Mathematics Worksheet cbse
Samadhan Academy
Mathematics Problem Worksheet
Class X | Quadratic Equations & Applications
Instructions: Attempt each problem independently first. Click the "Show Solution" button to verify your answer. Diagrams are provided where applicable.
Finding Two Positive Numbers
The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number.
Find the two numbers.
★ Solution
Let the greater number be x and the smaller number be y.
Given: x² - y² = 180 ... (1)
Also: y² = 8x ... (2)
Substituting (2) in (1):
x² - 8x = 180
x² - 8x - 180 = 0
(x - 18)(x + 10) = 0
x = 18 or x = -10 (rejected, as x is positive)
From (2): y² = 8 × 18 = 144
y = 12
Answer: The two numbers are 18 and 12.
Quadratic Equation Coefficients
Express x + 1x = 3 as a quadratic equation in the form ax² + bx + c = 0.
Find the value of a - b + c.
★ Solution
Given: x + 1x = 3
Multiplying both sides by x:
x² + 1 = 3x
x² - 3x + 1 = 0
Comparing with ax² + bx + c = 0:
a = 1, b = -3, c = 1
Therefore: a - b + c = 1 - (-3) + 1 = 1 + 3 + 1 = 5
Answer: a - b + c = 5
Rectangular Field Dimensions
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side.
Find the length of the sides.
๐ Diagram: Rectangular Field
★ Solution
Let the shorter side = x m
Then: Longer side = (x + 30) m, Diagonal = (x + 60) m
By Pythagoras theorem:
x² + (x + 30)² = (x + 60)²
x² + x² + 60x + 900 = x² + 120x + 3600
x² - 60x - 2700 = 0
(x - 90)(x + 30) = 0
x = 90 (rejecting negative value)
Shorter side = 90 m, Longer side = 90 + 30 = 120 m
Answer: Shorter side = 90 m, Longer side = 120 m
Fraction Problem
The denominator of a fraction is 2 more than the numerator. If 2 is added to both the numerator and denominator, the sum of the new and original fractions is 4635.
Find the original fraction.
★ Solution
Let the numerator = x
Then denominator = x + 2
Original fraction = xx + 2
New fraction = x + 2x + 4
Given: xx + 2 + x + 2x + 4 = 4635
Solving: 35[x(x+4) + (x+2)²] = 46(x+2)(x+4)
35[x² + 4x + x² + 4x + 4] = 46[x² + 6x + 8]
70x² + 280x + 140 = 46x² + 276x + 368
24x² + 4x - 228 = 0
6x² + x - 57 = 0
(6x + 19)(x - 3) = 0
x = 3 (rejecting negative value)
Answer: The original fraction is 35
Age Problem
Sourav's age is 3 years more than the square of Ravi's age. When Ravi reaches Sourav's current age, Sourav will be 6 years less than 13 times Ravi's current age.
Find their present ages.
★ Solution
Let Ravi's present age = x years
Then Sourav's present age = x² + 3 years
Years until Ravi reaches Sourav's current age = (x² + 3) - x = x² - x + 3
Sourav's age at that time = (x² + 3) + (x² - x + 3) = 2x² - x + 6
Given: 2x² - x + 6 = 13x - 6
2x² - 14x + 12 = 0
x² - 7x + 6 = 0
(x - 6)(x - 1) = 0
x = 6 or x = 1
If x = 1, Sourav = 4, but Ravi reaching age 4 in 3 years doesn't fit context.
So x = 6: Ravi = 6 years, Sourav = 36 + 3 = 39 years
Answer: Ravi is 6 years old and Sourav is 39 years old.
Weaving Sweaters (LCM Problem)
Ranjita, Neha, and Salma take 15, 18, and 20 days respectively to weave a sweater. They all start on the same day.
When will they all start a new sweater at the same time, and how many sweaters will they have completed in total?
★ Solution
To find when they all start a new sweater together, we need LCM of 15, 18, and 20.
Prime factorization:
15 = 3 × 5
18 = 2 × 3²
20 = 2² × 5
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 days
Sweaters completed:
Ranjita: 180 ÷ 15 = 12 sweaters
Neha: 180 ÷ 18 = 10 sweaters
Salma: 180 ÷ 20 = 9 sweaters
Total = 12 + 10 + 9 = 31 sweaters
Answer: After 180 days, total sweaters completed = 31
2-Digit Number Problem
A 2-digit number (where tens digit > unit digit) is obtained by:
- Multiplying the sum of digits by 7 and adding 3, OR
- Multiplying the difference of digits by 19 and subtracting 1
Find the 2-digit number.
★ Solution
Let tens digit = x, units digit = y (where x > y)
The number = 10x + y
From condition 1: 10x + y = 7(x + y) + 3
10x + y = 7x + 7y + 3
3x - 6y = 3
x - 2y = 1 ... (1)
From condition 2: 10x + y = 19(x - y) - 1
10x + y = 19x - 19y - 1
-9x + 20y = -1
9x - 20y = 1 ... (2)
From (1): x = 1 + 2y
Substituting in (2): 9(1 + 2y) - 20y = 1
9 + 18y - 20y = 1
-2y = -8
y = 4
Therefore: x = 1 + 2(4) = 9
Answer: The 2-digit number is 94
Ratio of Areas: Square and Circle
If the perimeter of a square is equal to the circumference of a circle.
Find the ratio of their areas.
๐ Diagram: Square and Circle Comparison
★ Solution
Let side of square = a, radius of circle = r
Given: Perimeter of square = Circumference of circle
4a = 2ฯr
a = ฯr2
Area of square = a² = ฯ²r²4
Area of circle = ฯr²
Ratio = Area of SquareArea of Circle = ฯ²r²/4ฯr² = ฯ4
Answer: Ratio of areas (Square : Circle) = ฯ : 4
Circle and Semicircle Radii
The area of a circle is equal to the perimeter of a semicircular disc of equal radius.
Find the radius.
๐ Diagram: Circle and Semicircular Disc
★ Solution
Let the common radius = r
Area of circle = ฯr²
Perimeter of semicircular disc = Half circumference + Diameter
= ฯr + 2r = r(ฯ + 2)
Given: Area of circle = Perimeter of semicircular disc
ฯr² = r(ฯ + 2)
ฯr = ฯ + 2 (dividing by r, assuming r ≠ 0)
r = ฯ + 2ฯ = 1 + 2ฯ
≈ 1 + 0.637 ≈ 1.637 units
Answer: r = ฯ + 2ฯ ≈ 1.64 units
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