CTET 2011 Question Paper-2 with Answer

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CTET 2011 Question Paper-2 with Answer

Child DevelopmentMathematics and ScienceSocial studiesLanguage-I(Eng)Language-II(Hindi)

Q41. A square sheet ABCD, when rotated on its diagonal AC at its axis of rotation, sweeps a

CTET_P2_MS_Q41

 

(a) cone

(b) spindle

(c) cylinder

(d) trapezium

 Answer: (b) spindle

Solution: When a square is rotated on it diagonal as its axis of rotation, the spindle shape is formed.

Q42. The area of a triangle with base x units is equal to the area of a square with side x units. Then, the altitude of the triangle is

(a) x/2 units

(b) x units

(c) 2x units

(d) 3x units

 Answer: (c) 2x units

Solution: Let altitude of the triangle is h units.

Now, ½ × base × altitude = (side)2

⇒ x × h = 2x2

⇒ h = 2x

Q43. Which is greatest among $ 33\frac{1}{2}\%, \frac{4}{15} $ and 0.35?

(a) 33½%

(b) 4/15

(c) 0.35

(d) cannot be compared

 Answer: (c) 0.35

Solution: $ 33\frac{1}{2}\% =\,\,\frac{67}{200}=0.335, \frac{4}{15}=0.266667 $

So, greatest 0.35

Q44. The factorization of 25 – p2 – q2 – 2pq is

(a) (5 + p + q) (5 – p + q)

(b) (5 + p +q) (5 – p – q)

(c) (5 + p – g) (5 – p + q)

(d) (5+ p – q ) (5 – p – q)

 Answer: (b) (5 + p +q) (5 – p – q)

Solution: 25 – p2 – q2 – 2pq = (5)2 – (p +q)2 = (5 + p + q)(5 – p – q)

Q45. A rectangle is divided horizontally into two equal parts. The upper part is further divided into three equal parts and the lower part is divided into four equal parts. Which fraction of the original rectangle is the shaded part?

CTET_P2_Q45

(a) 3/5

(b) 2/7

(c) 4/7

(d) 7/12

 Answer: (d) 7/12

Solution: fraction of upper part = ½ and shaded area = 1/2 × 2/3 = 1/3

fraction of lower part = ½ and shaded area = 1/2 × 2/4 = 1/4

Total shaded area = 1/3 + 1/4 = 7/12

Q46. In the given figure, PS = SQ = SR and ∠SPQ= 54°. Find the measure of ∠x.

CTET_P2_Q46

 

(a) 54°

(b) 72°

(c) 108°

(d) 36°

 Answer: (d) 36°

Solution: SP = SQ, ∴ ∠SQP = ∠SPQ = 54°, ∴ ∠PSQ = 180° – (54° + 54°)= 72°

∴ ∠QSR = 180° – 72° = 108° , now ∠SRQ + ∠SQR = 180° – 108° = 72°

∴ x° = 72°/2 = 36° ( ∠SRQ = ∠SQR)

Q47. 2x – 13, 2x – 11, 2x – 9, 2x – 7 are consecutive

(a) prime numbers

(b) even numbers

(c) odd numbers

(d) natural numbers

 Answer: (c) odd numbers

Solution: let x is integer, then 2x is even number

∴ 2x – 13, 2x – 11, 2x – 9, 2x – 7 are consecutive  odd numbers

Q48. The fractional equivalent of 57.12% (approx.) is

(a) 349/625

(b) 359/625

(c) 357/625

(d) 347/625

 Answer: (c) 357/625

Solution: 57.12% = 57.12/100 = 5712/10000 = 357/625

Q49. The ratio of the side and height of an equilateral triangle is

(a) 2 : 1

(b) 1 : 1

(c) $ \text{2:}\sqrt{3} $

(d) $ \sqrt{3}:2 $

 Answer: (c)$ \text{2:}\sqrt{3} $

Solution: height of an equilateral triangle  (h) = $ \frac{\sqrt{3}}{2}\times \,\,side\,\,\Rightarrow \,\,side\,\,: (h)\,\,=\,\,\text{2 : }\sqrt{3} $

Q50. If two adjacent sides of a square paper are decreased by 20% and 40% respectively, by what percentage does the new area decrease?

(a) 48%

(b) 50%

(c) 52%

(d) 60%

 Answer: (c) 52%

Solution: let side of square = 100x

Then area = 10000x2

New side are 80x and 60x and area = 4800x2

Area decrease = 10000x2 – 4800x2 = 5200x2

Percentage decrease = (5200x2/10000x2) × 100 = 52%

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