(A) an integer.
\n(B) a natural number.
\n(C) an odd integer.
\n(D) an even integer.
\nAnswer:
\n(C) an odd integer.<\/p>\n
Explanation: Question 2.<\/p>\n (A) 13. Explanation: Question 3.<\/p>\n (A) xy. Explanation: Question 4.<\/p>\n (A) 4. Explanation: Question 5.<\/p>\n (A) ab Explanation: Question 6.<\/p>\n (A) 10. Explanation: Question 7.<\/p>\n (A) always irrational. Explanation: Question 8.<\/p>\n (A) one decimal place. Explanation: Question 9.<\/p>\n (A) fraction Explanation: Question 10.<\/p>\n (A) an integer Explanation: Question 11.<\/p>\n (A) an irrational number Explanation: Question 12.<\/p>\n (A) an integer Explanation: Question 13.<\/p>\n \\(\\text { (A) } \\frac{2027}{625}\\) Explanation: Question 14.<\/p>\n (A) one decimal place Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: \u21d2 LCM = \\(\\frac{336 \\times 54}{6}\\)<\/p>\n = 3024 Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: So, assertion is incorrect: Now, in case of Reason Question 6.<\/p>\n Answer: Explanation: Case -Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. (A) 144 Explanation: Question 2.<\/p>\n (A) 2 Question 3. (A) 22<\/sup> \u00d7 32<\/sup> Question 4.<\/p>\n (A) Prime number Question 5.<\/p>\n (A) ab Explanation: Given,p = ab2<\/sup> = a \u00d7 b\u00d7b II. Read the following text and answer the following questions on the basis of the same: A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60,84 and 108 respectively.<\/p>\n Question 1. (A) 14 Explanation: Question 2.<\/p>\n (A) 11 Explanation: Question 3.<\/p>\n (A) 3780 Question 4.<\/p>\n (A) 55360 Question 5.<\/p>\n (A) 23<\/sup> \u00d7 32<\/sup> III. Read the following text and answer the following questions on the basis of the same: A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns. Question 1.<\/p>\n (A) 45 Explanationi: Question 2.<\/p>\n (A) 135 Explanation: Question 3.<\/p>\n (A) 5 Explanation: Question 4.<\/p>\n (A) 3 Explanation: Question 5.<\/p>\n (A) 2 Explanation: IV Read the following text and answer the following questions on the basis of the same: A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience. (A) 15005 Explanation: Question 2.<\/p>\n (A) 23 Explanation: Question 3.<\/p>\n (A) 22 Explanation: 253 = 11 x z Sum of exponents = 3 + 2 + 1 = 6.<\/p>\n IV Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) 15005 Explanation: Question 2.<\/p>\n (A) 23 Explanation: Question 3.<\/p>\n (A) 22 Explanation: Question 4.<\/p>\n (A) Composite number Question 5.<\/p>\n (A) 5 \u00d7113<\/sup> \u00d7 13,2<\/sup> Real Numbers Class 10 MCQ Questions with Answers Question 1. n2– 1 is divisible by 8, if n is: (A) an integer. (B) a natural number. (C) an odd integer. (D) an even integer. Answer: (C) an odd integer. Explanation: Any odd integer can be written as 2m + 1. put n = 2m + …<\/p>\n
\nAny odd integer can be written as 2m + 1.
\nput n = 2m + 1 in n2<\/sup>– 1
\nn2<\/sup>– 1 =(n + 1)(n-1) (2m + 2)(2m) = 4m(m + 1 ) The product of two consecutive numbers is divisible by 2. Thus, m(m + 1) is divisible bv 2. Let m(m + 1) = 2k
\nn2<\/sup>– 1 =(n + 1)(n-1)= (2m + 2)(2m)
\n4m(m + 1) = 4 \u00d7 2k =8k
\n1Thus, if n is an odd integer then n2<\/sup>-1 is divisible by 8.<\/p>\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nThe largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is:<\/h2>\n
\n(B) 65.
\n(C) 875.
\n(D) 1,750.
\nAnswer:
\n(A) 13.<\/p>\n
\nRequired largest number = HCF 1 of (70 – 5) and (125 – 8) = HCF of 65 and I 117 = 13<\/p>\nIf two positive integers a and b are written as a = x3<\/sup>y2<\/sup> and b = xy2<\/sup>; x, y are prime numbers, then HCF (a, b) is:<\/h2>\n
\n(B) xy2<\/sup>
\n(C) x3<\/sup>y3<\/sup>
\n(D) x2<\/sup>y2<\/sup>
\nAnswer:
\n(B) xy2<\/sup><\/p>\n
\nSince n-x3<\/sup>y3<\/sup>x \u00d7 x \u00d7 x \u00d7 y \u00d7 y and b – xy3<\/sup>=x \u00d7 y \u00d7 y \u00d7 y, Thus, HCF: of a and b = x \u00d7 y \u00d7 y =x y2<\/sup><\/p>\nIf the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is<\/h2>\n
\n(B) 2.
\n(C) 1.
\n(D) 3.
\nAnswer:
\n(B) 2.<\/p>\n
\nBy the Ludid’s division algorithm,
\nHCF of (65, 117) = 13
\nSince 65m – 117 = 13 \u203a m = 2<\/p>\nIf two positive integers p and q can be expressed as p = ab2<\/sup> and q = a3<\/sup>b; a, b being prime numbers, then LCM (p, q) is<\/h2>\n
\n(B) a2<\/sup>b2<\/sup>
\n(C) a3<\/sup>b3<\/sup>
\n(D) a2<\/sup>b2<\/sup>
\nAnswer:
\n(C) a3<\/sup>b3<\/sup><\/p>\n
\nSince p = ab2<\/sup>=a x b x b
\nq=a3<\/sup>b = a x a x a x b x b =
\nThus, LCM of p and q = a x a x a x b x b = a3<\/sup>b2<\/sup><\/p>\n<\/p>\nThe least number that is divisible by all the numbers from 1 to 10 (both inclusive) is<\/h2>\n
\n(B) 100.
\n(C) 504.
\n(D) 2,520.
\nAnswer:
\n(D) 2,520.<\/p>\n
\nRequired number = LCM( 1,2, 3,4, 5, 6,8, 9,10)
\n=1 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 7 = 2,520<\/p>\nThe product of a non-zero rational and an irrational number is :<\/h2>\n
\n(B) always rational.
\n(C) rational or irrational.
\n(D) one.
\nAnswer:
\n(A) always irrational.<\/p>\n
\nThe product of a non-zo rational with and an irrational number alwavs irrational.<\/p>\nThe decimal expansion of the rational number 14587 wyj terminal after : 1250<\/h2>\n
\n(B) two decimal places.
\n(C) three decimal places.
\n(D) four decimal places.
\nAnswer:
\n(D) four decimal places.<\/p>\n
\n\\(\\frac{14587}{1250}=\\frac{14587}{2 \\times 5^{4}}=\\frac{14587}{2 \\times 5^{4}} \\times \\frac{2^{4}}{2^{4}}\\)
\n=\\(\\frac{14587 \\times(2)^{3}}{10^{4}}=\\frac{116,696}{10000}=11.6696\\)<\/p>\n\\(\\frac{1}{\\sqrt{2}}\\)= is a\/an<\/h2>\n
\n(B) rational number
\n(C) irrational number
\n(D) none of these
\nAnswer:
\n(C) irrational number<\/p>\n
\n\\(\\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{2}}{\\sqrt{2} \\cdot \\sqrt{2}}=\\frac{1}{2} \\cdot \\sqrt{2} \\)
\nWe have \\(\\frac{1}{2}\\) is a rational but ?2 is irrational. So, the product of a rational and an irrational is irrational.
\nHence, \\(\\frac{1}{\\sqrt{2}}\\) = = \\(\\frac{1}{2} \\cdot \\sqrt{2}\\) is irrational.<\/p>\n<\/p>\n(3 + \u221a5) is:<\/h2>\n
\n(B) a rational number
\n(C) an irrational number
\n(D) none of these
\nAnswer:
\n(C) an irrational number<\/p>\n
\nHere 3 is a rational and \u221a5 is an irrational.
\n?The sum (3 + \u221a5) i.e., rational and irrational is irrational.Hence (3 + \u221a5) is an irrational number.<\/p>\nThe number 1.732 is:<\/h2>\n
\n(B) a rational number
\n(C) an integer
\n(D) a whole number
\nAnswer:
\n(B) a rational number<\/p>\n
\nWe have \\(1.732=\\frac{1732}{1000}=\\frac{433}{250}\\) Which is a rational number.<\/p>\n2.1311 3111 311113…………is<\/h2>\n
\n(B) a rational number
\n(C) an irrational number
\n(D) none of these.
\nAnswer:
\n(C) an irrational number<\/p>\n
\nGiven number is non-terminating, non-repeating decimal, So, it is an irrational number.<\/p>\nWhich of the following rational numbers is expressible as a terminating decimal?<\/h2>\n
\n\\(\\text { (B) } \\frac{1625}{462}\\)
\n\\(\\text { (C) } \\frac{131}{35}\\)
\n\\(\\text { (D) } \\frac{124}{165}\\)
\nAnswer:
\n\\(\\text { (A) } \\frac{2027}{625}\\)<\/p>\n
\nGiven,\\(\\frac{2027}{625}=\\frac{2027}{5 \\times 5 \\times 5 \\times 5} \\times \\frac{2^{4}}{2^{4}}\\)
\n\\(=\\frac{2027 \\times 2^{4}}{5^{4} \\times 2^{4}}=\\frac{32,432}{10^{4}}\\)
\n=3.2432
\nThus, it is expressible as a terminating decimal.<\/p>\nThe decimal expansion of the sum of rational numbers\\(\\frac{15}{4}\\) and\\(\\frac{5}{40}\\) will terminate after.<\/h2>\n
\n(B) two decimal places
\n(C) three decimal places
\n(D) four decimal places
\nAnswer:
\n(C) three decimal places<\/p>\n
\nThe sum of rational numbers
\n\\(\\begin{aligned} &=\\frac{15}{4}+\\frac{5}{40} \\\\
\n&=\\frac{15 \\times 25}{4 \\times 25}+\\frac{5 \\times 25}{40 \\times 25} \\\\
\n&=\\frac{375}{100}+\\frac{125}{1000}
\n\\end{aligned}\\)
\n= 3.75 + 0.125 = 3.875.
\nSo, it will terminate after 3 decimal places.<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): (7 \u00d7 13 \u00d7 11) + 11 and (7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1) + 3 have exactly composite numbers.
\nReason (R): (3 \u00d7 12 \u00d7 101) + 4 is not a composite number.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nfirstly consider the assertion, Since (7 \u00d7 13 \u00d7 11) + 11 = 11 \u00d7 (7 \u00d7 13 + 1)
\n= 11\u00d7 (91 + 1)
\n=11\u00d792 \u21d211 \u00d7 2 \u00d7 2 \u00d7 23 and (7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 l) + 3
\n= 3(7 x 6 x 5 x 4 x 2 x l + l)
\n= 3 x (1681) \u21d2 3 \u00d7 41 \u00d7 41 Given numbers have more than two prime factors. So, both the numbers are composite. Hence, assertion is correct.
\nNow let us consider the reason:
\n3\u00d7 12\u00d7 101 + 4 = 4(3 \u00d7 3 \u00d7 101 + 1)
\n= 4(909 + 1)
\n= 4(910)
\n= 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 7 \u00d7 13 = a composite number
\n[\u2234 Product of more than two prime factors] Thus, reason is not correct. Thus, assertion is correct but reason is incorrect.<\/p>\n<\/p>\nAssertion (A): If HCF (336,54) = 6, then LCM (336,54) = 3000.
\nReason (R): The sum of exponents of prime factors in the prime factorisation of 196 is 4.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nLet us consider the assertion,
\n\u2234 HCF \u00d7 LCM = Product of numbers
\n\u2234 6 x LCM = 336 x 54<\/p>\n
\nThus, the assertion is incorrect:
\nNow, let us consider the reason:
\nPrime factors of 196 = 22<\/sup> x 72<\/sup>
\n\u2234 The sum of exponents of prime factors = 2 + 2 = 4.
\nSo, the reason is correct: Thus, assertion is incorrect but reason is correct.<\/p>\nAssertion (A): HCF of two or more numbers = Product of the smallest power of each common prime factor, involved in the numbers.
\nReason (R): The HCF of 12,21 and 15 is 3.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A<\/p>\n
\nLet a, a2<\/sup> and a2 be three numbers, then we have the smallest power of a1<\/sup>, a2<\/sup> and a3<\/sup> is 1. So, HCF is a. Now, let us consider the reason:
\nPrime factors of 12 = 22 \u00d7 3
\nPrime factors of 21 =3\u00d77
\nPrime factors of 15 =3\u00d75
\n?HCF of 12,21 and 15 = 3, which is a common prime factor. Thus both assertion and reason are correct and reason is the correct explanation for assertion.<\/p>\nAssertion (A): The product of two consecutive positive integers is divisible by 2.
\nReason (R): 13233343563715 is a composite number.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion. Since, in the product of two consecutive positive integers, p = n(n +1), one of n or (n + 1) is an even number. Hence, the product of two consecutive positive integers is divisible by 2. So, it is correct. Now, let us consider the reason: Since, the given number ends in 5. It is a multiple of 5. Therefore, it is a composite number. Thus, both assertion and reason are correct and reason is not the correct explanation for assertion.<\/p>\nAssertion (A): The decimal expansion of assertion: \\(\\frac{15}{16000}\\) is 0.09375
\nReason (R): The decimal expansion of \\(\\frac{23}{2^{3} 5^{2}}\\)<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\n\\(\\frac{15}{1600}\\) = \\(\\begin{aligned}
\n&\\frac{15}{2^{4} \\times 100} \\\\ &\\frac{15 \\times 5^{4}}{2^{4} \\times 5^{4} \\times 100} \\\\ &\\frac{9375}{(2 \\times 5)^{4} \\times 100} \\\\ &\\frac{9375}{1000000}=0.009375
\n\\end{aligned}\\)<\/p>\n
\n\\(\\frac{23}{2^{3} 5^{2}}\\) = \\(\\begin{aligned} &\\frac{23 \\times 5}{2^{3} \\times 5^{2} \\times 5} \\\\ &\\frac{115}{2^{3} \\times5^{3}}=\\frac{115}{(2 \\times 5)^{3}} \\\\
\n&\\frac{115}{1000}=0.115
\n\\end{aligned}\\)
\nSo, reason is correct. Thus, asset tion incorrect but reason is correct.<\/p>\nAssertion (A): The decimal expansion of the rational number \\(\\frac{57}{2^{2} \\times 5}\\) will terminate after 2 decimal places. 2 x 5
\nReason (R): The decimal expansion of the rational number \\(\\frac{13}{3125}\\)will terminate after 3 decimal places.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\n\\(\\frac{57}{2^{2} \\times 5}\\)
\n\\(=\\frac{57 \\times 5}{2^{2} \\times 5^{2}}\\)
\n\\(=\\frac{285}{(2 \\times 5)^{2}}\\)
\n\\(=\\frac{285}{100}\\)
\nSo, U will terminate after 2 decimal places.
\n\u2234 Assertion is correct.
\nIn case of Reason:
\n\\(\\frac{13}{3125}=\\frac{13}{5^{5}}\\)
\n[\u2234 prim factor of 3125 = 5 \u00d7 5 \u00d7 5 \u00d7 5]
\n\\(=\\frac{13 \\times 2^{5}}{5^{5} \\times 2^{5}}\\)
\n\\(=\\frac{416}{(5 \\times 2)^{5}}=\\frac{416}{100000}\\)
\n= 0.00416.
\nSo. it will terminate after 5 decimal places.
\n\u2234Reason is incorrect.
\nThus, assertion is correct but reason is incorrect.<\/p>\n
\nI. Read the following text and answer the following questions on the basis of the same:
\nTo enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 32 students in section A and 36 students in section B.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Computer-Applications-Chapter-1-Real-Numbers.png?resize=215%2C147&ssl=1\")
\nQuestion 1.<\/p>\nWhat is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?<\/h2>\n
\n(B) 128
\n(C) 288
\n(D) 272
\nAnswer:
\n(C) 288<\/p>\n
\nWe have to find the LCM of 32 and 36.
\nLCM(32, 36) = 25<\/sup> \u00d7 9 = 288
\nHence, the minimum number of books required to distribute equally among students of section A and section B are 288.<\/p>\n<\/p>\nIf the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32,36) is<\/h2>\n
\n(B) 4
\n(C) 6
\n(D) 8
\nAnswer:
\n(B) 4<\/p>\n
\n36 can be expressed as a product of its primes as<\/span><\/p>\n
\n(B) 21<\/sup> \u00d7 35<\/sup>
\n(C) 23<\/sup> \u00d731<\/sup>
\n(D) 2\u00b0 x 3\u00b0
\nAnswer:
\n(A) 22 x 32<\/p>\n7 \u00d711\u00d7 13\u00d7 15 + 15 is a<\/h2>\n
\n(B) Composite number
\n(C) Neither prime nor composite
\n(D) None of the above
\nAnswer:
\n(B) Composite number<\/p>\nIf p and q are positive integers such that p = ab2<\/sup> and q = a2<\/sup>b, where a, b are prime numbers, then the LCM (p, q) is<\/h2>\n
\n(B) a2<\/sup>b2<\/sup>
\n(C) a3<\/sup>b2<\/sup>
\n(D)a3<\/sup>b3<\/sup>
\nAnswer:
\n(D)a3<\/sup>b3<\/sup><\/p>\n
\nq = a2<\/sup>b= a\u00d7a\u00d7b
\nLCM of (p, q) =a2<\/sup>b2<\/sup><\/p>\n
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Computer-Applications-Chapter-1-Real-Numbers-2-1.png?resize=295%2C159&ssl=1\")
<\/p>\nIn each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are<\/h2>\n
\n(B) 12
\n(C) 16
\n(D) 18
\nAnswer:
\n(B) 12<\/p>\n
\nNo. of participants seated in each room would be HCF of all the three values above.
\n60 = 2 \u00d7 2 \u00d7 3 \u00d7 5
\n84 = 2 \u00d7 2 \u00d7 3 \u00d7 7
\n108 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3
\nHence, HCF = 12.<\/p>\nWhat is the minimum number of rooms required during the event?<\/h2>\n
\n(B) 31
\n(C) 41
\n(D) 21
\nAnswer:
\n(D) 21<\/p>\n
\nMinimum no. of rooms required are total number of students divided by number of students in each robin.
\nNo. of rooms \\(\\begin{aligned} &=\\frac{60+84+108}{12} \\\\ &=21 \\end{aligned}\\)<\/p>\n<\/p>\nThe LCM of 60, 84 and 108 is<\/h2>\n
\n(B) 3680
\n(C) 4780
\n(D) 4680
\nAnswer:
\n(A) 3780<\/p>\nThe product of HCF and LCM of 60,84 and 108 is<\/h2>\n
\n(B) 35360
\n(C) 45500
\n(D) 45360
\nAnswer:
\n(D) 45360<\/p>\n108 can be expressed as a product of its primes as<\/h2>\n
\n(B) 23<\/sup> \u00d7 33<\/sup>
\n(C) 22<\/sup> \u00d7 32<\/sup>
\n(D) 22<\/sup> \u00d7 32<\/sup>
\nAnswer:
\n(D) 22<\/sup> \u00d7 32<\/sup><\/p>\n
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Computer-Applications-Chapter-1-Real-Numbers-3.png?resize=255%2C194&ssl=1\")
\nRead carefully the above paragraph and answer the following questions:<\/p>\nWhat is the maximum number of columns in which they can be planted?<\/h2>\n
\n(B) 40
\n(C) 15
\n(D) 35
\nAnswer:
\n(A) 45<\/p>\n
\nNo. of rose plants = 135
\nNo. of marigold plants =225
\nThe maximum number of columns in which they can be planted = HCF of 135 and 225
\n\u2234 Prime factors of 135 = 3 x 3 x 3 x 5 and 225 = 3 \u00d7 3 \u00d7 5 \u00d7 5
\n\u2234 Prime factors of 135 = 3 x 3 x 5 = 45.<\/p>\nFind the total number of plants<\/h2>\n
\n(B) 225
\n(C) 360
\n(D) 45
\nAnswer:
\n(C) 360<\/p>\n
\nTotal number of plants 135 + 225 = 360 plants<\/p>\nFind the sum of exponents of the prime factors of the maximum number of columns in which they can be planted.<\/h2>\n
\n(B) 3
\n(C) 4
\n(D) 6
\nAnswer:
\n(B) 3<\/p>\n
\nWe have proved that the maximum number of columns = 45
\nSo, prime factors of 45 = 3 \u00d7 3 \u00d7 5
\n= 32<\/sup> \u00d7 51<\/sup>
\n?Sum of exponents =2 + 1 = 3<\/p>\n<\/p>\nWhat is total numbers of row in which they can be planted<\/h2>\n
\n(B) 5
\n(C) 8
\n(D) 15
\nAnswer:
\n(C) 8<\/p>\n
\nNumber of rows of Rose plants \\(=\\frac{135}{45}=3\\)
\nNumber of rows of marigold plants = \\(=\\frac{225}{45}=5\\)
\nTotal numbers of rows = 3 + 5 = 8<\/p>\nFind the sum of exponents of the prime factors of total number of plants<\/h2>\n
\n(B) 3
\n(C) 5
\n(D) 6
\nAnswer:
\n(D) 6<\/p>\n
\nTotal number of plants = 135 + 225 =360
\nThe prime factors of 360 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5
\n= 23 <\/sup>\u00d7 32 <\/sup>\u00d7 51<\/sup>
\nSum of exponents = 3 + 2 + 1 = 6.<\/p>\n
\nig4
\nQuestion 1.<\/p>\nWhat will be the value of x?<\/h2>\n
\n(B) 13915
\n(C) 56920
\n(D) 17429
\nAnswer:
\n(B) 13915<\/p>\n
\nx = 2783 \u00d7 5
\nx = 13915<\/p>\nWhat will be the value of y?<\/h2>\n
\n(B) 22
\n(C) 11
\n(D) 19
\nAnswer:
\n(C) 11<\/p>\n
\n2783 = y \u00d7253
\n\\(\\begin{aligned} &y=\\frac{2783}{253} \\\\ &y=11 \\end{aligned}\\)<\/p>\nWhat will be the value of z?<\/h2>\n
\n(B) 23
\n(C) 17
\n(D) 19
\nAnswer:
\n(B) 23<\/p>\n
\n\\(\\begin{aligned} &z=\\frac{253}{11} \\\\ &z=23 \\end{aligned}\\)<\/p>\n
\nA Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.<\/p>\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Computer-Applications-Chapter-1-Real-Numbers-4.png?resize=252%2C197&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Computer-Applications-Chapter-1-Real-Numbers-5.png?resize=243%2C310&ssl=1\")
<\/p>\nWhat will be the value of x?<\/h2>\n
\n(B) 13915
\n(C) 56920
\n(D) 17429
\nAnswer:
\n(B) 13915<\/p>\n
\nx = 2783 \u00d7 5
\nx = 13915<\/p>\n<\/p>\nWhat will be the value of y?<\/h2>\n
\n(B) 22
\n(C) 11
\n(D) 19
\nAnswer:
\n(C) 11<\/p>\n
\n2783 = y \u00d7 253
\n\\(\\begin{aligned} &y=\\frac{2783}{253} \\\\ &y=11 \\end{aligned}\\)<\/p>\nWhat will be the value of z?<\/h2>\n
\n(B) 23
\n(C) 17
\n(D) 19
\nAnswer:
\n(B) 23<\/p>\n
\n253 = 11 \u00d7 z
\n\\(\\begin{aligned} &z=\\frac{253}{11} \\\\ &z=23 \\end{aligned}\\)<\/p>\nAccording to Fundamental Theorem of Arithmetic 13915 is a<\/h2>\n
\n(B) Prime number
\n(C) Neither prime nor composite
\n(D) Even number
\nAnswer:
\n(A) Composite number<\/p>\n<\/p>\nThe prime factorisation of 13915 is<\/h2>\n
\n(B) 5 \u00d7 113<\/sup> \u00d7 232<\/sup>
\n(C) 5 \u00d7 112<\/sup> \u00d7 23
\n(D) 5 \u00d7 112<\/sup> \u00d7 232<\/sup>
\nAnswer:
\n(C) 5 \u00d7 112<\/sup> \u00d7 23<\/p>\nMCQ Questions for Class 10 Maths with Answers <\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"