Class 10 Real Numbers


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1 ID : in10realnumbers [1] Class 10 Real Numbers For more such worksheets visit Answer t he quest ions (1) The LCM of two numbers is 760 and their product is Find their HCF. (2) Prove that n 2 n is even f or every positive integer n. (3) Three people go f or a morning walk together. Their steps measure 45 cm, 70 cm and 80 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? (4) Lovleen and Aditya are racing on a circular track. If Lovleen takes 20 minutes and Aditya takes 24 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? (5) Prove that is an irrational number. (6) Find the LCM of 986 and 374. Choose correct answer(s) f rom given choice (7) T wo tankers contain 522 litres and 252 litres of petrol respectively. Find the maximum capacity of the container which can measure the petrol of either tanker in exact number of litres. a. 17 b. 20 c. 18 d. None of above (8) How many prime f actors are there in the prime f actorization of 18150? a. 6 b. 5 c. 7 d. 8 (9) Which of the f ollowing is not an irrational number.. a b c d (10) The HCF of 7792 and 288 is a. 17 b. 16 c. 15 d. None of above a (11) Euclid's Division Lemma states that if a and b are any two positive integers, then there exist unique integers c and d such that a. a = bc + d, 0 d < b b. a = bc + d, 0 c < b c. a = bc + d, 0 < c b d. a = bc + d, 0 < d b
2 ID : in10realnumbers [2] (12) In a school annual day f unction parade, a group of 232 students need to march behind the band of 92 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? a. 3 b. 5 c. 4 d. None of above (13) 1 3 is number a. an irrational b. a whole c. a natural d. a rational (14) 7 is number a. a natural b. an irrational c. whole d. a rational (15) is number a. a whole b. a natural c. a rational d. an irrational 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at
3 Answers ID : in10realnumbers [3] (1) 8 It is given that the LCM of two numbers is 760 and their product 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 = 6080/760 = 8 Theref ore, the HCF of the numbers is 8. (2) 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 Step 4 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.
4 (3) 50.4 m ID : in10realnumbers [4] 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 45, 70 and 80 cm. The LCM of 45, 70 and 80 is Thus, the number of steps af ter which their steps will meet f or the f irst time af ter starting the walk are 5040 cm. Step 4 On converting centimeters to meters, answer will be 50.4 meters. (4) 120 Time af ter which they meet again at the starting point should be multiple of 20 and 24. Theref ore, number of minutes af ter which they will be meet again at starting point will be equal to the LCM of 20 and 24. Let us f ind the LCM of 20 and 24. All prime f actors of 20 = 2 2 5, and, 24 = Thus, the LCM of 20 and 24 = = 120. Theref ore, they will meet again af ter 120 minutes.
5 (5) ID : in10realnumbers [5] Let's assume that is a rational number. Theref ore, we can f ind two integers and b, such that, = a b [where, b is not equal to zero.] Now, 11 = Now, 11 = 6b + a 10b a 10b Since a and b are integers, (6b + a) and (10b) are also integers Step 4 Theref ore 6b + a is rational and hence, 11 should also be rational. T his contradicts the 10b f act that 11 is irrational.theref ore, our assumption is f alse and hence, is irrational. Step 5 Theref ore, is an irrational number.
6 (6) ID : in10realnumbers [6] Let us f ind the LCM of 986 and 374. All prime f actors of 986: is a factor of is a factor of is a factor of 29 1 Thus, 986 = All prime f actors of 374: is a factor of is a factor of is a factor of 17 1 Thus, 374 = Thus, the LCM of 986 and 374 = = (7) c. 18 It is given that, the two tankers contain 522 litres and 252 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 522 and 252. Let's f ind the HCF of 522 and 252, using Euclid's Division Algorithm, 522 = , 252 = Thus, the HCF of 522 and 252 is 18. T heref ore, the maximum capacity of the container which can measure the petrol of either tanker in exact number of litres is 18 litres.
7 (8) a. 6 ID : in10realnumbers [7] We have been asked to f ind the all prime f actors of All prime f actors of 18150: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 11 1 Thus, = Theref ore, the number of prime f actor of is 6. (9) d We know that an irrational number can not be written in the f orm p q. If we look at all the options, we notice that can be written as, = = 25 = 25 1 Hence, is not an irrational number.
8 (10) b. 16 ID : in10realnumbers [8] If we look at the question, we notice that 7792 > 288. Theref ore, a = 7792 and b = 288, By applying the Euclid's Division Algorithm(a = bq + r) we get, 7792 = Since, r 0, by applying the Euclid's Division Algorithm by taking a = 288 and b = 16, we get, 288 = Since, r = 0, the process stops. The divisor at this stage is 16. Thus, the HCF of the pair 7792 and 288 is 16. (11) a. a = bc + d, 0 d < b Euclid s division lemma: Euclid s division lemma, states that f or any two positive integers a and b we can f ind two whole numbers c and d such that a = b c + d where 0 d < b. Let us take an example to understand it. Suppose we have to divide 77 by = where, a = 77, b = 14, c = 5 and d = 7 This type of division is known as Euclid's division lemma. One important point to note here is that d is the remainder when a is divided by b. Remainder can be greater than or equal to 0, but should be less than divisor (i.e. 0 d < b). Theref ore, the correct answer is a = bc + d where 0 d < b.
9 (12) c. 4 ID : in10realnumbers [9] 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 232 students can march behind the band of 92 members are equal to the HCF of 232 and 92. Let's f ind the HCF of 232 and 92 using Euclid's Division Algorithm, 232 = , 92 = , 48 = , 44 = Thus, the HCF of 232 and 92 is 4. Step 4 Theref ore, the maximum number of columns in which they can march are 4.
10 (13) a. an irrational ID : in10realnumbers [10] Let's assume 1 3 is a rational number. Theref ore, we can f ind two integers, a and b, such that, 1 3 = a b [where, b is not equal to zero.] Now, 3 = b a b is a rational number as a and b are integers. a Theref ore, 3 is a rational which contradicts to the f act that 3 is Irrational. Hence, our assumption is f alse and 1 is irrational. 3 Theref ore, 1 3 is an irrational number.
11 (14) b. an irrational ID : in10realnumbers [11] Let's assume that 7 is a rational number. i.e. 7 = a/b (a and b are the two integers, such that, b is not equal to zero(0), and a and b do not have common f actor other than 1) a = 7b Squaring both side, a 2 = 7b 2 Theref ore, a 2 is divisible by 7 and it can be said that a is divisible by 7. Let a = 7m, where m is an integer. Squaring both side, a 2 = (7m) 2 Now, (7m) 2 = 7b 2 b 2 = 7m 2 This means that b 2 is divisible by 7 and hence, b is divisible by 7. This implies that a and b have a common f actor and this is contradiction to the f act that a and b are a coprime. Theref ore, 7 is an irrational number.
12 (15) d. an irrational ID : in10realnumbers [12] Let's assume that this sum is rational, that is, = a b [where, and b are integers with b is not equal to zero.] On squaring both sides of above equation we get, 3 = a b = a2 b a b + 11 Now solve this equation f or 11, 2 11 a b = a2 b 2 11 = 8 2 a b + a 2 b 2 2 a b 11 = 8b 2a + a 2b This implies that 11 is a rational number which is a contradiction. Step 4 Theref ore, is an irrational number.
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