ID : ww-10-real-numbers [1] Grade 10 Real Numbers For more such worksheets visit www.edugain.com Answer t he quest ions (1) In a seminar, the number of participants in German, English and French are 378, 210 and 154 respectively. Find the numbers of rooms required to house them if in each room, the same number of participants are to be accommodated and all of them must belong to the same language. (2) Find the largest positive integer that will divide 1664 and 593 leaving remainders 8 and 17 respectively. (3) Prove that 1 5 is an irrational number. (4) Using Euclid's Division Algorithm, f ind the HCF of 8490 and 525 (5) The LCM of two numbers is 1560 and their product is 20280. Find their HCF. (6) Three people go f or a morning walk together. Their steps measure 66 cm, 72 cm and 90 cm respectively. What is the minimum distance traveled when their steps will exactly match af ter starting the walk assuming that their walking speed is same? (7) Find the LCM of 690 and 300. (8) In a school annual day f unction parade, a group of 954 students need to march behind the band of 612 members. The two groups have to march in the same number of columns. What is the maximum number of columns in which they can march? (9) Sarah and Ronald are racing on a circular track. If Sarah takes 54 minutes and Ronald takes 60 minutes to complete the round. If they both start at the same point at the same time and go in same direction, af ter how many minutes will they meet again at the start point? Choose correct answer(s) f rom given choice (10) n 2 -n is f or every positive integer n. a. can't say b. even c. odd d. negative (11) The HCF of 828 and 324 is a. 35 b. 41 c. 36 d. None of above r
ID : ww-10-real-numbers [2] (12) Euclid's Division Lemma states that if r and s are any two positive integers, then there exist unique integers t and u such that a. r = st + u, 0 < u s b. r = st + u, 0 u < s c. r = st + u, 0 t < s d. r = st + u, 0 < t s (13) 3 + 2 is number a. an irrational b. a rational c. a whole d. a natural (14) Find the smallest number which when increased by 6 is exactly divisible by both 150 and 130. a. 1956 b. 1950 c. 10 d. 1944 (15) T wo tankers contain 649 litres and 297 litres of petrol respectively. A container with maximum capacity is used which can measure the petrol of either tanker in exact number of litres. How many containers of petrol are there in the f irst tanker. a. 59 b. 60 c. 56 d. 11 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ww-10-real-numbers [3] (1) 53 It is given that, in the seminar, the number of participants in German, English and French are 378, 210 and 154 respectively. The total number of participants = 378 + 210 + 154 = 742 T he number of participants accommodated in each room belong to the same language is equal to the HCF of the number of participants in German, English and French. Let's f ind the HCF of 378 and 210, using Euclid's Division Algorithm, 378 = 210 1 + 168, 210 = 168 1 + 42, 168 = 42 4 + 0. The HCF of 378 and 210 is 42. Now, f ind the HCF of 42 and 154, using Euclid's Division Algorithm, 154 = 42 3 + 28, 42 = 28 1 + 14, 28 = 14 2 + 0. The HCF of 42 and 154 is 14. Theref ore, the HCF of 378, 210 and 154 is 14. Step 5 The numbers of rooms required to house them = Total total number participants / HCF of the number of participants in German, English and French. = 742/14 = 53 Step 6 Thus, the number of rooms required to house them is 53.
(2) 72 ID : ww-10-real-numbers [4] We have to f ind the largest positive integer that will divide 1664 and 593 leaving remainders 8 and 17 respectively. In other words, we have to f ind the largest number that will divide (1664-8) and (593-17) leaving no remainder. Such number is the Highest Common Factor (HCF) of : 1656 [i.e., 1664-8] and 576 [i.e., 593-17]. Let's f ind the HCF of 1656 and 576 using Euclid's Division Algorithm, 1656 = 576 2 + 504, 576 = 504 1 + 72, 504 = 72 7 + 0. Thus, the HCF of 1656 and 576 is 72. Theref ore, 72 is the largest positive integer that will divide 1664 and 593 leaving the remainders 8 and 17 respectively. (3) Let's assume 1 5 is a rational number. Theref ore, we can f ind two integers, a and b, such that, 1 5 = a b [where, b is not equal to zero.] Now, 5 = b a b is a rational number as a and b are integers. a Theref ore, 5 is a rational which contradicts to the f act that 5 is Irrational. Hence, our assumption is f alse and 1 is irrational. 5 Theref ore, 1 5 is an irrational number.
(4) 15 ID : ww-10-real-numbers [5] If we look at the question, we notice that 8490 > 525. Theref ore, a = 8490 and b = 525, By applying the Euclid's Division Algorithm(a = bq + r) we get, 8490 = 525 16 + 90 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 525 and b = 90, we get, 525 = 90 5 + 75 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 90 and b = 75, we get, 90 = 75 1 + 15 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 75 and b = 15, we get, 75 = 15 5 + 0 Step 5 Since, r = 0, the process stops. The divisor at this stage is 15. Thus, the HCF of the pair 8490 and 525 is 15. (5) 13 It is given that the LCM of two numbers is 1560 and their product 20280. We know that the product of two numbers is equal to the product of HCF and LCM of the numbers. HCF of the numbers LCM of the numbers = Product of the numbers HCF of the numbers = Product of the numbers/lcm of the numbers = 20280/1560 = 13 Theref ore, the HCF of the numbers is 13.
(6) 39.6 m ID : ww-10-real-numbers [6] Distance af ter which their steps match should be multiple of their steps. Also since we want to f ind the miniumum distance, the distance af ter which their steps will match f or the f irst time af ter starting the walk is equal to the LCM of the steps measured f or the three persons i.e 66, 72 and 90 cm. The LCM of 66, 72 and 90 is 3960. Thus, the number of steps af ter which their steps will meet f or the f irst time af ter starting the walk are 3960 cm. On converting centimeters to meters, answer will be 39.6 meters.
(7) 6900 ID : ww-10-real-numbers [7] Let us f ind the LCM of 690 and 300. All prime f actors of 690: 2 690 2 is a factor of 690 3 345 3 is a factor of 345 5 115 5 is a factor of 115 23 23 23 is a factor of 23 1 Thus, 690 = 2 3 5 23. All prime f actors of 300: 2 300 2 is a factor of 300 2 150 2 is a factor of 150 3 75 3 is a factor of 75 5 25 5 is a factor of 25 5 5 5 is a factor of 5 1 Thus, 300 = 2 2 3 5 5. Thus, the LCM of 690 and 300 = 2 2 3 5 5 23 = 6900.
(8) 18 ID : ww-10-real-numbers [8] Since number of students in each line (or column) should be same, number of columns should f ully divide number of students. Similarly number of band members should also be f ully divisible by number of columns Theref ore number of columns should be highest possible number which f ully divides both number of students and number of band members. T heref ore, maximum of number of columns in which 954 students can march behind the band of 612 members are equal to the HCF of 954 and 612. Let's f ind the HCF of 954 and 612 using Euclid's Division Algorithm, 954 = 612 1 + 342, 612 = 342 1 + 270, 342 = 270 1 + 72, 270 = 72 3 + 54, 72 = 54 1 + 18, 54 = 18 3 + 0. Thus, the HCF of 954 and 612 is 18. Theref ore, the maximum number of columns in which they can march are 18. (9) 540 Time af ter which they meet again at the starting point should be multiple of 54 and 60. Theref ore, number of minutes af ter which they will be meet again at starting point will be equal to the LCM of 54 and 60. Let us f ind the LCM of 54 and 60. All prime f actors of 54 = 2 3 3 3, and, 60 = 2 2 3 5. Thus, the LCM of 54 and 60 = 2 2 3 5 3 3 = 540. Theref ore, they will meet again af ter 540 minutes.
(10) b. even ID : ww-10-real-numbers [9] We have been asked to prove that n 2 - n is even f or every positive integer n. Bef ore beginning, we have to understand the f ollowing Even Even = Even Even - Even = Even Odd - Odd = Even Odd Odd = Odd First, suppose n is odd: Now, n 2 - n = (odd) 2 - (odd) = odd odd - odd = odd - odd = Even Second, suppose n is even: Now, n 2 - n = (even) 2 - (even) = even even - even = even - even = even Step 5 Theref ore, n 2 - n is even f or every positive integer n.
(11) c. 36 ID : ww-10-real-numbers [10] If we look at the question, we notice that 828 > 324. Theref ore, a = 828 and b = 324, By applying the Euclid's Division Algorithm(a = bq + r) we get, 828 = 324 2 + 180 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 324 and b = 180, we get, 324 = 180 1 + 144 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 180 and b = 144, we get, 180 = 144 1 + 36 Since, r 0, by applying the Euclid's Division Algorithm by taking a = 144 and b = 36, we get, 144 = 36 4 + 0 Step 5 Since, r = 0, the process stops. The divisor at this stage is 36. Thus, the HCF of the pair 828 and 324 is 36. (12) b. r = st + u, 0 u < s Euclid s division lemma: Euclid s division lemma, states that f or any two positive integers r and s we can f ind two whole numbers t and u such that r = s t + u where 0 u < s. Let us take an example to understand it. Suppose we have to divide 77 by 14. 77 = 14 5 + 7 where, r = 77, s = 14, t = 5 and u = 7 This type of division is known as Euclid's division lemma. One important point to note here is that u is the remainder when r is divided by s. Remainder can be greater than or equal to 0, but should be less than divisor (i.e. 0 u < s). Theref ore, the correct answer is r = st + u where 0 u < s.
(13) a. an irrational ID : ww-10-real-numbers [11] Let's assume that this sum is rational, that is, 3 + 2 = a b [where, and b are integers with b is not equal to zero.] On squaring both sides of above equation we get, 2 = a b - 3 2 = a2 b 2-2 3 a b + 3 Now solve this equation f or 3, 2 3 a b = 3-2 + a2 b 2 3 = 1 2 a b + a 2 b 2 2 a b 3 = 1b 2a + a 2b This implies that 3 is a rational number which is a contradiction. Theref ore, 3 + 2 is an irrational number. (14) d. 1944 The smallest number which is exactly divisible by 150 and 130 is the LCM of 150 and 130. The LCM of 150 and 130 is 1950. According to the question, the required number is 6 less than the smallest number which is exactly divisible by 150 and 130. Hence, the required number = 1950-6 = 1944.
(15) a. 59 ID : ww-10-real-numbers [12] It is given that, the two tankers contain 649 litres and 297 litres of petrol respectively. T he maximum capacity of the container which can measure the petrol of either tanker in exact number of litres is equal to the HCF of 649 and 297. Let's f ind the HCF of 649 and 297, using Euclid's Division Algorithm, 649 = 297 2 + 55, 297 = 55 5 + 22, 55 = 22 2 + 11, 22 = 11 2 + 0. Thus, the HCF of 649 and 297 is 11. T heref ore, the maximum capacity of the container which can measure the petrol of either tanker in exact number of litres is 11 litres. Now, the number of containers of petrol in the f irst tanker = Petrol contained by the f irst tanker in litres/t he maximum capacity of the container which can measure the petrol of either tanker in exact number of litres = 649/11 = 59 Step 5 Thus, the number of containers of petrol in the f irst tanker are 59.