{"id":27178,"date":"2020-06-29T21:14:49","date_gmt":"2020-06-29T15:44:49","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27178"},"modified":"2021-07-02T15:07:10","modified_gmt":"2021-07-02T09:37:10","slug":"kseeb-solutions-for-class-8-maths-chapter-1-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-1-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 1 Playing with Numbers Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus\u00a0Class 8 Maths Chapter 1 Playing with Numbers Additional Questions<\/h2>\n

Question 1.
\nChoose the correct option.
\n(a) The general form of 456 is
\nA. (4 \u00d7 100) + (5 \u00d7 10) + (6 \u00d7 l)
\nB. (4 \u00d7 100) + (6 \u00d7 10) + (5 \u00d7 1)
\nC. (5 \u00d7 100) + (4 \u00d7 10) + (6 \u00d7 1)
\nD. (6 \u00d7 100) + (5 \u00d7 10) + (4 \u00d7 1)
\nAnswer:
\nA. (4 \u00d7 100) + (5 \u00d7 10) + (6 \u00d7 1)<\/p>\n

(b) Computers use
\nA. decimal system
\nB. binary system
\nC. base 5 system
\nD. base 6 system
\nAnswer:
\nB. binary system<\/p>\n

(c) If \\(\\overline{a b c}\\) is a 3 digit number, then the number.
\n\\(n=\\overline{a b c}+\\overline{a c b}+\\overline{b a c}+\\overline{b c a}+\\overline{c a b}+\\overline{c b a}\\) is always divisible by
\nA. 8
\nB. 7
\nC. 6
\nD. 5
\nAnswer:
\nC. 6<\/p>\n

(d) If \\(\\overline{a b c}\\) is a 3-digit number then
\n\\(n=\\overline{a b c}-\\overline{a c b}+\\overline{b a c}-\\overline{b c a}+\\overline{c a b}-\\overline{c b a}\\) is always divisible by
\nA. 12
\nB. 15
\nC. 18
\nD. 21
\nAnswer:
\nC. 18<\/p>\n

(e) If IK \u00d7 K1 = K2K, then the letter K stands for the digit.
\nA. 1
\nB. 2
\nC. 3
\nD. 4
\nAnswer:
\nA. 1<\/p>\n

(f) The number 345111 is divisible by
\nA. 15
\nB. 12
\nC. 9
\nD. 3
\nAnswer:
\nD. 3<\/p>\n

(g) The number of integers of the form 3AB4, where A, B denote some digits, which are divisible by 11 is
\nA. 0
\nB. 4
\nC. 7
\nD. 9
\nAnswer:
\nD. 9<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nWhat is the smallest 5 digit number divisible by 11 and containing each of the digits 2, 3, 4, 5, 6?
\nAnswer:
\nThe number is 24365
\n(\u2235 (2 + 3 + 5) – (4 + 6) = 0)<\/p>\n

Question 3.
\nHow many 5 – digit numbers divisible by 11 are there containing each of the digits 2, 3, 4, 5, 6?
\nAnswer:
\nThe number having the digits 2, 3, 4, 5, 6 is divisible by 11 if 2, 3 and 5 are in the odd places and 4 and 6 are in the even places as (2 + 3 + 5) – (4 + 6) = 0 which is divisible by 11.<\/p>\n

If 2 in the first place the numbers can be 24365, 26345, 24563, 26543 (4 numbers)<\/p>\n

Similarly if 3 is in the first place we get 4 numbers and 5 is in the first place we get 4 numbers.
\nTotally there are 12 numbers.<\/p>\n

Question 4.
\nIf 49A and A49 where A > 0, have a common factor, find all possible values of A.
\nAnswer:
\nIf A = 1 the numbers are 491 and 149 which do not have a common Tactor. If A = 2, the numbers are 492 and 249 which have 3 as a common factor.
\nIf A = 3 the numbers are 493 and 349 which do not have a common factor.
\nIf A = 4, the numbers are 494 and 449 which do not have a common factor.
\nIf A = 5, the numbers are 495 and 549 which have a common factor 3.
\nIf A = 6, the numbers are 496 and 649 8 which do not have a comon factor.
\nIf A = 7 the numbers are 497 and 749 which have a common factor 7.
\nIf A = 8, the numbers are 498 and 849 which have a common factor 3.
\nIf A = 9, the numbers are 499 and 949 which do not have a. common factor,
\n\u2234 the values of K are 2, 5, 7 or 8.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nWrite 1 to 10 using 3 and 5, each atleast once and using addition and subtraction
\n(For example 7 = 5 + 5 – 3)
\nAnswer:
\n1 = 3 + 3 – 5
\n2 = 5 – 3
\n3 = 5 + 3 – 5
\n4 = 3 + 3 + 3 – 5
\n5 = 3 + 5 – 3
\n6 = 5 + 5 + 5 – 3 – 3 – 3
\n7 = 5 + 5 – 3
\n8 = 5 + 3
\n9 = 5 + 5 + 5 – 3 – 3
\n10 = 5 + 5 + 3 – 3<\/p>\n

Question 6.
\nFind all 2-digit numbers each of which is divisible by the sum of its digits.
\nAnswer:
\nThe numbers are 10, 12, 18, 21,
\n24, 27, 36, 42, 45, 54, 63, 72 and 81.<\/p>\n

Question 7.
\nThe page numbers of a book written in a row gives a 216 digit number.
\nHow many pages are there in the book?
\nAnswer:
\nPage numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number.
\nThe remaining number digits i.e., (216 – (9 + 180)) = 27
\n27 is formed by pages 100 to 108 (9 pages)
\n\u2234 The total number of pages = 9 + ’90 + 9 = 108<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nLook at the following pattern.
\n\"KSEEB<\/p>\n

This is called Pascal\u2019s triangle. What is the middle number in the 9th row?
\nAnswer:
\nComplete the pascal\u2019s triangle
\n\"KSEEB
\nThe middle number in the 9th row is 70.<\/p>\n

Question 9.
\nComplete the adjoining magic square (Hint. In a 3 \u00d7 3 magic square the magic sum is 3 . times the central number)
\n\"KSEEB
\nAnswer:
\nCentral number is 7. The magic sum is 3 \u00d7 7 = 21
\n\"KSEEB<\/p>\n

Question 10.
\nFind all 3 digit natural numbers which are 12 times as large as the sum of their digits.
\nAnswer:
\nThe number 108. Because 1 + 0 + 8 = 9 and 12 \u00d7 9 = 108<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind all digits x, y such that \\(\\overline{34 \\times 5 y}\\) is divisible by 36.
\nAnswer:
\nA number is divisible by 36 if it is divisible by both 4 and 9 as 4 \u00d7 9 = 36.
\nIf the last two digits number is divisible by 4 then the number is divisible by 4. So in the given number y can be 2 or 6 as 52 and 56 are divisible by 4.<\/p>\n

If the sum of all the digits of a number is divisible by 9 then the number is di-visible by 9. In the given number 34x5y if y = 2 and x = 4 the sum of the digits . 3 + 4 + 4 + 5 + 2 = 18 which is
\ndivisible by 9<\/p>\n

\u2234 34452 is divisible by 4, 9 and 36
\nSimilarly if y = 6 and x = 0 and y = 6 and x = 9 the number is divisible by 36<\/p>\n

Question 12.
\nCan you divide the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 into two groups
\nsuch that the product of numbers in one group divides the product of numbers in the other group and the quotient is minimum.
\nAnswer:
\nGroup 1 = {1, 2, 3, 4, 5, 6, 7} the product is 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
\nGroup 2 = {8,9,10} the product is 8 \u00d7 9 \u00d7 10 = 720
\n\\(\\frac{\\text { Group } 1}{\\text { Group } 2}=\\frac{5040}{720}=7\\)<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nFind all 8 digit numbers 273A49B5
\nwhich are divisible by 11 as well as 25.
\nAnswer:
\nThe number is divisible by 25 if the number formed by last 2 digits is divisible by 5. In the number 273A49B5, B is either 2 or 7.
\nLet B is 2 then the number is divisible by 11 if (sum of oddly placed digits) (sum of evenly place ! digits) = multiple of 11
\n\u2234 (2 + 3 + 4 + 2) – (7 + A + 9 + 5) => 11 – (21 + A) = multiple of 11. It is possible when A = 1
\n\u2234 The number is 27314925.
\nSimilarly if B = 7 then A must be 6
\n\u2234 The number is 27364975<\/p>\n

Question 14.
\nSuppose a, b are integers such that 2 + a and 35 – b are divisible by 11 prove that a + b is divisible by 11.
\nAnswer:
\n2 + a and 35 – b
\nare divisible by 11
\n\u2234 2 + a – (35 – b)
\nis also divisible by 11
\n2 + a – 35 + b = a + b – 33 is divisible by 11
\n\u2234 – 33 is divisible by 11 Hence a + b is divisible by 11<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nIn the multiplication table A8 x 3B = 2730, A and B represnet distinct digits different from 0. Find A + B.
\nAnswer:
\nA8 \u00d7 3B = 2730 The units place of the product is 0 B must be 5.
\nThe product is A8 \u00d7 35. 35 \u00d7 8 = 280.
\nWe write 0 in the units place and 28 in carried
\n\u2234 35 x A + 28 = 273 \u21d2 35 x A = 273 – 28
\n= 245 => A = \\(\\frac{245}{35}\\) \u21d2 A = 7
\n\u2234 A = 7 and B = 5
\nA + B = 7 + 5 = 12<\/p>\n

Question 16.
\nFind the least natural number which leaves the remainder 6 and 8 when divided by 7 and 9 respectively.
\nAnswer:
\nSince 7 – 6 = 1, 9 – 8 = 1
\nThe remainder in each case in less than the devisor by 1.
\nHence if 1 is added to the required number it becomes exactly divisible by 7 and 9.<\/p>\n

The least number divisible by 7 and 9 is_ in LCM of 7 and 9 which is equal to 63
\nThe required number is 63 – 1 = 62<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nProve that the sum of the cubes of three consecutive natural numbers is always divisible by 3.
\nAnswer:
\nLet the 3 consecutive natural num-bers be x, x + 1 and x + 2.
\nSum of their cubes
\n= x + (x + l)3 + (x + 2) 3<\/sup>= x3<\/sup> + x3<\/sup> + 13<\/sup> + 3x(x +1) + x3<\/sup> + 23<\/sup> + 6x(x + 2)= x3<\/sup> + x3<\/sup> +1 + 3x2<\/sup> + 3x + x3<\/sup> + 8 + 6x2<\/sup> + 12x= 3x3<\/sup> + 9x2<\/sup> +15x + 9 .= 3(x3<\/sup> + 3x2<\/sup> +5x + 3)
\nSum of their cubes is a multiple of 3 which means that it is always divisible by 3.<\/p>\n

Question 18.
\nWhat is the smallest number you have to add to 100000 to get a multiple of 1234?
\nAnswer:
\n\"KSEEB
\n\u2234 1234 – 46 = 1188 should be added to 100000 to get a multiple of 1234.<\/p>\n

Question 19.
\nUsing the digits 4,5,6,7,8, each once, construct a 5-digit number which is divisible by 264.
\nAnswer:
\nUsing divisibility by 4 and 11, we get – 4 numbers.
\n58476, 48576, 57684, 67584.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Playing with Numbers Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus\u00a0Class 8 Maths Chapter 1 Playing with Numbers Additional Questions Question 1. Choose the correct option. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27178"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27178"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27178\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27178"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27178"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27178"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}