{"id":26947,"date":"2021-06-28T18:20:18","date_gmt":"2021-06-28T12:50:18","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26947"},"modified":"2022-03-02T10:47:26","modified_gmt":"2022-03-02T05:17:26","slug":"ncert-solutions-for-class-7-maths-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-4-ex-4-1\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1"},"content":{"rendered":"
Put m = 5 \nL.H.S = 3(5) – 14 \n= 15 – 14 \n= 1 \u2260 R.H.S \nL.H.S \u2260 R.H.S<\/p>\n
Put m = 6 \nL.H.S = 3(6) – 14 \n= 18 – 14 \n= 4 = R.H.S \nL.H.S = R.H.S \nm = 6 is the solution to 3m – 14 = 4<\/p>\n
<\/p>\n
Question 4. \nWrite equations for the following statements: \n(i) The sum of numbers x and 4 is 9. \n(ii) 2 subtracted from y is 8. \n(iii) Ten times a is 70. \n(iv) The number b divided by 5 gives 6. \n(v) Three-fourth of t is 15. \n(vi) Seven times m plus 7 gets you 77. \n(vii) One-fourth of a number x minus 4 gives 4. \n(viii) If you take away 6 from 6 times y, you get 60. \n(ix) If you add 3 to one-third of z, you get 30. \nAnswer: \n(i) x + 4 = 9 \n(ii) y – 2 = 8 \n(iii) 10 a = 70 \n(iv) \\(\\frac { b }{ 5 }\\) = 6 \n(v) \\(\\frac { 3t }{ 4 }\\) = 15 \n(vi) 7m + 7 = 77 \n(vii) \\(\\frac { x }{ 4 }\\) x – 4 = 4 \n(viii) 6 y – 6 = 60 \n(ix) \\(\\frac { z }{ 3 }\\) + 3 = 30<\/p>\n
Question 5. \nWrite the following equations in statement forms: \n(i) p + 4 = 15 \n(ii) m – 7 = 3 \n(iii) 2m = 7 \n(iv) \\(\\frac { m }{ 5 }\\) = 3 \n(v) \\(\\frac { 3m }{ 5 }\\) = 6 \n(vi) 3p + 4 = 25 \n(vii) 4p – 2 = 18 \n(viii) \\(\\frac { p }{ 2 }\\) + 2 = 8 \nAnswer: \n(i) The sum of p and 4 is 15. \n(ii) 7 subtracted from m is 3. \n(hi) Twice a number m is 7. \n(iv) One-fifth of a number m is 3. \n(v) Three-fifth of a number m is 6. \n(vi) Three times a number p when added to 4 gives 25. \n(vii) 2 subtracted from four times a number p is 18. \n(viii) 2 added to half of a number p is 8.<\/p>\n
<\/p>\n
Question 6. \nSet up an equation in the following cases: \n(i) Irfan says that he has 7 marbles more than live times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit\u2019s marbles.)<\/p>\n
(ii) Laxmi\u2019s father is 49 years old. He is 4 years older than three times Laxmi\u2019s age. (Take Laxmi\u2019s age to be y years.)<\/p>\n
(iii) 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. (Take the lowest score to be 1.)<\/p>\n
(iv) In an isosceles triangle, the verte x angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees). \nAnswer: \n(i) Let Parmit has m marbles. \nThen, five time the marbles Parmit has = 5 m. \nIrfan has 7 marbles more than five times the marbles parmit has \nSo, Irfan has (5m + 7) marbles \nBut it is given that Irfan has 37 marbles. \n5m + 7 = 37<\/p>\n
(ii) Let Laxmi\u2019s age = y years \n3 times Laxmi\u2019s age = 3y years. \nAge of Laxmi\u2019s father = 3 times \nLaxmi\u2019s age + 4 years \n= 3y + 4 years \nBut Laxmi\u2019s father is 49 years old. \n3y + 4 = 49<\/p>\n
(iii) Let the lowest score (marks) = 1 \nTwice the lowest marks = 2 l \nSince highest marks = (twice the lowest marks) + 7 = 2l + 7 \nBut the highest marks = 87 \n2l + 7 = 87<\/p>\n
(iv) Let the base angle be b degrees \nThe base angle of an isosceles triangle are equal. \nThe other base angle = b degrees \nSince the vertex angle = Twice either base angle = 2b degrees. \nAlso, the sum of three angles of triangle = 180\u00b0 \nb + b + 2b = 180\u00b0 \n(OR) \n4b = 180\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 Question 1. Complete the last column of the table. Answer: Equation Value Say, whether the Equation is Satisfied, (Yes\/No) …<\/p>\n