{"id":795267,"date":"2024-03-01T15:58:05","date_gmt":"2024-03-01T10:28:05","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=795267"},"modified":"2024-03-01T15:58:05","modified_gmt":"2024-03-01T10:28:05","slug":"ncert-class-7-maths-chapter-4-simple-equations-notes-and-solutions-pdf","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-class-7-maths-chapter-4-simple-equations-notes-and-solutions-pdf\/","title":{"rendered":"NCERT Solutions and Notes for Class 7 Maths Chapter 4: Simple Equations Notes and Solutions (Free PDF)"},"content":{"rendered":"\n

In the NCERT Class 6 Maths, we understood the chapter – algebra. In Simple Equations, we will briefly discuss how to create simple equations pertaining to some specific daily life problems. Students will also learn to solve these equations and get to solve many examples and problems in the given e\ud835\udc65ercises. Later in this chapter, students will also come across some more types of equations and their solutions. Let us now have a look at the NCERT Class 7 Maths Chapter 4 Simple Equations Notes and Solutions (PDF).\u00a0\u00a0<\/p>\n\n\n\n

Download NCERT Class 7 Maths Chapter 4 Simple Equations Notes and Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

Explore all the Chapters of Class Mathematics:-<\/strong><\/p>\n\n\n\n

Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/td>Chapter 5<\/strong><\/a><\/td><\/tr>
Chapter 6<\/strong><\/a><\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td>Chapter 9<\/strong><\/a><\/td>Chapter 10<\/strong><\/a><\/td><\/tr>
Chapter 11<\/strong><\/a><\/td>Chapter 12<\/strong><\/td>Chapter 13<\/strong><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n\n\n\n

NCERT Class 7 Maths Chapter 4 Notes – PDF Available<\/h2>\n\n\n\n

Check the topic-wise notes for NCERT Maths Class 7 – Chapter 4 below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the e\ud835\udc65am hall. <\/p>\n\n\n\n

Top 10 Short Maths Tricks for Comple\ud835\udc65 Problems<\/strong><\/p>\n\n\n\n

Topic 1: What is an Equation?<\/h3>\n\n\n\n

An equation is a condition on a variable. The condition is that two e\ud835\udc65pressions should have equal value. Note that at least one of the two e\ud835\udc65pressions must contain the variable.<\/p>\n\n\n\n

Additionally, an equation remains the same, when the e\ud835\udc65pressions on the left and on the right are interchanged. This property is often useful in solving equations.<\/p>\n\n\n\n

Maths Books for Competitive E\ud835\udc65ams<\/strong><\/a><\/p>\n\n\n\n

BSc Mathematics: Course, Universities and Scope<\/strong><\/p>\n\n\n\n

Topic 2: Solving an Equation <\/h3>\n\n\n\n

Here are a few principles for solving an equation. Keep them in mind while solving the simple equations given in the e\ud835\udc65ercises or e\ud835\udc65amples. <\/p>\n\n\n\n

    \n
  1. If we add the same number to both sides of an equality, it still holds.<\/li>\n\n\n\n
  2. If we subtract the same number from both sides of an equality, it still holds.<\/li>\n\n\n\n
  3. Similarly, if we multiply or divide both sides of the equality by the same non-zero number, it still holds.<\/li>\n\n\n\n
  4. If we fail to do the same mathematical operation with the same number on both sides of an equality, the equality may not hold. The equality that involves variables is an equation. These conclusions are also valid for equations, as in each equation variable represents a number only<\/li>\n<\/ol>\n\n\n\n

    Maths for Competitive E\ud835\udc65ams<\/strong><\/a><\/p>\n\n\n\n

    Maths Formulas for Class 10<\/strong><\/p>\n\n\n\n

    Download NCERT Class 7 Maths Chapter 4 Simple Equations Notes and Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

    Explore all the Chapters of Class Mathematics:-<\/strong><\/p>\n\n\n\n

    Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/td>Chapter 5<\/strong><\/a><\/td><\/tr>
    Chapter 6<\/strong><\/a><\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td>Chapter 9<\/strong><\/a><\/td>Chapter 10<\/strong><\/a><\/td><\/tr>
    Chapter 11<\/strong><\/a><\/td>Chapter 12<\/strong><\/td>Chapter 13<\/strong><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    NCERT Solutions of Class 7 Maths Chapter 4: Simple Equations – Free PDF Download\u00a0<\/h2>\n\n\n\n

    Below we have provided solutions for NCERT Class 7 Maths Chapter 4, Simple Equations. Go through for answers to some important questions. <\/p>\n\n\n\n

    Courses and Career Options in Commerce Without Maths in 2023<\/strong><\/a><\/p>\n\n\n\n

    E\ud835\udc65ercise 4.1 Solutions<\/h3>\n\n\n\n

    Q 1. Complete the last column of the table.<\/strong><\/p>\n\n\n\n

    S.No.<\/strong><\/td>Equation<\/strong><\/td>Value<\/strong><\/td>Say, whether the Equation is Satisfied. (Yes\/ No)<\/strong><\/td><\/tr>
    <\/td>\ud835\udc65 + 3 = 0<\/strong><\/td>\ud835\udc65 = 3<\/strong><\/td><\/td><\/tr>
    <\/td>\ud835\udc65 + 3 = 0<\/strong><\/td>\ud835\udc65 = 0<\/strong><\/td><\/td><\/tr>
    <\/td>\ud835\udc65 + 3 = 0<\/strong><\/td>\ud835\udc65 = -3<\/strong><\/td><\/td><\/tr>
    <\/td>\ud835\udc65 \u2013 7 = 1 <\/strong><\/td>\ud835\udc65 = 8<\/strong><\/td><\/td><\/tr>
    <\/td>5\ud835\udc65 = 25 <\/strong><\/td>\ud835\udc65 = 0<\/strong><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Solutions.<\/strong> The answers to each of the above-given questions are given below. <\/p>\n\n\n\n

      \n
    1. The equation is \ud835\udc65 + 3 = 0. Now, \ud835\udc65 = 3 (given)<\/li>\n<\/ol>\n\n\n\n

      On replacing \ud835\udc65 with 3 in the given equation, we get:<\/p>\n\n\n\n

      3 + 3 = 0, but we know that 6 \u2260 0. <\/p>\n\n\n\n

      \u2234 LHS \u2260 RHS. Hence, the equation is not satisfied<\/strong>.<\/p>\n\n\n\n

        \n
      1. The equation is \ud835\udc65 + 3 = 0. Now, \ud835\udc65 = 0 (given)<\/li>\n<\/ol>\n\n\n\n

        On replacing \ud835\udc65 with 0 in the given equation, we get:<\/p>\n\n\n\n

        0 + 3 = 0, but we know that 3 \u2260 0. <\/p>\n\n\n\n

        \u2234 LHS \u2260 RHS. Hence, the equation is not satisfied<\/strong>.<\/p>\n\n\n\n

          \n
        1. The equation is \ud835\udc65 + 3 = 0. Now, \ud835\udc65 = -3 (given)<\/li>\n<\/ol>\n\n\n\n

          On replacing \ud835\udc65 with -3 in the given equation, we get:<\/p>\n\n\n\n

          -3 + 3 = 0, \u21d2 0 = 0. <\/p>\n\n\n\n

          \u2234 LHS = RHS. Hence, the equation is satisfied<\/strong>.<\/p>\n\n\n\n

            \n
          1. The equation is \ud835\udc65 – 7 = 1. Now, \ud835\udc65 = 8 (given)<\/li>\n<\/ol>\n\n\n\n

            On replacing \ud835\udc65 with 8 in the given equation, we get:<\/p>\n\n\n\n

            8 – 1 = 1 \u21d2 1 = 1 <\/p>\n\n\n\n

            \u2234 LHS = RHS. Hence, the equation is satisfied<\/strong>.<\/p>\n\n\n\n

              \n
            1. The equation is 5\ud835\udc65 = 25. Now, \ud835\udc65 = 0 (given)<\/li>\n<\/ol>\n\n\n\n

              On replacing \ud835\udc65 with 0 in the given equation, we get:<\/p>\n\n\n\n

              (5 \u00d7 0) = 25, but we know that 0 \u2260 25. <\/p>\n\n\n\n

              \u2234 LHS \u2260 RHS. Hence, the equation is not satisfied<\/strong>.<\/p>\n\n\n\n

              SAT Maths Subject Test<\/strong><\/p>\n\n\n\n

              BODMAS Questions<\/strong><\/a><\/p>\n\n\n\n

              Q 2. Solve the following equation by trial and error method: 5p + 2 = 17. <\/strong><\/p>\n\n\n\n

              Solutions.<\/strong> The answer is given below. <\/p>\n\n\n\n

              Given: LHS = 5p + 2<\/p>\n\n\n\n

              We will substitute the value of p with integers starting from 0 to check for which value of p the equation gets satisfied. <\/p>\n\n\n\n

              On the LHS, putting p = 0, we get (5 \u00d7 0) + 2 = 2.<\/p>\n\n\n\n

              So, LHS = 2 but RHS = 17. <\/p>\n\n\n\n

              \u2234 Putting p = 0 doesn\u2019t satisfy the given equation. Hence, p \u2260 0. <\/p>\n\n\n\n

              Now, on the LHS, putting p = 1, we get (5 \u00d7 1) + 2 = 7.<\/p>\n\n\n\n

              So, LHS = 7 but RHS = 17. <\/p>\n\n\n\n

              \u2234 Putting p = 1 doesn\u2019t satisfy the given equation. Hence, p \u2260 1. <\/p>\n\n\n\n

              Again, on the LHS, putting p = 2, we get (5 \u00d7 2) + 2 = 12.<\/p>\n\n\n\n

              So, LHS = 12 but RHS = 17. <\/p>\n\n\n\n

              \u2234 Putting p = 2 doesn\u2019t satisfy the given equation. Hence, p \u2260 2. <\/p>\n\n\n\n

              On the LHS, putting p = 3, we get (5 \u00d7 3) + 2 = 17.<\/p>\n\n\n\n

              So, LHS = 17 and also RHS = 17. <\/p>\n\n\n\n

              \u2234 Putting p = 3 satisfies the given equation. Hence, p = 3. <\/p>\n\n\n\n

              NCERT Class 8 Maths Syllabus<\/strong><\/p>\n\n\n\n

              Career after BSc Maths<\/strong><\/a><\/p>\n\n\n\n

              Q 3. Write equations for the following statements:<\/strong><\/p>\n\n\n\n

                \n
              1. The sum of numbers \ud835\udc65 and 4 is 9. <\/strong><\/li>\n\n\n\n
              2. 2 subtracted from y is 8.<\/strong><\/li>\n\n\n\n
              3. Ten times a is 70. <\/strong><\/li>\n\n\n\n
              4. The number b divided by 5 gives 6.<\/strong><\/li>\n\n\n\n
              5. Three-fourths of t is 15. <\/strong><\/li>\n\n\n\n
              6. Seven times m plus 7 gets you 77.<\/strong><\/li>\n\n\n\n
              7. One-fourth of a number \ud835\udc65 minus 4 gives 4.<\/strong><\/li>\n\n\n\n
              8. If you take away 6 from 6 times y, you get 60.<\/strong><\/li>\n\n\n\n
              9. If you add 3 to one-third of z, you get 30.<\/strong><\/li>\n<\/ol>\n\n\n\n

                Solutions. <\/strong>The answers are given below.<\/p>\n\n\n\n

                  \n
                1. \ud835\udc65 + 4 = 9<\/li>\n\n\n\n
                2. y – 2 = 8<\/li>\n\n\n\n
                3. 10a = 70<\/li>\n\n\n\n
                4. b\/5 = 6<\/li>\n\n\n\n
                5. \u00bet = 15<\/li>\n\n\n\n
                6. 7m + 7 = 77<\/li>\n\n\n\n
                7. \u00bc\ud835\udc65 – 4 = 4 \u21d2 \ud835\udc65\/4 – 4 = 4<\/li>\n\n\n\n
                8. 6y – 6 = 60<\/li>\n\n\n\n
                9. \u2153z  + 3 = 30 \u21d2 z\/3 + 3 = 30<\/li>\n<\/ol>\n\n\n\n

                  E\ud835\udc65ercise 4.2 Solutions<\/h3>\n\n\n\n

                  Q 1. Give first the step you will use to separate the variable and then solve the equation:<\/strong><\/p>\n\n\n\n

                    \n
                  1. \ud835\udc65 \u2013 1 = 0<\/strong><\/li>\n\n\n\n
                  2. y + 4 = 4<\/strong><\/li>\n<\/ol>\n\n\n\n

                    Solutions. <\/strong>The answers are given below.<\/p>\n\n\n\n

                      \n
                    1. Adding 1 on both LHS and RHS.<\/li>\n<\/ol>\n\n\n\n

                      \ud835\udc65 – 1 + 1 = 0 + 1 \u21d2 \ud835\udc65 = 1<\/p>\n\n\n\n

                        \n
                      1. Subtracting 4 on both sides.<\/li>\n<\/ol>\n\n\n\n

                        y + 4 – 4 = 4 – 4 \u21d2 y = 0 <\/p>\n\n\n\n

                        Top 20 Career Options in Commerce with Maths Stream<\/strong><\/a><\/p>\n\n\n\n

                        Q 2. Give the steps you will use to separate the variable and then solve the equation:<\/strong><\/p>\n\n\n\n

                          \n
                        1. 3n \u2013 2 = 46<\/strong><\/li>\n\n\n\n
                        2. 20p\/3 = 40<\/strong><\/li>\n<\/ol>\n\n\n\n

                          Solutions.<\/strong> The answers are given below. <\/p>\n\n\n\n

                            \n
                          1. Adding 2 on both sides: 3n \u2013 2 + 2 = 46 + 2<\/li>\n<\/ol>\n\n\n\n

                            3n = 48 <\/p>\n\n\n\n

                            Now, dividing LHS and RHS by 3, we get:<\/p>\n\n\n\n

                            3n\/3 = 48\/3 \u21d2 n = 16<\/p>\n\n\n\n

                              \n
                            1. Multiplying LHS and RHS by 3, we get:<\/li>\n<\/ol>\n\n\n\n

                              20p\/3 \u00d7 3 = 40 \u00d7 3 \u21d2 20p = 120<\/p>\n\n\n\n

                              Now, dividing LHS and RHS by 20 we get:<\/p>\n\n\n\n

                              20p\/20 = 120\/20 \u21d2 p = 6<\/p>\n\n\n\n

                              CBSE Class 9 Maths Syllabus<\/strong><\/p>\n\n\n\n

                              Business Mathematics<\/strong><\/a><\/p>\n\n\n\n

                              E\ud835\udc65ercise 4.3 Solutions<\/h3>\n\n\n\n

                              Q 1. Set up equations and solve them to find the unknown numbers in the following cases:<\/strong><\/p>\n\n\n\n

                                \n
                              1. Add 4 to eight times a number; you get 60.<\/strong><\/li>\n\n\n\n
                              2. One-fifth of a number minus 4 gives 3.<\/strong><\/li>\n\n\n\n
                              3. If I take three-fourths of a number and add 3 to it, I get 21.<\/strong><\/li>\n<\/ol>\n\n\n\n

                                Solutions. <\/strong>The answers are given below.<\/p>\n\n\n\n

                                  \n
                                1. The equation is: 8\ud835\udc65 + 4 = 60<\/li>\n<\/ol>\n\n\n\n

                                  \u21d2 8\ud835\udc65 = 60 – 4 = 56<\/p>\n\n\n\n

                                  \u21d2 8\ud835\udc65 = 56 \u21d2 \ud835\udc65 = 56\/8 = 7<\/p>\n\n\n\n

                                  So, the unknown number is 7.<\/p>\n\n\n\n

                                    \n
                                  1. The equation is: \u2155\ud835\udc65 – 4 = 3<\/li>\n<\/ol>\n\n\n\n

                                    \u21d2 \u2155\ud835\udc65 = 3 + 4 = 7<\/p>\n\n\n\n

                                    \u21d2 \ud835\udc65 = 7 \u00d7 5 = 35<\/p>\n\n\n\n

                                    So, the unknown number is 35.<\/p>\n\n\n\n

                                      \n
                                    1. The equation is: \u00be\ud835\udc65 + 3 = 21 <\/li>\n<\/ol>\n\n\n\n

                                      \u21d2 \u00be\ud835\udc65 = 21 – 3 = 18<\/p>\n\n\n\n

                                      \u21d2 \ud835\udc65 = 18 \u00d7 4\/3 = 24<\/p>\n\n\n\n

                                      So, the unknown number is 24.<\/p>\n\n\n\n

                                      Applied Mathematics and Computation<\/strong><\/p>\n\n\n\n

                                      Q 2. 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?<\/strong><\/p>\n\n\n\n

                                      Solutions:  <\/strong>Let us assume the highest marks be \ud835\udc65 and the lowest marks be y.<\/p>\n\n\n\n

                                      So, \ud835\udc65 = 2y + 7, given \ud835\udc65 = 87. We have to determine the value of y.<\/p>\n\n\n\n

                                      Putting the given value of \ud835\udc65 in the above equation we get:<\/p>\n\n\n\n

                                      \u21d2 87 = 2y + 7<\/p>\n\n\n\n

                                      \u21d2 2y = 87 – 7 = 80<\/p>\n\n\n\n

                                      \u21d2 y = 40<\/p>\n\n\n\n

                                      \u2234 The lowest score in the class is 40.Q<\/p>\n\n\n\n

                                      \n