{"id":25573,"date":"2021-06-23T13:52:16","date_gmt":"2021-06-23T08:22:16","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25573"},"modified":"2021-06-25T15:30:35","modified_gmt":"2021-06-25T10:00:35","slug":"ncert-solutions-for-class-6-maths-chapter-11-ex-11-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-11-ex-11-5\/","title":{"rendered":"NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5"},"content":{"rendered":"

These NCERT Solutions for Class 6 Maths<\/a> Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 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.
\n(a) 17 = x + 7
\n(b) (t – 7) > 5
\n(c) \\(\\frac{4}{2}\\) = 2
\n(d) (7 \u00d7 3) – 19 = 8
\n(e) 5 \u00d7 4 – 8 = 2x
\n(f) x – 2 = 0
\n(g) 2m < 30
\n(h) 2n + 1 = 11
\n(i) 7 = (11 \u00d7 5) – (12 \u00d7 4)
\n(j) 7 = (11 \u00d7 2) + p
\n(k) 20 = 5y
\n(l) \\(\\frac{3 \\mathrm{q}}{2}\\) < 5
\n(m) z + 12 > 24
\n(n) 20 – (10 – 5) = 3 \u00d7 5
\n(o) 7 – x = 5
\nAnswer:
\n(a) It is an equation of variable as both the sides are equal. The variable is x.
\n(b) It is not an equation as L.H.S. is greater than R.H.S.
\n(c) It is an equation with no variable. But it is a false equation.
\n(d) It is an equation with no variable. But it is a false equation.
\n(e) It is an equation of variable as both the sides are equal. The variable is x.
\n(f) It is an equation of variable x.
\n(g) It is not an equation as L.H.S. is less than R.H.S.
\n(h) It is an equation of variable as both the sides are equal. The variable is n.
\n(i) It is an equation with no variable as its both sides are equal.
\n(j) It is an equation of variable p.
\n(k) It is an equation of variable y.
\n(l) It is not an equation as L.H.S. is less than R.H.S.
\n(m) It is not an equation as L.H.S. is greater than R.H.S.
\n(n) It is an equation with no variable.
\n(o) It is an equation of variable x<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nComplete the entries in the third column of the table.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Equation<\/td>\nValue of variable<\/td>\nEquation satisfied Yes\/No<\/td>\n<\/tr>\n
(a) 10y = 80<\/td>\ny = 10<\/td>\n–<\/td>\n<\/tr>\n
(b) 10y = 80<\/td>\ny = 8<\/td>\n–<\/td>\n<\/tr>\n
(c) 10y = 80<\/td>\ny = 5<\/td>\n–<\/td>\n<\/tr>\n
(d) 4l = 20<\/td>\nl = 20<\/td>\n–<\/td>\n<\/tr>\n
(e) 4l = 20<\/td>\nl = 80<\/td>\n–<\/td>\n<\/tr>\n
(f) 4l = 20<\/td>\nl = 5<\/td>\n–<\/td>\n<\/tr>\n
(g) b + 5 = 9<\/td>\nb = 5<\/td>\n–<\/td>\n<\/tr>\n
(h) b + 5 = 9<\/td>\nb = 9<\/td>\n–<\/td>\n<\/tr>\n
(i) b + 5 = 9<\/td>\nb = 4<\/td>\n–<\/td>\n<\/tr>\n
(j) h – 8 = 5<\/td>\nh = 13<\/td>\n–<\/td>\n<\/tr>\n
(k) h – 8 = 5<\/td>\nh = 8<\/td>\n–<\/td>\n<\/tr>\n
(l) h – 8 = 5<\/td>\nh = 0<\/td>\n–<\/td>\n<\/tr>\n
(m) p + 3 = 1<\/td>\nP = 3<\/td>\n–<\/td>\n<\/tr>\n
(n) p + 3 = 1<\/td>\nP = 1<\/td>\n–<\/td>\n<\/tr>\n
(o) p + 3 = 1<\/td>\nP = 0<\/td>\n–<\/td>\n<\/tr>\n
(p) p + 3 = 1<\/td>\nP = -1<\/td>\n<\/td>\n<\/tr>\n
(q) p + 3 = 1<\/td>\nP = -2<\/td>\n–<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"NCERT<\/p>\n

\"NCERT<\/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.
\n(a) 5m = 60 (10,5,12,15)
\n(b) n + 12= 20(12,8,20,0)
\n(c) p – 5 = 5 (0, 10, 5,-5)
\n(d) \\(\\frac{\\mathrm{q}}{2}\\) = 7 (7, 2, 10, 14)
\n(e) r – 4 = 0 (4,-4, 8,0)
\n(f) x + 4 = 2 (-2, 0, 2, 4)
\nAnswer:
\n(a) 5m = 60
\nPutting the given values in L.H.S.,
\n5 \u00d7 10 = 50
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 m = 10 is not the solution
\n5 \u00d7 5 = 25
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 m = 5 is not the solution
\n5 \u00d7 12 = 60
\n\u2235 L.H.S. = R.H.S
\n\u2234 m = 12 is a solution
\n5 \u00d7 15 = 75
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 m = 15 is not the solution<\/p>\n

(b) n + 12 = 20
\nPutting the given values in L.H.S.,
\n12 + 12 = 24
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 n = 12 is not the solution
\n8 + 12 = 20
\n\u2235 L.H.S. = R.H.S
\n\u2234 n = 8 is a solution
\n20 + 12 = 32
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 n = 20 is not the solution
\n0 + 12 = 12
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 n = 0 is not the solution<\/p>\n

(c) p – 5 = 5
\nPutting the given values in L.H.S.,
\n0 – 5 = -5
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 p = 0 is not the solution
\n10 – 5 = 5
\n\u2235 L.H.S. = R.H.S
\n\u2234 p = 10 is a solution
\n5 – 5 = 0
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 p = 5 is not the solution
\n-5 -5 = -10
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 p = -5 is not the solution<\/p>\n

\"NCERT<\/p>\n

(d) \\(\\frac{\\mathrm{q}}{2}\\) = 7
\nPutting the given values in L.H.S.,
\n\\(\\frac{\\mathrm{7}}{2}\\)
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 q = 7 is not the solution
\n\\(\\frac{\\mathrm{2}}{2}\\) = 1
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 q = 2 is not the solution
\n\\(\\frac{\\mathrm{10}}{2}\\) = 5
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 q = 10 is not the solution
\n\\(\\frac{\\mathrm{14}}{2}\\) = 7
\n\u2235 L.H.S. = R.H.S
\n\u2234 q = 14 is a solution<\/p>\n

(e) r – 4 = 0
\nPutting the given values in L.H.S.,
\n4 – 4 = 0
\n\u2235 L.H.S. = R.H.S
\n\u2234 r = 4 is a solution
\n– 4 -4 = -8
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 r = – 4 is not the solution
\n8 – 4 = 4
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 r = 8 is not the solution
\n0 – 4 = – 4
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 r = 0 is not the solution<\/p>\n

(f) x + 4 = 2
\nPutting the given values in L.H.S.,
\n-2 + 4 = 2
\n\u2235 L.H.S. = R.H.S
\n\u2234 x = -2 is a solution
\n0 + 4 = 4
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 x = 0 is not the solution
\n2 + 4 = 6
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 x = 0 is not the solution
\n4 + 4 = 8
\n\u2235 L.H.S. \u2260 R.H.S
\n\u2234 x = 4 is not the solution<\/p>\n

\"NCERT<\/p>\n

Question 4.
\n(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.
\n\"NCERT<\/p>\n

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.
\n\"NCERT<\/p>\n

(c) Complete the table and find the solution of the equation z\/3 = 4 using the table.
\n\"NCERT<\/p>\n

(d) Complete the table and find the solution to the equation m – 7 = 3
\n\"NCERT
\nAnswer:
\n(a)
\n\"NCERT
\n\u2235 At m = 6, m + 10 = 16 \u2234 m = 6 is the solution.<\/p>\n

(b)
\n\"NCERT
\n\u2235 At t = 7, 5t = 35 \u2234 t = 7 is the solution.<\/p>\n

(c)
\n\"NCERT
\n\u2235 At z = 12, z\/3 = 4 \u2234 z = 12 is the solution.<\/p>\n

(d)
\n\"NCERT
\n\u2235 At m = 10, m- 7= 3 \u2234 m = 10 is the solution.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nSolve the following riddles, you may yourself construct such riddles.
\nWho am I?
\n(i) Go round a square
\nCounting every corner
\nThrice and no more!
\nAdd the count to me
\nTo get exactly thirty four!<\/p>\n

(ii) For each day of the week
\nMake an upcount from me
\nIf you make no mistake
\nYou will get twenty three!<\/p>\n

(iii) I am a special number
\nTake away from me a six!
\nA whole cricket team
\nYou will still be able to fix!<\/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!
\nAnswer:
\n(i) According to the condition,
\n1 + 12 = 34 or \u00d7 + 12 = 34
\n\u2234 By inspection, we have
\n22 + 12 = 34
\nSo, I am 22.<\/p>\n

(ii) Let I am \u2018\u00d7\u2019.
\nWe know that there are 7 days in a week.
\n\u2234 up-counting from \u00d7 for 7, the sum = 23
\nBy inspections, we have
\n16 + 7 = 23
\n\u2234 \u00d7 = 16
\nThus I am 16.<\/p>\n

(iii) Let the special number be x and there are 11 players in cricket team.
\n\u2234 Special Number -6 = 11
\n\u00d7 – 6 = 11
\nBy inspection, we get
\n17 – 6 = 11
\n\u2234 \u00d7 = 17
\nThus I am 17.<\/p>\n

(iv) Suppose I am \u2018\u00d7\u2019
\n\u2234 22 – 1 = 1
\nor 22 – \u00d7 = \u00d7
\nBy inspection, we have
\n22 -11 = 11
\n\u2234 \u00d7 = 11
\nThus I am 11.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5 Question 1. State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from …<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"yoast_head":"\nNCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-11-ex-11-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5 Question 1. State which of the following are equations (with a variable). Give reason for your answer. 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