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NCERT Solutions for Class 9th Maths Chapter 2 Exercise Set 2.2 — Linear Polynomials

Book page 21 Updated on2026-09-08

Q1.
Find the value of the linear polynomial 5x – 3 if: (i) x = 0   (ii) x = –1   (iii) x = 2
Answer

Substitute each value in turn.

(i)   5(0) − 3 = 0 − 3 = −3
(ii)   5(−1) − 3 = −5 − 3 = −8
(iii)   5(2) − 3 = 10 − 3 = 7
Why the answers rise in equal steps: from x = −1 to x = 0 the value goes −8 → −3, a rise of 5; from x = 0 to x = 1 it would go −3 → 2, again 5. In general 5(x + 1) − 3 − (5x − 3) = 5, the coefficient of x. Each one-unit step in the input changes the output by exactly the coefficient — this is the number we will soon call the slope.
Tip: Put the substituted value in brackets, especially when it is negative. Writing 5−1 instead of 5(−1) is the commonest slip here.
Q2.
Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0   (ii) s = –3   (iii) s = 4
Answer
(i)   7(0)² − 4(0) + 6 = 0 − 0 + 6 = 6
(ii)   7(−3)² − 4(−3) + 6
    = 7(9) + 12 + 6
    = 63 + 12 + 6 = 81
(iii)   7(4)² − 4(4) + 6
    = 7(16) − 16 + 6
    = 112 − 16 + 6 = 102
Why the two minus signs in (ii) do different things: (−3)² = +9, because squaring a negative gives a positive — so the first term is +63, not −63. But −4(−3) = +12, because a minus times a minus is a plus. Handling these separately is the whole difficulty of the question.
Check it yourself: the outputs 6, 81, 102 are not in a linear pattern — unlike Q1, equal steps in s do not give equal steps here. That is the visible signature of the s² term.
Q3.
The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.
Answer

Let Salil’s present age be x years. Then his mother’s present age is 3x years.

After 5 years: Salil = x + 5,   mother = 3x + 5
(x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15

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

Check: in 5 years they will be 20 and 50, and 20 + 50 = 70. ✓

Why one letter is enough: the first sentence ties the two ages together, so the mother’s age is not a second unknown — it is 3x. That leaves one unknown and one condition, and 4x + 10 is a linear polynomial; setting it equal to 70 gives a linear equation with exactly one solution. Note also that 5 is added twice, once for each person; forgetting that gives 4x + 5 = 70 and a wrong answer.
Tip: the age gap 45 − 15 = 30 years never changes. You can use that as a second check on any age problem.
Q4.
The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Answer

The ratio 2 : 5 lets both numbers be written with one letter. Let them be 2k and 5k.

5k − 2k = 63
3k = 63
k = 21
2k = 42,   5k = 105

Check: 105 − 42 = 63 ✓  and  42 : 105 = 2 : 5 (divide both by 21). ✓

Why the “k” trick is legitimate: saying the ratio is 2 : 5 means the two numbers are 2 and 5 parts of the same size. Calling that common part k captures the ratio exactly, and turns two unknowns into one. Every pair in the ratio 2 : 5 — 4 and 10, 6 and 15, 42 and 105 — is of the form (2k, 5k); the difference condition then picks out the single pair we want.
Check it yourself: the difference of 2k and 5k is 3k, so any such difference must be a multiple of 3. 63 = 3 × 21 is, which is why whole numbers come out.
Q5.
Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹88, how many coins does she have of each type?
Answer

Let the number of five-rupee coins be f. Then the number of two-rupee coins is 3f.

Value of the two-rupee coins = 2 × 3f = 6f
Value of the five-rupee coins = 5f
6f + 5f = 88
11f = 88
f = 8

Ruby has 8 five-rupee coins and 24 two-rupee coins (32 coins in all).

Check: 24 × ₹2 = ₹48 and 8 × ₹5 = ₹40; ₹48 + ₹40 = ₹88. ✓

Why you must not add the coins directly: the ₹88 is a total of value, not of coins, so each count has to be multiplied by what that coin is worth before adding. The neat consequence is that every “bundle” of 3 two-rupee coins and 1 five-rupee coin is worth ₹11, and 88 ÷ 11 = 8 tells you there are 8 such bundles.
Tip: Choose the letter for the smaller count (here the five-rupee coins). Then the other count is 3f, a whole number automatically.
Q6.
A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Answer

Let the shorter piece be x feet. Then the longer piece is 4x feet.

x + 4x = 300
5x = 300
x = 60

The pieces are 60 feet and 240 feet.

Check: 60 + 240 = 300 ✓ and 240 = 4 × 60. ✓

Why 5 appears: the whole fence is made of the shorter piece plus four more of the same size — 5 equal parts in all. So the shorter piece is one-fifth of 300 and the longer piece is four-fifths. Reading the ratio 1 : 4 as “5 equal parts” is a faster route to the same linear equation.
Tip: The question says the pieces are of different sizes, which is consistent: 60 ≠ 240. If a problem ever gives you two equal answers where it promised different ones, re-read the set-up.
Q7.
If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Answer

Let the width be w cm. Then the length is (2w + 3) cm.

Perimeter = 2(length + width)
2[(2w + 3) + w] = 24
2(3w + 3) = 24
6w + 6 = 24
6w = 18
w = 3

Width = 3 cm, length = 2(3) + 3 = 9 cm.

Check: perimeter = 2(9 + 3) = 24 cm ✓ and 9 is three more than twice 3. ✓

Why the bracket matters: “three more than twice the width” is 2w + 3, and the perimeter formula doubles the whole of (length + width). Writing 2 × 2w + 3 + w instead of 2(2w + 3 + w) leaves the 3 undoubled and gives w = 21/5, which is not a sensible answer here. The perimeter 6w + 6 is itself a linear polynomial in w, so setting it to 24 has exactly one solution.
Check it yourself: the area is 27 cm². Now try the same shape rule with perimeter 36 cm — you should get w = 5, length 13.
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