Home · Schooling · CBSE · Class 9 · Mathematics · Introduction to Linear Polynomials · Exercise Set 2.2

NCERT Class 9 Mathematics Exercise Set 2.2 Solutions: Introduction to Linear Polynomials

Step-by-step working for every question of Exercise Set 2.2 in NCERT's Class 9 Mathematics book, chapter “Introduction to Linear Polynomials”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 2.2 Q1

   Value of 5x − 3 at (i) x = 0, (ii) x = −1, (iii) x = 2.

   1. (i) Substitute x = 0 into the expression 5x − 3.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (i) Calculate 5 × 0 − 3.
   3. (i) The result is 0 − 3 = −3.
   4. (ii) Substitute x = −1 into the expression 5x − 3.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   5. (ii) Calculate 5 × (−1) − 3.
   6. (ii) The result is −5 − 3 = −8.
   7. (iii) Substitute x = 2 into the expression 5x − 3.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. (iii) Calculate 5 × 2 − 3.
   9. (iii) The result is 10 − 3 = 7.

   Answer: (i) −3; (ii) −8; (iii) 7

   Report an error in this solution

2. Ex 2.2 Q2

   Value of 7s² − 4s + 6 at (i) s = 0, (ii) s = −3, (iii) s = 4.

   1. (i) Substitute s = 0 into the expression 7s² − 4s + 6.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (i) Calculate 7(0)² − 4(0) + 6, which simplifies to 0 − 0 + 6.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   3. (i) The value is 6.
   4. (ii) Substitute s = −3 into the expression 7s² − 4s + 6.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   5. (ii) Calculate 7(−3)² − 4(−3) + 6, which is 7(9) + 12 + 6.
   6. (ii) Add the terms to get 63 + 12 + 6 = 81.
   7. (iii) Substitute s = 4 into the expression 7s² − 4s + 6.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. (iii) Calculate 7(4)² − 4(4) + 6, which is 7(16) − 16 + 6.
   9. (iii) Add and subtract the terms to get 112 − 16 + 6 = 102.

   Answer: (i) 6; (ii) 81; (iii) 102

   Report an error in this solution

3. Ex 2.2 Q3

   Salil's mother is now three times as old as Salil; in 5 years their ages will total 70. Find both present ages.

   1. Let Salil's present age be x years.
   2. Then his mother's present age is 3x years.
   3. In 5 years, Salil's age will be (x + 5) years.
   4. In 5 years, his mother's age will be (3x + 5) years.
   5. The sum of their ages after 5 years is 70, so (x + 5) + (3x + 5) = 70.
   6. Simplifying gives 4x + 10 = 70.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Subtracting 10 from both sides gives 4x = 60.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   8. Dividing by 4 gives x = 15.
   9. So Salil's present age is 15 years.
   10. His mother's present age is 3 × 15 = 45 years.

   Answer: Salil is 15 years old and his mother is 45 years old.

   Report an error in this solution

4. Ex 2.2 Q4

   Two positive integers differ by 63 and are in the ratio 2 : 5. Find them.

   1. Let the two positive integers be 2x and 5x, since they are in the ratio 2 : 5.
   2. The difference between the larger and smaller integer is 63, so 5x − 2x = 63.
   3. Simplify the equation to get 3x = 63.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   4. Divide both sides by 3 to find x = 21.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   5. Calculate the first integer as 2 × 21 = 42.
   6. Calculate the second integer as 5 × 21 = 105.

   Answer: 42, 105

   Report an error in this solution

5. Ex 2.2 Q5

   Ruby has three times as many ₹2 coins as ₹5 coins, ₹88 in all. How many coins of each kind?

   1. Let the number of ₹5 coins be x.
   2. Since Ruby has three times as many ₹2 coins as ₹5 coins, the number of ₹2 coins is 3x.
   3. The total value of the ₹5 coins is 5x and the total value of the ₹2 coins is 2(3x) = 6x.
   4. The sum of the values is given as ₹88, so 5x + 6x = 88.
   5. Simplifying the equation gives 11x = 88.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Dividing both sides by 11 gives x = 8.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. The number of ₹5 coins is x = 8.
   8. The number of ₹2 coins is 3x = 3 × 8 = 24.

   Answer: 24 two-rupee coins, 8 five-rupee coins

   Report an error in this solution

6. Ex 2.2 Q6

   A 300 ft fence is cut into two pieces, the longer four times the shorter. Find both lengths.

   1. Let the length of the shorter piece be x feet.
   2. The length of the longer piece is four times the shorter, so it is 4x feet.
   3. The total length of the fence is the sum of both pieces, so x + 4x = 300.
   4. Combine the terms to get 5x = 300.
   5. Divide both sides by 5 to find x = 60.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   6. The shorter piece is 60 feet.
   7. The longer piece is 4 × 60 = 240 feet.

   Answer: 60 feet, 240 feet

   Report an error in this solution

7. Ex 2.2 Q7

   A rectangle's length is 3 more than twice its width and its perimeter is 24 cm. Find its length and width.

   1. Let the width of the rectangle be x cm.
   2. Then the length is (2x + 3) cm.
   3. The perimeter of a rectangle is 2(length + width).
   4. So, 2((2x + 3) + x) = 24.
   5. Simplify inside the bracket: 2(3x + 3) = 24.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Divide both sides by 2: 3x + 3 = 12.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. Subtract 3 from both sides: 3x = 9.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   8. Divide by 3: x = 3.
   9. So, width = 3 cm.
   10. Length = 2(3) + 3 = 9 cm.

   Answer: width 3 cm, length 9 cm

   Report an error in this solution

← Introduction to Linear Polynomials: notes, practice and all exercises