CBSE 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.

How Does Class 10 Maths Important Questions Help You with Exams?

• Important questions often include application-based problems, which require students to think critically and apply theoretical concepts in real-life scenarios. 

• This not only enhances problem-solving skills but also deepens the understanding of mathematical concepts.

• These questions are carefully curated to reflect the type of problems examiners may focus on. 

• Practising them gives students insights into how examiners test understanding, helping them better anticipate what to expect.

• Important questions often highlight recurring patterns in past exams, helping students prioritise topics that have higher weight or are frequently asked. This ensures strategic and efficient revision.

• These questions cover a range of difficulty levels, allowing students to prepare for both basic and advanced problems. This balanced exposure equips them to handle the variety of questions in the exam confidently.

• Important questions often come with detailed solutions that guide students through each step. This not only teaches them the correct approach but also helps them avoid common errors that could lead to losing marks.

Focusing on important questions in Class 10 Maths plays a vital role in exam preparation. These questions not only help students identify and understand key concepts but also enhance their problem-solving and critical-thinking skills. By practising these targeted questions, students can improve their confidence, manage their time effectively, and pinpoint areas that require further attention.

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