# CTET-Mumbai November 2012 Maths Question Paper I CTET-Mumbai November 2012 Maths Question Paper I.

CTET-Mumbai November 2012 Maths Question Paper I I For (Class I to V CTET Mathematics Paper I in Hindi and English Language

31. Ten thousands + ten ones + ten tens equals
(1) 10110
(2) 11010
(3) 10011
(4) 101010

32. Which one of the following is not correct?
(1) 7 metres 10 centimetres   = 710 centimetres
(2) Half of one half  = one-third of three quarters
(3) 2kgll2g=2112g
(4) 3L6OmL=360mL

33. Number of minutes in 10 days is equal to the number of seconds in
(1) 2hours
(2) 3 hours
(3) 4 hours
(4) 5 hours

34. Perimeter of a square and a rectangle arc equal. If the perimeter of the square is 40 cm and length of rectangle is 2 cm more than the side of the square, then the area, in square cm, of the rectangle is
(1) 48
(2) 72
(3) 96
(4) 100

35. Sania bought a battery. She read on it “Life : 4200 hours”. She uses it throughout day and night. How many weeks will the battery run?
(1) 28
(2) 25
(3) 24
(4) 21

36. Internal length, breadth and height of  a rectangular box are 15 cm, 9 cm  and 10 cm respectively. How many  such boxes are needed to pack 6750  centimetre cubes?
(1) 6
(2) 5
(3) 4
(4) 3
37. The weight of 18 gulabjamun is one kilogram. If 16 gulabjamun can be packed in one box, then the number of boxes needed to pack 40 kg gulabjamun is
(1) 40
(2) 42
(3) 45
(4) 48

38. In a hockey match between School  A and School B, School A scored 11  goals and School B scored 3 goals.
What fraction of the total goals did  School A score?
(1) 3/14
(2) 8/11
(3) 11/14
(4) 11/3

39. Number of degrees in five and twot hird of a right angle is
(1) 510
(2) 490
(3) 486
(4) 480

40. Rohan of Class II can do the skip counting correctly. In which developmental phase of number is he?
(1) Emergent phase
(2) Matching phase
(3) Quantifying phase
(4) Partitioning phase

41. A Class IV Maths Test has a question “Convert the fraction into decimal.” The question aims to test the students’
(I) knowledge of decimal facts.
(2) understanding of concepts of fraction and decimal.
(3) procedural skill of converting fraction into decimal.
(4) skill of exploring the decimal equivalent of given fraction by drawing paper grid of 10 x 10.
42.. it is observed that children use various methods of problem-solving strategies like
(a) Counting
(b) Direct Modelling
(c) Number facts
Sequence from which children pass through while developing problem solving strategies is
(1) (a)—(b)–÷(c)
(2) (b) — (c) —* (a)
(3) (c) — (a) — (b)
(4) (b) —+ (a) —+ (c)

43. Number sense in the primary school is always accompanied with developing operation sense. Which of the following statement cannot be referred as component of operation sense?
(1) Determining the operation to be used in given situation.
(2) Recognizing that the same operation can be applied in different problem situations.
(3) Realizing the operations’ effect on numbers.
(4) Comparing the numbers and quantities.

( 44) Mr. Jam draws the following figure on the blackboard: Then he posed a problem to the students ‘if  1/2 paint is required to  paint 1/8 of a wall, how much paint
would be requ red to paint complete wall ?“
Mr. Jam wants to expose the students to problems based on
(1) product of fractions.
(2) division of whole number by a fraction.
(3) division of fraction by a fraction.
(4) fraction divided by a whole number.

45. A child studying in Class IV exhibits difficulty in sorting, recognizing patterns, orientating numbers and shapes, telling time, measuring etc. The child may be
(1) suffering from dyslexia
(2) suffering from dyscalculia
(3) suffering from dysgraphia
(4) suffering from attention deficit disorder

46. A teacher introduced the concept of perimeter in Class V. She used lots of geometrical cut-outs and asked the children to measure all around. To assess the understanding of concept she asked the following question:
Ravi is running on footpath of a square shaped park of length 10 metres. How much distance Ravi covered in one round?
Children were not able to respond to this question. The reason may be
(1) children have not understood at all the concept of perimeter.
(2) there is gap in instructional objective and assessment objective.
(3) children were not able to understand the language of question.
(4) children of Class V are too young to understand this type of problem.

47. Students struggle with estimation as

(1) they are not able to measure accurately.
(2) they are not able to differentiate between two units.
(3) they are not having much experience with different units.
(4) they are not skilled enough to use ruler.

48. Teacher conducted an oral assessment in class and found that Harish can speak definition of all type of numbers — odd, even, prime and composite accurately but not able to identify the numbers accurately, when given a set of numbers. Teacher reports that Harish
(1) has good memory, but lacks practice.
(2) has good memory, but lacks concentration.
(3) has good memory, but lacks conceptual understanding.
(4) has good memory, but lacks analytical ability.

49. Rubrics of assessment to check the concept of shapes in Class II, shall be
(1) can draw shapes accurately, knows the number of sides in each polygon.
(2) can draw shapes accurately and can name them correctly.
(3) can draw the right shape, can sort out the shape, can tell its number of sides, edges and vertices accurately.
(4) can tell the name of given shape, differentiate between a circle and a polygon, can identify its parallel or nonp arallel sides.

50. Major aspect of inquiry based lesson plan is
(1) Exposition
(4) Cross-curricular integration

51. Following ICT tool can be effectively used for exploratory tasks in geometry class
(1) Geo-gebra software
(2) Calculator
(4) Cartoon strip software

52. ‘Algebra Tiles’ are used to teach
(2) Exponents
(3) Square roots and cube roots
(4) Graphing of linear equation

53. Teacher gave following sorting activity in class: This sorting activity is

55. What should be subtracted from the  product 101 x 101 to get 10101?
(1) 102
(2) 101
(3) 100
(4) 99

56. The number of factors of 18 is
(1) 4
(2) 5
(3) 6
(4) 7

57. The sum of place values of 4 in 6403  and 3640 is
(1) 8
(2) 404
(3) 443
(4) 440

58. (Smallest common multiple of 6, 9, 12)  — (Smallest common multiple of 4, 6, 8) is equal to
(1) 6
(2) 12
(3) 18
(4) 24

59. How many 1/10 are there in 3/5
(1) 8
(2) 6
(3) 4
(4) 2

60. When 71777 is divided by 7, the remainder is
(1) 1
(2) 3
(3) 5
(4) 6