(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Let the number of boys be x and the number of girls be y.
According to the problem:
x + y = 10 (Total students)
y = x + 4 (Girls are 4 more than boys)
Substituting y in the first equation: x + (x + 4) = 10 → 2x + 4 = 10 → 2x = 6 → x = 3
Then y = 3 + 4 = 7
So, number of boys = 3, number of girls = 7
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Let the cost of one pencil be ₹x and the cost of one pen be ₹y.
According to the problem:
5x + 7y = 50
7x + 5y = 46
Solving these equations:
Multiply first equation by 7 and second by 5:
35x + 49y = 350
35x + 25y = 230
Subtracting: 24y = 120 → y = 5
Substituting in first equation: 5x + 35 = 50 → 5x = 15 → x = 3
So, cost of one pencil = ₹3, cost of one pen = ₹5
(i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0
a₁/a₂ = 5/7, b₁/b₂ = -4/6 = -2/3, c₁/c₂ = 8/-9 = -8/9
Since a₁/a₂ ≠ b₁/b₂, the lines intersect at a point.
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2
Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.
(iii) 6x - 3y + 10 = 0 and 2x - y + 9 = 0
a₁/a₂ = 6/2 = 3, b₁/b₂ = -3/-1 = 3, c₁/c₂ = 10/9
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
(i) 3x + 2y = 5; 2x - 3y = 7
a₁/a₂ = 3/2, b₁/b₂ = 2/-3 = -2/3
Since a₁/a₂ ≠ b₁/b₂, the equations are consistent.
(ii) 2x - 3y = 8; 4x - 6y = 9
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = -3/-6 = 1/2, c₁/c₂ = 8/9
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the equations are inconsistent.
(iii) (3/2)x + (5/3)y = 7; 9x - 10y = 14
Rewriting first equation: 9x + 10y = 42 (multiplying by 6)
a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/-10 = -1
Since a₁/a₂ ≠ b₁/b₂, the equations are consistent.
(iv) 5x - 3y = 11; -10x + 6y = -22
a₁/a₂ = 5/-10 = -1/2, b₁/b₂ = -3/6 = -1/2, c₁/c₂ = 11/-22 = -1/2
Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are consistent.
(v) (4/3)x + 2y = 8; 2x + 3y = 12
Rewriting first equation: 4x + 6y = 24 (multiplying by 3)
a₁/a₂ = 4/2 = 2, b₁/b₂ = 6/3 = 2, c₁/c₂ = 24/12 = 2
Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are consistent.
(i) x + y = 5, 2x + 2y = 10
These equations are consistent (dependent) as the second equation is just a multiple of the first. They have infinitely many solutions.
(ii) x - y = 8, 3x - 3y = 16
a₁/a₂ = 1/3, b₁/b₂ = -1/-3 = 1/3, c₁/c₂ = 8/16 = 1/2
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the equations are inconsistent.
(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0
These equations are consistent. Solving them:
From first equation: y = 6 - 2x
Substitute in second: 4x - 2(6 - 2x) - 4 = 0 → 4x - 12 + 4x - 4 = 0 → 8x = 16 → x = 2
Then y = 6 - 2(2) = 2
Solution: x = 2, y = 2
(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = -2/-4 = 1/2, c₁/c₂ = -2/-5 = 2/5
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the equations are inconsistent.
Let the width be x meters and length be y meters.
According to the problem:
y = x + 4 (length is 4m more than width)
Half perimeter = 36 → x + y = 36
Substituting: x + (x + 4) = 36 → 2x + 4 = 36 → 2x = 32 → x = 16
Then y = 16 + 4 = 20
So, width = 16m, length = 20m
(i) intersecting lines: 3x + 2y - 7 = 0 (different ratio of coefficients)
(ii) parallel lines: 4x + 6y - 5 = 0 (same ratio of x and y coefficients but different constant)
(iii) coincident lines: 4x + 6y - 16 = 0 (same ratio for all coefficients)
First equation: x - y + 1 = 0 → y = x + 1
Second equation: 3x + 2y - 12 = 0 → 2y = 12 - 3x → y = 6 - 1.5x
Finding intersection point:
x + 1 = 6 - 1.5x → 2.5x = 5 → x = 2
Then y = 2 + 1 = 3
So intersection point is (2, 3)
First line intersects x-axis when y=0: x - 0 + 1 = 0 → x = -1 → point (-1, 0)
Second line intersects x-axis when y=0: 3x + 0 - 12 = 0 → 3x = 12 → x = 4 → point (4, 0)
Vertices of triangle: (-1, 0), (4, 0), (2, 3)
(i) x + y = 14, x - y = 4
From second equation: x = 4 + y
Substitute in first: (4 + y) + y = 14 → 4 + 2y = 14 → 2y = 10 → y = 5
Then x = 4 + 5 = 9
Solution: x = 9, y = 5
(ii) s - t = 3, s/3 + t/2 = 6
From first equation: s = 3 + t
Substitute in second: (3+t)/3 + t/2 = 6
Multiply by 6: 2(3+t) + 3t = 36 → 6 + 2t + 3t = 36 → 5t = 30 → t = 6
Then s = 3 + 6 = 9
Solution: s = 9, t = 6
(iii) 3x - y = 3, 9x - 3y = 9
From first equation: y = 3x - 3
Substitute in second: 9x - 3(3x - 3) = 9 → 9x - 9x + 9 = 9 → 9 = 9
This is always true, so the equations have infinitely many solutions.
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
From first equation: 0.2x = 1.3 - 0.3y → x = (1.3 - 0.3y)/0.2 = 6.5 - 1.5y
Substitute in second: 0.4(6.5 - 1.5y) + 0.5y = 2.3 → 2.6 - 0.6y + 0.5y = 2.3 → -0.1y = -0.3 → y = 3
Then x = 6.5 - 1.5(3) = 6.5 - 4.5 = 2
Solution: x = 2, y = 3
(v) √2x + √3y = 0, √3x - √8y = 0
From first equation: √2x = -√3y → x = -√(3/2)y
Substitute in second: √3(-√(3/2)y) - √8y = 0 → -3/√2 y - 2√2 y = 0
Multiply by √2: -3y - 4y = 0 → -7y = 0 → y = 0
Then x = 0
Solution: x = 0, y = 0
(vi) (3x)/2 - (5y)/3 = -2, x/3 + y/2 = 13/6
Multiply first equation by 6: 9x - 10y = -12
Multiply second equation by 6: 2x + 3y = 13
From second equation: 2x = 13 - 3y → x = (13 - 3y)/2
Substitute in first: 9((13 - 3y)/2) - 10y = -12
Multiply by 2: 9(13 - 3y) - 20y = -24 → 117 - 27y - 20y = -24 → -47y = -141 → y = 3
Then x = (13 - 3(3))/2 = (13 - 9)/2 = 4/2 = 2
Solution: x = 2, y = 3
Solving the equations:
2x + 3y = 11
2x - 4y = -24
Subtract second from first: (2x + 3y) - (2x - 4y) = 11 - (-24) → 7y = 35 → y = 5
Substitute in first: 2x + 3(5) = 11 → 2x + 15 = 11 → 2x = -4 → x = -2
Now, y = mx + 3 → 5 = m(-2) + 3 → 5 = -2m + 3 → 2m = -2 → m = -1
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Let the numbers be x and y.
x - y = 26
x = 3y
Substitute: 3y - y = 26 → 2y = 26 → y = 13
Then x = 3(13) = 39
Numbers are 39 and 13
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let the angles be x and y, with x > y.
x + y = 180 (supplementary)
x = y + 18
Substitute: (y + 18) + y = 180 → 2y + 18 = 180 → 2y = 162 → y = 81
Then x = 81 + 18 = 99
Angles are 99° and 81°
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
Let cost of one bat be ₹x and cost of one ball be ₹y.
7x + 6y = 3800
3x + 5y = 1750
From second equation: 3x = 1750 - 5y → x = (1750 - 5y)/3
Substitute in first: 7((1750 - 5y)/3) + 6y = 3800
Multiply by 3: 7(1750 - 5y) + 18y = 11400 → 12250 - 35y + 18y = 11400 → -17y = -850 → y = 50
Then x = (1750 - 5(50))/3 = (1750 - 250)/3 = 1500/3 = 500
Cost of one bat = ₹500, cost of one ball = ₹50
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Let fixed charge be ₹x and charge per km be ₹y.
x + 10y = 105
x + 15y = 155
Subtract first from second: (x + 15y) - (x + 10y) = 155 - 105 → 5y = 50 → y = 10
Then x + 10(10) = 105 → x + 100 = 105 → x = 5
Fixed charge = ₹5, charge per km = ₹10
For 25 km: 5 + 25(10) = 5 + 250 = ₹255
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Let the fraction be x/y.
(x+2)/(y+2) = 9/11 → 11(x+2) = 9(y+2) → 11x + 22 = 9y + 18 → 11x - 9y = -4
(x+3)/(y+3) = 5/6 → 6(x+3) = 5(y+3) → 6x + 18 = 5y + 15 → 6x - 5y = -3
From first equation: 11x = 9y - 4 → x = (9y - 4)/11
Substitute in second: 6((9y - 4)/11) - 5y = -3
Multiply by 11: 6(9y - 4) - 55y = -33 → 54y - 24 - 55y = -33 → -y = -9 → y = 9
Then x = (9(9) - 4)/11 = (81 - 4)/11 = 77/11 = 7
Fraction is 7/9
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
Let Jacob's present age be x years and his son's present age be y years.
Five years hence: x + 5 = 3(y + 5) → x + 5 = 3y + 15 → x - 3y = 10
Five years ago: x - 5 = 7(y - 5) → x - 5 = 7y - 35 → x - 7y = -30
Subtract second from first: (x - 3y) - (x - 7y) = 10 - (-30) → 4y = 40 → y = 10
Then x - 3(10) = 10 → x - 30 = 10 → x = 40
Jacob's present age = 40 years, son's present age = 10 years
(i) x + y = 5 and 2x - 3y = 4
Elimination method:
Multiply first equation by 3: 3x + 3y = 15
Add to second: (3x + 3y) + (2x - 3y) = 15 + 4 → 5x = 19 → x = 19/5
Substitute in first: 19/5 + y = 5 → y = 5 - 19/5 = 6/5
Solution: x = 19/5, y = 6/5
Substitution method:
From first equation: y = 5 - x
Substitute in second: 2x - 3(5 - x) = 4 → 2x - 15 + 3x = 4 → 5x = 19 → x = 19/5
Then y = 5 - 19/5 = 6/5
(ii) 3x + 4y = 10 and 2x - 2y = 2
Elimination method:
Multiply second equation by 2: 4x - 4y = 4
Add to first: (3x + 4y) + (4x - 4y) = 10 + 4 → 7x = 14 → x = 2
Substitute in first: 3(2) + 4y = 10 → 6 + 4y = 10 → 4y = 4 → y = 1
Solution: x = 2, y = 1
Substitution method:
From second equation: 2x - 2y = 2 → x - y = 1 → x = 1 + y
Substitute in first: 3(1 + y) + 4y = 10 → 3 + 3y + 4y = 10 → 7y = 7 → y = 1
Then x = 1 + 1 = 2
(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
Elimination method:
Rewrite second equation: 9x - 2y = 7
Multiply first equation by 3: 9x - 15y = 12
Subtract from second: (9x - 2y) - (9x - 15y) = 7 - 12 → 13y = -5 → y = -5/13
Substitute in first: 3x - 5(-5/13) = 4 → 3x + 25/13 = 4 → 3x = 4 - 25/13 = 27/13 → x = 9/13
Solution: x = 9/13, y = -5/13
Substitution method:
From first equation: 3x = 5y + 4 → x = (5y + 4)/3
Substitute in second: 9((5y + 4)/3) = 2y + 7 → 3(5y + 4) = 2y + 7 → 15y + 12 = 2y + 7 → 13y = -5 → y = -5/13
Then x = (5(-5/13) + 4)/3 = (-25/13 + 52/13)/3 = (27/13)/3 = 9/13
(iv) x/2 + 2y/3 = -1 and x - y/3 = 3
Elimination method:
Multiply first equation by 6: 3x + 4y = -6
Multiply second equation by 3: 3x - y = 9
Subtract second from first: (3x + 4y) - (3x - y) = -6 - 9 → 5y = -15 → y = -3
Substitute in second: x - (-3)/3 = 3 → x + 1 = 3 → x = 2
Solution: x = 2, y = -3
Substitution method:
From second equation: x = 3 + y/3
Substitute in first: (3 + y/3)/2 + 2y/3 = -1 → 3/2 + y/6 + 2y/3 = -1
Multiply by 6: 9 + y + 4y = -6 → 5y = -15 → y = -3
Then x = 3 + (-3)/3 = 3 - 1 = 2
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let the fraction be x/y.
(x+1)/(y-1) = 1 → x + 1 = y - 1 → x - y = -2
x/(y+1) = 1/2 → 2x = y + 1 → 2x - y = 1
Subtract first from second: (2x - y) - (x - y) = 1 - (-2) → x = 3
Substitute in first: 3 - y = -2 → y = 5
Fraction is 3/5
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Let Nuri's present age be x years and Sonu's present age be y years.
Five years ago: x - 5 = 3(y - 5) → x - 5 = 3y - 15 → x - 3y = -10
Ten years later: x + 10 = 2(y + 10) → x + 10 = 2y + 20 → x - 2y = 10
Subtract first from second: (x - 2y) - (x - 3y) = 10 - (-10) → y = 20
Substitute in first: x - 3(20) = -10 → x - 60 = -10 → x = 50
Nuri's present age = 50 years, Sonu's present age = 20 years
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let the tens digit be x and units digit be y.
x + y = 9
The number is 10x + y
Reversed number is 10y + x
9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x - 11y = 0 → 8x - y = 0
Add to first: (x + y) + (8x - y) = 9 + 0 → 9x = 9 → x = 1
Then 1 + y = 9 → y = 8
The number is 18
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
Let number of ₹50 notes be x and number of ₹100 notes be y.
x + y = 25
50x + 100y = 2000 → x + 2y = 40
Subtract first from second: (x + 2y) - (x + y) = 40 - 25 → y = 15
Then x + 15 = 25 → x = 10
₹50 notes = 10, ₹100 notes = 15
(v) 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 seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Let fixed charge for first 3 days be ₹x and charge per extra day be ₹y.
For Saritha (7 days): x + 4y = 27
For Susy (5 days): x + 2y = 21
Subtract second from first: (x + 4y) - (x + 2y) = 27 - 21 → 2y = 6 → y = 3
Then x + 2(3) = 21 → x + 6 = 21 → x = 15
Fixed charge = ₹15, charge per extra day = ₹3