Extra Practice Questions
Pair of Linear Equations (R.D. Sharma Level)
1. Solve for x and y: 2x + 3y = 13 and 5x – 4y = -2.
Let u = 1/x and v = 1/y. The equations become:
2u + 3v = 13 —(1)5u - 4v = -2 —(2)
Multiply (1) by 4 and (2) by 3 to eliminate v:8u + 12v = 5215u - 12v = -6
Add the two new equations: 23u = 46 => u = 2.
Substitute u=2 into (1): 2(2) + 3v = 13 => 4 + 3v = 13 => 3v = 9 => v = 3.
Since u = 1/x, x = 1/u = 1/2. Since v = 1/y, y = 1/v = 1/3.
Answer: x = 1/2, y = 1/3.
2. For what value of k will the following pair of linear equations have no solution? 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1.
For no solution, the condition is a1a2 = b1b2 ≠c1c2.
Here, a1=3, b1=1, a2=2k-1, b2=k-1.
Set a1a2 = b1b2: 32k-1 = 1k-1.
Cross-multiply: 3(k - 1) = 1(2k - 1) => 3k - 3 = 2k - 1 => k = 2.
Check the second condition for k=2: b1/b2 = 1/1 = 1 and c1/c2 = 1/(2(2)+1) = 1/5. Since 1 ≠1/5, the condition holds.
Answer: k = 2.
3. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Find the fraction.
Let the fraction be x/y.
Eq 1: x + y = 2x + 4 => y - x = 4.
Eq 2: (x+3)/(y+3) = 2/3 => 3(x+3) = 2(y+3) => 3x - 2y = -3.
From Eq 1, y = x + 4. Substitute into Eq 2:
3x - 2(x + 4) = -3 => 3x - 2x - 8 = -3 => x = 5.
Substitute x back: y = 5 + 4 = 9.
Answer: The fraction is 5/9.
4. 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 stream and that of the boat in still water.
Let boat speed = x km/hr, stream speed = y km/hr.
Upstream speed = x-y, Downstream speed = x+y.
Eq 1: 30/(x-y) + 44/(x+y) = 10.
Eq 2: 40/(x-y) + 55/(x+y) = 13.
Let u=1/(x-y), v=1/(x+y).
30u + 44v = 10 => 15u + 22v = 5.
40u + 55v = 13.
Multiply first new eq by 8, second by 3: 120u + 176v = 40, 120u + 165v = 39.
Subtracting gives 11v = 1 => v = 1/11.
Substitute v: 15u + 22(1/11) = 5 => 15u + 2 = 5 => 15u = 3 => u = 1/5.
x-y = 1/u = 5 and x+y = 1/v = 11.
Adding these gives 2x = 16 => x = 8. Then y = 3.
Answer: Speed of boat = 8 km/hr, Speed of stream = 3 km/hr.
… (16 more questions follow)
5. For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 and (a-b)x + (a+b)y = 3a+b-2.
Condition for infinite solutions: a1a2 = b1b2 = c1c2.
2/(a-b) = 3/(a+b) => 2a+2b = 3a-3b => a = 5b.
3/(a+b) = 7/(3a+b-2) => 9a+3b-6 = 7a+7b => 2a-4b = 6 => a-2b=3.
Substitute a=5b into the second simplified eq: 5b-2b=3 => 3b=3 => b=1.
Then a = 5(1) = 5.
Answer: a = 5, b = 1.
6. The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number.
Let number be 10x+y. Reversed number is 10y+x.
Eq 1: (10x+y) + (10y+x) = 66 => 11x+11y=66 => x+y=6.
Eq 2: x-y=2 or y-x=2.
Case 1: Add x+y=6 and x-y=2 => 2x=8 => x=4, y=2. Number is 42.
Case 2: Add x+y=6 and y-x=2 => 2y=8 => y=4, x=2. Number is 24.
Answer: The number is 42 or 24.
7. Solve: ax + by = a – b and bx – ay = a + b.
Multiply first eq by a, second by b:
a2x + aby = a2 - abb2x - aby = ab + b2
Add them: (a2+b2)x = a2+b2 => x = 1.
Substitute x=1 into first eq: a(1) + by = a-b => by = -b => y = -1.
Answer: x = 1, y = -1.
8. The area of a rectangle gets reduced by 9 sq. units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 sq. units. Find the dimensions.
Let length=l, breadth=b. Area=lb.
Eq 1: (l-5)(b+3) = lb - 9 => lb + 3l - 5b - 15 = lb - 9 => 3l - 5b = 6.
Eq 2: (l+3)(b+2) = lb + 67 => lb + 2l + 3b + 6 = lb + 67 => 2l + 3b = 61.
Multiply Eq 1 by 2, Eq 2 by 3: 6l - 10b = 12, 6l + 9b = 183.
Subtracting gives -19b = -171 => b = 9.
Substitute b=9 into 3l - 5b = 6: 3l - 45 = 6 => 3l = 51 => l = 17.
Answer: Length = 17 units, Breadth = 9 units.
9. Solve: 10x+y + 2x-y = 4 and 15x+y – 5x-y = -2.
Let u=1/(x+y), v=1/(x-y).
10u + 2v = 4 => 5u + v = 2.
15u - 5v = -2.
From first new eq, v = 2-5u. Substitute: 15u - 5(2-5u) = -2.
15u - 10 + 25u = -2 => 40u = 8 => u = 1/5.
v = 2 - 5(1/5) = 2 - 1 = 1.
x+y = 1/u = 5 and x-y = 1/v = 1.
Adding gives 2x=6 => x=3. Then y=2.
Answer: x = 3, y = 2.
10. For what value of k will the equations kx + 3y = k-3 and 12x + ky = k have a unique solution?
Condition for a unique solution: a1a2 ≠b1b2.
k/12 ≠3/k => k2 ≠36 => k ≠±6.
Answer: The system has a unique solution for all real values of k except 6 and -6.
11. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹1000, whereas a student B, who takes food for 26 days, pays ₹1180. Find the fixed charges and the cost of food per day.
Let fixed charge = x, charge per day = y.
Eq 1: x + 20y = 1000.
Eq 2: x + 26y = 1180.
Subtract Eq 1 from Eq 2: 6y = 180 => y = 30.
Substitute y=30 into Eq 1: x + 20(30) = 1000 => x + 600 = 1000 => x = 400.
Answer: Fixed charge = ₹400, Cost per day = ₹30.
12. Father’s age is three times the sum of the ages of his two children. After 5 years, his age will be twice the sum of the ages of his two children. Find the present age of the father.
Let father’s age = x, sum of children’s ages = y.
Eq 1: x = 3y.
After 5 years, father’s age = x+5. Sum of children’s ages = y + 5 + 5 = y+10.
Eq 2: x+5 = 2(y+10) => x+5 = 2y+20 => x-2y = 15.
Substitute x=3y: 3y-2y = 15 => y = 15.
x = 3(15) = 45.
Answer: The father’s present age is 45 years.
13. Solve: (a – b)x + (a + b)y = a2 – 2ab – b2 and (a + b)(x + y) = a2 + b2.
Eq 1: (a-b)x + (a+b)y = a2-2ab-b2.
Eq 2 simplifies to: (a+b)x + (a+b)y = a2+b2.
Subtract Eq 1 from Eq 2:
[(a+b) - (a-b)]x = (a2+b2) - (a2-2ab-b2)
2bx = 2ab + 2b2 => 2bx = 2b(a+b) => x = a+b.
Substitute x in Eq 2: (a+b)(a+b) + (a+b)y = a2+b2.
a2+2ab+b2 + (a+b)y = a2+b2 => (a+b)y = -2ab => y = -2ab/(a+b).
Answer: x = a+b, y = -2ab/(a+b).
14. 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 the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Let speed of car from A = x km/hr, from B = y km/hr. Assume x > y.
Same direction: Relative speed = x-y. Distance = 100 km. Time = 5 hr.
100 = (x-y) * 5 => x-y = 20.
Opposite direction: Relative speed = x+y. Distance = 100 km. Time = 1 hr.
100 = (x+y) * 1 => x+y = 100.
Add the two equations: 2x = 120 => x = 60.
Then y = 40.
Answer: Speed of cars are 60 km/hr and 40 km/hr.
15. In a cyclic quadrilateral ABCD, ∠A = (2x – 1)°, ∠B = (y + 5)°, ∠C = (2y + 15)° and ∠D = (4x – 7)°. Find the four angles.
In a cyclic quadrilateral, opposite angles are supplementary (∠A+∠C=180, ∠B+∠D=180).
Eq 1: (2x-1) + (2y+15) = 180 => 2x+2y = 166 => x+y = 83.
Eq 2: (y+5) + (4x-7) = 180 => 4x+y = 182.
Subtract Eq 1 from Eq 2: 3x = 99 => x = 33.
Substitute x: 33+y=83 => y=50.
Angles: ∠A = 2(33)-1=65°, ∠B = 50+5=55°, ∠C = 2(50)+15=115°, ∠D = 4(33)-7=125°.
Answer: ∠A=65°, ∠B=55°, ∠C=115°, ∠D=125°.
16. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Let numbers be x and y.
Eq 1: x+y=16.
Eq 2: 1/x + 1/y = 1/3 => (x+y)/(xy) = 1/3.
Substitute x+y=16 into the simplified Eq 2: 16/(xy) = 1/3 => xy = 48.
We need two numbers that sum to 16 and have a product of 48. These are 12 and 4.
Answer: The numbers are 12 and 4.
17. Solve: xa – yb = 0 and ax + by = a2 + b2.
From first eq, x/a = y/b => x = ay/b.
Substitute into second eq: a(ay/b) + by = a2+b2.
a2y/b + by = a2+b2. Multiply by b:
a2y + b2y = b(a2+b2) => y(a2+b2) = b(a2+b2) => y = b.
Substitute y back: x = a(b)/b = a.
Answer: x = a, y = b.
18. A man has only 20 paise coins and 25 paise coins in his purse. If he has 50 coins in all totaling ₹11.25, how many coins of each kind does he have?
Let number of 20p coins = x, 25p coins = y. Total money in paise = 11.25 * 100 = 1125.
Eq 1: x+y = 50.
Eq 2: 20x + 25y = 1125 (divide by 5) => 4x + 5y = 225.
Multiply Eq 1 by 4: 4x + 4y = 200.
Subtract this from the simplified Eq 2: y = 25.
Substitute y: x+25=50 => x=25.
Answer: He has 25 coins of each kind.
19. If x, y are real numbers such that 4x + 3y = 41 and 3x – y = 3, find the value of x – y.
We have 4x + 3y = 41 and 3x - y = 3.
From second eq, y = 3x - 3. Substitute into first eq:
4x + 3(3x-3) = 41 => 4x + 9x - 9 = 41 => 13x = 50 => x = 50/13.
y = 3(50/13) - 3 = 150/13 - 39/13 = 111/13.
x - y = 50/13 - 111/13 = -61/13.
Answer: x – y = -61/13.
20. Solve for x and y: xa + yb = a+b and xa2 + yb2 = 2.
Eq 1: bx + ay = ab(a+b).
Eq 2: b2x + a2y = 2a2b2.
From Eq 1, x = (ab(a+b) - ay)/b. Substitute into Eq 2 (this is complex). Let’s use elimination.
Multiply Eq 1 by b: b2x + aby = ab2(a+b).
Subtract this from Eq 2:
(a2y - aby) = 2a2b2 - ab2(a+b)
ay(a-b) = ab2[2a - (a+b)] = ab2(a-b).
ay = ab2 => y = b2.
Substitute y into simplified Eq 1: x/a + b2/b = a+b => x/a + b = a+b => x/a = a => x = a2.
Answer: x = a2, y = b2.
