CBSE Important Questions for Class 8 Maths - 2025-26

1. Tell what property allows you to compute 

$\dfrac{1}{3}\times(6 \times \dfrac{4}{3})$ as $(\dfrac{1}{3}\times 6)\times \dfrac{4}{3} $

Ans: Since, $a\times(b\times c)= (a \times b)\times c$.

2. Solve and check result: $5t-3=3t-5$

Ans: 

\[5t-3=3t-5\]

On Transposing \[3t\] to L.H.S and \[-3\] to R.H.S, we obtain 

\[5t-3=-5-\left( -3 \right)\]

\[2t=-2\]

On dividing both sides by\[2\], we obtain 

\[t=-1\]

L.H.S \[=5t-3=5\times \left( -1 \right)-3=-8\]

R.H.S \[=3t-5=3\times \left( -1 \right)-5=-3-5=-8\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

3. Find the measure of each exterior angle of a regular polygon of 9 Sides.

Ans:

Given: a regular polygon with $9$ sides

We need to find the measure of each exterior angle of the given polygon.

We know that all the exterior angles of a regular polygon are equal.

The sum of all exterior angles of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Therefore,

The sum of all angles of a given regular polygon $ = {360^ \circ }$

Number of sides $ = 9$

Therefore, a measure of each exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{9} $

$  = {40^ \circ } $

4. The following number is not a perfect square. Give reason. $\text{1057}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it.

We can see that $1057$has its unit place digit as $7$.

Hence, $1057$cannot be a perfect square.

5. Find the smallest number by which the following numbers must be divided to obtain a perfect cube. The number is 81.

Ans: $ 81\text{ }=\text{ }\underline{3\text{ }\times \text{ }3\text{ }\times \text{ }3}\text{ }\times \text{ }3$ . Here, one 3 is extra which is not in a triplet. Dividing 81 by 3, will make it a perfect cube. 

Thus,  $ 81\text{ }\div \text{ }3\text{ }=\text{ }27\text{ }=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3$  is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3. 

6. $72\% $ of  $25$ students are good in mathematics. How many are not good in mathematics?

Ans: Total number of students $ = $ 25.

Percentage of students are good in mathematics  $ = $$72\% $

Percentage of students who are not good in mathematics  $ = $ $(100 - 72)\% $

$ \Rightarrow 28\% $

$\therefore $ Number of students who are not good in mathematics $ = $$28\%  \times 25$

$ \Rightarrow \dfrac{{28}}{{100}} \times 25$

$ \Rightarrow \dfrac{{28}}{4}$

$ \Rightarrow 7$

Students are not good in mathematics $ = 7$

7. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$\left( {{\text{p, q}}} \right){\text{; }}\left( {{\text{10m, 5n}}} \right){\text{; }}\left( {{\text{20}}{{\text{x}}^{\text{2}}}{\text{ , 5}}{{\text{y}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{4x, 3}}{{\text{x}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{3mn, 4np}}} \right){\text{ }}$

Ans:  We know that,

Area of rectangle = length x breadth

Area of 1st rectangle = p x q = pq

Area of 2nd rectangle = ${{10m  \times  5n  =  10  \times  5  \times  m  \times  n   =  50mn}}$

Area of 3rd rectangle = ${\text{20}}{{\text{x}}^{\text{2}}}{{  \times  5}}{{\text{y}}^{\text{2}}}{{ =  20 \times 5 \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = 100}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

Area of 4th rectangle = ${{4x }} \times {\text{ 3}}{{\text{x}}^{\text{2}}}{{  =  4 \times 3}} \times {{x}} \times {{\text{x}}^2}{\text{ = 12}}{{\text{x}}^3}$

Area of 5th rectangle ${{ =  3mn  \times  4np  =  3  \times  4  \times  m  \times  n  \times  n  \times  p  =  12m}}{{\text{n}}^{\text{2}}}{\text{p}}$

8. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of the garden (Length of rectangle is ${\text{20 - (3}}{\text{.5 + 3}}{\text{.5)m}}$)?

Ans: As we have given that length of rectangle = ${\text{[20 - (3}}{\text{.5 + 3}}{\text{.5)]m}}$

=  ${\text{[20 - 7]m}}$= ${\text{13m}}$

Breadth = \[{\text{7m}}\].

Now, we have to find the circumference of both semi-circles.

As, Diameter = ${\text{7m}}$, so, Radius(r) = $\dfrac{7}{{\text{2}}}{\text{m}}$= ${\text{3}}{\text{.5m}}$

Circumference of one semicircle = ${\text{$\pi$ r}}$= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(3}}{\text{.5)m}}$= ${\text{11m}}$

Circumference of both circles = ${\text{2 x 11m}}$= ${\text{22m}}$

Now, the Perimeter of the garden = AB + CD + Length of both semi-circular regions AD & BC

= ${\text{(13 + 13 + 22)m}}$

=  ${\text{48m}}$

Area of the garden = Area of the rectangle + $2 x $Area of two semi-circular regions

= (Length x Breadth) + ${\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{$\pi$ }}{{\text{r}}^{\text{2}}}$

= ${\text{[(13 x 7) + 2 x (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)]}}{{\text{m}}^{\text{2}}}$

= ${\text{(91 + 38}}{\text{.5)}}{{\text{m}}^{\text{2}}}$

= \[{\text{129}}{\text{.5}}{{\text{m}}^{\text{2}}}\]

9. The following graph shows the temperature forecast and the actual temperature for each day of the week. 

(a) On which days was the forecast temperature the same as the actual temperature? 

(b) What was the maximum forecast temperature during the week? 

(c) What was the minimum actual temperature during the week? 

(d) On which day did the actual temperature differ the most from the forecast temperature? 

Graph Temperature  Vs Days, Image Credits-NCERT

Ans: (a) On Tuesday, Friday, and Sunday, the forecast temperature matched the actual temperature.

(b)The maximum forecast temperature during the week was $35{}^\circ C$. 

(c) The minimum actual temperature during the week was $15{}^\circ C$.

(d)On Thursday, the actual temperature differs the most from the forecast temperature. 

10. Carry out the following divisions. 

$28{{x}^{4}}\div 56x$

Ans: Write the numerator and denominator in its factors and divide. 

$28{{x}^{4}}=2\times 2\times 7\times x\times x\times x\times x$

$56x=2\times 2\times 2\times 7\times x$

$28{{x}^{4}}\div 56x=\dfrac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}$

$=\dfrac{{{x}^{3}}}{2}$

$=\dfrac{1}{2}{{x}^{3}}$

Here are some important questions for Class 8 Maths. For a deeper 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 prepare effectively for your exams.

How do Class 8 Maths Important Questions Help you with Exams?

• These important questions highlight the major themes and ideas from each chapter, ensuring students concentrate on the most critical areas.

• The questions are carefully chosen based on previous exam trends, helping students get familiar with the types of questions that are likely to appear in their exams.

• Working through these questions deepens students' understanding of the chapters, allowing them to connect with the underlying themes and meanings more effectively.

• Consistent practice with these questions trains students to manage their time efficiently during the exam by answering in a well-organised manner.

• By regularly practising these essential questions, students build confidence as they become more comfortable with the exam format and answering techniques, improving their overall performance.

The Class 8 Maths Important Questions offer a valuable tool for exam preparation. They help students focus on key themes, understand important concepts, and practice answering exam-oriented questions. By regularly solving these questions, students can improve their comprehension, time management, and overall confidence. This targeted preparation ensures a better grasp of the chapters and enhances their ability to perform well in exams.

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