{"id":18239,"date":"2020-01-06T11:33:32","date_gmt":"2020-01-06T06:03:32","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18239"},"modified":"2021-07-02T15:31:09","modified_gmt":"2021-07-02T10:01:09","slug":"kseeb-solutions-for-class-6-maths-chapter-11-ex-11-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-6-maths-chapter-11-ex-11-5\/","title":{"rendered":"KSEEB Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5"},"content":{"rendered":"

Students can Download Chapter 11 Algebra Ex 11.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 6 Maths Chapter 11 Algebra Ex 11.5<\/h2>\n

Question 1.
\nState which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
\nSolution:
\na) 17 = x + 7
\nAn equation with variable x<\/p>\n

b) (t – 7) > 5
\nAn inequality<\/p>\n

c) \\(\\frac{4}{2}\\) = 2
\nNo, it is a numerical equation<\/p>\n

d) (7 \u00d7 3) – 19 = 8
\nNo, it is a numerical equation<\/p>\n

e) 5 \u00d7 4 – 8 = 2x
\nAn equation with variable x<\/p>\n

\"KSEEB<\/p>\n

f) x – 2 = 0
\nAn equation with variable x<\/p>\n

g) 2m < 30
\nAn inequality<\/p>\n

h) 2n + 1 = 11
\nAn equation with variable n<\/p>\n

i) 7 = (11 \u00d7 5) – (12 \u00d7 4)
\nNo, it is a numerical equation<\/p>\n

j) 7 = (11 \u00d7 2 ) + p
\nAn equation with variable p<\/p>\n

k) 20 = 5y
\nAn equation with variable y<\/p>\n

l) \\(\\frac{3 p}{2}\\) < 5 An inequality m) z + 12 > 24
\nAn inequality<\/p>\n

n) 20 – (10 – 5) = 3 \u00d7 5
\nNo, it is a numerical equation<\/p>\n

o) 7 – x = 5
\nAn equation with variable x.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nComplete the entire in the third column of the table.
\n\"KSEEB
\nSolution:
\na) 10y = 80
\ny = 10 is net a solution to the given equation because for y = 10
\n10y = 10 \u00d7 10 = 100, and net 80<\/p>\n

b) 10y = 80
\ny = 8 is a solution to the given equation because for y = 8
\n10y = 10 \u00d7 8 = 80 and hence, the equation is satisfied<\/p>\n

c) 10y = 80
\ny = 5 is net a solution to the given equation because for y = 5
\n10y = 10 \u00d7 5 = 50, and net 80.<\/p>\n

d) 4l = 20
\nl = 20 is net a solution to the given equation because for l = 20
\n4l = 4 \u00d7 20 = 80, and net 20<\/p>\n

e) 4l = 80
\nl = 80 is net a solution to the given equation because for l = 80
\n4l = 4 \u00d7 80 = 320, and net 20<\/p>\n

f) 4l = 20
\n4l = 4 \u00d7 5 = 20 and hence, the equation because for l = 5
\n4l = 4 \u00d7 5 = 20 and hence, the equation is satisfied<\/p>\n

g) b + 5 = 9
\nb = 5 is net a solution to the given equation because for b = 5<\/p>\n

h) b + 5 = p
\nb = 9 is net a solution to the given equation because for b = 9
\nb + 5 = 9 + 5 = 14, and net 9<\/p>\n

i) b + 5 = 9
\nb + 5 = 4 + 5 = 9 and hence, the equation is satisfied<\/p>\n

\"KSEEB<\/p>\n

j) h – 8 = 5
\nh = 13 is a solution to the given equation because for h = B
\nh – 8 = 13 – 8 = 5 and hence, the equation is satisfied<\/p>\n

k) h – 8 = 5
\nh = 8 is net a solution to the given equation because for h = 8
\nh = 8 = -8 = 0, and net 5<\/p>\n

l) h – 8 = 5
\nh = 0 is net a solution to the given equation because for h = 0,
\nh – 8 = 0 – 8 = -8, and net 5<\/p>\n

m) p + 3 = 1
\np + 3 = 3 + 3 = 6, and net 1<\/p>\n

n) p + 3 = 1
\np + 3 = 1 + 3 = 4 , and net 1<\/p>\n

o) p + 3 = 1
\np + 3 = 0 + 3 = 3, and net 1<\/p>\n

p) p + 3 = 1
\np + 3 = -1 + 3 = 2, and net 1
\np + 3 = -2 + 3 = 1 and hence, the equation is satisfied.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nPick out the solution from the values given in the bracket next to each equation. show that the other values do not satisfy the equation.
\nSolution:
\na) 5m = 60 (10, 5, 12, 15)
\nm = 12 a solution to the given equation because form = 12
\n5m = 5 \u00d7 12 = 60 and hence, the equation is satisfied
\nm = 10 is not a solution to the given equation because for m = 10.
\n5m = 5 \u00d7 10 = 50, and net 60
\nm = 5 is net a solution to the given equation because for m = 5,
\n5m = 5 \u00d7 5 = 25, and net 60
\nm = 15 is net a solution to the given equation because for m = 15
\n5m = 5 \u00d7 15 = 75, and net 6.<\/p>\n

b) n + 12 = 20 (12, 8, 20, 0)
\nn = 8 is a solution to the given equation because for n = 8
\nn + 12 = 8 + 12 = 20 and hence, the equation is satisfied
\nn +12 = 12 + 12 = 24, and net 20
\nn = 20 is net a solution to the given equation because for n = 20
\nn + 12 = 20 + 12 = 32, and net 20<\/p>\n

c) p – 5 = 5 (0, 10, 5 – 5)
\np = 10 is a solution to the given equation because for p = 10, p – 5 = 10 – 5 = 5 and hence, the equation is satisfied.
\np = 0 is net a solution to the given equation because for p = 0 .
\np – 5 = 5 – 5 = 0, and net 5
\np = -5 is net s solution to the given equation because for p = -5
\np = 5 = -5 – 5 = -10 and net 5<\/p>\n

d) \\(\\frac{q}{2}\\) = 7 (7, 2, 10, 14)
\nq = 14 is a solution to the given equation because for q = 7 \\(\\frac{q}{2}\\) = \\(\\frac{7}{2}\\) and not 7
\n\\(\\frac{q}{2}\\) = \\(\\frac{7}{2}\\) and net 7
\nq = 2 is not a solution to the given equation because for q = 7
\n\\(\\frac{q}{2}\\) = \\(\\frac{7}{2}\\) and net 7
\nq = 2 is net a solution to the given equation because for q = 2
\n\\(\\frac{q}{2}\\) = \\(\\frac{2}{2}\\) = 1, and net 7
\nq = 10 is net a solution to the given equation because for qn = 10,
\n\\(\\frac{q}{2}\\) = \\(\\frac{10}{2}\\) = 5, and net 7<\/p>\n

e) r – 4 = 0 (4, -4, 8, 0)
\nr = 4 is a solution to the given equation because for r = 4
\nr = 4 = 4 – 4 = 0 and hence, the equation is satisfied
\nr = -4 is net a solution to-the given equation because for r = -4
\nr – 4 = -4 – 4 = -8 and net 0
\nr = 8 is net a solution to the given equation because for r = 8
\nr – 4 = 8 – 4 = 4 and net 0
\nr = 0 is net a solution to the given equation because for r = 0
\nr – 4 = 0 – 4 = -4 and net 0 .<\/p>\n

f) x + 4 = 2 (-2, 9, 2, 4)
\nx = -2 is a solution to the given equation because for x – 2
\nx + 4 = -2 + 4 = 2 and hence, the equation is satisfied
\nx = 0 is net a solution to the given equation because for x = 0
\nx + 4 = 0 + 4 = 4 and net 2
\nx = 2 is net a solution to the given equation because for x = 2
\nx + 4 = 2 + 4 = 6 and net 2
\nx = 4 is net a solution to the given equation because for x = 4
\nx + 4 = 4 + 4 = 8, and net 2.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\na) Complete the table and by inspection of the table find the solution to the equation m +10 = 6
\n\"KSEEB
\nSolution:
\nBy inspection, we can find that m = 6 is the solution of the above equation as for m = 6, m + 10 = 6 + 10 = 16<\/p>\n

b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35
\n\"KSEEB
\nSolution:
\nFor 5t, the table can be constructed as follows
\n\"KSEEB
\n\"KSEEB
\nBy inspection, we can find that t = 7 is the solution of the above equation as for t = 7, 5t = 5 \u00d7 7 = 35<\/p>\n

c) Complete the table and find the solution of the equation x\/3 = 4 using the table.
\n\"KSEEB
\nSolution:
\nFor, the table can be constructed as follows
\n\"KSEEB
\n\"KSEEB
\nBy inscription, we can find that z = 12 is the solution of the above equation of the above equation as for z = 12, \\(\\frac{z}{3}\\) = 4<\/p>\n

\"KSEEB<\/p>\n

d) Complete the table and find the solution to the equation m – 7 = 3
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nBy inspection, we can find that m = 10 is the solution of the above equation as for m = 10, m – 7 = 10 – 7 = 3<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nSolve the following riddles, you may yourself construct such riddles. Who am I?
\n\"KSEEB
\ni) Go round a square
\nCounting every corner
\nThrice and no more!
\nAdd the count to me
\nTo get exactly thirty four!
\nSolution:
\nThere are 4 comers in a square
\nThrice the number of comers in the square will be 3 \u00d7 4 = 12
\nWhen this result result i,e 12, is added to the number if Comes to be 34. therefore the number will be the difference of 34 and 12 i.e 34 – 12 = 22<\/p>\n

ii) For each day of the week
\nMake an upcount from me
\nIf you make mo mistake
\nyou will get twenty three!
\nSolution:
\n23 was the result when the old number was up Counted on Saturday
\n22 was the result when the old number was up counted on Saturday
\n20 was the result ‘when the old number was up counted on Thursday
\n19 was the result when the old number was up counted on tuesday
\n18 was the result when the old number was up counted on Tuesday
\n17 was the result when the old number was up counted on Tuesday<\/p>\n

iii) I am a special number
\nTake away from me a six!
\nA whole cricket term
\nyou will still be able to fix!
\n\"KSEEB
\nSolution:
\nIn a cricket taken at the 11 players Hence the number is such that when 6 is subtracted from it, the result is 11, Therefore, the number is 11 + 6 = 17<\/p>\n

iv) Tell me who I am
\nI shall give a pretty clue!
\nyou will get me back
\nIf you take me out of twenty two!
\nSolution:
\nThe number is such that when it is subtracted from 22 the result is again the number it self the number is 11, when again gives 11, when it is subtracted from 22.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 11 Algebra Ex 11.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 6 Maths Chapter 11 Algebra Ex 11.5 Question 1. State which of the following are …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[83],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18239"}],"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=18239"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18239\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}