🎯 Questions 91-100
Final Set – Most Expected Maths Questions for Class 10
Real Numbers – HCF and LCM
Find the HCF and LCM of 96 and 404 using prime factorization method. Verify that HCF × LCM = Product of the two numbers.
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✅ Solution
Step 1: Prime factorization
96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12
= 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3
= 2⁵ × 3¹
404 = 2 × 202 = 2 × 2 × 101
= 2² × 101¹
Step 2: Find HCF
HCF = Product of lowest powers of common prime factors
Common prime factor: 2
Lowest power: 2² (from 404)
HCF = 2² = 4
Step 3: Find LCM
LCM = Product of highest powers of all prime factors
Prime factors: 2, 3, 101
Highest powers: 2⁵, 3¹, 101¹
LCM = 2⁵ × 3 × 101 = 32 × 3 × 101 = 9696
Step 4: Verification
HCF × LCM = 4 × 9696 = 38784
Product of numbers = 96 × 404 = 38784
Since HCF × LCM = Product of numbers, verified.
Answer: HCF = 4, LCM = 9696
Polynomials – Zeroes and Coefficients
If α and β are zeroes of the polynomial x² – 6x + k, and α – β = 2, find the value of k.
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✅ Solution
Step 1: Use relationships between zeroes and coefficients
For quadratic polynomial x² – 6x + k:
Sum of zeroes: α + β = -(-6)/1 = 6
Product of zeroes: αβ = k/1 = k
Step 2: Use given condition α – β = 2
We have: α + β = 6 …(1)
And: α – β = 2 …(2)
Adding (1) and (2):
2α = 8 ⇒ α = 4
From (1): 4 + β = 6 ⇒ β = 2
Step 3: Find k
k = αβ = 4 × 2 = 8
Answer: k = 8
Pair of Linear Equations – Word Problem
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number.
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✅ Solution
Step 1: Define variables
Let the two-digit number be 10x + y
Where x = tens digit, y = units digit
Reversed number = 10y + x
Step 2: Form equations
Given: (10x + y) + (10y + x) = 66
11x + 11y = 66
x + y = 6 …(1)
Digits differ by 2: |x – y| = 2 …(2)
Step 3: Solve equations
Case 1: x – y = 2 …(3)
From (1) and (3):
(x + y) + (x – y) = 6 + 2 ⇒ 2x = 8 ⇒ x = 4
Then y = 6 – 4 = 2
Number = 42
Case 2: x – y = -2 ⇒ y – x = 2 …(4)
From (1) and (4):
(x + y) + (y – x) = 6 + 2 ⇒ 2y = 8 ⇒ y = 4
Then x = 6 – 4 = 2
Number = 24
Both numbers satisfy the conditions.
Answer: The number can be 42 or 24
Quadratic Equations – Nature of Roots
Find the values of k for which the quadratic equation 2x² + kx + 3 = 0 has equal roots.
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✅ Solution
Step 1: Condition for equal roots
For quadratic equation ax² + bx + c = 0 to have equal roots:
Discriminant (D) = b² – 4ac = 0
Given equation: 2x² + kx + 3 = 0
Here: a = 2, b = k, c = 3
Step 2: Apply condition
D = b² – 4ac = 0
k² – 4(2)(3) = 0
k² – 24 = 0
k² = 24
k = ±√24 = ±2√6
Answer: k = 2√6 or k = -2√6
Arithmetic Progression – Word Problem
The sum of first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term of this AP.
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✅ Solution
Step 1: Find sum of first 14 terms
Sum of first 7 terms = 63
Sum of next 7 terms (8th to 14th) = 161
Sum of first 14 terms = 63 + 161 = 224
Using sum formula: Sₙ = \(\frac{n}{2}[2a + (n-1)d]\)
Step 2: Form equations
For n = 7: \(\frac{7}{2}[2a + 6d] = 63\)
\(\frac{7}{2} × 2[a + 3d] = 63\)
7(a + 3d) = 63
a + 3d = 9 …(1)
For n = 14: \(\frac{14}{2}[2a + 13d] = 224\)
7[2a + 13d] = 224
2a + 13d = 32 …(2)
Step 3: Solve for a and d
From (1): a = 9 – 3d
Substitute in (2):
2(9 – 3d) + 13d = 32
18 – 6d + 13d = 32
18 + 7d = 32
7d = 14 ⇒ d = 2
Then a = 9 – 3(2) = 9 – 6 = 3
Step 4: Find 28th term
a₂₈ = a + 27d
a₂₈ = 3 + 27(2)
a₂₈ = 3 + 54 = 57
Answer: 28th term = 57
Coordinate Geometry – Section Formula
Find the coordinates of the point which divides the line segment joining A(-2,2) and B(2,8) into four equal parts.
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✅ Solution
Step 1: Understand the division
We need to divide AB into 4 equal parts, so we need 3 division points.
Let these points be P, Q, R such that:
AP = PQ = QR = RB
Step 2: Find point P (divides AB in 1:3)
Using section formula:
P = \(\left(\frac{mx₂ + nx₁}{m+n}, \frac{my₂ + ny₁}{m+n}\right)\)
For ratio 1:3 (m=1, n=3)
P = \(\left(\frac{1×2 + 3×(-2)}{1+3}, \frac{1×8 + 3×2}{1+3}\right)\)
= \(\left(\frac{2 – 6}{4}, \frac{8 + 6}{4}\right)\)
= \(\left(\frac{-4}{4}, \frac{14}{4}\right)\) = (-1, 3.5)
Step 3: Find point Q (midpoint, divides AB in 1:1)
Q = midpoint = \(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\)
= \(\left(\frac{0}{2}, \frac{10}{2}\right)\) = (0, 5)
Step 4: Find point R (divides AB in 3:1)
For ratio 3:1 (m=3, n=1)
R = \(\left(\frac{3×2 + 1×(-2)}{3+1}, \frac{3×8 + 1×2}{3+1}\right)\)
= \(\left(\frac{6 – 2}{4}, \frac{24 + 2}{4}\right)\)
= \(\left(\frac{4}{4}, \frac{26}{4}\right)\) = (1, 6.5)
Answer: Division points are:
P(-1, 3.5), Q(0, 5), R(1, 6.5)
Trigonometry – Height and Distance
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
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✅ Solution
Step 1: Draw diagram
Let AB be the tower, height = h m
Point C is 30 m away from foot B
∠ACB = 30°
Step 2: Apply trigonometric ratio
In right triangle ABC:
tan 30° = AB/BC
\(\frac{1}{\sqrt{3}} = \frac{h}{30}\)
h = \(\frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\) m
Using √3 ≈ 1.732:
h ≈ 10 × 1.732 = 17.32 m
Answer: Height of tower = \(10\sqrt{3}\) m ≈ 17.32 m
Circles – Tangent Theorem
Prove that the lengths of tangents drawn from an external point to a circle are equal.
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✅ Proof
Step 1: Draw figure and label
Let P be an external point to circle with centre O.
Let PA and PB be two tangents from P to the circle.
A and B are points of contact.
Join OA, OB, and OP.
Step 2: Identify right triangles
In triangles OAP and OBP:
1. OA = OB (radii of same circle)
2. OP = OP (common side)
3. ∠OAP = ∠OBP = 90° (radius ⟂ tangent at point of contact)
Step 3: Prove congruence
By RHS congruence rule:
Right angle – Hypotenuse – Side
∠OAP = ∠OBP = 90° (right angles)
OP = OP (common hypotenuse)
OA = OB (equal sides)
∴ ΔOAP ≅ ΔOBP
Corresponding parts of congruent triangles are equal:
PA = PB
Hence proved that lengths of tangents from an external point are equal.
Surface Areas & Volumes – Combination of Solids
A solid is in the shape of a cone surmounted on a hemisphere. The radius of each is 3.5 cm and total height of solid is 9.5 cm. Find the volume of the solid. [Use π = 22/7]
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✅ Solution
Given: Radius r = 3.5 cm
Total height = 9.5 cm
Step 1: Find height of cone
Height of hemisphere = radius = 3.5 cm
Height of cone = Total height – Height of hemisphere
= 9.5 – 3.5 = 6 cm
Step 2: Volume of hemisphere
Volume of hemisphere = \(\frac{2}{3}πr³\)
= \(\frac{2}{3} × \frac{22}{7} × (3.5)^3\)
= \(\frac{2}{3} × \frac{22}{7} × 42.875\)
= \(\frac{2}{3} × 22 × 6.125\)
= \(\frac{2}{3} × 134.75 = 89.83 \text{ cm}^3\)
Step 3: Volume of cone
Volume of cone = \(\frac{1}{3}πr²h\)
= \(\frac{1}{3} × \frac{22}{7} × (3.5)^2 × 6\)
= \(\frac{1}{3} × \frac{22}{7} × 12.25 × 6\)
= \(\frac{1}{3} × 22 × 10.5\)
= \(\frac{231}{3} = 77 \text{ cm}^3\)
Step 4: Total volume
Total volume = Volume of hemisphere + Volume of cone
= 89.83 + 77 = 166.83 cm³
Answer: Volume of solid ≈ 166.83 cm³
Probability – Combined Events
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
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✅ Solution – Final Question!
Step 1: Define variables
Let number of blue balls = x
Total balls = red balls + blue balls = 5 + x
Step 2: Write probabilities
P(red ball) = \(\frac{5}{5 + x}\)
P(blue ball) = \(\frac{x}{5 + x}\)
Step 3: Use given condition
Given: P(blue) = 2 × P(red)
\(\frac{x}{5 + x} = 2 × \frac{5}{5 + x}\)
Step 4: Solve equation
Since denominator (5 + x) is same and non-zero:
x = 2 × 5
x = 10
Answer: Number of blue balls = 10
🎉 Congratulations!
You have completed all 100 Most Expected Maths Questions for Class 10 CBSE.
Keep revising and practicing for best results in your exams!