{"id":27139,"date":"2021-06-29T15:44:08","date_gmt":"2021-06-29T10:14:08","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27139"},"modified":"2021-06-29T15:44:08","modified_gmt":"2021-06-29T10:14:08","slug":"ncert-solutions-for-class-7-maths-chapter-4-ex-4-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-4-ex-4-4\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4<\/h2>\n

Question 1.
\nSet up equations and solve them to find the unknown numbers in the following cases:
\n(a) Add 4 to eight times a number; you get 60.
\n(b) One-fifth of a number minus 4 gives 3.
\n(c) If I take three-fourths of a number and add 3 to it, I get 21.
\n(d) When I subtracted 11 from twice a number, the result was 15.
\n(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
\n(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
\n(g) Anwar thinks of a number. If he takes away 7 from \\(\\frac{5}{2}\\) of the number, the result is 23.
\nAnswer:
\n(a) Let the required number be x
\n\u2234 Eight times of the number = 8x
\nAccording to the given question, we get
\n8x + 4 =60
\nTransposing 4 to R.H.S
\n8x = 60 – 4
\n8x = 56
\nDividing both sides by 8, we get
\n\\(\\frac{8 x}{8}=\\frac{56}{8}\\)
\nx = 7
\n\u2234 The required number = 7<\/p>\n

(b) Let the number be x
\none-fifth of the number = \\(\\frac{1}{5}\\) x
\nAccording to the given question, we get
\n\\(\\frac{1}{5}\\) x – 4 = 3
\nTransposing -4 to R.H.S
\n\\(\\frac{1}{5}\\)x = 3 + 4
\n\\(\\frac{1}{5}\\)x = 7
\nMultiplying both sides by 5, we get 1
\n\\(\\frac{1}{5}\\)x \u00d7 5 = 7 \u00d7 5
\nx = 35
\n\u2234 The required number = 35<\/p>\n

(c) Let the number be x
\nThree-fourths of the number = \\(\\frac{3}{4}\\) x
\nAccording to the given question, we get
\n\\(\\frac{3}{4}\\) x + 3 = 21
\nTransposing 3 to R.H.S
\n\\(\\frac{3}{4}\\) x = 21 – 3
\n\\(\\frac{3}{4}\\) x = 18
\nMultiplying both sides by 4, we get
\n\\(\\frac{3}{4}\\) x \u00d7 4 = 18 \u00d7 4
\n3x = 72
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{3}=\\frac{72}{3}\\)
\nx = 24
\n\u2234 Thus, the required number = 24<\/p>\n

(d) Let the number be x
\nTwice a number = 2x
\nAccording to the given question
\n2x – 11 = 15
\nTransposing -11 to R.H.S
\n2x = 15 + 11
\n2x = 26
\nDividing both sides by 2, we get
\n\\(\\frac{2 \\mathrm{x}}{2}=\\frac{26}{2}\\)
\nx = 13
\n\u2234 The required number 13.<\/p>\n

(e) Let the number of notebooks with Munna be \u2018x\u2019
\nThrice the number of notebooks = 3x
\nAccording to the given question, we get
\n50 – 3x = 8
\nTransposing 50 to R.H.S
\n-3x = 8 – 50
\n-3x = -42
\nDividing both sides by (-3), we get
\n\\(\\frac{-3 x}{-3}=\\frac{-42}{-3}\\)
\nx = 14
\n\u2234 The required number of notebooks is 14.<\/p>\n

(f) Let the number be x
\nAccording to the given question, we get
\n\\(\\frac{x+19}{5}\\) = 8
\nMultiplying both sides by 5, we get
\n\\(\\frac{x+19}{5}\\) \u00d7 5 = 8 \u00d7 5
\nx + 19 = 40
\nTransposing 19 to R.H.S
\nx = 40 – 19
\nx = 21
\n\u2234 The required number = 21<\/p>\n

(g) Let the number be x
\n\\(\\frac{5}{2}\\) of the number = \\(\\frac{5}{2}\\) x
\nAccording to the given question, we get
\n\\(\\frac{5}{2}\\) x – 7 = 23
\nTransposing -7 to R.H.S
\n\\(\\frac{5}{2}\\)x = 23 + 7
\n\\(\\frac{5}{2}\\)x = 30
\nMultiplying both sides by 2, we get
\n\\(\\frac{5}{2}\\) x \u00d7 2 = 30 \u00d7 2
\n5x = 60
\nDividing both sides by 5, we get
\n\\(\\frac{5 x}{5}=\\frac{60}{5}\\)
\nx = 12
\nThe required number = 12.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nSolve the following:
\n(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?<\/p>\n

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40\u00b0. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180\u00b0).<\/p>\n

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
\nAnswer:
\n(a) Let the lowest marks scored be x
\nTwice the lowest marks = 2x
\nAccording to the given question, we get
\n2x + 7 = 87
\nTransposing 7 to R.H.S
\n2x = 87 – 7
\n2x = 80
\nDividing both sides by 2, we get
\n\\(\\frac{2 x}{2}=\\frac{80}{2}\\)
\nx = 40
\n\u2234 The lowest marks = 40<\/p>\n

(b) Let the base angle be x\u00b0
\n\u2234 The other base angle = x\u00b0
\nThe vertex angle = 40\u00b0
\n\u2234 Sum of the angles = x + x + 40
\n= 2x + 40
\nAccording to the given question, we get
\n2x + 40 = 180\u00b0
\n(Sum of the three angles of a triangle)
\n2x = 180\u00b0 – 40\u00b0
\n2x = 140\u00b0
\nDividing both sides by 2, we get
\n\\(\\frac{2 x}{2}=\\frac{140}{2}\\)
\nx = 70\u00b0
\n\u2234 The base angle of the triangle
\n= 70\u00b0<\/p>\n

(c) Let the runs scored by Rahul be x
\n\u2234 Sachins runs = 2x
\nAccording to the given question, we get
\nx + 2x =2(100) -2
\n3x = 200 – 2
\n3x = 198
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{x}=\\frac{198}{3}\\)
\nx = 66
\n\u2234 Rahul score = 66 runs and Sachins score = 2 \u00d7 66 = 132 runs<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nSolve the following:
\n(i) Ifran says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?<\/p>\n

(ii) Laxmi\u2019s father is 49 years old. He is 4 years older than three times Laxmi\u2019s age. What is Laxmi\u2019s age?<\/p>\n

(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
\nAnswer:
\n(i) Let the number of marbles with Parmit be x
\nNumber of marbles with Irfan = 37
\nAccording to the given question, we get
\n5x + 7 = 37
\nTransposing 7 to R.H.S
\n5x = 37 – 7
\n5x = 30
\nDividing both sides by 5, we get
\n\\(\\frac{5 x}{5}=\\frac{30}{5}\\)
\n\\(\\frac{5 x}{5}=\\frac{30}{5}\\) \u21d2 x = 6
\n\u2234 Parmit has 6 marbles.<\/p>\n

(ii) Let Laxmi\u2019s age be x years
\nThree times Laxmi s age = 3x
\nAccording to the given question, we get
\n3x + 4 = 49
\nTransposing 4 to R.H.S
\n3x = 49 – 4
\n3x = 45
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{3}=\\frac{45}{3}\\)
\nx = 15
\n\u2234 Laxmi\u2019s age= 15 years<\/p>\n

(iii) Let the number of fruit trees be x
\nThree times fruit trees = 3x
\nNumber of non-fruit trees = 2 + 3x
\nAccording to the given question, we get
\n2 + 3 x = 77
\nTransposing 2 to R.H.S
\n3x = 77 – 2
\n3x = 75
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{3}=\\frac{75}{3}\\)
\n\u21d2 x = 25
\n\u2234 The number of fruit trees = 25<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nSolve the following riddle :
\nI am a number,
\nTell my identity!
\nTake me seven times over
\nAnd add a fifty!
\nTo reach a triple century
\nYou still need forty!
\nAnswer:
\nLet the required number be x
\n7 times the number = 7x
\neach triple century = 3 \u00d7 100 = 300
\nAccording to the given riddle, we get
\n7x + 50 = 300 – 40
\n7x + 50 = 260
\nTransposing 50 to R.H.S
\n7x = 260 – 50
\n7x = 210
\nDividing both sides by 7, we get
\n\\(\\frac{7 \\mathrm{x}}{7}=\\frac{210}{7}\\)
\nx = 30
\n\u2234 The required number is 30.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4 Question 1. Set up equations and solve them to find the unknown numbers in the following cases: (a) Add …<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4<\/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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 - 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-7-maths-chapter-4-ex-4-4\/\" \/>\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 7 Maths Chapter 4 Simple Equations Ex 4.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4 Question 1. 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