{"id":10668,"date":"2024-01-18T15:04:15","date_gmt":"2024-01-18T09:34:15","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=10668"},"modified":"2024-01-19T17:49:29","modified_gmt":"2024-01-19T12:19:29","slug":"ap-board-6th-class-maths-solutions-chapter-2-unit-exercise","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-6th-class-maths-solutions-chapter-2-unit-exercise\/","title":{"rendered":"AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Unit Exercise"},"content":{"rendered":"

AP State Syllabus AP Board 6th Class Maths Solutions<\/a> Chapter 2 Whole Numbers Unit Exercise Textbook Questions and Answers.<\/p>\n

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Unit Exercise<\/h2>\n

Question 1.
\nChoose the appropriate symbol from < or > and place it in the blanks.
\ni) 8 ……. 7
\nii) 5 ……. 2
\niii) 0 ……. 1
\niv) 10 ……. 5
\nAnswer:
\ni) 8 …. > …. 7
\nii) 5 …. > …. 2
\niii) 0 …. < …. 1
\niv) 10 …. > …. 5<\/p>\n

\"AP<\/p>\n

Question 2.
\nPresent the successor of 11 and predecessor of 5 on the number line.
\nAnswer:
\ni)
\n\"AP
\nSuccessor of 11 is 12.<\/p>\n

ii)
\n\"AP
\nPredecessor of 5 is 4.<\/p>\n

Question 3.
\nWhich of the statements are true ( T ) and which are false ( F ). Correct the false statements.
\ni) There is a natural number that has no predecessor. ( T )
\nii) Zero is the smallest whole number. ( T )
\niii) A whole number on the left of another number on the number line, is greater than that number. (F)
\nAnswer:
\ni) There is a natural number that has no predecessor. ( T )
\nii) Zero is the smallest whole number. ( T )
\niii) A whole number on the left of another number on the number line, is greater than that number. ( F )<\/p>\n

\"AP<\/p>\n

Question 4.
\nGive the results without actually performing the operations, using the given information.
\ni) 28 \u00d7 19 = 532 , then 19 \u00d7 28 =
\nii) a \u00d7 b = c , then b \u00d7 a =
\niii) 85 + 0 = 85 , then 0 + 85 =
\nAnswer:
\ni) 28 \u00d7 19 = 532 , then 19 \u00d7 28 = 532
\nii) a \u00d7 b = c , then b \u00d7 a = c
\niii) 85 + 0 = 85 , then 0 + 85 = 85<\/p>\n

Question 5.
\nFind the value of the following:
\ni) 368 \u00d7 12 + 18 \u00d7 368
\nii) 79 \u00d7 4319 + 4319 \u00d7 11
\nAnswer:
\ni) 368 \u00d7 12 + 18 \u00d7 368 – 368 \u00d7 12 + 368 \u00d7 18
\nDistributive property of multiplication over addition.
\n= 368 \u00d7 (12 + 18)
\n= 368 \u00d7 30 = 11040<\/p>\n

ii) 79 \u00d7 4319 + 4319 \u00d7 11
\n= 79 \u00d7 4319 + 11 \u00d7 4319
\nDistributive property of multiplication over addition.
\n= (79 + 11) \u00d7 4319
\n= 90 \u00d7 4319;
\n= 388710<\/p>\n

Question 6.
\nChandana and Venu purchased 12 note books and 10 note books respectively. The cost of each note book is Rs. 15. Then how much amount should they pay to the shopkeeper?
\nAnswer:
\nNumber of note books purchased by Chandana = 12
\nNumber of note books purchased by Venu = 10
\nTotal number of note books purchased together = 12 + 10
\nCost of each note book = Rs. 15
\nCost of (12 + 10) note books = (12 + 10) \u00d7 15
\n= 22 \u00d7 15
\nThe amount paid to the shopkeeper = Rs. 330<\/p>\n

\"AP<\/p>\n

Question 7.
\nMatch the following.
\ni) 3 + 1991 + 7 = 3 + 7 + 1991\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [ ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 A) Additive identity
\nii) 2 \u00d7 68 \u00d7 50 = 2 \u00d7 50 \u00d7 68\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[ ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B) Multiplicative identity
\niii) 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [ ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C) Commutative under addition
\niv) 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [ ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D) Distributive property of multiplication over addition
\nv) 879 \u00d7 (100 + 30) = 879 \u00d7 100 + 879 \u00d7 30\u00a0 [ ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 E) Commutative under multiplication
\nAnswer:
\ni) C
\nii) E
\niii) B
\niv) A
\nv) D<\/p>\n

Question 8.
\nStudy the pattern:
\n91 \u00d7 11 \u00d7 1 = 1001
\n91 \u00d7 11 \u00d7 2 = 2002
\n91 \u00d7 11 \u00d7 3 = 3003
\nWrite next seven steps. Check, whether the result is correct.
\nAnswer:
\n91 \u00d7 11 \u00d7 1 = 1001
\n91 \u00d7 11 \u00d7 7 = 7007
\n91 \u00d7 11 \u00d7 8 = 8008
\n91 \u00d7 11 \u00d7 9 = 9009
\n91 \u00d7 11 \u00d7 10 = 10010<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Unit Exercise Textbook Questions and Answers. AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Unit Exercise Question 1. Choose the appropriate symbol from < or > and place it in the blanks. i) 8 ……. 7 ii) 5 ……. … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":38522,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/04\/Chapter-2-Whole-Numbers-Unit-Exercise.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10668"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=10668"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10668\/revisions"}],"predecessor-version":[{"id":38523,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10668\/revisions\/38523"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/38522"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=10668"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=10668"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=10668"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}