Bihar School Examination Board (BSEB) is responsible for conducting matrics and intermediate examinations across the state. The Bihar board class 10 examination started on 17 February with two shifts (morning and evening), which will conclude on 25 February 2025.
This Story also Contains
The Bihar board class 10 mathematics (subject code- 110) examination is scheduled for 18 February (Tuesday), today in both the shifts from 9:30 am to 12:45 pm and 2:00 pm to 5:15 pm, respectively. The Bihar board matric mathematics examination 2025 both morning and evening shifts is over now. The level of difficulty of the Bihar board class 10 mathematics exam was moderate. Arithmetic Progression questions were dominating in the paper. Download the Bihar board class 10 maths question paper, answer key with details explanation and details analysis of BSEB matric maths exam 2025 scoring down to this page.
Below is the Bihar Board matric examination overview, which contains exam information, the number of questions, and all the necessary details.
Exam Details |
Information |
---|---|
Exam Name |
BSEB Class 10th Mathematics Exam 2025 |
Conducting Body |
Bihar School Examination Board (BSEB) |
Exam Mode |
Offline (Pen & Paper) |
Total Marks |
100 |
Section A (MCQ) |
50 |
Section B |
50 |
Exam Duration |
3 Hours 15 Minutes |
Total Questions |
138 (100+30+8) |
Question Booklet Sections |
Section A & Section B |
Passing Marks |
30% (24/80 in Theory & 6/20 in Internal) |
The BSEB Class 10 Mathematics Exam 2025 was moderate-level, with almost 45% of the questions requiring two-step calculations and formula application.
Let's have a look at the level of difficulty of secondary school examination 2025 math from the table given below:
Difficulty Level | Percentage | Description |
Easy | 40% |
Formula-based, direct substitution, minimal calculation. |
Moderate | 45% |
Two-step calculations, direct application of formulas. |
Challenging | 15% |
Concept-heavy, multi-step problem-solving required. |
The stacked bar chart below represents the number of questions asked in the BSEB Class 10 Mathematics Examination 2025, categorized by difficulty levels across different topics.
BSEB secondary school examination 2025 math is successfully over for shift 1 now. Click on the link below to download the BSEB class 10 maths question paper PDF 2025.
BSEB class 10 Mathematics Question paper |
Download PDF |
Bihar Board Class 10 Mathematics Question Paper |
Question 1. $\sin \left(90^{\circ}-A\right)=\cos A$
(A) $\sin A$
(B) $\cos A$
(C) $\tan A$
(D) $\sec A$
Question 2. If $\alpha=\beta=60^{\circ}$ then the value of $\cos (\alpha-\beta)$ is
(A) $\frac{1}{2}$
(B) 1
(C) 0
(D) 2
Question 3. If $\theta=45^{\circ}$ then the value of $\sin \theta+\cos \theta$ is
(A) $\frac{1}{\sqrt{2}}$
(B) $\sqrt{2}$
(C) $\frac{1}{2}$
(D) 1
Question 4. If $A=30^{\circ}$ then the value of $\frac{2 \tan A}{1-\tan ^2 A}$ is
(A) $2 \tan 30^{\circ}$
(B) $\quad \tan 60^{\circ}$
(C) $2 \tan 60^{\circ}$
(D) $\tan 30^{\circ}$
Question 5. If $\tan \theta=\frac{12}{5}$ then the value of $\sin \theta$ is
(A) $\frac{5}{12}$
(B) $\frac{12}{13}$
(C) $\frac{5}{13}$
(D) $\frac{12}{5}$
Question 6. $
\frac{\cos 59^{\circ}}{\sin 31^{\circ}} \times \frac{\tan 80^{\circ}}{\cot 10^{\circ}}=
$
(A) $\frac{1}{\sqrt{2}}$
(B) 1
(C) $\frac{\sqrt{3}}{2}$
(D) $\frac{1}{2}$
Question 7. If $\tan 25^{\circ} \times \tan 65^{\circ}=\sin A$ then the value of $A$ is
(A) $25^{\circ}$
(B) $65^{\circ}$
(C) $90^{\circ}$
(D) $45^{\circ}$
Question 8. If $\cos \theta=x$ then $\tan \theta=$
(A) $\frac{\sqrt{1+x^2}}{x}$
(1) $\frac{\sqrt{1-x^2}}{x}$
(C) $\sqrt{1-x^2}$
(D) $\frac{x}{\sqrt{1-x^2}}$
Question 9. $\quad\left(1-\cos ^4 \theta\right)=$
(A) $\cos ^2 \theta\left(1-\cos ^2 \theta\right)$
(B4) $\sin ^2 \theta\left(1+\cos ^2 \theta\right)$
(C) $\sin ^2 \theta\left(1-\sin ^2 \theta\right)$.
(D) $\sin ^2 \theta\left(1+\sin ^2 0\right)$
Question 10. What is the form of a point lying on $y$-axis ?
(A) $(y, 0)$
(B) $(2, y)$
(C) $(0, x)$
(D) None of these
Check the Bihar Board Matric previous year's question paper, which is one of the essential resources to score good marks in future Bihar Board matric exams.
BSEB Class 10 Maths Previous year Question Paper | Download PDF |
BSEB Class 10 Maths Question Paper 2024 | |
BSEB Class 10 Maths Question Paper 2023 | |
BSEB Class 10 Maths Question Paper 2022 | |
BSEB Class 10 Maths Question Paper 2021 | |
BSEB Class 10 Maths Question Paper 2020 |
Bihar Board class 10 maths shift 1 paper was concluded and shift 2 is going according to the scheduled time. Bihar Board class 10 mathematics Answer key of question paper set F is given below.
Bihar Board Class 10 Maths Answer Key 2025 (SET- F): Section A | |||
Question Number | Correct Option | Question Number |
Correct Option |
1 | B | 11 | C |
2 | B | 12 | A |
3 | B | 13 | C |
4 | B | 14 | B |
5 | B | 15 | B |
6 | B | 16 | A |
7 | C | 17 | D |
8 | B | 18 | B |
9 | B | 19 | C |
10 | C | 20 | C |
(Q1)
We use the complementary angle identity of trigonometry:
$
\sin \left(90^{\circ}-A\right)=\cos A
$
Comparing with the given expression, we see that:
$
\sin \left(90^{\circ}-A\right)=\cos A
$
Hence, the answer is the option (2).
(Q2.)
Given:
$
\alpha=\beta=60^{\circ}
$
We use the trigonometric identity:
$
\cos (\alpha-\beta)=\cos 0^{\circ}
$
Since $\cos 0^{\circ}=1$, we get:
$
\cos \left(60^{\circ}-60^{\circ}\right)=\cos 0^{\circ}=1
$
Hence, the answer is the option (2).
(Q3.)
Given:
$
\theta=45^{\circ}
$
We calculate:
$
\sin 45^{\circ}+\cos 45^{\circ}
$
Since,
$
\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \cos 45^{\circ}=\frac{1}{\sqrt{2}}
$
Adding these values,
$
\sin 45^{\circ}+\cos 45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}
$
Hence, the answer is the option (2),
Ans.4)
We need to evaluate:
$
\frac{2 \tan A}{1-\tan ^2 A}
$
Given $A=30^{\circ}$, we substitute $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ :
$
\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}
$
Calculating the denominator:
$
1-\frac{1}{3}=\frac{3}{3}-\frac{1}{3}=\frac{2}{3}
$
Thus,
$
\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}
$
Since $\tan 60^{\circ}=\sqrt{3}$, we conclude:
$
\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=\tan 60^{\circ}
$
Hence, the answer is the option (2).
Ans.5)
We are given:
$
\tan \theta=\frac{12}{5}
$
Using the identity:
$
\tan \theta=\frac{\sin \theta}{\cos \theta}
$
we consider a right-angled triangle where the opposite side is 12 and the adjacent side is 5 . Using the Pythagorean theorem, the hypotenuse is:
$
r=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13
$
Hence, the answer is the option (2).
Ans 6)
We simplify the given expression:
$
\frac{\cos 59^{\circ}}{\sin 31^{\circ}} \times \frac{\tan 80^{\circ}}{\cot 10^{\circ}}
$
Using Complementary Angle Identities
We know that:
$
\cos 59^{\circ}=\sin \left(90^{\circ}-59^{\circ}\right)=\sin 31^{\circ}
$
Thus,
$
\frac{\tan 80^{\circ}}{\cot 10^{\circ}}=\tan 80^{\circ} \times \tan 10^{\circ}
$
Using the identity:
$
\tan \left(90^{\circ}-x\right)=\cot x
$
we get:
$
\tan 80^{\circ}=\cot 10^{\circ}
$
So,
$
\tan 80^{\circ} \times \tan 10^{\circ}=\cot 10^{\circ} \times \tan 10^{\circ}=1
$
Hence, the answer is the option (2).
Ans.7)
We are given:
$
\tan 25^{\circ} \times \tan 65^{\circ}=\sin A
$
We know that:
$
\tan \left(90^{\circ}-x\right)=\cot x
$
Thus,
$
\tan 65^{\circ}=\cot 25^{\circ}
$
$A=90^{\circ}$
Hence, the answer is the option (3).
Ans.8)
We are given:
$
\cos \theta=x
$
From the fundamental identity of trigonometry:
$
\sin ^2 \theta+\cos ^2 \theta=1
$
Substituting $\cos \theta=x$ :
$
\begin{aligned}
& \sin ^2 \theta+x^2=1 \\
& \sin ^2 \theta=1-x^2 \\
& \sin \theta=\sqrt{1-x^2}
\end{aligned}
$
We use the definition of tangent:
$
\tan \theta=\frac{\sin \theta}{\cos \theta}
$
Substituting values:
$
\tan \theta=\frac{\sqrt{1-x^2}}{x}
$
Hence, the answer is the option (2).
Ans.9)
We start with the given expression:
$
1-\cos ^4 \theta
$
We use the identity:
$
a^2-b^2=(a-b)(a+b)
$
Rewriting $\cos ^4 \theta$ as $\left(\cos ^2 \theta\right)^2$, we get:
$
1-\cos ^4 \theta=\left(1-\cos ^2 \theta\right)\left(1+\cos ^2 \theta\right)
$
Hence, the answer is the option (2).
Ans.10)
A point lying on the $y$-axis means that its $x$-coordinate is always 0 because any point on the $y$-axis has no horizontal displacement.
Thus, the general form of a point on the $y$-axis is:
$
(0, y)
$
Hence, the answer is the option (3).
To access the complete Bihar board class 10 mathematics answer key with a proper explanation, click on the link below, which is free of charge.
BSEB Class 10 Mathematics Paper Answer Key 2025 | Download PDF |
BSEB Class 10 Mathematics Answer key with Solution |
Also Check:
Frequently Asked Questions (FAQs)
A total of 100 MCQs are present for Bihar board class 10.
A total of 6 subjects are present in the Bihar board class 10 examination.
A total of 138 questions were present from both sections with 100 multiple-choice questions.
On Question asked by student community
Hello,
The NMMS Bihar 2025 application form is open. The last date to submit the form is 7th December 2024 .
Hope it helps !
Hello! If you are looking for the Bihar Board Class 9 half-yearly exam 2025, here is the link provided by Careers360. Going through this paper will help you understand the exam pattern, important chapters, and types of questions asked. I’ll be attaching it for your reference.
https://school.careers360.com/boards/bseb/bihar-board-class-9-half-yearly-question-paper-2025-26
Hello! If you are looking for the BSEB Class 9 half yearly Science question paper 2025-26, here is the link provided by Careers360. Going through this paper will help you understand the exam pattern, important chapters, and types of questions asked. I’ll be attaching it for your reference so you can practice effectively.
https://school.careers360.com/boards/bseb/bihar-board-class-9-half-yearly-question-paper-2025-26
Hello! If you are looking for the Bihar Board NCERT Class 8 Science book, here is the link provided by Careers360. I’ll be attaching it for your reference so you can access the textbook easily and use it for your exam preparation.
https://school.careers360.com/ncert/ncert-books-for-class-8-science
https://school.careers360.com/ncert/ncert-books-for-class-8
Hey dear Ashwini!
You can download the bihar board question papers in hindi medium for all subjects from this link https://school.careers360.com/boards/bseb/bihar-board-10th-last-5-years-question-papers
For each and every subject there are question papers for nearly last 5 years which will be a great aid in exam preparation. You just need to click on the download link and then download the PDF and solve the questions from it.
Thankyou !
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters