Important Questions For Class 10 Maths - 2025-26

1. Use Euclid’s division lemma to show that the square of any positive integer is either of form $3m$ or $3m+1$ for some integer $m$.

(Hint: Let $x$ be any positive integer then it is of the form $3q,3q+1$ or $3q+2$. Now square each of these and show that they can be rewritten in the form $3m$ or $3m+1$.)

Ans: Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$  …..(1)

In the expression (1), we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.    ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.    ……(3)

Squaring both sides of equation (3) we get,

${{a}^{2}}={{(3q)}^{2}}$ or ${{(3q+1)}^{2}}$ or ${{(3q+2)}^{2}}$ 

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$  …..(4)

Taking $3$ common from LHS of equation (4) we get,

${{a}^{2}}=3\times (3{{q}^{2}})$ or $3(3{{q}^{2}}+2q)+1$ or $3(3{{q}^{2}}+4q+1)+1$.   …..(5)

Equation (5) could be written as 

${{a}^{2}}=3{{k}_{1}}$ or $3{{k}_{2}}+1$ or $(3{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $3m$ or $3m+1$.

2. Find the value of ‘$k$ ’ such that the quadratic polynomial ${{x}^{2}}-(k+6)x+2(2k+1)$ has sum of the zeros is half of their product.

Ans: It is given that, Sum of the zeros $=\dfrac{1}{2}\times $ Product of the zeros.

From the given quadratic polynomial, ${{x}^{2}}-(k+6)x+2(2k+1)$, the sum of the zeros is $(k+6)$ and product of the zeros is $2(2k+1)$.

Hence,

$\Rightarrow (k+6)=\dfrac{1}{2}[2(2k+1)]$

$\Rightarrow k+6=2k+1$

$\Rightarrow k=5$.

3. The sum of the areas of two squares is 468m2. If their perimeters differ from 24cm, find the sides of the two squares. 

Ans:

Let, the side of the larger square be x. 

Let, the side of the smaller square be y. 

${{x}^{2}}+{{y}^{2}}=468 $ 

Cond. II 4x-4y = 24 

$ \Rightarrow xy=6 $ 

$ \Rightarrow x=6+y $ 

$ \Rightarrow {{x}^{2}}+{{y}^{2}}=468 $ 

$ \Rightarrow {{\left( 6+y \right)}^{2}}+{{y}^{2}}=\text{ }468 $ 

on solving we get 

y = 12 

⇒ x = (12+6) = 18 m 

∴ The lengths of the sides of the two squares are 18m and 12m. 

4.If the \[{{p}^{th}}\] , \[{{q}^{th}}\] and \[{{r}^{th}}\] term of an AP is \[x,y\] and \[z\] respectively, show that \[x\left( q-r \right)+y\left( r-p \right)+z\left( p-q \right)=0\] .

Ans: \[{{\text{p}}^{th}}\text{term}\Rightarrow x=A+\left( p-1 \right)D\]

\[{{q}^{th}}\text{term}\Rightarrow y=A+\left( q-1 \right)D\]

\[{{r}^{th}}\text{term}\Rightarrow z=A+\left( r-1 \right)D\]

We have to prove:

\[x\left( q-r \right)+y\left( r-p \right)+z\left( p-q \right)=0\] 

\[\Rightarrow \left\{ A+\left( p-1 \right)D \right\}\left( q-r \right)+\left\{ A+\left( q-1 \right)D \right\}\left( r-p \right)~+\left\{ A+\left( r-1 \right)D \right\}\left( p-q \right)~=0\]

\[\Rightarrow A\left\{ \left( q-r \right)+\left( r-p \right)+\left( p-q \right) \right\}+D\{\left( p-1 \right)\left( q-r \right)~+\left( r-1 \right)\left( r-p \right)+\left( r-1 \right)\left( p-q \right)\}=0\]

\[\Rightarrow A\cdot 0+D\{p\left( q-r \right)+q\left( r-p \right)+r\left( p-q \right)-\left( q-r \right)\left( r-p \right)-\left( p-q \right)\}=0\]

\[\Rightarrow A\cdot 0+D\cdot 0=0\]

\[\therefore LHS=RHS\]

Hence proved. 

5. If  $ \mathbf{\vartriangle ABC\sim\vartriangle DEF} $ and  $ \mathbf{AB=5~\text{cm}, }$ area  $ \mathbf{(\Delta ABC)=20~\text{c}{{\text{m}}^{2}}, }$ area  $\mathbf{ (\Delta DEF)=45~\text{c}{{\text{m}}^{2}}, }$ then  $ \mathbf{\text{DE}= }$ 

a. $ \mathbf{\dfrac{4}{5}~\text{cm}} $ 

b. $ \mathbf{7.5~\text{cm} }$ 

c. $ \mathbf{8.5~\text{cm}} $ 

d. $ \mathbf{7.2~\text{cm}} $ 

Ans: (b)  $ 7.5~\text{cm} $ 

The ratio of the area of a triangle is equal to the square the ratio of the corresponding sides

Therefore,  $ \sqrt{\dfrac{(\Delta ABC)}{(\Delta DEF)}}=\dfrac{AB}{DE} $ 

$ \sqrt{\dfrac{20}{45}}=\dfrac{5}{DE}\Rightarrow DE=7.5 $ 

$ \dfrac{(\Delta ABC)}{(\Delta DEF)}=\dfrac{20}{45}=\dfrac{5}{7.5} $ 

6. Evaluate the following equation:                                                                                                       

\[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Ans:

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then,

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]                   

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$

7. The perimeter of a sector of a circle of radius $8$cm is $25$m, what is area of sector?

a. $50c{{m}^{2}}$ 

b. $42c{{m}^{2}}$ 

c. $52c{{m}^{2}}$

d. none of these

Ans: Given radius= 8 cm and perimeter of sector=25 cm

$ \text{Perimeter of a sector of circle=}\left( \dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r \right)+2r $

$ \Rightarrow 25=\left[ \dfrac{\theta }{{{360}^{\circ }}}\times 2\pi \left( 8 \right) \right]+2\left( 8 \right) $

$ \Rightarrow 25=\dfrac{\theta }{{{360}^{\circ }}}\pi \times 16+16 $

$ \Rightarrow 25-16=\dfrac{\theta }{{{360}^{\circ }}}\pi \times 16 $ 

$ \Rightarrow \dfrac{9}{16}=\dfrac{\theta }{{{360}^{\circ }}}\pi $

$ \text{Area of a sector of a circle=}\dfrac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}}$

$ =\left( \dfrac{\theta }{{{360}^{\circ }}}\pi \right)\times {{r}^{2}} $

$ =\dfrac{9}{16}\times {{\left( 8 \right)}^{2}} $

$ =\dfrac{9}{16}\times 64 $

$ =36 $

Hence, (d) none of these.

8. A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is $\sqrt{6}:\sqrt{\pi }$.

Ans: Given that a sphere and cube have equal surface areas.

We get

$\Rightarrow 4\pi {{r}^{2}}=6{{a}^{2}}$

$\Rightarrow r=\sqrt{\dfrac{6{{a}^{2}}}{4\pi }}$ 

Now, the ratio of the volumes is

$\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}^{3}}}{{{a}^{3}}}$ 

$\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{4}{3}\pi {{\left( \sqrt{\dfrac{6{{a}^{2}}}{4\pi }} \right)}^{3}}}{{{a}^{3}}}$

$\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{4}{3}\pi \times {{a}^{3}}{{\left( \sqrt{\dfrac{6}{4\pi }} \right)}^{3}}}{{{a}^{3}}}$

$\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{4}{3}\pi {{\left( \sqrt{\dfrac{6}{4\pi }} \right)}^{3}}$

\[\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{4}{3\times {{2}^{3}}}\pi {{\left( \sqrt{\dfrac{6}{\pi }} \right)}^{3}}\]

\[\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{3\times 2}\pi {{\left( \sqrt{\dfrac{6}{\pi }} \right)}^{3}}\]

\[\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\pi }{6}{{\left( \dfrac{6}{\pi } \right)}^{\dfrac{3}{2}}}\]

\[\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\dfrac{6}{\pi }}\]

Therefore, the volume of the sphere to that of the cube is $\sqrt{6}:\sqrt{\pi }$.

9. In the fig points ${A},{B},{C},$ and ${D}$ are the centres of four circles. each having a radius of ${1}$ unit. If a point is chosen at random from the interior of a square \[{ABCD}\]. What is the probability that the point will be chosen from the shaded region?

Ans: It is given that the radius of the circle is $1$ unit.

Therefore, the area of the circle $=$ Area of $4$ sector.

Thus, the side of the square \[ABCD\] is \[2\] units.

Therefore, the area of square \[=\text{2 }\!\!\times\!\!\text{ 2}=\text{4}\] units.

So, the area of the shaded region

$=$ area of square $-\text{ }4\times $ area of the sectors.

$=4-\pi $.

Hence, the required probability $=\frac{4-\pi }{4}$.

10. A game consists of tossing a one rupee coin $3$  times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans: Hanif will win if he gets 3 heads and 3 tails consecutively.

The probability of Hanif losing the game, = The probability of not getting 3 heads and 3 tails.

The possible outcomes of the tosses, $(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)$ 

The total number of outcomes is 8.

Thus, the probability of not getting 3 heads and 3 tails, 

$=1-\frac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}$.

Here are some important questions for Class 10 Maths. For better understanding of each chapter, please go through the Chapter-wise Important Questions table. This resource will assist you in understanding the key concepts and important questions in each chapter and preparing effectively for your exams.

Why Should You Practise Vedantu’s Important Questions for Class 10 Maths?

Scoring high in Maths is not about studying more; it’s about practising smartly. The Important Questions for Class 10 Maths help you focus on high-weightage concepts, repeated problem types, and exam-oriented questions from every chapter. By solving these Maths Important Questions Class 10, you get clarity on formulas, stepwise marking, and common mistakes students usually make in board exams.

How Does It Help to Improve Your Class 10 Math Preparation?

Practising PYQ Class 10 Maths gives you real exam exposure. When you solve these questions, you understand question patterns, frequently tested topics, and the level of difficulty expected in the board exam. Combining Class 10 Maths Important Questions with previous year questions strengthens your confidence and improves your speed and accuracy.

Advantages of Solving Maths Important Questions Class 10 Regularly

• Improves conceptual clarity and problem-solving skills

• Enhances time management for the 3-hour exam

• Builds confidence through repeated practice

• Helps in understanding stepwise answer presentation

• Covers frequently asked and high-scoring topics

Regular practice of Important Questions for Class 10 Maths alone ensures complete preparation. 

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