{"id":42475,"date":"2023-04-09T01:58:08","date_gmt":"2023-04-08T20:28:08","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=42475"},"modified":"2023-04-10T09:51:00","modified_gmt":"2023-04-10T04:21:00","slug":"ml-aggarwal-class-6-solutions-for-icse-maths-chapter-4-objective-type-questions","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-4-objective-type-questions\/","title":{"rendered":"ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Objective Type Questions"},"content":{"rendered":"
Mental Maths Question 2. Question 3. (ii) The sum of two prime numbers is always a prime number. False (iii) The sum of two prime numbers can never (iv) No odd number can be written as the sum of two prime numbers. False (v) If two numbers are co-prime, then atleast one of them must be prime. False (vi) If a number is divisible by 18, it must be divisible by 3 and 6 both. True<\/p>\n (vii) If a number is divisible by 2 and 4 both, it (viii)If a number is divisible by 3 and 6 both, it must be divisible by 18. False (ix) HCF of an even number and an odd number is always 1. False Multiple Choice Questions<\/strong> Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Question 25. Question 26. Question 27. Question 28. Value Based Questions<\/strong> (ii) Maximum number of cards = 540 (iii) Yes. Higher Order Thinking Skills (Hots)<\/strong> Question 2. Question 3. ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Objective Type Questions Mental Maths Question 1. Fill in the blanks: (i) The only natural number which has exactly one factor is ………. (ii) The only prime number which is even is ………. (iii) The HCF of two co-prime numbers is ………. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42475"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=42475"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42475\/revisions"}],"predecessor-version":[{"id":159017,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42475\/revisions\/159017"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=42475"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=42475"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=42475"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n<\/strong>Question 1.
\nFill in the blanks:
\n(i) The only natural number which has exactly one factor is ……….
\n(ii) The only prime number which is even is ……….
\n(iii) The HCF of two co-prime numbers is ……….
\n(iv) Two perfect numbers are ………. and ……….
\n(v) The only prime-triplet is ……….
\nSolution:
\n(i) The only natural number which has exactly one factor is 1.
\n(ii) The only prime number which is even is 2.
\n(iii) The HCF of two co-prime numbers is 1.
\n(iv) Two perfect numbers are 6 and 28.
\n(v) The only prime-triplet is 3, 5, 7.<\/p>\n
\nState whether the following statements are true (T) or false (F):
\n(i) Every natural number has a finite number of factors.
\n(ii) Every natural number has an infinite number of its multiples.
\n(iii) There are infinitely many prime numbers.
\n(iv) If two numbers are separately divisible by a number, then their difference is also divisible by that number.
\n(v) LCM of two prime numbers equals their product.
\n(vi) LCM of two co-prime numbers equals their product.
\nSolution:
\n(i) Every natural number has a finite number of factors. True<\/strong>
\n(ii) Every natural number has an infinite number of its multiples. True<\/strong>
\n(iii) There are infinitely many prime numbers. True<\/strong>
\n(iv) If two numbers are separately divisible by a number, then their difference is also divisible by that number. True<\/strong>
\n(v) LCM of two prime numbers equals their product. True<\/strong>
\n(vi) LCM of two co-prime numbers equals their product. True<\/strong><\/p>\n
\nState whether the following statements are true or false. If a statement is false, justify your answer.
\n(i) The sum of two prime numbers is always an even number.
\n(ii) The sum of two prime numbers is always a prime number.
\n(iii) The sum of two prime numbers can never be a prime number
\n(iv) No odd number can be written as the sum of two prime numbers.
\n(v) If two numbers are co-prime, then atleast one of them must be prime.
\n(vi) If a number is divisible by 18, it must be divisible by 3 and 6 both.
\n(vii) If a number is divisible by 2 and 4 both, it must be divisible by 8.
\n(viii) If a number is divisible by 3 and 6 both, it must be divisible by 18.
\n(ix) HCF of an even number and an odd number is always 1.
\nSolution:
\n(i) The sum of two prime numbers is always an even number. False
\nCorrect:
\n2 and 7 both are prime numbers but their sum = 2 + 7 = 9, which is an odd number.<\/p>\n
\nCorrect:
\n3 and 5 both are prime numbers but their sum = 3 + 5 = 8, which is a composite number.<\/p>\n
\nbe a prime number. False
\nCorrect:
\n2 and 5 both are prime numbers but their sum = 2 + 5 = 7, which is a prime number.<\/p>\n
\nCorrect:
\n13 is an odd number and 13 = 2+11, which is the sum of two prime numbers.<\/p>\n
\nCorrect :
\n8 and 15 are co-prime numbers but neither 8 is prime nor 15 is prime.<\/p>\n
\nmust be divisible by 8. False
\nCorrect :
\n20 is divisible by 2 and 4 both but 20 is not divisible by 8.<\/p>\n
\nCorrect:
\n12 is divisible by 3 and 6 both bu 12 is not divisible by 18.<\/p>\n
\nCorrect:
\n6 is even and 9 is odd but HCF of 6 and 9 is 3.<\/p>\n
\nChoose the correct answer from the given four options (4 to 28):<\/strong>
\nQuestion 4.
\nAll factors of 6 are
\n(a) 1, 6
\n(b) 2, 3
\n(c) 1, 2, 3
\n(d) 1, 2, 3, 6
\nSolution:
\nThe factors of 6 are 1, 2, 3, 6 (d)<\/p>\n
\nWhich of the following is an odd composite number?
\n(a) 7
\n(b) 9
\n(c) 11
\n(d) 12
\nSolution:
\n9, is an odd composite number. (c)<\/p>\n
\nThe number of even numbers between 68 and 90 is
\n(a) 10
\n(b) 11
\n(c) 12
\n(d) 31
\nSolution:
\nThe even numbers between 68 and 90 is 70, 72, 74, 76, 78, 80, 82, 84, 86, 88 = 10 numbers (a)<\/p>\n
\nWhich of the following is a prime number?
\n(a) 69
\n(b) 87
\n(c) 91
\n(d) 97
\nSolution:
\nSince, the factors of 97 are 1 and 97
\n97 is a prime number. (d)<\/p>\n
\nWhich of the following is a pair of twin- prime number?
\n(a) 19, 21
\n(b) 43, 47
\n(c) 59, 61
\n(d) 73, 79
\nSolution:
\n59, 61
\nPairs of prime numbers whose difference is 2 are called twin-prime numbers. (c)<\/p>\n
\nThe number of distinct prime factors of the largest 4-digit number is
\n(a) 2
\n(b) 3
\n(c) 5
\n(d) none of these
\nSolution:
\nLargest 4 digit number = 9999
\n
\n3 is prime factor. (b)<\/p>\n
\nThe number of distinct prime factors of the smallest 5-digit number is
\n(a) 2
\n(b) 4
\n(c) 6
\n(d) 8
\nSolution:
\nSmallest 5-digit number = 10000
\n
\nNumber of distinct prime factors of smallest 5-digit number = 2 (a)<\/p>\n
\nThe sum of the prime factors of 1729 is
\n(a) 13
\n(b) 19
\n(c) 32
\n(d) 39
\nSolution:
\n
\nPrime factors of 1729 are 7, 13 and 19 Sum of prime factors = 7 + 13 + 19 = 39 (d)<\/p>\n
\nWhich of the following is a pair of co-prime numbers?
\n(a) 8, 45
\n(b) 3, 18
\n(c) 5, 35
\n(d) 6, 39
\nSolution:
\n8, 15
\nThe factors of 8 are 1, 2, 4, 8 The factors of 15 are 1, 3, 5, 15 The common factors of 8 and 15 is 1 They are co-prime. (a)<\/p>\n
\nEvery natural number has an infinite number of
\n(a) prime factors
\n(b) factors
\n(c) multiples
\n(d) none of these
\nSolution:
\nMultiples
\ne.g. Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, …….. (c)<\/p>\n
\nWhich of the following numbers is divisible by 4?
\n(a) 308594
\n(b) 506784
\n(c) 732106
\n(d) 9301538
\nSolution:
\n506784
\nBecause the number formed by tens and ones digits is divisible by 4 i.e. 84 \u00f7 4 = 21 (b)<\/p>\n
\nWhich of the following numbers is divisible by 8?
\n(a) 503786
\n(b) 505268
\n(c) 305678
\n(d) 703568
\nSolution:
\n703568
\nBecause the number formed by hundred, tens
\nand ones digit is divisible by 8
\ni. e. 568 – 8 = 71 (d)<\/p>\n
\nWhich of the following numbers is divisible by 3?
\n(a) 50762
\n(b) 42063
\n(c) 52871
\n(d) 37036
\nSolution:
\n42063
\nBecause sum of its digits is = 4 + 2 + 0 + 6 + 3 = 15 Which is divisible by 3 (b)<\/p>\n
\nWhich of the following numbers is divisible by 9?
\n(a) 972063
\n(b) 730542
\n(c) 785423
\n(d) 5612844
\nSolution:
\n972063
\nBecause sum of digits
\n= 9 + 7 + 2 + 0 + 6 + 3 = 27 Which is divisible by 9 (a)<\/p>\n
\nWhich of the following numbers is divisible by 6?
\n(a) 560324
\n(b) 650374
\n(c) 798653
\n(d) 750972
\nSolution:
\n750972
\nBecause sum of its digit
\n= 7 + 5 + 0 + 9 + 7 + 2 = 30 Which is divisible by 3.
\nHence it is divisible by 6. (d)<\/p>\n
\nThe digit by which ‘*’ should be replaced in 54 * 281 so that the number formed is divisible by 9 is
\n(a) 6
\n(b) 7
\n(c) 8
\n(d) 9
\nSolution:
\nFor a number to be divisible by 9, sum of its digits should be divisible by 9.
\nSum of given digits in 54 * 281
\n= 5 + 4 + 2 + 8 + 1 = 20.
\nIf we add 7, it becomes 27, which is divisible by 9.
\n\u2234 * is to be replaced by 7. (b)<\/p>\n
\nThe digit by which should be replaced in 7254 * 98 so that the number formed is divisible by 22 is
\n(a) 0
\n(b) 1
\n(c) 2
\n(d) 6
\nSolution:
\nFor a number to be divisible by 22, sum of its digits should be divisible by 2 and by 11.
\nSince, the last digit of 7254 * 98 is 8, which is divisible by 2.
\nNow,
\nSum of the digits at odd places = 7 + 5 + 8 = 20
\nSum of the digits at even places = 9 + 4 + 2 = 15
\n\u2234 Their Difference = 20 – 15 = 5
\nSince, 5 is not divisble by 11, so to make a number divisible by 11 we must add 6.
\n\u2234 * is to be replaced by 6 (d)<\/p>\n
\nIf a number is divisible by 5 and 6 both, then it may not be divisible by
\n(a) 10
\n(b) 15
\n(c) 30
\n(d) 60
\nSolution:
\n60 (d)<\/p>\n
\nThe number of common prime factors of 60, 75 and 105 is
\n(a) 2
\n(b) 3
\n(c) 4
\n(d) 5
\nSolution:
\n60, 75 and 105
\n
\n=2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 7 = 2 (a)<\/p>\n
\nThe H.C.F. of 144 and 198 is
\n(a) 6
\n(b) 9
\n(c) 12
\n(d) 18
\nSolution:
\nH.C.F. of 144 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3
\nH.C.F. of 198 = 2 \u00d7 3 \u00d7 3 \u00d7 11
\n
\n\u2234 H. C. F of 144 and 198
\n= 2 \u00d7 3 \u00d7 3 = 18 (d)<\/p>\n
\nThe L.C.M. of 30 and 45 is
\n(a) 15
\n(b) 30
\n(c) 45
\n(d) 90
\nSolution:
\nL.C.M. of 30 and 45 is 90
\n
\n\u2234 L.C.M. = 3 \u00d7 5 \u00d7 2 \u00d7 3 = 90 (d)<\/p>\n
\nThe L.C.M. of 4 and 44 is
\n(a) 4
\n(b) 11
\n(c) 44
\n(d) 176
\nSolution:
\nLCM of 4 and 44 is 44
\n
\n\u2234 L.C.M = 4 \u00d7 11 =44 (c)<\/p>\n
\nThe LCM of 7 and 13 is 1
\n(a) 1
\n(b) 7
\n(c) 13
\n(d) 91
\nSolution:
\nLCM of 7 and 13 is
\n
\nLCM of 7 and 13 is 91
\nL.C.M. = 7 \u00d7 13 = 91 (d)<\/p>\n
\nIf H.C.F. of two numbers is 15 and their product is 1575, then their L.C.M. is
\n(a) 15
\n(b) 105
\n(c) 525
\n(d) 1575
\nSolution:
\nProduct of numbers =1575
\nH.C.F. = 15
\nWe know,
\nL.C. M = \\(\\frac{\\text { Product of numbers }}{\\text { H.C.F. }}\\)
\n\\(=\\frac{1575}{15}=105\\) (b)<\/p>\n
\nIf the LCM of two natural numbers is 180, then which of the following is not the HCF of the numbers?
\n(a) 45
\n(b) 60
\n(c) 75
\n(d) 90
\nSolution:
\nL.C.M. of 2 natural numbers = 180
\nWe know that,
\nL.C.M. of 2 numbers is always exactly divisible by their H.C.F.
\n\u2234 Taking (a) 45 as H.C.F.
\n
\nHere remainder = 0
\n\u2234 45 is H.C.F.
\nNow, taking (b) 60 as H.C.F.
\n
\nHere remainder = 0
\n\u2234 60 is also H.C.F.
\nNow, taking 75 as H.C.F.
\n
\nHere, remainder = 30
\ni. e. remainder \u2260 0
\nHence, 75 is not the H.C.F. of two natural numbers whose L.C.M. is 180
\nHence, answer is (c).<\/p>\n
\nQuestion 1.
\nTo teach the value of gratitude and appreciation to the students, a school organised a ‘Card Making’ activity in which the students were asked to make “THANK YOU CARDS” for the people who helped them in some way. Assorted cards were made with different titles. Their numbers are given below:
\nT cards for teachers = 120
\nF cards for friends = 540
\nS cards for servants = 90
\nP cards for parents = 240 and
\nG cards for grandparents = 150
\n(i) Find the HCF and LCM of all the different number of cards.
\n(ii) Find HCF and LCM of maximum and minimum number of cards.
\n(iii) Is the number of T-cards is a factor of number of P-cards?
\nSolution:
\n(i) HCF of 120, 540, 90, 240, 150
\n
\n
\n\u2234 HCF = 30
\nNow, LCM of 120, 540, 90, 240 and 150 is
\n
\n\u2234 LCM = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5
\n= 10800<\/p>\n
\nMinimum number of cards = 90
\n\u2234 HCF is as follow :
\n
\nHence, HCF of 90 and 540 is 90
\nLCM of 90 and 540 is as follow :
\n
\nLCM = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 = 540<\/p>\n
\nT cards = 120
\nP cards = 240
\n240 = 120 \u00d7 2<\/p>\n
\nQuestion 1.
\nWrite 2-digit odd numbers whose sum of digits is 8.
\nSolution:
\n17, 71, 35, 53<\/p>\n
\nWrite all pairs of 2-digit twin primes such that on changing the places of their digits, they still remain prime numbers.
\nSolution:
\n11, 13, 71, 73<\/p>\n
\nThere are just four natural numbers less than 100, which have exactly three factors. One of them is 25, what are the other three? What can be said about these numbers?
\nSolution:
\nFour natural numbers less than 100 which have three factors :
\nOne of them is 25 = 1, 5, 25
\nSecond is 49 = 1, 7, 49
\nThird is 9 = 1, 3, 9
\nFourth is 4 = 1, 2, 4<\/p>\nML Aggarwal Class 6 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"