Lotus Valley every 5 days to deliver coaching classes. If they both delivered the coaching sessions today, how many days in the next 60 days they both would take the session on the same day? (2) Two brands of chocolates are available in packs of 49 and 21 respectively. If Adila wants to buy equal number of chocolates of both brands, what is the least number of boxes of each chocolate brand that she will need to buy? (3) A rectangular courtyard with length 4 m 26 cm and breadth 2 m 46 cm is to be paved with square stones of the same size. Find the least number of such stones required. (4) 119 cola cans and 140 f ruit juice cans need to be stacked in a school canteen. If each stack is to be of the same height and is to contain cans of the same type, what would be the greatest number of cans each stack can have? Choose correct answer(s) f rom given choice (5) Alf erno-princip is a comet that orbits around the sun once in 54 years, and Wallaby-Grove is a comet that orbits around the sun once in 252 years. The last time they were seen together in the sky was in In which year will they be seen together in the sky next? a b c d (6) If a number is divisible by 8 and 24, it will be necessarily divisible by a. 28 b. 24 c. 26 d. 22 (7) The lowest natural number which when divided by 16, 24, 20 leaves the remainder of 4 in each case is a. 247 b. 244 c. 243 d. none of these Fill in the blanks (8) The largest number which divides 186 and 158 leaving remainder 4 and 2 respectively is.

2 (9) The product of any two numbers is equal to product of their and. ID : ae-6-lcm-and-hcf [2] (10) The greatest number which divides 553, 290 and 478 leaving remainder 3, 4 and 5 respectively is. (11) The length, breadth and height of a room are 16 m 83 cm, 5 m 61 cm and 11 m 90 cm respectively. The length of the longest rod which can measure the dimensions of the room exactly is cm (12) Find the Least Common Multiple (LCM) of f ollowing: A) LCM of 27, 360, 72, 45, 72 is. B) LCM of 18, 192, 60, 16 is. C) LCM of 18, 64, 18, 80 is. D) LCM of 10, 30, 6, 25 is. E) LCM of 9, 84, 18 is. F) LCM of 32, 128, 32, 8, 36 is. (13) The least common multiple (LCM) of 70 and 14 is. (14) T wo tankers contain 1560 litres and 320 litres of petrol respectively. T he largest measuring container which can measure the petrol of either tanker exactly will have a capacity of litres. (15) The largest number that divides 1447 and 382 leaving remainder 7 is Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : ae-6-lcm-and-hcf [3] (1) 6 Farah goes to Lotus Valley every 2 nd days and Hadil goes to Lotus Valley every 5 th days. So like this they will both will take session on every 10 days. In order to f ind the number of days af ter which both of them goes together, we need to f ind a number which is as small as possible and divisible by both 2 and 5. Theref ore we need to f ind LCM of 2 and 5 which is 10 days. Now since both of them goes to school every 10 th days. Theref ore number of session in next 60 days they will take = 60 / 10 = 6 (2) 3 boxes of f irst brand, 7 boxes of second brand. Adila wants to buy equal number of chocolates of two brands, but she needs to buy them in multiples of 49 and 21 respectively. Also, she wants to buy the least number of boxes. the number of chocolates of each brand she will need to buy = L.C.M of 49 and 21. L.C.M of 49 and 21 = 147. Number of boxes of the f irst brand Adila 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 Adila 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 = = 7 she will need to buy 3 boxes of f irst brand and 7 boxes of second brand.

4 (3) 2911 ID : ae-6-lcm-and-hcf [4] According to the question, the length and breadth of the courtyard is 4 m 26 cm and 2 m 46 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 426 cm and 246 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 426 cm and 246 cm, the length and breadth of the courtyard in centimeters. The HCF of 426 and 246 is 6. the needed size of square stones = 6 cm The number of stones needed = Area of the couryard Area of a single stone = 6 6 = 2911 The least number of such stones required are 2911.

5 (4) 7 ID : ae-6-lcm-and-hcf [5] 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 119 and 140. Since the number of cans in each stack should additionally be greatest possible, it should be the Highest Common Factor(HCF) of 119 and 140. Lets f ind the HCF of 119 and 140. All prime f actors of 119: is a factor of is a factor of = 7 17 All prime f actors of 140: is a factor of is a factor of is a factor of is a factor of = The HCF of 119 and 140 is = 7 = 7 Step 6 The greatest number of cans each stack can have 7.

6 (5) d ID : ae-6-lcm-and-hcf [6] We have been told that the comet Alf erno-princip orbits around the sun once in 54 years, and the comet Wallaby-Grove orbits around the sun once in 252 years. Once the two comets are seen together, the number of years af ter which they will be seen together in the sky is equal to the LCM of 54 and 252. Let us now calculate the LCM of 54 and 252. All prime f actors of 54: is a factor of is a factor of is a factor of is a factor of 3 54 = All prime f actors of 252: is a factor of is a factor of is a factor of is a factor of is a factor of = the LCM of 54 and 252 is = = 756. Step 6 The year when they were least seen together in the sky = Theref ore, the year in which they will be seen together in the sky next = = 2757.

7 (6) b. 24 ID : ae-6-lcm-and-hcf [7] If a number is divisible by two dif f erent numbers, it is necessarily divisible by their L.C.M. L.C.M of 8 and 24 = 24. if a number is divisible by 8 and 24, it will be necessarily divisible by 24. (7) b. 244 The lowest number which is divisible by 16, 24, 20, is the Least Common Multiple (LCM) of 16, 24, 20 The LCM of 16, 24, 20 = 240. Now we need to f ind the number which leaves the remainder of 4 when divided by these numbers. This number should be 4 more than the LCM. Theref ore, the required number = LCM (16, 24, 20) + 4 = = 244

8 (8) 26 ID : ae-6-lcm-and-hcf [8] We need to f ind the largest number that divides 186 leaving remainder 4 and divides 158 leaving remainder 2. Such number is the H.C.F of (186-4) and (158-2), that is, the HCF of 182 and 156. Let us now f ind the H.C.F of 182 and 156. All prime f actors of 182: is a factor of is a factor of is a factor of = All prime f actors of 156: is a factor of is a factor of is a factor of is a factor of = Step 6 The H.C.F of 182 and 156 is = 2 13 = 26. Step 7 Theref ore, the largest number which divides 186 and 158 leaving remainders 4 and 2 respectively is 26. (9) HCF LCM

9 (10) 11 ID : ae-6-lcm-and-hcf [9] The biggest number that divides 553, 290 and 478 leaving remainder 3, 4 and 5 respectively is the same number that divides the numbers (553-3), (290-4) and (478-5) leaving no remainder. In other words, we need to f ind the HCF of the f ollowing three numbers: 550 [Simplify 553-3],, 286 [Simplify 290-4],, 473 [Simplify 478-5]. All prime f actors of 550: is a factor of is a factor of is a factor of is a factor of = All prime f actors of 286: is a factor of is a factor of is a factor of = All prime f actors of 473: is a factor of is a factor of = Step 6 The HCF of 550, 286 and 473 is = 11 = 11.

10 Step 7 ID : ae-6-lcm-and-hcf [10] Theref ore, the greatest number which divides 553, 290 and 478 leaving remainder 3, 4 and 5 respectively is 11.

11 (11) 17 ID : ae-6-lcm-and-hcf [11] According to the question, the length, breadth and height of the room are 16 m 83 cm, 5 m 61 cm and 11 m 90 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 1683 cm, 561 cm and 1190 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 1683, 561 and All prime f actors of 1683: is a factor of is a factor of is a factor of is a factor of 17 Theref ore, 1683 = All prime f actors of 561: is a factor of is a factor of is a factor of 17 Theref ore, 561 = Step 6 All prime f actors of 1190: is a factor of is a factor of is a factor of is a factor of 17

12 ID : ae-6-lcm-and-hcf [12] Theref ore, 1190 = Step 7 The HCF of 1683, 561 and 1190 is = 17 = 17. Step 8 Hence, the length of the longest rod which can measure the dimensions of the room exactly is 17 cm. (12) A) B) C) D) E) F)

13 (14) 40 ID : ae-6-lcm-and-hcf [13] The container which can measure petrol of both tanks, should be such that its volume in litres should f ully divide 1560 and 320. T heref ore, capacity of the largest measuring container which can measure the petrol of either tanker exactly is the HCF of 1560 and 320. Let us f ind 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 = Let us now f ind all prime f actors of 320: 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 = The HCF of 1560 and 320 is = = 40 T heref ore, the largest measuring container which can measure the petrol of either tanker exactly will have a capacity of 40 liters.

14 (15) 15 ID : ae-6-lcm-and-hcf [14] We have to f ind the largest number that divides 1447 and 382 leaving remainder 7. In other words, we have to f ind the largest number that divides (1447-7) and (382-7) leaving no remainder. Such number is the Highest Common Factor (HCF) of : 1440 [i.e., ] and 375 [i.e., 382-7]. Let us now f ind the HCF of 1440 and 375. All prime f actors of 1440: 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 5 Theref ore, 1440 = All prime f actors of 375: is a factor of is a factor of is a factor of is a factor of 5 Theref ore, 375 = The HCF of 1440 and 375 is = 3 5 = 15. Step 6 the largest number which divides 1447 and 382 leaving remainder 7 is 15.