Maths, Science and Pedagogy MCQ Questions | ||||
Quiz-1 | Quiz-2 | Quiz-3 | Quiz-4 | Quiz-5 |
Quiz-6 | Quiz-7 | Quiz-8 | Quiz-9 | Quiz-10 |
Quiz-11 | Quiz-12 | Quiz-13 | Quiz-14 | Quiz-15 |
Directions (Q.1 – 60): Answer the following questions by selecting the most appropriate option.
Q1. 2^{14} – 1 is divisible by
(a) 13
(b) 14
(c) 3
(d) 4
Answer: (c) Explanation: a^{n} – b^{n} is always divisible by (a + b) if n is even. 2^{14} – 1= 2^{14} – 1^{14} ; As 14 is even, 2^{14} – 1^{14} is divisible by 2 + 1= 3 |
Q2. If the prime factorisation of three numbers x, y and z is
X = 2^{3} × 3^{2} × 5^{2}
y=2^{2} × 3 × 5^{2}
z = 2^{4} × 3^{3} × 5,
(a) 2 × 3 × 5
(b) 2^{2} × 3 × 5
(c) 2^{2} × 3^{2} × 5
(d) 2^{3} × 3^{2} × 5^{2}
Answer: (b) Explanation: x = 2^{3} x 3^{2} x 5^{2} y = 2^{2} x 3 x 5^{2} z = 2^{4} x 3^{3} x 5 2 – Lowest occurrence is 2 times. 3 – Lowest occurrence is 1 time. 5 – Lowest occurrence is 1 time. H.C.F. = 2^{2} x 3 x 5 |
Q3. If $\frac{\frac{1946}{-1a}}{19b7}$ then the sum of the digits a and b is
(a) 15
(b) 14
(c) 13
(d) 11
Answer: (d) Explanation: $\frac{\frac{1946}{-1a}}{19b7}$ Here, 7 is greater than 6. There must be a carry-over of digits from 4 to 6. If 6 becomes 16 and ‘a’ is subtracted from 16, we get 7. ∴ a = 16-7= 9 Now, 4 becomes 3 and 1 is subtracted from it. ∴ b= 3-1 = 2 Hence, a + b = 9+ 2 = 11 |
Q4. Find the values of x and y.
331x + 337y=360 and 337x + 331y=308
(a) $x=\,\,-\frac{23}{6}\,\,and\,\,y\,\,=\,\,\frac{29}{6}$
(b) $x=\,\,-\frac{29}{6}\,\,and\,\,y\,\,=\,\,-\frac{23}{6}$
(c) $x=\,\,\frac{29}{6}\,\,and\,\,y\,\,=\,\,\frac{23}{6}$
(d) $x=\,\,\frac{23}{6}\,\,and\,\,y\,\,=-\,\,\frac{29}{6}$
Answer: (a) Explanation: On adding 331x + 337y = 360 and 337x + 331y = 308, We get 668x + 668y = 668 ⇒ x + y = 1 … (a) On subtracting both equations, we get –6x +6y = 52 ⇒ –3x + 3y = 26 … (b) From (a) × 3 + (b), we get 3x + 3y = 3 ⇒ y = 29/6 Now, x + y = 1 ⇒ x = $1-\frac{29}{6}=-\frac{23}{6}$ |
Q5. Find the zero of the polynomial p(x) = 2x^{2} – 8.
(a) ±2
(c) 2
(b) 3
(d) –2
Answer: (a) Explanation: Putting p(x) = 0, we have 2x^{2}-8=0 ⇒ 2x^{2} = 8 ⇒ x² = 4 ⇒ x = ±2 |
Q6. x- 2 is a factor of
(a) 2x^{2} + 3x + 2
(c) 2x^{2} –3x – 2
(b) 2x^{2} – 3x + 2
(d) 2x^{2} + 4x +2
Answer: (c) Explanation: Since x – 2 is a factor of p(x), p(b) should be equal to zero. Here, for p(x) = 2x^{2} – 3x – 2 p(b) = 0 As 2(b)^{2} – 3(b)-2 = 8-6-2 = 0 |
Q7. Find the remainder on dividing
p(x)=x^{4} + 3x^{2} – 2x + 7 by x – 3.
(a) 109
(c)-110
(b)-109
(d) -108
Answer: (a) Explanation: If p(x) is divided by x – 3, the remainder is p(c). p(c) = (c)^{4} + 3(c)^{2} – 2(c)+7 = 81 +27 – 6 +7 = 109 |
Q8. Which of the following pairs of angles has complementary angles?
(a) 47 ad 46
(c) 72 and 18
(b) 43 and 37
(d) 53 and 47
Answer: (c) Explanation: Two angles are complementary if they add up to 90 degrees. 72° + 18° = 90° |
Q9. The number of degrees in 5 right angles is
(a) 90
(c) 450
(b) 180
(d) 360
Answer: (c) Explanation: 1 right angle = 90° ∴ 5 right angles = 90° 5 = 450° |
Q10. The marks of 20 students of Class X are given below 42, 45, 50, 52, 55, 56, 60, 62, 64, 66, 68, 70, 71, 72, 74, 80, 82, 85, 90, and 92
Calculate the mean.
(a) 59.4
(c) 96.2
(b) 66.8
(d) 64.5
Answer: (b) Explanation: Mean =$\frac{\sum{x}}{N}$ = $\frac{42+45+50+52+55+56+60+62+64+66+68+70+71+72+74+80+82+85+90+92}{20}$ = $\frac{1336}{20}=66.8$ |
Q11. The following series show, marks in the economics of 11 students of Class X:
17 32 35 33 15 21 41 32 11 10 20
Find the median marks.
(a) 21
(c) 63.3
(b) 52
(d) 60
Answer: (a) Explanation: Marks of 11 students in ascending order are: 10, 11, 15, 17, 20, 21, 32, 32, 33, 35, 41 Median = Size of $\left( \frac{N+1}{2} \right) ^{th}$item Here, N = 11 M= $\left( \frac{11+1}{2} \right) ^{th}=6^{th}$ item = Size of 6th item, i.e. 21 Therefore, median = 21 |
Q12. The following graph shows that the students of a particular class watched television during holidays (in the number of hours):
For how many hours did the maximum number of students watch television?
(a) 4-5 hours
(c) 3-4 hours
(b) 6-7 hours
(d) 2-3 hours
Answer: (a) Explanation: The height of the bar corresponding to 4-5 is the maximum So, the maximum number of students watched TV for 4-5 hours |
Q13. If the area of an equilateral triangle is $4\sqrt{3}$ cm^{2}, then its perimeter is
(a) 16 cm
(c) 12 cm
(b) 14 cm
(d) 15 cm
Answer: (c) Explanation: Area of equilateral triangle = $\frac{\sqrt{3}}{4}\left( side \right) ^2$ ⇒ $4\sqrt{3}=\frac{\sqrt{3}}{4}\left( side \right) ^2$ ⇒ (Side)^{2} = 16 ⇒ Side = 4 Perimeter = 3 x Side= 12 cm |
Q14. If the length of a rectangle is decreased by x% and breadth is increased by (x+5)%, find the value of x if the area of the rectangle is not changed.
(a) 10
(c) 25
(b) 20
(d) 30
Answer: (b) Explanation: Let the original length and breadth be L and B Area of original rectangle = LB Length of new rectangle = $\frac{100-x}{100}L$ Breadth of new rectangle = $\frac{105+x}{100}B$ Area of new rectangle = $\left( \frac{100-x}{100}L \right) \left( \frac{105+x}{100}B \right) $ = $\frac{10500-5x-x^2}{10000}LB$ Now, $\frac{10500-5x-x^2}{10000}LB$ = LB ⇒ 10500 – 5x – x^{2} = 10000 ⇒ x^{2} + 5x -500 = 0 ⇒ (x + 25)(x-20) = 0 ⇒ x = 20 |
Q15. Find the volume of a cone, if its height and radius are 24 m and 14 m respectively
(a) 4928 m^{3}
(c) 4289 m^{3}
(b) 4298 m^{3}
(d) 2948 m^{3}
Answer: (a) Explanation: Volume of cone = $\frac{1}{3}\pi r^2h$ = $\frac{1}{3}\times \frac{22}{7}\times 14\times 14\times 24=4928$ m^{3} |
Q16. The value of $\frac{2}{5}\times 5\frac{1}{4}$
(a) $\frac{1}{2}$
(b) $\frac{21}{10}$
(c) $\frac{11}{10}$
(d) $\frac{15}{10}$
Answer: (b) Explanation: $\frac{2}{5}\times 5\frac{1}{4}=\frac{2}{5}\times \frac{21}{4}=\frac{21}{10}$ |
Q17. What is the value of “a'” on the given number line?
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{4}$
(d) 1
Answer: (b) Explanation: The portion between 0 and 1 is divided into 3 equal parts Therefore, a = 1/3 |
Q18. The additive inverse of S, where S = 2–3+4 –5+6 – 7+8–9 +……+ 50–51, is
(a) 0
(c)-25
(b) 50
(d) 25
Answer: (d) Explanation: We have, S= 2-3+4-5+ 6-7+8-9+…+50-51 = (2-3)+(4-5)+(6 – 7)+(8-9) +…+(50-51) = (-1)+(-1)+(-1) + up to 25^{th} term = -25 Therefore, additive inverse of S is 25. |
Q19. Which of the following are not prime factors?
(a) 2 × 4 × 5
(b) 2 × 3 × 5
(c) 7 × 5
(d) 11 × 5 × 13
Answer: (a) Explanation: Since 4 is a composite number, 2 × 4 × 5 are not prime factors. |
Q20. Which one of the following is true?
(a) $\sqrt[3]{0.8}$= 0.2
(b) $\sqrt{0.9}$ = 0.3
(c) Square root of an odd number is always an odd number.
(d) Square of an even number is always an even number.
Answer: (d) Explanation: The Square of an even number is even and the square of an odd number is odd. However, in the case of square root, it may not be true. For example, the square root of 3 is a fraction but not an odd number, and the square root of 6 is a decimal number but not an even number. |
Q21. A student is able to solve all addition and subtraction sums, but he often forgets to carry over numbers while adding and to deduct the borrowed number while subtracting. The best remedial strategy to correct these errors is
(a) Punishing him for every mistake he commits
(b) Rewarding him for every correct answer
(c) Giving him a long worksheet to practice the sums of carrying over and borrowing
(d) Asking him to continue while carrying over and borrowing and do self-checking to ensure that his answer is correct
Answer: (c) Explanation: The student knows the regrouping process after carrying over or borrowing, but he is careless, and most of the mistakes he makes are due to his absentmindedness. Though punishments and rewards may help sometimes, a habit of working carefully and ensuring that the sums are correctly solved can be developed through practice. The habit of self-checking answers helps minimise such errors. |
Q22. The nature of mathematics is
(a) Ornamental
(b) Difficult
(c) Logical
(d) Uncommon
Answer: (c) Explanation: The discipline of mathematics has a logical structure because its most fundamental concepts include numbers, symbols, shapes, algorithms, axioms, theorems, proofs, etc. The understanding of mathematics requires critical thinking and reasoning. |
Q23. Which of the following is not a central theme as stated in the position paper of the National Focus Group on “Teaching of Mathematics”?
(a) Shifting the focus of mathematics education from achieving ‘narrow’ goals to ‘higher goals
(b) Mathematisation of the student’s mind
(c) Changing the modes of assessment to examine students’ mathematisation abilities rather than procedural knowledge
(d) Enabling teachers by providing them with a variety of mathematical resources
Answer: (b) Explanation: NCERT has suggested some directions in which action needs to be taken to overcome the problems of teaching. It has grouped these directions into four central themes. The Mathematisation of a learner’s mind is not stated as a central theme in the NCF document on teaching mathematics. |
Q24. “Students’ ability to come up with a formula is more important than being able to correctly use well-known formulae”. Which learning approach do you think is best suited for this?
(a) Contextual learning approach
(b) Constructivist approach
(c) Cooperative learning approach
(d) Mastery learning approach
Answer: (b) Explanation: The given approach is a constructivist approach. Under constructivism, students construct their own knowledge by testing ideas and approaches, based on their prior knowledge and experience. In contextual learning, students plan and use their own experiences and process new knowledge with reference to their memory and everyday life experiences. In cooperative learning, students carry out activities in groups under the supervision and guidance of a teacher. They exchange ideas through discussions in order to solve a problem. In mastery learning, the whole curriculum content is broken down into small units and each unit is mastered one by one. The teacher makes sure that all students have mastered a unit taught before proceeding to the next unit. |
Q25. NCF 2005 considers that mathematics involves ‘a certain way of thinking and reasoning’. The vision can be realised by
(a) Adopting exploratory approach, use of manipulative, connecting concepts to real-life and involving students in discussions
(b) Rewriting all textbooks of mathematics to make them suitable for teaching in classrooms
(c) Giving lots of problem-solving worksheets to students for practice
(d) Giving special coaching to students in order to develop their thinking and reasoning skills
Answer: (a) Explanation: Adopting an exploratory approach, i.e. providing opportunities to explore problem situations using various manipulative and discussions with teachers and the peer group, provides a certain way of thinking and reasoning’ to the students. These not only help in developing manipulative skills and achieving narrow aims of education but also help students move towards achieving the higher aim of education. Rewriting a hundred textbooks or worksheets or providing special coaching is of little use if the students are not given hands-on experiences to which they can connect with real-life situations. |
Q26. Which of the following should be avoided in a good mathematics textbook?
(a) Linking the mathematics that children do (in their textbooks) to the mathematics they see and experience all around
(b) Introducing concepts through situations of life in which they are placed
(c) Beginning concept formation with definitions and mathematical terminology
(d) Arriving at concepts and ideas by observing patterns, exploring them and providing children opportunities to define them in their own words
Answer: (c) Explanation: Inductive method of teaching is always more helpful than the deductive method. Rather than loading and burdening a student with heavy terms or concepts, which they are not aware of, a teacher should try to make them draw conclusions by observing and exploring the situation. |
Q27. Roona was asked to add 89 to 354. This is how she did it:
1
354
+89
———-
1244
What is the problem with her?
(a) Roona has written numbers carelessly, bothering about the place values.
(b) There is a lack of alignment in her work while writing not numbers in columns.
(c) Roona has problems with the basic number facts.
(d) Roona has a visual processing disability.
Answer: (b) Explanation: Roona has misaligned the numerals in columns for calculation. She seems to have problems with place value, which requires the understanding of the base ten systems, i.e. aligning numbers correctly before performing a mathematical operation. |
Q28. Which level of thinking skill (refer to Bloom’s cognitive domain) is exhibited if a student is able to arrange a set of numbers in ascending and descending orders?
(a) Knowledge
(c) Application
(b) Comprehension
(d) Analysis
Answer: (b) Explanation: The competence of arranging numbers in ascending or descending order comes when students understand the concept of greater and smaller numbers. So, a student answering this question exhibits his/her “comprehension’ skill. |
Q29. For teaching multiplication, a teacher re-introduced multiplication as repeated addition. Then, she grouped the same number of objects taken multiple times and introduced the mathematical operator for multiplication. Further, she conducted a small activity of finding products using criss-cross lines or matchsticks. Here, the teacher is
(a) Using multiple representations to make the class interesting
(b) Developing a lesson and taking the students from concrete to abstract concept
(c) Catering to the learners with different learning styles
(d) Providing remedial strategies for low achievers in mathematics
Answer: (b) Explanation: The teacher is trying to develop an understanding of multiplication in her students. Multiplication as repeated addition is shown using a number of objects and activities. The students can learn by doing the operations on their own. The class in which students are engaged in activities is surely going to be interesting. Since only one type of activity is being done using different objects, we cannot say that the need of students for different learning styles is being catered to. Since the diagnosis of students’ problems will come after this phase is over, remedial teaching can be planned after that. |
Q30. Find the odd one out.
(a) Rapport
(c) Involvement
(b) Individuality
(d) Defective speech
Answer: (d) Explanation: Defective speech is the odd one among the given options because it is a subtype of physically exceptional children, while rapport building, individual traits and involvement or participation are attributes used during remedial teaching. |