New : 10 most repeated math question types in CTET
2. What mathematics educationcan do to improve communication skills of every child and how it can make them employable agter school
3. What mathematics education can do to engage the mind of every student and how it can strengthen the student’s resources
4. What mathematics can do to retain every child in school and how it can help them to be self-dependent
2. 2 x 33 x 5 x 11
3. 22 x 32 x 5 x 11
4. 23 x 32 x 5 x 11
2. Level 3 – deduction
3. Level 0 – visualization
4. Level 1 – analysis
2. All types of materials including language , written symbols, meaningful instructions to establish their purpose
3. Physical material like geo-board and 3D models, cubic rods etc
4. Charts based on formulae and concepts, graph paper and dotted sheets
2. Everyday language - > language of mathematics problem-> mathematized situatin language -> symbolic language
3. Everyday language - > language of mathematics problem-> -> symbolic language mathematized situation language
4. Everyday language - > symbolic language-> language of mathematics problem-> mathematized situation language
2. Visual –motor coordination
3. Visual-spatial skills
4. Visual-memory
2. Growing ability in numerical innovations and training
3. Growing aptitude in numerical innovations and techniques
4. Growing aptitude in numerical innovations and training
2. Learning lots of new formulae and algorithms
3. Remembering solutions or methods of various types of mathematical problems
4. Learning problem solving techniques only
2. Multi- disciplinary question
3. Open-ended question
4. Close-ended question
2. -120
3. 840
4. -360
2. A composite number can be odd
3. There is no even prime number
4. ‘1’ is the smallest prime number
2. Assess the student’s understanding and ability to combine two skills learnt to accomplish the given task
3. Engage every student in some work
4. Give the exposure of experiential learning
2. 2.3321
3. 2.245
4. 2.432
2. 5
3. 2
4. 3
(11 numbers in all. the number lying in the 6th place is the median )
2. 72
3. 120
4. 135
1080 = (n-2)180
(n-2) = 1080/180
(n-2) = 6
ð n= 8
2. 100
3. 60
4. 70
2. ∆ DEF ≡ ∆ QRP
3. ∆ DEF ≡ ∆ PQR
4. ∆ DEF ≡ ∆ QPR
Only option 4 could satisfy this condition.
2. 512
3. 320
4. 640
104= sum of two lengths of parallel sides + sum lengths of non-parallel sides
104= sum of two lengths of parallel sides + 18 + 12
104= sum of two lengths of parallel sides + 40
ð sum of two lengths of parallel sides = 104 -40 = 64
2. 625
3. 150
4. 225
Therefore, the % increase would be {(37.5 - 6)/6} x 100 ={ 31.5/6}x100 = 5.25 x 100 = 525
2. 4:9
3. 20:27
4. 10:9
2. 2x + y - 4
3. 2x - y + 3
4. 2x - y - 4
2. -29/21
3. 1/21
4. -1/21
2. -3.722
3. 4.722
4. -3.278
2. 1612
3. 945
4. 1042
2. 73
3. -37
4. 42
2. 0 or 3
3. 3 or 9
4. 0 or 9
2. 6
3. 2
4. 3
Q58. The scale of a map is 1:3x10 6 . Two cities are 9 cm apart on the map. The actual distance (in km) between the cities is
1. 180
2. 360
3. 135
4. 270
2. 12500
3. 10200
4. 11350.50
And after the end of second year = 0.88x (1-0.12) = 0.88 x 0.88 x = 0.7744x
But it is given that the value at the end of second year = 0.7744 x = 9680
Q31. Place of
mathematics education in the curricular framework is positioned on twin
concerns:
1. What mathematics education can do to improve the score of
students summative examination and how it can help to choose right stream in
higher classes2. What mathematics educationcan do to improve communication skills of every child and how it can make them employable agter school
3. What mathematics education can do to engage the mind of every student and how it can strengthen the student’s resources
4. What mathematics can do to retain every child in school and how it can help them to be self-dependent
Q32. LCM of 22,
54,135 and 198 is
1. 22 x 33 x 5 x 112. 2 x 33 x 5 x 11
3. 22 x 32 x 5 x 11
4. 23 x 32 x 5 x 11
Solution:
2
|
22
|
54
|
135
|
198
|
3
|
11
|
27
|
135
|
99
|
3
|
11
|
9
|
45
|
11
|
3
|
11
|
3
|
15
|
11
|
5
|
11
|
1
|
5
|
11
|
11
|
11
|
1
|
1
|
11
|
1
|
1
|
1
|
1
|
Q33. In class
VI, in the unit of ‘ Understanding Quadrilateral’ , important result related to
angle-sum property of quadrilaterals are introduced using paper folding
activity followed by the exercise based on these properties.
At these levels
proof of the angle property is not given, as the students of class VI are at Van Hiele level of
1.
Level 2 – informal deduction2. Level 3 – deduction
3. Level 0 – visualization
4. Level 1 – analysis
Q34. The term
‘Mathematical tools’ refers to
1. Calculators, rulers, tape measures,
protractors, compass etc2. All types of materials including language , written symbols, meaningful instructions to establish their purpose
3. Physical material like geo-board and 3D models, cubic rods etc
4. Charts based on formulae and concepts, graph paper and dotted sheets
Q35. Four stages
of language developments in mathematics classroom in order are
1.
Everyday language -> mathematized
situatin language -> language of mathematics problem solving -> symbolic
language2. Everyday language - > language of mathematics problem-> mathematized situatin language -> symbolic language
3. Everyday language - > language of mathematics problem-> -> symbolic language mathematized situation language
4. Everyday language - > symbolic language-> language of mathematics problem-> mathematized situation language
Q36. A learner exhibiting
difficulties in sorting , recognizing pattern,
orienting numbers and shapes, telling time and measurement may have
dyscalculia with difficulty in
1. Language processing 2. Visual –motor coordination
3. Visual-spatial skills
4. Visual-memory
Q37. CBSE announced
the celebration of GANIT WEEK in schools to commemorate the birth anniversary
of the legendary mathematician Srinivasa Ramanujam . GANIT stands for
1. Growing ability in numerical innovations
and techniques2. Growing ability in numerical innovations and training
3. Growing aptitude in numerical innovations and techniques
4. Growing aptitude in numerical innovations and training
Q38. Learning
mathematics in upper primary level is about
1.
Gaining understanding of mathematical
concepts and their application in solveing problems logically2. Learning lots of new formulae and algorithms
3. Remembering solutions or methods of various types of mathematical problems
4. Learning problem solving techniques only
Q39. Read the
following question from class VI textbook:
“Write a pair of
integers whose sum gives a negative integer”
The above
question refers to
1. Reflective question2. Multi- disciplinary question
3. Open-ended question
4. Close-ended question
Q40. The product
of integers between -7 and -3 is
1. 1202. -120
3. 840
4. -360
Answer:
Integers
between -7 and -3 are -6,-5,-4. Multiply these numbers to get -120
Q41. Which one
of the following statement is correct?
1. Sum of two prime numbers is always a
prime number2. A composite number can be odd
3. There is no even prime number
4. ‘1’ is the smallest prime number
Q42. In geometry
class of VI grade students , the teacher explained the construction of angles
measuring 30, 60 and 90 with the help of demonstration of construction and
bisector of an angle. Then she asked the students to construct an angle of 15
and an angle of 46
1. Assess the learner’s performance in
summative assessment2. Assess the student’s understanding and ability to combine two skills learnt to accomplish the given task
3. Engage every student in some work
4. Give the exposure of experiential learning
Q43. The value
of 1 + (11/10) + (11/100) + (111/1000) + (111/10000) is
1. 3.32212. 2.3321
3. 2.245
4. 2.432
Answer:
1
+ 1.1 + 0.11 + 0.111 + 0.0111
=
2.3321
Q44. The mean of
range, mode and median of the data
4,3,2,2,7,2,2,0,3,4,4
is
1.
42. 5
3. 2
4. 3
Solution:
Mean
= sum of all number/(number of number)
The
range of the above number is (largest number – smallest number) = 7-0 = 7
The
mode of the above series is 2 (mode is the number that is repeated the most )
And
the median is the number that lies exactly in the middle for a series with odd
number of elements when arranged in increasing order.
0,2,2,2,2,3,3,4,4,4,7(11 numbers in all. the number lying in the 6th place is the median )
Here
the number is 3
Therefore
the mean of range, mode and median is = (7+3+2)/3 = 12/3 = 4
Q45. The sum of
all interior angles of a regular convex polygon is 1080. The measure of each of
its interior angles is
1. 1082. 72
3. 120
4. 135
Solution:
The
sum of all angles of a regular convex polygon = (n-2)180
(where n is the number of sides of a regular
hexagon)1080 = (n-2)180
(n-2) = 1080/180
(n-2) = 6
ð n= 8
Hence,
the polygon is regular octagon and the measure of each angle is 1080/8 = 135
Q46. In a
quadrilateral ABCD, < D = = 60 and <C= 100. The bisectors of angle A and
D meet at point P. the measure of angle < APB is
1.
802. 100
3. 60
4. 70
Solution:
Angle
A is opposite to Angle C. Since angle C is 100 then angle A would be 80
Bisector
of angle A and D meets at P. Then the points A,P and D forms a triangle. Thus the
angle
angle APB
would be equal to
<APB
= 180 – 60 – 40 = 80
Q47. In ∆ DEF and ∆ PQR, PQ = DE, EF = PR and FD =
QR then
1. ∆ DEF ≡ ∆ RPQ2. ∆ DEF ≡ ∆ QRP
3. ∆ DEF ≡ ∆ PQR
4. ∆ DEF ≡ ∆ QPR
Solution:
Given
that,
PQ
= DE and EF = PR and FD = QROnly option 4 could satisfy this condition.
Q48. The
perimeter of trapezium is 104 cm, the length of its non-parallel sides is 18 cm
and 22 cm and its altitude is 16 cm. the area (in cm2) of the
trapezium
1. 10242. 512
3. 320
4. 640
Solution:
Perimeter
of trapezium = length of all 4 sides 104= sum of two lengths of parallel sides + sum lengths of non-parallel sides
104= sum of two lengths of parallel sides + 18 + 12
104= sum of two lengths of parallel sides + 40
ð sum of two lengths of parallel sides = 104 -40 = 64
Now, the area of trapezium = (1/2)( sum of two lengths of parallel
sides) x altitude
= (1/2)(64)(16) = 512
Q49. If each
edge of a solid cube is increased by 150 %, the percentage increase in the
surface area is
1.
5252. 625
3. 150
4. 225
Solution:
Surface
area of cube = 6 x L x L
If
the edge is increased by 150% then the new edge = 1L + 1.5 L = 2.5L
Therefore,
the surface area = 6 x 2.5L x 2.5L = Therefore, the % increase would be {(37.5 - 6)/6} x 100 ={ 31.5/6}x100 = 5.25 x 100 = 525
Q50. The radii
of the bases of two cylinders are in the ratio of 2:3 and their heights are in
the ratio of 5:3 . The ratio of their volumes is
1. 7:62. 4:9
3. 20:27
4. 10:9
Solution:
Let the radii be
2x and 3x and the heights would be 5h and 3h.
Then the ratio
of volumes would be (2x) 2
(5h) / (3x) 2 (3h) = 4 x 5 / 9 x 3 = 20/27
Q51. One of the
factors of
4x2 + y2 + 14x –
7y -4xy + 12 is
1. 2x + y + 42. 2x + y - 4
3. 2x - y + 3
4. 2x - y - 4
Q52. What should
be subtracted from (-5/7) to get (-2/3)?
1. 29/212. -29/21
3. 1/21
4. -1/21
Solution:
Subtract
(-2/3) from (-5/7)
ð
(-5/7) -(-2/3)
ð
2/3 - 5/7
ð
(14 - 15)/21
ð
-1/21
Q53. In standard
form, 0.00001278 is expressed as k x 10n. The value of (k + n)
1. 3.2782. -3.722
3. 4.722
4. -3.278
Solution:
K would
be 1.278 and then n would be -5
And
k + n would be -5 + 1.278 = -3.722
Q54. The least
number which must be added to 893304 to obtain a perfect square is
1. 15212. 1612
3. 945
4. 1042
Q55. The value of
3√ -91125 - 3√512 is
1.
-532. 73
3. -37
4. 42
Solution:
We know that
cubic root of 512 is 8
And the cube root
of -991125 is -45
Thus 3√
-91125 - 3√512 = -45 – 8 =
-53
Q56. The values
of y for which the four digit number 51y3 is divisible by
1. 2 or 32. 0 or 3
3. 3 or 9
4. 0 or 9
Solution:
Use
the rules of divisibility to check for the right answer.
Q57. In the
product (x2 - 2) (1-3x+2x2) the sum of coefficients of x
2 and x is
1. 52. 6
3. 2
4. 3
Solution:
(x2
- 2) (1-3x+2x2) = x 2 – 3x3 + 2x4 -2 + 6x
- 4x2 = 2x4
– 3x3 - 3x2
+4
It
is clear that the coefficient of x is 6 and the coefficient of x 2
is -3 and their sum is 3.Q58. The scale of a map is 1:3x10 6 . Two cities are 9 cm apart on the map. The actual distance (in km) between the cities is
1. 180
2. 360
3. 135
4. 270
Answer:
9
cm translates to 9 x 3 x106 = 27 x 106 cm = 27 x 104
m = 270 kilometer
Q59. The value
of a machine which was purchased two years ago, depreciates at 12 % per annum.
If it’s present value is 9680 for how much was it purchased?
1. 12142.502. 12500
3. 10200
4. 11350.50
Answer:
Let the initial
price was x. Then the price after first year is x (1-0.12) = And after the end of second year = 0.88x (1-0.12) = 0.88 x 0.88 x = 0.7744x
But it is given that the value at the end of second year = 0.7744 x = 9680
ð
X =
12500
Q60. As per NCF
2005, the goal of mathematics teaching in school curriculum is that children
learn important mathematics” . Important
mathematics implies
1.
Understanding appropriate use of learnt
mathematical techniques
2. Verifying geometrical theorems in Maths
lab
3. Knowing mathematical procedures and algorithms
4. Solving mathematical games and puzzles
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