NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 3.1
Form the pair of linear equations in the following problems, and find their solutions graphically:
Answer
(i) 10 students of Class X took part in a Mathematics quiz. The number of girls is 4 more than the number of boys.
Let boys = x, girls = y. Equations: x + y = 10 and y – x = 4.
Solution: x = 3, y = 7. 3 boys and 7 girls.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46.
5x + 7y = 50 and 7x + 5y = 46. Solution: x = 3 (pencil), y = 5 (pen).
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0: a₁/a₂ = 5/7, b₁/b₂ = –4/6. Since 5/7 ≠ –2/3, lines intersect (unique solution).
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. All equal → lines are coincident (infinite solutions).
(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0: a₁/a₂ = 3, b₁/b₂ = 3, c₁/c₂ = 10/9. a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → lines are parallel (no solution).
Answer
On comparing the ratios, find out whether the following pairs of equations are consistent or inconsistent:
(i) 3x + 2y = 5, 2x – 3y = 7: 3/2 ≠ 2/–3 → Consistent (unique solution)
(ii) 2x – 3y = 8, 4x – 6y = 9: 2/4 = –3/–6 ≠ 8/9 → Inconsistent (no solution)
(iii) 3/2x + 5/3y = 7, 9x – 10y = 14: (3/2)/9 = 1/6, (5/3)/–10 = –1/6 → 1/6 ≠ –1/6 → Consistent
(iv) 5x – 3y = 11, –10x + 6y = –22: 5/–10 = –3/6 = 11/–22 = –1/2 → Consistent (infinite solutions, dependent)
Answer
Exercise 3.2 (Algebraic Methods)
Solve the following pairs of linear equations by the substitution method:
(i) x + y = 14, x – y = 4: From eq 1: x = 14 – y. Substituting: 14 – y – y = 4 → y = 5, x = 9. (9, 5)
(ii) s – t = 3, s/3 + t/2 = 6: From eq 1: s = t + 3. Sub: (t+3)/3 + t/2 = 6 → 5t/6 = 5 → t = 6, s = 9. (9, 6)
(iii) 3x – y = 3, 9x – 3y = 9: From eq 1: y = 3x – 3. Sub: 9x – 3(3x–3) = 9 → 9 = 9. Infinite solutions (coincident lines).
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3: From eq 1: x = (1.3 – 0.3y)/0.2. Solving: x = 2, y = 3
Answer
Solve 2x + 3y = 11 and 2x – 4y = –24 by substitution, and find the value of 'm' for which y = mx + 3.
Answer
Subtracting: 7y = 35
⇒ y = 5, x = –2. For y = mx + 3: 5 = m(–2) + 3
⇒ m = –1
Exercise 3.3 (Elimination Method)
Solve the following pairs by the elimination method and the substitution method:
(i) x + y = 5, 2x – 3y = 4: Multiply eq 1 by 3: 3x + 3y = 15. Add to eq 2: 5x = 19 → x = 19/5, y = 6/5. (19/5, 6/5)
(ii) 3x + 4y = 10, 2x – 2y = 2: Multiply eq 2 by 2: 4x – 4y = 4. Add to eq 1: 7x = 14 → x = 2, y = 1. (2, 1)
(iii) 3x – 5y – 4 = 0, 9x = 2y + 7: Multiply eq 1 by 3: 9x – 15y = 12. Subtract eq 2: –15y + 2y = 12 – 7 → –13y = 5 → y = –5/13, x = 9/13. (9/13, –5/13)
Answer
(iv) x/2 + 2y/3 = –1, x – y/3 = 3: Multiply eq 2 by 2: 2x – 2y/3 = 6. Subtract from eq 1 multiplied by 2: gives y = –3, x = 2. Wait — let me redo: eq 1: 3x + 4y = –6 (multiply by 6), eq 2: 3x – y = 9 (multiply by 3). Subtract: 5y = –15
⇒ y = –3, x = 2. (2, –3)
Q2. Form the pair of linear equations and solve
(i) Meena went to a bank to withdraw ₹2000. She asked the cashier to give ₹50 and ₹100 notes only. She got 25 notes in all. Find the number of notes of each denomination.
50x + 100y = 2000 and x + y = 25. From eq 2: x = 25 – y. Sub: 50(25–y) + 100y = 2000 → 1250 + 50y = 2000 → y = 15, x = 10.
10 notes of ₹50 and 15 notes of ₹100.
(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for 7 days, while Susy paid ₹21 for a book kept for 5 days.
x + 4y = 27, x + 2y = 21. Subtracting: 2y = 6 → y = 3, x = 15.
Fixed charge = ₹15, Additional charge per day = ₹3.
Important Extra Questions
For what values of k will the following pair of linear equations have infinitely many solutions?
Answer
kx + 3y – (k – 3) = 0, 12x + ky – k = 0
Ans. For infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
k/12 = 3/k = (k–3)/k
From k/12 = 3/k: k² = 36 → k = ±6
Check k = 6: 3/6 = 1/2, (6–3)/6 = 1/2, 6/12 = 1/2. All equal ✓
Check k = –6: 3/–6 = –1/2, –6/12 = –1/2, (–6–3)/–6 = 3/2 ≠ –1/2 ✗
k = 6
Solve using cross-multiplication method: 2x + 3y = 46, 3x + 5y = 74.
Answer
x/(3×–74 – 5×–46) = y/(–46×3 – (–74)×2) = 1/(2×5 – 3×3)
x/(–222 + 230) = y/(–138 + 148) = 1/(10 – 9)
x/8 = y/10 = 1/1 → x = 8, y = 10
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the boat in still water and the speed of the stream.
Answer
Let speed of boat = x, speed of stream = y.
Upstream speed = x – y, Downstream speed = x + y.
30/(x–y) + 44/(x+y) = 10 ... (1)
40/(x–y) + 55/(x+y) = 13 ... (2)
Let 1/(x–y) = u, 1/(x+y) = v: 30u + 44v = 10, 40u + 55v = 13
Solving: u = 1/5, v = 1/11. So x – y = 5, x + y = 11.