Are you looking for solution of Class 10 Important Maths Questions for 2026 Exams? Here you’ll find complete, step-by-step solutions for every question. Designed for both students and teachers, these answers make board exam revision easier with clear explanations, formula highlights, and downloadable PDF options.
Class 10 Important Maths Questions for 2026 Exams ⇒
Q 1: Which of the following cannot be the unit digit of 8^n where n is a natural number?
a) 4 b) 2 c) 0 d) 6
Solution:
To solve this, we need to check the cyclicity of the powers of 8. This means we look at the pattern of the last digit (unit digit) as we increase the power n.
Step 1: Calculate the powers of 8 for natural numbers
n = 1, 2, 3, 4, \dots
- For n = 1: 8^1 = 8 (Unit digit is 8)
- For n = 2: 8^2 = 64 (Unit digit is 4)
- For n = 3: 8^3 = 512 (Unit digit is 2)
- For n = 4: 8^4 = 4096 (Unit digit is 6)
- For n = 5: 8^5 = 32768 (Unit digit repeats to 8)
Step 2: Analyze the pattern.
The cycle of unit digits for powers of 8 is: 8, 4, 2, 6.
This pattern repeats after every 4th power.
Step 3: Compare with the given options.
- Option (a) is 4: This is possible (e.g., 8^2).
- Option (b) is 2: This is possible (e.g., 8^3).
- Option (d) is 6: This is possible (e.g., 8^4).
- Option (c) is 0: The unit digit 0 is never obtained in powers of 8. For a number to end in 0, it must have prime factors of both 2 and 5. The number 8 has only 2 as a prime factor (2^3).
Final Answer: The correct option is (c). The unit digit cannot be 0.
Q2: If the zeros of the polynomial ax^{2}+bx+\frac{2a}{b} are reciprocal of each other, then find b.
Solution:
We use the relationship between the coefficients and the product of zeros.
Step 1: Identify the coefficients of the quadratic polynomial.
Comparing the given polynomial P(x) = ax^2 + bx + \frac{2a}{b} with the standard form Ax^2 + Bx + C:
Coefficient of x^2 (A) = a
Constant term (C) = \frac{2a}{b}
Step 2: Apply the property of reciprocal zeros.
Let one zero be \alpha.
Since the other zero is its reciprocal, it will be \frac{1}{\alpha}.
We know that: \text{Product of zeros} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}
Step 3: Substitute the values.
\alpha \cdot \frac{1}{\alpha} = \frac{\frac{2a}{b}}{a}
Step 4: Simplify the equation.
1 = \frac{2a}{b \cdot a}
1 = \frac{2}{b}
b = 2
Final Answer: The value of \boxed{ b = 2}.
Q3: Find the value of k such that the polynomial x^{2}-(k+6)x+2(2k+1) has the sum of its zeros equal to half of their product.
Solution:
We need to use the relationship between coefficients and zeros of a quadratic polynomial.
Step 1: Compare with standard form ax^2 + bx + c.
- a = 1
- b = -(k+6)
- c = 2(2k+1)
Step 2: Write the formulas for sum and product of zeros.
\text{Sum of zeros} (\alpha + \bets) = \frac{-b}{2} = \frac{-(-(k+6))}{1} = k+6
\text{Product of zeros} (\alpha \beta) = \frac{c}{a} = \frac{2(2k+1)}{1} = 4k+2
Step 3: Apply the given condition.
The problem states: Sum of zeros = \frac{1}{2} \times Product of zeros.
k+6 = \frac{1}{2} (4k+2)
Step 4: Solve for k.
k+6 = 2k+1
6 - 1 = 2k - k
k = 5
Final Answer: The value of \boxed{k = 5}.
Q4: The perimeter of a sector of a circle of radius 8 cm is 25 cm. What is the area of the sector?
Solution:
We are given the perimeter of the sector, which includes the two radii and the arc length.
Step 1: Use the perimeter formula to find Arc Length (l).
\text{Perimeter} = 2r + l
Given r = 8 and \text{Perimeter} = 25.
25 = 2(8) + 1
25 = 16 + l
l = 9 \text{ cm}
Step 2: Calculate the Area using the arc length formula.
\text{Area} = \frac{1}{2} \times l \times r
\text{Area} = \frac{1}{2} \times 9 \times 8
\text{Area} = 9 \times 4
\text{Area} = 36 \text{ cm}^2
Final Answer: The area of the sector is \boxed{36 \text{ cm}^2}.
Q5: If x(\frac{2 \tan 30^{\circ}}{1+\tan^{2}30^{\circ}}) = y(\frac{2 \tan 30^{\circ}}{1-\tan^{2}30^{\circ}}), then find x : y.
Solution:
We will solve this by substituting the standard value of \tan 30^{\circ} directly into the equation.
Step 1: Recall the trigonometric value.
We know that \tan 30^{\circ} = \frac{1}{\sqrt{3}}.
Therefore, \tan^2 30^{\circ} = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}.
Step 2: Simplify the Left Hand Side (LHS) term attached to x.
x(\frac{2 \tan 30^{\circ}}{1+\tan^{2}30^{\circ}}) = \frac{2(\frac{1}{\sqrt{3}})}{1+\frac{1}{3}}
= \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}
= \frac{2}{\sqrt{3}} \times \frac{3}{4}
= \frac{3}{2\sqrt{3}} Rationalizing the denominator (multiplying top and bottom by \sqrt{3}):
\frac{3\sqrt{3}}{2(3)} = \frac{\sqrt{3}}{2}
Step 3: Simplify the Right Hand Side (RHS) term attached to y.
\frac{2 \tan30^{\circ}} {1-\tan^{2} 30^{\circ}} = \frac{2(\frac{1}{\sqrt{3}})}{1-\frac{1}{3}}
= \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}
= \frac{2}{\sqrt{3}} \times \frac{3}{2}
= \frac{3}{\sqrt{3}} Rationalizing the denominator:
= \sqrt{3}
Step 4: Equate the simplified terms.
The original equation becomes:
x (\frac{\sqrt{3}}{2}) = y(\sqrt{3})
Step 5: Find the ratio x : y.
Divide both sides by \sqrt{3}:
\frac{x}{2} = y
x = 2y
\frac{x}{y} = \frac{2}{1}
Final Answer: The ratio x : y is \boxed{2 : 1}
Q6: In the adjoining figure, PQ||XY||BC, AP=2 \text{ cm}, PX=1.5 \text{ cm} and BX=4 \text{ cm}. If QY=0.75 \text{ cm} then find AQ+CY.
Solution:
Since the lines are parallel, we can apply the Basic Proportionality Theorem (Thales Theorem).
Step 1: Find AQ using PQ || XY.
In \Delta AXY, since PQ || XY:
\frac{AP}{PX} = \frac{AQ}{QY}
\frac{2}{1.5} = \frac{AQ}{0.75}
AQ = \frac{2 \times 0.75}{1.5} = \frac{1.5}{1.5} = 1 \text{ cm}
Step 2: Find CY using XY || BC.
In \Delta ABC, since XY || BC:
\frac{AX}{XB} = \frac{AY}{YC}
Note: AX = AP + PX = 2 + 1.5 = 3.5 \text{ cm}.
Also, AY = AQ + QY = 1 + 0.75 = 1.75 \text{ cm}.
\frac{3.5}{4} = \frac{1.75}{YC}
YC = \frac{4 \times 1.75}{3.5}
YC = \frac{7}{3.5} = 2 \text { cm}
Step 3: Calculate AQ + CY.
AQ + CY = 1 + 3 = 3 \text{ cm}
Final Answer: The value of AQ + CY is \boxed{3 \text{ cm}}.
Q7: If the mode of some observations is 10 and sum of mean and median is 25, then find the mean and median.
Solution:
We use the empirical relationship: \text{Mode} = 3(\text{Median}) - 2(\text{Mean}).
Step 1: Set up the equations.
Eq 1: \text{Mode} = 10
Eq 2: \text{Mean} + \text{Median} = 25 \Rightarrow \text{Mean} = 25 - \text{Median}
Step 2: Substitute into the empirical formula.
10 = 3(\text{Median}) - 2(25 - \text{Median})
10 = 3(\text{Median}) - 50 + 2(\text{Median})
60 = 5(\text{Median})
\text{Median} = 12
\text {Mean} = 25 - 12 = 13
Final Answer: Mean is \boxed{13} and Median is \boxed{12}.
Q8: Solve the following pair of equation algebraically:
101x + 102y = 304
102x+101y=305
Solution:
Since the coefficients of x and y are interchanged, we use the addition and subtraction method.
Step 1: Add the two equations.
(101x + 102y) + (102x + 101y) = 304 + 305
203x + 203y = 609
Divide by 203: x + y = 3 — (i)
Step 2: Subtract the first equation from the second.
(102x + 101y) - (101x + 102y) = 305 - 304
x - y = 1 — (ii)
Step 3: Solve (i) and (ii).
Add (i) and (ii):
2x = 4 \Rightarrow x =2
Substitute x=2 in (i):
2+ y = 3 \Rightarrow y = 1
Final Answer: \boxed{x = 2, y = 1}.
Q9: The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random is \frac{12}{23}, find the total number of balls.
Solution:
Let the number of black balls be x.
Step 1: Define the variables.
Black balls = x Red balls = x + 3 Total balls = x + (x + 3) = 2x + 3
Step 2: Form the probability equation.
P(\text{Red}) = \frac{\text{Number of Red balls}}{\text{Total balls}}
\frac{12}{23} = \frac{x + 3}{2x + 3}
Step 3: Solve for x.
12(2x+3) = 23(x+3)
24x + 36 = 23x + 69
24 - 23x = 69 -36
x = 33 (Black balls)
Step 4: Calculate total balls.
\text{Total Balls} = 2(33) + 3 = 66 + 3 = 69
Final Answer: The total number of balls is \boxed{69}.
Q10: If points A(6,1), B(p,2), C(9,4) and D(7,q) are vertices of a parallelogram ABCD, find p and q. Check if it is a rectangle.
Solution:
In a parallelogram, diagonals bisect each other.
Thus, Midpoint of AC = Midpoint of BD.
Step 1: Find Midpoint of AC.
M_{AC} = (\frac{6+9}{2}, \frac{1+4}{2}) = (\frac{15}{2}, \frac{5}{2})
Step 2: Find Midpoint of BD.
M_{BD} = (\frac{p+7}{2}, \frac{2+p}{2})
Step 3: Equate coordinates to find p and q.
For x: \frac{p+7}{2} = \frac{15}{2} \Rightarrow p+7 = 15 \Rightarrow p=8
For y: \frac{2+q}{2} = \frac{5}{2} \Rightarrow 2+q = 5 \Rightarrow q=3
So, coordinates are A(6,1), B(8,2), C(9,4), D(7,3).
Step 4: Check for Rectangle (Diagonals must be equal).
AC^2 = (9-6)^2 + (4-1)^2 = 3^2 + 3^2 = 18
BD^2 = (7-8)^2 + (3-2)^2 = (-1)^2 + 1^2 = 2
Since AC \neq BD, it is not a rectangle.
Final Answer: \boxed{p=8, q=3}. It is not a rectangle.
Q11: Find the smallest value of p for which x^{2}-2(p+1)x+p^{2}=0 has real roots. Hence find the roots.
Solution:
For real roots, the discriminant D \ge 0.
Step 1: Calculate Discriminant (D).
D = b^2 - 4ac
D = [-2(p+1)]^2 - 4(1)(p^2)
D = 4(p^2 + 2p +1) - 4p^2
D = 4p^2 + 8p + 4 - 4p^2
D = 8p + 4
Step 2: Set inequality D \ge 0.
8p + 4 \ge 0
8p \ge -4
p \ge -0.5
The smallest value is p = -0.5.
Step 3: Find roots for p = -0.5.
Equation becomes: x^2 - 2(0.5)x + 0.25 = 0
x^2 - x + 0.25 = 0
(x - 0.5)^2 = 0
x = 0.5, 0.5
Final Answer: Smallest p = -0.5. Roots are \boxed{0.5}.
Q12: From a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Solution:
The largest cone will have a base diameter equal to the side of the cube and height equal to the side of the cube.
Step 1: Dimensions.
Cube side a = 14.
Cone radius r = 7, height h = 14.
Slant height l = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5}.
Using \sqrt{5}=2.2, l = 7(2.2) = 15.4 \text{ cm}.
Step 2: Volume of Remaining Solid.
V = V_{\text{Cube}} - V_{\text{cone}}
V = (14)^3 - \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 14
V = 2744 - \frac{1}{3} (22 \times 7 \times 14)
V = \frac{6067}{3} \text{ cm}^3
Step 3: Surface Area of Remaining Solid.
SA = SA_{\text{cube}} - \text{Area}<em>{\text{base}} + CSA</em>{\text{cone}}
SA = 6a^2 - \pi r^2 + \pi rl
SA = 6(196) - \frac{22}{7}(49) + \frac{22}{7}(7)(15.4)
SA = 1176 - 154 + 338.8
SA = 1360.8 \text{ cm}^2
Final Answer: Volume \boxed{\frac{6067}{3} \text{ cm}^3}, Surface Area \boxed{1360.8 \text{ cm}^2}.
Q13: A boy whose eye level is 1.35 m from the ground spots a balloon from his eyes. At an instant, the angle of elevation is 60 degrees. After 12 seconds, the angle of elevation changes to 30 degrees. If the speed of the wind is 3 m/sec, find the height of the balloon from the ground.
Solution:
This is a standard double triangle problem.
Let the height of the balloon be H.
Let effective height above eye level be h = H - 1.35.
Step 1:
Calculate distance travelled by balloon.
\text{Distance} (d) = \text{Speed} \times \text{Time} = 3 \times 12 = 36 \text{ m}.
Step 2: Apply trigonometry.
In first triangle (60°): \tan 60^{\circ} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}.
In second triangle (30°): \tan 30^{\circ} = \frac{h}{x+36}.
\frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 36}
Step 3: Solve for h.
\frac{h}{\sqrt{3}} + 36 = h\sqrt{3}
36 = h\sqrt{3} - \frac{h}{\sqrt{3}}
36 = h(\frac{3-1}{\sqrt{3}})
36 = h(\frac{2}{\sqrt{3}})
h = 18 \sqrt{3}
Step 4: Total Height.
H = 18(1.73) + 1.35 = 31.14 + 1.35 = 32.49 \text{ m}.
Final Answer: The height of the balloon is \boxed{32.49 \text{ m}}.
Q14: Two dice are rolled bearing numbers 4,6,7,9,11,12. Find probability that product is odd.
Solution:
The product of two numbers is odd ONLY if both numbers are odd.
Step 1: Identify Odd and Even numbers on the die.
Numbers: {4, 6, 7, 9, 11, 12} Odd numbers: {7, 9, 11} (Count = 3)
Even numbers: {4, 6, 12} (Count = 3)
Total outcomes when rolling two dice = 6 \times 6 = 36.
Step 2: Find favorable outcomes.
For the product to be odd, we must pick an odd number from the first die AND an odd number from the second.
Favorable outcomes = 3 \times 3 = 9.
Step 3: Calculate probability. P(\text{Odd Product}) = \frac{9}{36} = \frac{1}{4}.
Final Answer: The probability is \boxed{\frac{1}{4}}.
Q15: Prove that the lengths of tangents drawn from an external point to a circle are equal.
Solution:
Step 1: Given and To Prove.
Given: A circle with center O and an external point P. Two tangents PQ and PR are drawn.
To Prove: PQ = PR.
Step 2: Construction. Join OP, OQ, and OR.
Step 3: Proof in \Delta OQP and \Delta ORP.
\angle OQP = \angle ORP = 90^{\circ} (Radius is perpendicular to tangent).
OP = OP (Common hypotenuse).
OQ = OR (Radii of the same circle).
By RHS Congruence Rule, \Delta OQP \cong \Delta ORP.
By CPCT, PQ = PR.
Final Answer: Hence Proved.
Q16: Prove that abscissa of a point P which is equidistant from points A(7,1) and B(3,5) is 2 more than its ordinate.
Solution:
Let point P be (x, y). Abscissa is x, Ordinate is y.
We need to prove x = y + 2.
Step 1: Apply distance formula PA = PB.
\sqrt{(x-7)^2 + (y-1)^2} = \sqrt{(x-3)^2 + (y-5)^2}
Square both sides:
(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2
Step 2: Expand the identity.
(x^2 - 14x + 49) + (y^2 -2y +1) = (x^2 - 6x + 9) + (y^2 - 10y + 25)
Step 3: Simplify.
x^2 and y^2 cancel out.
- 14x - 2y + 50 = - 6x - 10y + 34
10y - 2y + 50 - 34 = 14x - 6x
8y + 16 = 8x
y + 2 = x
Final Answer: Since x = y + 2, the abscissa is 2 more than the ordinate. Hence Proved.
Q17: If 1+\sin^{2}\theta = 3 \sin\theta \cos\theta, then prove that \tan \theta=\frac{1}{2} or 1.
Solution:
We need to convert the equation into terms of \tan \theta.
Step 1: Divide the entire equation by \cos^2 \theta.
\frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{3 \sin \theta \cos \theta}{\cos^2 \theta}
\sec^2 \theta + \tan^2 \theta = 3 \tan \theta
Step 2: Use identity \sec^2 \theta = 1 + \tan^2 \theta.
(1 + \tan^2 \theta) + \tan^2 \theta = 3 \tan \theta
2 \tan^2 \theta - 3 \tan \theta + 1 = 0
Step 3: Solve the quadratic equation for \tan \theta.
Let u = \tan \theta.
2 u^2 - 3 u + 1 = 0
2u^2 - 2u - u + 1 = 0
2u(u-1) - 1(u-1) = 0
(2u - 1)(u - 1) = 0
u = 1 or u = \frac{1}{2}.
Therefore, \tan \theta = 1 or \tan \theta = \frac{1}{2}.
Final Answer: Hence Proved.