Preparing for the Odisha Teacher Eligibility Test (OTET) Paper-II (Science) requires a strong grasp of mathematics concepts relevant to the secondary school curriculum. Here, we present a curated selection of 30 multiple-choice questions (MCQs) covering various topics in mathematics. These questions are designed to help you assess and reinforce your understanding in preparation for the exam.
Top 30 Mathematics MCQS For OTET Paper-II(Science) Exam
- What is the main purpose of evaluation in the mathematics classroom?
(a) To grade students.
(b) To identify weak students.
(c) To improve instruction.
(d) To keep a record of student progress.
Ans. (c) To improve instruction. - Which type of evaluation is done at the end of a unit or term?
(a) Diagnostic evaluation
(b) Formative evaluation
(c) Summative evaluation
(d) Continuous evaluation
Ans. (c) Summative evaluation - What does high performance through rote memorization indicate?
(a) Good understanding of concepts
(b) Effective learning
(c) Superficial learning
(d) Strong problem-solving skills
Ans. (c) Superficial learning - What kind of evaluation is used to diagnose learning difficulties?
(a) Formative evaluation
(b) Summative evaluation
(c) Diagnostic evaluation
(d) Continuous evaluation
Ans. (c) Diagnostic evaluation - Which activity is not included in evaluation?
(a) Giving tests
(b) Asking questions in class
(c) Checking homework
(d) Only grading assignments
Ans. (d) Only grading assignments - What should a teacher assess before planning a lesson?
(a) The content difficulty
(b) Student attendance
(c) Previous knowledge and learning levels of students
(d) Available teaching aids
Ans. (c) Previous knowledge and learning levels of students - Why is formative evaluation important during instruction?
(a) To finalize grades
(b) To diagnose learning difficulties
(c) To teach new concepts
(d) To evaluate the final learning outcome
Ans. (b) To diagnose learning difficulties - How does evaluation reflect on teaching?
(a) It shows how much content was covered.
(b) It indicates the effectiveness of the teaching methods.
(c) It records student attendance.
(d) It monitors homework completion.
Ans. (b) It indicates the effectiveness of the teaching methods. - What does evaluation for enhancement of learning involve?
(a) Only grading tests
(b) Reinforcing earlier learning
(c) Ignoring informal evaluations
(d) Focusing solely on new concepts
Ans. (b) Reinforcing earlier learning - How can students participate in the evaluation process?
(a) By preparing test items
(b) By skipping difficult questions
(c) By only attending tests
(d) By memorizing answers
Ans. (a) By preparing test items - If l, m and n are prime numbers, such that l<m<n<13. In how many cases (l + m + n) is also a prime number?
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: The prime numbers less than 13 are 2, 3, 5, 7 and 11.
Also, l<m<n<13 and (l+m+n) is a prime number.
Hence, only two possible pairs exist i.e., (3, 5, 11) and (5, 7, 11). - If A, A+2 and A+4 are prime numbers, then how many possible solutions do A have?
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: When A is a prime number, there is only one possible case that A, A+2 and A+4 are prime numbers, i.e., 3,5,7. Therefore, number of possible solutions is 1. - Every prime number of the form 3x + 1 can be represented in the form 6y + 1 where, x and y are integers, then what should be the values of x?
a) Odd number
b) Perfect square
c) Natural number
d) Even number
Answer: d
Explanation: Every prime number of the form 3x + 1 can be represented in the form 6y + 1, only when x is even.
For example, when x = 2, 3x + 1 i.e., 7 which is a prime number can also be represented by 6y + 1, where y=1.
When x = 4, 3x + 1 i.e., 13 which is a prime number can also be represented by 6y + 1, where y=2.
advertisement - In how many ways the number 3234 can be written as the product of two numbers which are co-prime to each other?
a) 8
b) 4
c) 16
d) 32
Answer: a
Explanation: Formula is 2n-1, where n is the distinct primes in the factorized form of the number.
3234 = 2*3*72*11.
Here, n=4.
So, 3234 can be written as the product of 2 co-prime numbers in 24-1 = 8 ways. - How many integers between 201 and 401 are co-prime to 20?
a) 70
b) 60
c) 80
d) 90
Answer: c
Explanation: 20 = 22 * 5
So, we need to find the numbers which do not have 2 or 5 as factors.
Numbers between 201 and 401 which are multiple of 2 = 200/2 = 100.
Numbers between 201 and 401 which are multiple of 5 = 200/5 = 40.
Now we have to find numbers which are multiples of both 2 and 5, i.e., multiple of 10, as these numbers would be counted twice while counting the numbers divisible by 2 and 5.
Numbers between 201 and 401 which are multiple of 10 = 200/10 = 20.
So, total number of integers between 201 and 401 are co-prime to 20 is = 200 – (100 + 40 – 20) = 80. - Find the sum of first 75 odd natural numbers.
a) 5625
b) 6425
c) 6525
d) 7225
Answer: a
Explanation: We know that, sum of first n odd natural number is given by n2.
Therefore, the sum of first 75 odd natural numbers = 752 = 5625. - Find the sum of first 90 even natural numbers.
a) 8110
b) 8100
c) 8190
d) 8180
Answer: c
Explanation: We know that, sum of first n even natural number is given by n2+n.
Therefore, the sum of first 90 even natural numbers = 902+90 = 8190. - Find the total number of three-digit numbers with unit digit 7 and divisible by 11.
a) 9
b) 8
c) 7
d) 6
Answer: b
Explanation: Three-digit numbers with unit digit 7 and divisible by 11 are 187, 297, 407, 517, 627, 737, 847 and 957.
Hence, the total number of three-digit numbers with unit digit 7 and divisible by 11 is 8. - If B is the set of squares of natural numbers and m & n are any two elements of B, then which of the following statement is correct?
a) m+n belongs to B
b) m-n belongs to B
c) m/n belongs to B
d) m*n belongs to B
Answer: d
Explanation: We know that product of two square numbers is also a square number.
For example, 25*36=90 and 81*100=8100 which are also square numbers.
Hence, m*n belongs to B. - If 2p+3q=25 and 2p+2-3q+1=37, then find the value of p and q?
a) 4,3
b) 4,2
c) 3,3
d) 3,5
Answer: b
Explanation: 2p+3q=25 …… (i)
2p+2-3q+1=37 = 4*2p – 3*3q …… (ii)
On multiplying equation (i) by 3 and adding with equation (2), we get,
7*2p = 112
2p = 16, hence, p=4.
Therefore, q = 2. - The product of two numbers is 10 more than 5 times the sum times the sum. Find the numbers if the numbers are in a ratio 5 : 6.
a) 10, 12
b) 5, 6
c) 15, 18
d) 20, 24
Answer: a
Explanation: Let the ratios be in terms of x.
The numbers will be 5x and 6x.
The product of the numbers = 30×2
The sum of the numbers = 6x + 5x = 11x
Now it is giving that 30×2 = 5 * 11x + 10 = 55x + 10
30×2 = 55x + 10
On solving the quadratic equation, we get x = 2
The numbers are 10 and 12. - Three numbers are in a ratio 8 : 9 : 10. Which of the following can be a sum of the numbers?
a) 2582
b) 2580
c) 2590
d) 2592
Answer: d
Explanation: The only possible sum of the numbers can be any digit which is a multiple of 27.
The sum of the terms of the ratio is 27 that means that the sum of the digits will be a multiple of 27 with x being the constant. Thus, the digits must be a multiple of 27. The only option multiple of 27 is 2592. - The difference of two numbers is 21 and the sum of the numbers is 402. Find the ratio of the numbers.
a) 190, 211
b) 190.2, 211.2
c) 190.4, 211.4
d) 190.5, 211.5
Answer: d
Explanation: Let the numbers be x and y.
X – y = 21
X + y = 402
On adding both the equations we get:
2x = 423
X = 211.5 and y = 402 – x = 190.5 - 4 numbers are in a ratio 4 : 5 : 6 : 7. Find the product of the unit digits of the numbers if their sum is 242.
a) 800
b) 840
c) 880
d) 960
Answer: b
Explanation: Let the ratios be in terms of x.
The numbers will be: 4x, 5x, 6x and 7x.
The sum of the numbers = 4x + 5x + 6x + 7x = 22x = 242
X = 242 / 22 = 11
4x = 44, 5x = 55, 6x = 66 and 7x = 77
The product = 4 * 5 * 6 * 7 = 840 - The product of two numbers is 10 times their sum. Find the sum of the numbers if they are in a ratio 1 : 1.
a) 20
b) 200
c) 40
d) 400
Answer: c
Explanation: Let the number be in terms of x.
The sum = x + x = 2x
The product = x * x = x2
X2 = 2x * 10 = 20x
The only possible value of x that satisfies the condition is 20.
So, the number is 20.
The sum = 20 + 20 = 40 - The zeroes of x2–2x –8 are:
(a) (2,-4)
(b) (4,-2)
(c) (-2,-2)
(d) (-4,-4)
Answer: (b) (4,-2)
Explanation: x2–2x –8 = x2–4x + 2x –8
= x(x–4)+2(x–4)
= (x-4)(x+2)
Therefore, x = 4, -2. - What is the quadratic polynomial whose sum and the product of zeroes is √2, ⅓ respectively?
(a) 3×2-3√2x+1
(b) 3×2+3√2x+1
(c) 3×2+3√2x-1
(d) None of the above
Answer: (a) 3×2-3√2x+1
Explanation: Sum of zeroes = α + β =√2
Product of zeroes = α β = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;
x2–(α+β)x +αβ
= x2 –(√2)x + (1/3)
= 3×2-3√2x+1 - If the zeroes of the quadratic polynomial ax2+bx+c, c≠0 are equal, then
(a) c and b have opposite signs
(b) c and a have opposite signs
(c) c and b have same signs
(d) c and a have same signs
Answer: (d) c and a have same signs
Explanation:
For equal roots, discriminant will be equal to zero.
b2 -4ac = 0
b2 = 4ac
ac = b2/4
ac>0 (as square of any number cannot be negative) - The degree of the polynomial, x4 – x2 +2 is
(a) 2
(b) 4
(c) 1
(d) 0
Answer: (b) 4
Explanation: Degree is the highest power of the variable in any polynomial. - If one of the zeroes of cubic polynomial is x3+ax2+bx+c is -1, then product of other two zeroes is:
(a) b-a-1
(b) b-a+1
(c) a-b+1
(d) a-b-1
Answer: (b) b-a+1
Explanation: Since one zero is -1, hence;
P(x) = x3+ax2+bx+c
P(-1) = (-1)3+a(-1)2+b(-1)+c
0 = -1+a-b+c
c=1-a+b
Product of zeroes, αβγ = -constant term/coefficient of x3
(-1)βγ = -c/1
c=βγ
βγ = b-a+1
