{"id":11389,"date":"2021-04-22T16:03:42","date_gmt":"2021-04-22T10:33:42","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=11389"},"modified":"2021-09-15T16:01:13","modified_gmt":"2021-09-15T10:31:13","slug":"ap-board-6th-class-maths-solutions-chapter-7-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-6th-class-maths-solutions-chapter-7-intext-questions\/","title":{"rendered":"AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions<\/a> Chapter 7 Introduction to Algebra InText Questions and Answers.<\/p>\n

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions<\/h2>\n

\"AP<\/p>\n

Let’s Explore<\/span> (Page No. 102)<\/p>\n

Question 1.
\nArrange 2 matchsticks to form the shape \"AP Continue the same shape for 2 times, 3 times and 4 times. Frame the rule for repeating the pattern.
\nSolution:
\nTo make the given shape 2 matchsticks are needed.
\nTo make the given 2 shapes 4 matchsticks are needed.
\nTo make the given 3 shapes 6 matchsticks are needed.
\nTo make the given 4 shapes 8 matchsticks are needed.
\nContinue and arrange the information in the following table.
\n\"AP
\n\"AP<\/p>\n

Number of matchsticks required = 2 \u00d7 Number of shapes to be formed
\n= 2 \u00d7 x = 2x<\/p>\n

Question 2.
\nRita took matchsticks to form the shape
\nShe repeated the pattern and gave a rule.\"AP
\nNumber of matchsticks needed = 6.y, where y is the number of shapes to be formed. Is it correct ? Explain.
\nWhat is the number of sticks needed to form 5 such shapes ?
\nSolution:
\n\"AP
\nTo make the given shape 6 matchsticks are needed.
\nTo make the given 2 shapes 12 matchsticks are needed.
\nTo make the given 3 shapes 18 matchsticks are needed.
\nContinue and arrange the information in the following table.
\n\"AP
\nYes, it is correct.
\nNumber of matchsticks required = 2 \u00d7 Number of shapes to be formed
\n= 2 \u00d7 y = 2y
\nNumber of matchsticks needed to form 5 such shapes = 6 \u00d7 5 = 30<\/p>\n

Let\u2019s Explore<\/span> (Page No. 103)<\/p>\n

Question 1.
\nA line of shapes is constructed using matchsticks.
\n\"AP
\nShape-1 Shape-2 Shape-3 Shape-4
\ni) Find the rule that shows how many sticks are needed to make a line of such shapes ?
\nii) How many matchsticks are needed to form shape -12 ?
\nSolution:
\n\"AP
\nNumber of matchsticks 3 5 7 9
\ni) Let us know the pattern
\nS1<\/sub> = 3 = 2 + 1 = (1 \u00d7 2) + 1
\nS2<\/sub> = 5 = 4 + 1 = (2 \u00d7 2) + 1
\nS3<\/sub> = 7 = 6 + 1 = (3 \u00d7 2) + 1
\nS4<\/sub> – 9 = 8 + 1 = (4 \u00d7 2) + 1
\nNow the rule for this pattern is number of matchsticks.<\/p>\n

ii) Used to make ‘n’ number of shapes is Sn = (n \u00d7 2) + 1 = 2n + 1
\nNumber of matchsticks needed to form shape – 12 is
\nS12<\/sub> = 2(12) + 1 = 24 + 1 = 25 sticks.<\/p>\n

Check Your Progress<\/span> (Page No. 105)<\/p>\n

Question 1.
\nFill the following table as instructed. One is shown for you.<\/p>\n\n\n\n\n\n\n\n
S.No.<\/td>\nExpression<\/td>\nVerbal Form<\/td>\n<\/tr>\n
1.<\/td>\ny + 3<\/td>\nThree more than y<\/td>\n<\/tr>\n
2.<\/td>\n2x – 1<\/td>\n<\/td>\n<\/tr>\n
3.<\/td>\n5z<\/td>\n<\/td>\n<\/tr>\n
4.<\/td>\n\\(<table border=”2″>\\)<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution:<\/p>\n\n\n\n\n\n\n\n
S.No.<\/td>\nExpression<\/td>\nVerbal Form<\/td>\n<\/tr>\n
1.<\/td>\ny + 3<\/td>\nThree more than y<\/td>\n<\/tr>\n
2.<\/td>\n2x – 1<\/td>\nOne less than the double of x<\/td>\n<\/tr>\n
3.<\/td>\n5z<\/td>\n5 times of z<\/td>\n<\/tr>\n
4.<\/td>\n\u00a0\\(<table border=”2″>\\)<\/td>\nHalf of the m<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let’s Explore ?<\/span> (Page No. 106)<\/p>\n

Question 1.
\nFind the general rule for the perimeter of a rectangle. Use variables T and ‘b’ for length and breadth of the rectangle respectively.
\nSolution:
\nGiven length of rectangle = l
\nbreadth of rectangle = b
\nWe know that the perimeter of rectangle is twice the sum of its length and breadth.
\nSum of length and breadth = l + b
\nTwice the sum of length and breadth = 2 \u00d7 (l + b)
\nRule for the perimeter of a rectangle = 2(l + b)<\/p>\n

Question 2.
\nFind the general rule for the area of a square by using the variable ‘s’ for the side of a square.
\nAnswer:
\nGiven side of a square = s
\nWe know that the area of a square is the product of side and side.
\nArea of a square = side \u00d7 side
\nRule for the area of a square = s.s<\/p>\n\n\n\n\n\n\n\n\n\n
Side<\/td>\nArea<\/td>\n<\/tr>\n
1<\/td>\n1 \u00d7 1<\/td>\n<\/tr>\n
2<\/td>\n2 \u00d7 2<\/td>\n<\/tr>\n
3<\/td>\n3 \u00d7 3<\/td>\n<\/tr>\n
4<\/td>\n4 \u00d7 4<\/td>\n<\/tr>\n
………..<\/td>\n…………<\/td>\n<\/tr>\n
s<\/td>\nS \u00d7 s<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(Page No. 107)<\/p>\n

Question 1.
\nFind the nth term in the following sequences.
\n0 3, 6, 9, 12, ii) 2, 5, 8, 11, iii) 1, 4, 9, 16,
\nSolution:
\ni) Given number pattern is 3, 6, 9, 12,……………..
\nTo find the nth term in the given pattern, we put the sequence in a table.
\n\"AP<\/p>\n

First number = 3 \u00d7 1
\nSecond number = 3 \u00d7 2
\nnth number = 3 \u00d7 n = 3n
\nSo, the nth term of the pattern 3, 6, 9, 12, is 3n.<\/p>\n

ii) Given number pattern is 2, 5, 8, 11,
\nTo find the nth term in the given pattern, we put the sequence in a table.
\n\"AP
\nFirst number = 2 = 3 \u00d7 1 – 1
\nSecond number = 5 = 3 \u00d7 2 – 1
\nThird number = 8 = 3 \u00d7 3 – 1
\nnth number = 3 \u00d7 n – 1 = 3n – 1
\nSo, the nth term of the pattern 2, 5, 8, 11 is 3n – 1.<\/p>\n

iii) Given number pattern is 1, 4, 9, 16,
\nTo find the nth term in the given pattern, we put the sequence in a table.
\n\"AP
\nFirst number =1 = 1 \u00d7 1
\nSecond number = 4 = 2 \u00d7 2
\nThird number =9 = 3 \u00d7 3
\nnth number = n \u00d7 n = n2<\/sup>
\nSo, the nth term of the pattern 1, 4, 9, 16 is n2<\/sup>.<\/p>\n

Check Your Progress<\/span> (Page No. 108)<\/p>\n

Question 1.
\nComplete the table and find the value of ‘p’ for the equation \\(\\frac{\\mathbf{p}}{\\mathbf{3}}\\) = 4<\/p>\n\n\n\n\n\n\n\n
p<\/td>\n\\(\\frac{\\mathbf{p}}{3}\\) = 4<\/td>\nCondition satisfied ? Yes\/ No<\/td>\n<\/tr>\n
3<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
6<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
9<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
12<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution:<\/p>\n\n\n\n\n\n\n\n
p<\/td>\n\\(\\frac{\\mathbf{p}}{3}\\) = 4<\/td>\nCondition satisfied ? Yes\/ No<\/td>\n<\/tr>\n
3<\/td>\n\\(\\frac{3}{3}\\) \u22601 \u2260 4<\/td>\nNo<\/td>\n<\/tr>\n
6<\/td>\n\u00a0\\(\\frac{6}{3}\\) \u22602 \u2260 4<\/td>\nNo<\/td>\n<\/tr>\n
9<\/td>\n\u00a0\\(\\frac{9}{3}\\) \u2260 3 \u2260 4<\/td>\nNo<\/td>\n<\/tr>\n
12<\/td>\n\u00a0\\(\\frac{12}{3}\\) \u2260 4 \u2260 4<\/td>\nYes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 2.
\nWrite LHS and RHS of following simple equations.
\ni) 2x + 1 = 10
\nii) 9 = y – 2
\niii) 3p + 5 = 2p + 10
\nSolution;
\ni) 2x+ 1 = 10
\nGiven equation is 2x + 1 = 10
\nL.H.S = 2x + 1
\nR.H.S = 10<\/p>\n

ii) 9 = y – 2
\nGiven equation is 9 = y – 2
\nLHS = 9
\nRHS = y – 2<\/p>\n

iii) 3p + 5 = 2p + 10
\nGiven equation is 3p + 5 = 2p + 10
\nLHS = 3p + 5
\nRHS = 2p + 10<\/p>\n

Question 3.
\nWrite any two simple equations and write their LHS and RHS.
\nSolution:
\ni) Consider 8x + 3 = 4 is a simple equation.
\nL.H.S = 8x + 3
\nRHS = 4<\/p>\n

ii) Consider 5a + 6 = 8a – 3 is a simple equation.
\nLHS = 5a + 6
\nRHS = 8a – 3<\/p>\n

Let\u2019s Explore<\/span> (Page No. 109)<\/p>\n

Observe for what value of m, the equation 3m = 15 has both LHS and RHS become equal.
\nSolution:
\nGiven equation is 3m = 15
\nIf m = 1, then the value of 3m = 3(1) = 3\u226015 \u2234 LHS \u2260RHS
\nIf m = 2, then the value of 3m = 3(2) = 6 \u2260 15 \u2234 LHS \u2260 RHS
\nIf m = 3, then the value of 3m = 3(3) = 9\u226015 \u2234 LHS \u2260 RHS
\nIf m = 4, then the value of 3m = 3(4) = 12 \u2260 15 \u2234 LHS \u2260 RHS
\nIf m = 5, then the value of 3m = 3(5) = 15 = 15 \u2234 LHS = RHS
\nFrom the above when m = 5 the both LHS and RHS are equal<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions and Answers. AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions Let’s Explore (Page No. 102) Question 1. Arrange 2 matchsticks to form the shape Continue the same shape for 2 times, 3 times and …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/11389"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=11389"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/11389\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=11389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=11389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=11389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}