Question Bank — Pair of Linear Equations

MCQ 1

The pair of equations x + 2y − 5 = 0 and 2x + 4y − 10 = 0 represents:

✓ Correct! a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 5/10 = 1/2. All ratios equal → coincident lines → infinitely many solutions.

MCQ 2

For the system 2x + 3y = 7 and 4x + 6y = 15 to be inconsistent, which condition holds?

✓ Correct! a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 7/15. First two ratios match, third differs → parallel lines → no solution (inconsistent).

MCQ 3

The solution of the pair x + y = 14 and x − y = 4 is:

✓ Correct! Adding: 2x = 18 → x = 9. Substituting: 9 + y = 14 → y = 5.

MCQ 4

Two lines represented by a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if:

✓ Correct! Parallel lines have same slope (a₁/a₂ = b₁/b₂) but different y-intercepts (≠ c₁/c₂), so they never meet.

MCQ 5

The sum of two numbers is 30 and their difference is 10. The smaller number is:

✓ Correct! x + y = 30, x − y = 10. Adding: 2x = 40 → x = 20 (larger). So y = 10 (smaller).

MCQ 6

The value of k for which the equations kx − y = 2 and 6x − 2y = 3 have no solution is:

✓ Correct! For no solution: k/6 = −1/−2 ≠ 2/3. k/6 = 1/2 → k = 3. Check: c-ratio = 2/3 ≠ 1/2 ✓ so lines are parallel.

MCQ 7

In the cross-multiplication method, the denominator for x is:

✓ Correct! x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁). The denominator for x uses b and c values.

MCQ 8

A two-digit number has its tens digit twice the units digit. If 18 is subtracted, the digits reverse. The number is:

✓ Correct! Let tens = a, units = b. a = 2b (Eq 1). 10a + b − 18 = 10b + a → 9a − 9b = 18 → a − b = 2 (Eq 2). From Eq 1 & 2: 2b − b = 2 → b = 2, a = 4. Number = 42.

MCQ 9

A consistent system of equations is one which has:

✓ Correct! A consistent system has one or infinitely many solutions. "Inconsistent" means no solution. Linear systems can never have exactly 2 or 3 solutions.

MCQ 10

If x = a, y = b is the solution of x − y = 2 and x + y = 4, then a and b are:

✓ Correct! Add: 2x = 6 → x = 3. Subtract: 2y = 2 → y = 1. So a = 3, b = 1. Verify: 3−1=2 ✓ and 3+1=4 ✓

AR 1

Assertion (A)The pair 2x + 3y = 5 and 4x + 6y = 10 has infinitely many solutions.

Reason (R)If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident and have infinite solutions.

✓ Correct! (A) 2/4 = 3/6 = 5/10 = 1/2 — all ratios equal, so infinitely many solutions. (R) correctly states the condition and is the exact explanation for (A).

AR 2

Assertion (A)The graphical method can always give exact answers for a pair of linear equations.

Reason (R)Two straight lines either intersect at one point, are parallel, or coincide.

✓ Correct! (A) is FALSE — graphical method gives approximate answers when the intersection is at non-integer coordinates. (R) is TRUE — the three cases for two lines are always one of those three.

AR 3

Assertion (A)The equations x + 2y = 3 and 2x + 4y = 6 cannot be solved by the substitution method to get a unique solution.

Reason (R)When one equation is a constant multiple of the other, the system has infinitely many solutions, not a unique one.

✓ Correct! (A) is TRUE — the second equation is exactly 2 × the first, so they are coincident. (R) is TRUE and correctly explains (A): a scalar multiple implies identical lines and infinite solutions.

SA 1

Solve by substitution: 2x + 3y = 11 and 2x − 4y = −24.

[3 marks]

Step 1: From Eq 1: 2x = 11 − 3y → x = (11 − 3y)/2

Step 2: Substitute in Eq 2:
2·(11 − 3y)/2 − 4y = −24
11 − 3y − 4y = −24
−7y = −35 → y = 5

Step 3: x = (11 − 15)/2 = −4/2 = x = −2

Verification: Eq 1: 2(−2)+3(5) = −4+15 = 11 ✓ Eq 2: 2(−2)−4(5) = −4−20 = −24 ✓
Solution: (−2, 5)

SA 2

Solve by elimination: 3x + 4y = 10 and 2x − 4y = 0.

[2 marks]

Coefficients of y are +4 and −4 (opposite signs) → ADD the equations:
(3x + 4y) + (2x − 4y) = 10 + 0
5x = 10 → x = 2

Substitute in Eq 2: 2(2) − 4y = 0 → 4y = 4 → y = 1

Verification: 3(2)+4(1) = 10 ✓ and 2(2)−4(1) = 0 ✓
Solution: (2, 1)

SA 3

Check whether the following pair is consistent or inconsistent: 3x + 2y = 8 and 6x + 4y = 9.

[2 marks]

Compute ratios: a₁/a₂ = 3/6 = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 8/9

a₁/a₂ = b₁/b₂ (both 1/2) ≠ c₁/c₂ (8/9)

Condition for no solution: ✓ matched.
∴ The lines are parallel and the system is inconsistent (no solution).

SA 4

Find the value of k so that the system 3x − y = 3 and 9x − ky = 9 has infinitely many solutions.

[2 marks]

For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

a₁/a₂ = 3/9 = 1/3 | b₁/b₂ = −1/(−k) = 1/k | c₁/c₂ = 3/9 = 1/3

Require: 1/k = 1/3 → k = 3

Check: c-ratio = 1/3 = 1/3 ✓ — all three ratios equal.
k = 3

SA 5

Solve by cross-multiplication: 2x + y = 5 and 3x + 2y = 8.

[3 marks]

Rewrite in ax + by + c = 0 form:
2x + y − 5 = 0 (a₁=2, b₁=1, c₁=−5)
3x + 2y − 8 = 0 (a₂=3, b₂=2, c₂=−8)

x/(b₁c₂ − b₂c₁) = x/[(1)(−8) − (2)(−5)] = x/(−8+10) = x/2
y/(c₁a₂ − c₂a₁) = y/[(−5)(3) − (−8)(2)] = y/(−15+16) = y/1
1/(a₁b₂ − a₂b₁) = 1/[(2)(2) − (3)(1)] = 1/1

x/2 = 1 → x = 2 | y/1 = 1 → y = 1

Verification: 2(2)+1=5 ✓ 3(2)+2(1)=8 ✓
Solution: (2, 1)

SA 6

The sum of a two-digit number and the number obtained by reversing its digits is 66. The units digit is 2 more than the tens digit. Find the number.

[3 marks]

Let tens digit = x, units digit = y. Number = 10x + y, Reversed = 10y + x.

Eq 1: (10x + y) + (10y + x) = 66 → 11x + 11y = 66 → x + y = 6
Eq 2: y = x + 2 → y − x = 2

Adding: 2y = 8 → y = 4, then x = 2.

Number = 10(2) + 4 = 24. Reversed = 42. Sum = 66 ✓ and 4 − 2 = 2 ✓

SA 7

5 pencils and 7 pens together cost ₹50, and 7 pencils and 5 pens cost ₹46. Find the cost of each.

[3 marks]

Let pencil = ₹x, pen = ₹y.
Eq 1: 5x + 7y = 50
Eq 2: 7x + 5y = 46

Multiply Eq 1 by 5: 25x + 35y = 250
Multiply Eq 2 by 7: 49x + 35y = 322
Subtract: 24x = 72 → x = 3

Sub in Eq 1: 15 + 7y = 50 → 7y = 35 → y = 5

Pencil = ₹3, Pen = ₹5 Verify: 5(3)+7(5)=50 ✓ 7(3)+5(5)=46 ✓

SA 8

Solve the reducible pair: 2/x + 3/y = 13 and 5/x − 4/y = −2 (x ≠ 0, y ≠ 0).

[3 marks]

Let u = 1/x, v = 1/y. System becomes:
2u + 3v = 13 … (1)
5u − 4v = −2 … (2)

Multiply (1) by 4: 8u + 12v = 52
Multiply (2) by 3: 15u − 12v = −6
Add: 23u = 46 → u = 2 → x = 1/2

Sub in (1): 4 + 3v = 13 → v = 3 → y = 1/3

x = 1/2, y = 1/3

SA 9

For what value of p does the pair px + 3y − (p − 3) = 0 and 12x + py − p = 0 have infinite solutions?

[3 marks]

For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

p/12 = 3/p = (p−3)/p

From p/12 = 3/p: p² = 36 → p = ±6
From 3/p = (p−3)/p: 3 = p − 3 → p = 6

Both conditions satisfied when p = 6.
Check: 6/12 = 3/6 = 3/6 = 1/2 ✓

SA 10

In a triangle ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also ∠A − ∠B = 10°. Find all three angles.

[3 marks]

Angle sum: x + (3x − 2) + y = 180 → 4x + y = 182 … (1)
Given: x − (3x − 2) = 10 → −2x + 2 = 10 → x = −4

Wait — let's re-read: ∠A − ∠B = 10 → x − (3x−2) = 10 → −2x = 8 → x = −4. That gives negative angle.
Correct reading: ∠B − ∠A = 10 → (3x−2) − x = 10 → 2x = 12 → x = 6

∠A = 6°, ∠B = 3(6)−2 = 16°
From (1): 24 + y = 182 → y = 158°
Check: 6 + 16 + 158 = 180 ✓
∠A = 6°, ∠B = 16°, ∠C = 158°

SA 11

Solve graphically: x + y = 6 and x − y = 2. Shade the region bounded by the two lines and the y-axis.

[3 marks]

Table for x + y = 6: (0,6), (6,0)
Table for x − y = 2: (0,−2), (2,0)

Plot both lines. They intersect when:
Adding: 2x = 8 → x = 4, y = 2 → Intersection at (4, 2)

The region bounded by the two lines and y-axis is the triangle with vertices:
(0, 6), (0, −2), (4, 2)

Area = ½ × base × height = ½ × 8 × 4 = 16 sq. units

SA 12

A fraction becomes 1/3 when 1 is subtracted from the numerator and 1 is added to the denominator. It becomes 1/2 when 1 is added to both. Find the fraction.

[3 marks]

Let fraction = x/y.
Eq 1: (x−1)/(y+1) = 1/3 → 3(x−1) = y+1 → 3x − y = 4
Eq 2: (x+1)/(y+1) = 1/2 → 2(x+1) = y+1 → 2x − y = −1

Subtract Eq 2 from Eq 1: x = 5
Sub in Eq 2: 10 − y = −1 → y = 11

Fraction = 5/11
Verify: (5−1)/(11+1) = 4/12 = 1/3 ✓ (5+1)/(11+1) = 6/12 = 1/2 ✓

LA 1

A father is 3 times as old as his son. After 12 years, the father will be twice as old as his son. Find their present ages. Also find how many years ago the father was 5 times as old as the son.

[5 marks]

Let father's age = x, son's age = y.
Eq 1: x = 3y … present ages
Eq 2: x + 12 = 2(y + 12) → x − 2y = 12

Substitute x = 3y: 3y − 2y = 12 → y = 12, x = 36

Father = 36 years, Son = 12 years

When was father 5× son's age?
Let t years ago: (36 − t) = 5(12 − t)
36 − t = 60 − 5t → 4t = 24 → t = 6 years ago

Verify: 6 years ago, father = 30, son = 6. 30 = 5 × 6 ✓
Present: father = 36 = 3 × 12 ✓ After 12 years: 48 = 2 × 24 ✓

LA 2

A boat can go 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.

[5 marks]

Let speed of boat = u km/h, speed of stream = v km/h.
Upstream speed = u − v, Downstream speed = u + v.

Let a = 1/(u+v), b = 1/(u−v). Then:
Eq 1: 44a + 30b = 10
Eq 2: 55a + 40b = 13

Multiply Eq 1 by 4: 176a + 120b = 40
Multiply Eq 2 by 3: 165a + 120b = 39
Subtract: 11a = 1 → a = 1/11 → u + v = 11

Sub in Eq 1: 4 + 30b = 10 → 30b = 6 → b = 1/5 → u − v = 5

Adding: 2u = 16 → u = 8 km/h
v = 11 − 8 = v = 3 km/h

Speed in still water = 8 km/h, Stream speed = 3 km/h

LA 3

Solve the following pair of equations using cross-multiplication method: ax + by = a − b and bx − ay = a + b.

[5 marks]

Rewrite in standard form:
ax + by − (a−b) = 0 → c₁ = −(a−b)
bx − ay − (a+b) = 0 → c₂ = −(a+b)

Apply cross-multiplication:
x/[b·(−(a+b)) − (−a)·(−(a−b))]
= x/[−b(a+b) − a(a−b)]
= x/[−ab − b² − a² + ab]
= x/(−a² − b²)

y/[(−(a−b))·b − (−(a+b))·a]
= y/[−b(a−b) + a(a+b)]
= y/[−ab + b² + a² + ab]
= y/(a² + b²)

1/(a·(−a) − b·b) = 1/(−a² − b²)

x/(−a²−b²) = 1/(−a²−b²) → x = 1
y/(a²+b²) = 1/(−a²−b²) → y = −1 → y = −1

Solution: x = 1, y = −1
Verify in Eq 1: a(1) + b(−1) = a − b ✓

LA 4

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. Find the speeds of the two cars.

[5 marks]

Let speed of car from A = x km/h, from B = y km/h. Assume x > y.

Condition 1 (same direction):
Car A overtakes Car B after 5 hours. In 5 hours A covers 5x, B covers 5y.
Distance A travels extra = 100 km: 5x − 5y = 100 → x − y = 20 … (1)

Condition 2 (opposite direction):
They cover combined distance of 100 km in 1 hour:
x + y = 100 … (2)

Add (1) and (2): 2x = 120 → x = 60 km/h
y = 100 − 60 = y = 40 km/h

Speed of car from A = 60 km/h, from B = 40 km/h
Verify: Same direction: 5(60−40) = 100 ✓ Opposite: 60+40=100 km in 1 h ✓

LA 5

A lending library charges fixed fee for first 3 days and then ₹ per day thereafter. Raju paid ₹27 for a book kept for 7 days. Tariq paid ₹21 for 5 days. Find the fixed charge and the charge per day.

[5 marks]

Let fixed charge = ₹x, charge per extra day = ₹y.

Raju: 7 days = 3 fixed + 4 extra days
x + 4y = 27 … (1)

Tariq: 5 days = 3 fixed + 2 extra days
x + 2y = 21 … (2)

Subtract (2) from (1): 2y = 6 → y = ₹3 per day
Substitute: x + 6 = 21 → x = ₹15 fixed charge

Fixed charge = ₹15, Charge per extra day = ₹3
Verify: 15 + 4(3) = 27 ✓ 15 + 2(3) = 21 ✓