JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

West Bengal 2020-21 NTSE Stage 1 Answer Key & Question Paper With Solutions

For students who are gearing up to prepare effectively and crack the NTSE exams, we at BYJU’S are providing an extensive set of West Bengal 2020-21 NTSE Stage 1 Answer Key & Question Paper With Solutions. The papers have been solved by our subject experts and they have taken extra care to deliver the accurate answers only. They have further designed the solutions to help each student get a comprehensive understanding of the various subjects and important topics covered in the West Bengal2020-21 NTSE Stage 1 Answer Key & Question Paper With Solutions.
Students will find the West Bengal 2020-21 NTSE Stage 1 Answer Key & Question Paper With Solutions here. Download the question paper to get familiar with the NTSE exam pattern, the marking scheme and the type of questions that are asked in the exam. Students can also start solving the different types of questions and work on improving their speed and accuracy. In essence, students will get to practice as per the actual exam format and this will make them feel motivated and well prepared.

West Bengal 2020-21 NTSE Stage 1 Answer Key & Question Paper With Solutions

Download PDF Download PDF

West Bengal 2020-21 NTSE Stage 1 – Question Paper With Solutions

Paper: MAT

Question 1. If there is a common roots of the equation x2 + ax + b = 0 and x2 + bx + a = 0 then the value of a + b is


a. 1
b. -1
c. 0
d. ½

Answer: (b)

Let the common root is ‘x’

x2 + ax + b = 0 ……(1)

x2 + bx + a = 0 ……(2)

Subtract equation (2) from (1), we get

(a – b)x + (b –a) = 0

⇒ x = (a-b)/(a-b) = 1

⇒ x = 1

⇒ {1 + a + b = 0} (From equation (1))

⇒ a + b = –1

Question 2. A principle becomes twice of its amount in 10 years at a certain rate of simple interest, at the same rate of simple interest, that principle becomes thrice of its amount in


a. 15 years
b. 20 years
c. 25 years
d. 30 years

Answer: (b)

Let, Principle = y

Amount = 2y

\(\begin{array}{l}2y = y + \frac{y\times R\times 10}{100}\end{array} \)
\(\begin{array}{l}y = y + \frac{y\times R\times 10}{100}\end{array} \)

⇒ R = 10%

Now, according to the question

\(\begin{array}{l}3y = y + \frac{y\times 10\times T’}{100}\end{array} \)

\(\begin{array}{l}2y = \frac{y\times T’}{10}\end{array} \)

⇒ T’ = 20 years

Question 3. The length of each of the two parallel chords AB and CD is 12 cm. If the length of the radius of the circle is 10 cm, then the distance between the two chords is


a. 12 cm
b. 14 cm
c. 16 cm
d. 18 cm

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 3 solution

Perpendicular line from the center bisects the chord, so it will bisect the chord in two equal parts of 6 cm each.

d.= OQ =

\(\begin{array}{l}\sqrt{10^{2} – 6^{2}}\end{array} \)
= 8 cm = OP

PQ = 8 + 8 = 16 cm

Question 4. The ratio of the volumes of two cubes is 1 : 27, the ratio of total surface areas of two cubes is


a.1 : 6
b.1 : 8
c. 1 : 9
d. 1 : 18

Answer: (c)

Let the sides of cubes a1 and a2 respectively

Now,

\(\begin{array}{l}\frac{V_{1}}{V_{2}} = \frac{a_{1}^{3}}{a_{2}^{3}} = \frac{1}{27}\end{array} \)
(Given)

\(\begin{array}{l}\frac{a_{1}}{a_{2}} = \frac{1}{3}\end{array} \)

Now, surface area =

\(\begin{array}{l}6a_{1}^{2} \: and \: 6a_{2}^{2}\end{array} \)
\(\begin{array}{l}\frac{S_{1}}{S_{2}} = \frac{6a_{1}^{3}}{6a_{2}^{2}} = (\frac{1}{3})^{2} = \frac{1}{9}\end{array} \)

Therefore, option (c) is correct.

Question 5. A box has 210 coins of denomination one rupee and fifty paise only. The ratio of their respective values is 13 : 11. The number of one-rupee is


a. 75
b. 76
c. 78
d. 87

Answer: (c)

Let numbers of 1 Rs coins = x ………..(1)

and, ½ Rs coins = (210 – x) ………..(2)

From (1) and (2)

\(\begin{array}{l}\frac{1\times x}{\frac{1}{2}(210-x)} = \frac{13}{11}\end{array} \)
(Given)

⇒ 22x = 2730 – 13x

⇒ 35x = 2730

⇒ x = 78

Question 6. If the volume of two solid right circular cylinders are same and their height are in the ratio 1:3, then the ratio of length of radii is


a. √3 : 1
b. 1 : √3
c. 1 : 3
d. 3 : 1

Answer: (a)

Volume of the cylinder = πR2h

πr21h1= πr22h2 (Given)

\(\begin{array}{l}\frac{h_{1}}{h_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2}\end{array} \)
…(1)

also given

\(\begin{array}{l}\frac{h_{1}}{h_{2}} = \frac{1}{3}\end{array} \)
,

\(\begin{array}{l}\frac{h_{1}}{h_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2} = \frac{1}{3}\end{array} \)

Take square root both side

\(\begin{array}{l}\frac{r_{1}}{r_{2}} = \frac{\sqrt{3}}{1}\end{array} \)

Question 7. The product of

\(\begin{array}{l}(2-\sqrt{3})\end{array} \)
\(\begin{array}{l}(2+\sqrt{5})\end{array} \)
\(\begin{array}{l}(4+2\sqrt{3})\end{array} \)
\(\begin{array}{l}(\sqrt{5}-2)\end{array} \)
is


a. 1
b. 2
c. 3
d. 4

Answer: (b)

= (2 – √3) (4 + 2√3) (2 + √5) (√5 – 2)

= (2 – √3) (2 + √3) (2) (2 + √5) (2 – √5) (-1)

= (4–3) (2) (4–5) (–1)

= (1) (2) (–1) (–1)

= 2

Question 8. If x – y ∝ 1/z, y – z ∝ 1/x, z – x ∝ 1/y then the sum of three variation constant is


a. –1
b. 0
c. 1
d. ±1

Answer: (b)

Given, x – y ∝ 1/z, y – z ∝ 1/x, z – x ∝ 1/y

Let x–y = k1 1/z , y–z = k2 1/x , z–x = k3 1/y

⇒ (x – y) z = k1 ……….(1)

⇒ (y – z) x = k2 ……….(2)

⇒ (z – x) y = k3 ……….(3)

By adding equation (1), (2) and (3), we get –

k1 + k2 + k3 = 0

Question 9. Keeping the radius of a right circular cone same. If the height of its increased thrice, the volume of it will be increased by


a. 100 %
b. 200 %
c. 300 %
d. 400 %

Answer: (b)

Let height for cones h1 and h2 respectively and h1 = h and h2 = 3h (Given)

Now, V1 = ⅓ πr2h1 = ⅓ πr2h …(1)

and, V2 = ⅓ πr2h2

⇒ V2 = ⅓ πr2(3h) ……..(2)

from (1) and (2), we get percentage increase as

(⅓ πr2(3h) – ⅓ πr2 h)/ ⅓ πr2 h x 100

2 x 100

⇒ Volume is increased by 200%

Therefore, option (b) is correct.

Question 10. In ΔABC and ΔPQR, if ∠A = ∠Q = 50°, AB : QP = AC : QR and ∠R = 60°, then ∠B is


a. 50°
b. 60°
c. 70°
d. 80°

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 10 solution

Given that, ∠A = ∠Q = 50° and ∠R = 60°

Also given, AB/QP = AC/QR

Now, by SAS Rule

ΔBAC ~ ΔPQR

⇒ ∠C = ∠R

⇒ ∠B = ∠P

∠P = 180° – (60 + 50)°

∠B = 70°

Question 11. If the measure of two angles of a triangle are 65°20’3” and 54°39’57”, then the circular value of the third angle is


a. πc
b. πc / 2
c. πc / 3
d. 2πc / 3

Answer: (c)

Given that,

∠A = 65° 20’ 3’’ ……….(1)

∠B = 54° 39’ 57’’ ……….(2)

adding equation (1) and (2) we get

∠A + ∠B = 120°

⇒ ∠C = 60°

⇒ ∠C = (π/3)c

Question 12. Three squares are based on the sides of a right triangle. The area of the two smaller ones are 225 sq.cm and 400 sq.cm. What is the area of the third one?


a. 484 sq.cm
b. 259 sq.cm
c. 576 sq.cm
d. 625 sq.cm

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 12 solution

Let the sides of Right angle triangle be a, b, c.

By Pythagoras theorem

a.sup>2 + b2 = c2

a.sup>2 + b2 = 225 + 400 (Given)

⇒ c2 = 625 sq cm

Question 13. If sinθ – cosθ = 0, 0° ≤ θ ≤ 90° and secθ + cosecθ = x then x will be


a. 1
b. 2
c. 2√2
d. √2

Answer: (c)

Given that sinθ – cosθ = 0, 0°≤θ≤90° and secθ – cosesθ = x

sin θ – cos θ = 0

⇒ sin θ = cos θ ⇒ tanθ = 1

⇒ θ = 45o

So, sec 45° + cosec 45°

= √2 + √2 = 2√2

Question 14. If 32 is removed from the date 32, 25, 23, 21, 17, 15, 13, 12, 10 then median will be


a. increased by 1. 5
b. decreased by 1
c. increase by 1
d. Same

Answer: (b)

Given 32, 25, 23, 21, 17, 15, 13, 12, 10

So, Median = 17 as it is the middle term

Now, 25, 23, 21, 17, 15, 13, 12, 10

(Median)new = 17 + 15/ 2= 16

⇒ Describe by 17 – 16 = 1

Question 15. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.7 m away from the wall. The length of the ladder is


a. 4.7 m
b. 9.4 m
c. 8.4 m
d. 9.7 m

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 15 solution

Cos 60° = 4.7 / x (cos = B/H)

⇒ 4.7 / x = ½

⇒ x = 2 x 4.7

⇒ x = 9.4 m

Direction: In each question 16 to 25 there are two words separated by ‘:’ and other two separated from the first two by the symbol ‘::’, find the relation between two sets of words and select one word from the right side of ‘:’ which have the same relation as left set of word of ‘::’. Fill the circle of the letter denoting your selected answer on the OMR sheet.

Question 16. Lion : Roar :: Ass : ?


a. Trumpet
b. Bray
c. Bark
d. Howl

Answer: (b)

As a lion is related to Roar. Similarly, Ass is related with Bray.

Question 17. Ocean : Water :: Glacier : ?


a. Mountain
b. Cave
c. Ice
d. Refrigerator

Answer: (c)

As the ocean consists of water. Similarly, glaciers consist of ice.

Question 18. Arc : Circle :: Line segment : ?


a. Sphere
b. Ellipse
c. Point
d. Square

Answer: (d)

Arc is a part of a circle. Similarly, the line segment is a part of the square.

Question 19. Court : Justice :: School : ?


a. Student
b. Teacher
c. Education
d. Building

Answer: (c)

As you get justice in court. Similarly, you get education in school.

Question 20. Protein : Growth :: Carbohydrates : ?


a. Immunity
b. Resistance
c. Diseases
d. Energy

Answer: (d)

Protein is needed for growth. Similarly. carbohydrates are needed for energy.

Question 21. USA : Congress :: Iran : ?


a. Cortes
b. Althing
c. Majlis
d. Storting

Answer: (c)

USA has congress. Similarly, Iran has Majlis.

Question 22. Country : President :: State : ?


a. Chief Minister
b. Minister of State
c. Speaker
d. Governor

Answer: (d)

As president is related to the country. Similarly, the Governor is related to the state.

Question 23. AB : ZY :: CD : ?


a. UV
b. WX
c. VU
d. XW

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 23 solution

Therefore, option (d) is correct.

Question 24. 42 : 56 :: 110 : ?


a. 148
b. 184
c. 132
d. 124

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 24 solution

Therefore, Option (c) is correct.

Question 25. 64 : 144 :: 256 : ?


a. 484
b. 412
c. 625
d. 402

Answer: (Bonus)

64 : 144 :: 256 : 400

↑ ↑ ↑ ↑

82 : 122 :: 162 : 202

But no option is given as 400

Direction 26-55: In question 26-55, numbers are placed in figure on the basis of some rules. One place in figure is indicated by the interrogation sign(?). Find out the correct alternative to replace the question mark and indicate your answer by filling the circle of the corresponding letter of alternatives in the OMR Sheet.

Question 26. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 26


a. 22
b. 32
c. 30
d. 23

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 25 solution

Logic: multiply all outside numbers and divide by 10.

Therefore, option (b) is correct.

Question 27. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 27


a. 120

b. 130

c. 140

d. 150

Answer: (b)

(56 – 21) × 2 = 70

(87 – 45) × 2 = 84

(180 – 115) × 2 = 130

Question 28. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 28


a. 20
b. 18
c. 16
d. 14

Answer: (c)

(7 × 4) ÷ 2 = 14

(4 × 9) ÷ 3 = 12

(6 × K) ÷ 4 = 24

K = 16

Question 29. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 29


a. 9
b. 8
c. 1
d. 7

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 29 solution

Question 30. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 30


a. 51
b. 61
c. 63
d. 54

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 30 solution

Therefore, Option (c) is correct

Question 31. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 31


a. 77
b. 78
c. 86
d. 88

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 31 solution

Therefore option (c) is correct.

Question 32. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 32


a. 185
b. 165
c. 175
d. 195

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 32 solution

Therefore, Option (a) is correct

Question 33. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 33


a. 46
b. 42
c. 43
d. 44

Answer: (a

WB NTSE 2021 Stage 1 MAT Paper Question 33 solution

Therefore, Option (a) is correct.

Question 34. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 34


a. 78
b. 80
c. 84
d. 82

Answer: Bonus

Image 1 and 2 does not tally for any relation as the only difference is 81 and 84 between them.

Question 35. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 35

a. J
b. K
c. S
d. P

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 35 solution

Therefore, option (b) is correct

Question 36. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 36


a. 146
b. 126
c. 175
d. 185

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 36 solution

Therefore, option (d) is correct

Question 37. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 37


a. 18
b. 33
c. 135
d. 145

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 37 solution

Therefore, option (c) is correct.

Question 38. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 38


a. 117
b. 115
c. 127
d. 112

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 38 solution

(2 × 1) + 1 = 3

(14 × 7) + 7 =105

(12 × 9) + 9 = 117

Therefore, Option (a) is correct

Question 39. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 39


a. 13
b. 16
c. 15
d. 17

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 39 solution

Therefore, option (b) is correct.

Question 40. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 40


a. 2
b. 3
c. 4
d. 5

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 40 solution

40 + (5 – 2) = 43

22 + (2 – 1) = 23

So, 21 + (1 – k) = 20

⇒ k = 2

Therefore, option (a) is correct.

Question 41. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 41


a. 1
b. 731
c. 1625
d. 2031

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 41 solution

Therefore, option (d) is correct.

Question 42. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 42


a. 6
b. 9
c. 12
d. 18

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 42 solution

Therefore, option (b) is correct.

Question 43. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 43


a. 72
b. 73
c. 74
d. 75

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 43 solution

Therefore, option (b) is correct.

Question 44. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 44 solution


a. 16A
b. 20B
c. 22D
d. 24C

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 44 solution

4 × 2 = 8 3 × 8 = 24 2 × 7 = 14

For numbers in column top number multiply by bottom number = middle number.

For alphabet, Alphabet A, B, C comes once in every row.

Therefore 24C is the correct answer.

Question 45. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 45 solution


a. 414
b. 424
c. 441
d. 484

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 45 solution

\(\begin{array}{l}\sqrt{(\sqrt{324\times 4})} = \sqrt{36}\: \: \: \sqrt{(\sqrt{9\times 225})} = \sqrt{45}\end{array} \)
\(\begin{array}{l}\sqrt{(\sqrt{36\times k^{2}})} = \sqrt{6k} = \sqrt{126}\end{array} \)

= 3√4 = 3√5 ⇒ 6k = 126

⇒ k = 21

⇒ k2 = 441

Question 46. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 46


a. 9
b. 7
c. 5
d. 3

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 46 solution

Therefore, Option (d) is correct

Question 47. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 47


a. 8
b. 9
c. 4
d. 7

Answer: Bonus

Question 48. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 48


a. K7
b. M8
c. K8
d. M7

Answer: (c)

For numbers in column 2+3 = 5 , 4 +5 = 9

Similarly, 6 + 8 = 14

For alphabets in row pattern +2, +2.

So, I, J , K

Therefore, Option (c) is correct

Question 49. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 49


a. 2
b. 3
c. 6
d. 4

Answer: (c)

 WB NTSE 2021 Stage 1 MAT Paper Question 49 solution

Therefore, option (c) is correct

Question 50. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 50


a. 10
b. 14
c. 8
d. 6

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 50 solution

Therefore, option (a) is correct.

Question 51. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 51


a. 8
b. 7
c. 9
d. 6

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 51 solution

Therefore, option (c) is correct.

Question 52. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 52 solution

a. 7
b. 6
c. 8
d. 5

Answer: (a)

We will denote the numerical order of the alphabet as their value.

WB NTSE 2021 Stage 1 MAT Paper Question 52 solution

Similarly,

T(20) + 7 = 27 [Reverse position]

Therefore, Option (a) is correct

Question 53. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 53


a. 4/3
b. 3/2
c. 5/4
d. 3/4

Answer: (c)

First column

= denominator of other term in Row

2 → ½ → 3/2 (2 = ½ + 3/2)

3 → 4/3 → 5/3 (3 = 4/3 + 5/3)

(4 = K/4 + 11/4)

k = 5

4 → 5/4 → 11/4

Hence ? = 5/4

Logic: in row first entry = sum of next two entries.

Therefore, Option (c) is correct

Question 54. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 54


a. 72
b. 68
c. 100
d. 120

Answer: (c)

1 × 0 = 0

2 × 2 = 4

3 × 6 = 18

4 × 12 = 18

5 × 20 = 100

WB NTSE 2021 Stage 1 MAT Paper Question 54 solution

Question 55. Find out the correct alternative.

WB NTSE 2021 Stage 1 MAT Paper Question 55


a. IJ18
b. JL12
c. JK24
d. JL22

Answer: (d)

Continuous series of alphabets.

1st letter

B → C → D

E → F → G

H → I → J

2nd letter

D → E → F

G → H → I

J → K → L

3rd number – Consecutive numbers have a difference of 2

6 → 8 → 10

12 → 14 → 16

18 → 20 → 22

So, JL22 will be the correct answer

Direction: In each of the following questions (56 to 70), a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and answer on the OMR Answer Sheet by filling the circle.

Question 56. 95, 94, 92, 89, 85, 80, ?


a. 78
b. 76
c. 74
d. 72

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 56 solution

Question 57. 0, 6, 24, 60, 120, 210, ?


a. 260
b. 275
c. 310
d. 336

Answer: (d)

0 = 13–1 = 0

6 = 23–2 = 6

24 = 33–3 = 24

60 = 43–4 = 60

120 = 53–5 = 120

210 = 66– 6 = 210

Similarly, 73–7 = 336

Question 58. 720, 360, 120, 30, 6, ?


a. 0
b. 1
c. 2
d. 3

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 58 solution

Therefore, option (b) is correct.

Question 59. 34, 18, 10, 6, 4, ?


a. 3
b. 2
c. 1
d. 0

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 59 solution

Therefore, Option (a) is correct

Question 60. 107, 97, 82, 62, ?


a. 42
b. 47
c. 37
d. 39

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 60

Therefore, option (c) is correct.

Question 61. 1, 2, 5, 12, 27, 58, ?


a. 116
b. 121
c. 125
d. 127

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 61

Therefore, option (b) is correct

Question 62. 2/√5, 3/5, 4/5√5, 5/25

a. 6 / 5√5

b. 6 / 25√5

c. 6 / 125

d. 7 / 25

Answer: (b)

Numerator

It is simply a consecutive number

2 → 3 → 4 → 5 → 6

Denominator

multiply by √5

WB NTSE 2021 Stage 1 MAT Paper Question 62

So, 6 / 25√5 is a required number.

Therefore, Option (b) is correct

Question 63. 3, 12, 27, 48, 75, 108, ?


a. 118
b. 135
c. 147
d. 152

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 63

Therefore, option (c) is correct

Question 64. 840, 168, 42, 14, ?


a. 2
b. 3
c. 5
d. 7

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 64

Therefore, option (d) is correct

Question 65. 4, 5, 7, 11, 19, 35, ?


a. 47
b. 57
c. 67
d. 77

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 65

Therefore, option (c) is correct

Question 66. 11, 10, 101, 100, 1001, 1000, ?


a. 10000
b. 10001

c. 11001
d. 10011

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 66

Therefore, option (b) is correct

Question 67. 2, 7, 27, 107, 427, ?


a. 1262
b. 1707
c. 4027
d. 4207

Answer: (b)

67

Therefore, Option (b) is correct

Question 68. 3, 8, 13, 24, 41, ?


a. 70
b. 75
c. 80
d. 85

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 68

Therefore, Option (a) is correct.

Question 69. 2, 8, 18, 32, 50, ?


a. 60
b. 66
c. 72
d. 82

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 69

Therefore, option (c) is correct.

Question 70. 4, 18, ?, 100, 180, 294


a. 38
b. 42
c. 48
d. 50

Answer: (c)

4 18 48 100 180 294

↑ ↑ ↑ ↑ ↑ ↑

22×1 32×2 42×3 52×4 62×5 72×6

Therefore, option (c) is correct.

Direction (71-80): In each of the questions 71 to 80 there are four items, three of which are alike by some means or other while one is out of the class. Find out the odd items and indicate your answer by filling the circle of the corresponding letter on the OMR Answer-Sheet.

Questions 71. Find out the odd items.


a. Bar
b. Pie
c. Rectangle
d. Pictogram

Answer: (c)

Pictogram, pictogram and bar all use to represent and analyse some set of data, but rectangle is a simple geometrical figure.

Question 72. Find out the odd items


a. Hygrometer
b. Hydrometer
c. Diameter
d. Barometer

Answer: (c)

Hygrometer, barometer and hydrometer are all measuring instruments except diameter. Diameter is the longest possible chord of any circle.

Questions 73. Find out the odd items


a. Sun
b. Earth
c. Moon
d. Universe

Answer: (d)

Sum, moon and earth are all parts of the universe itself.

Question 74. Find out the odd items


a. Iron
b. Mercury
c. Copper
d. Aluminum

Answer: (b)

Iron, copper and aluminium are all solid metals except mercury which is a liquid metal.

Question 75. Find out the odd items


a. Seismograph
b. Anemometer
c. Richter Scale
d. Epicentre

Answer: (b)

Anemometer is used to measure wind speed while the rest three are all related to earthquakes.

Question 76. Find out the odd items


a. Wood
b. Cork
c. Stone
d. Paper

Answer: (c)

Wood, cork and paper are all obtained from trees except stone.

Question 77. Find out the odd items


a. SARS
b. COVID-19
c. EBOLA
d. Typhoid

Answer: (d)

Typhoid is a bacterial infection whereas all the other options are viral infections.

Question 78. Find out the odd items


a. Herpes virus
b. Papilloma Virus
c. Coronavirus
d. Pox virus

Answer: (c)

Coronavirus moves down the respiratory tract. While all other infections affect the skin.

Question 79. Find out the odd items


a. Arc
b. Diagonal
c. Chord
d. Diameter

Answer: (a) and (b)

Arc, chord and diameter are related to circles whereas, Diagonal is related to polygon.

Therefore, Option (b) is correct

Arc is curve line and all other are straight line

Therefore, Option (a) can also be correct

Question 80. Find out the odd items


a. BCIUX
b. BDAYO
c. BEMUZ
d. BQMPX

Answer: (d)

BCIUX (2 + 3 + 9 + 21 + 24) = 59

BDAYO (2 + 4 + 1 + 25 + 15) = 47

BEMUZ (2 + 5 + 13 + 21 + 26) = 67

BQMPX (2 + 17 + 13 + 16 + 24) = 72

Sum of all letters is prime number except BQMPX

Direction (81-85): Choose the correct one and given the answer by filling the circle of the letter denoting your selected answer on the OMR Answer Sheet.

Question 81. Ibrahim ranks 8th in a class of 35 students. What is his rank from the last?


a. 26th
b. 27th
c. 29th
d. 28th

Answer: (d)

No. Of total students in class=35

Ibrahim secure rank = 8th

So, students after him = 35 – 8 = 27

Hence, Ibrahim have rank from last = 28th

Question 82. If the digit 12 of a clock is pointing towards the East, then in which direction will digit 3 point?


a. West
b. South
c. North
d. South-West

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 82

3 will point toward the south.

Question 83. Which one is the most accurate laboratory method for detecting COVID-19 coronavirus?


a. RT-PCR
b. RDT
c. PCR
d. Serology Test

Answer: (a)

Compared to other available virus isolation methods, real-time RT-PCR is significantly faster and has a lower potential for contamination or errors as the entire process can be carried out within a closed cube:

It continues to be the most accurate method available for the detection of the COVID-19 virus.

Question 84. Rahim walks 15 km towards North, From there he walks 9km towards South. Then he walks 8km towards the East. How far and in which direction is he now from his starting point?


a. 7 km North-East
b. 10 km North-East
c. 10 km South-West
d. 7 km South- East

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 84

Given that,

OA = 15, AB = 9, and BC = 8

We have to find OC = ?

WB NTSE 2021 Stage 1 MAT Paper Question 84 solution

OC2 = OB2 + BC2

⇒ OC2 = (6)2 + (8)2

⇒ OC = 10

Therefore he is 10 km in the North-East direction.

Question 85. If 1st January 2008 is Tuesday then what day of the week lies on 1st January, 2009?


a. Wednesday
b. Thursday
c. Sunday
d. Monday

Answer: (b)

1st Jan. 2008 ࢮ Tuesday

Since 2008 is leap year (364 + 2)

So, 1st Jan. 2009 will be Tuesday + 2 = Thursday

Directions (86-90): Here the four fundamental operations +, × and + are represented by symbols different from the usual one. You have to solve the problem by substituting the real symbol accordingly and indicate your answer by filling the Circle of the letter denoting your selected answer on the OMR Answer sheet.

Question 86. If L denotes ×, M denotes ÷, P denotes + and Q denotes, then 7P24M8Q6M2L3 =


a. 1
b. 2
c. 3
d. 4

Answer: (a)

L → ×

M → ÷

P → +

Q → –

7P24M8Q6M2L3

= 7 + 24 ÷ 8 – 6 ÷ 2 × 3

= 7 + – × 3

= 7 + – 9

= 7 + 3 – 9 = 1

Question 87. If A means ‘–’ B means ‘÷’ C means ‘+’ and D means ‘×’, then 15B3C24A12D2 =


a. 4
b. 2
c. 5
d. 3

Answer: (c)

A → –

B → ÷

C → +

D → ×

15B3C24A12D2

= 15 ÷ 3 + 24 – 12 × 2

= 5 + 24 – 24

= 5

Question 88. If A means +; M means ×; D means ÷, G means >, L means < , then which of the following will be logically correct?


a. 4A3M2L4D2M6
b. 4A3M2D3L4M6
c. 4A5M4L6D2A8
d. 4A5D3G6A2M3

Answer: (a, b)

A → +; M → ×; D → ÷, G → >, L → <

Option (a): 4A3M2L4D2M6

4 + 3 × 2 < 4 ÷ 2 × 6

= 10 < 12 [Accepted]

Option (b): 4A3M2D3L4M6

4 + 3 × 2 ÷ 3 < 4 × 6

= 6 < 24 [Accepted]

Option (c): 4A5M4L6D2A8

4 + 5 × 4 < 6 ÷ 2 + 8

= 24 < 11 [Rejected]

Option (d): 4A5D3G6A2M3

4 + 5 ÷ 3 > 6 + 2 × 3

= 17/3 > 12 [Rejected]

Therefore, Option (a, b) both are correct.

Question 89. Select the correct combination of mathematical signs to replace * signs and to balance the given equation:

(5 * 6) * 5 * 8 = 14


a. ×, +, ÷
b. ×, ÷ ,+
c. +, ×, ÷
d. +, ÷, ×

Answer: (b)

(5 * 6) * 5 * 8 = 14

Option (a): ×, +, ÷

(5 × 6) + 5 ÷ 8

= 30 + ⅝ ≠ 14 [Rejected]

Option (b): ×, ÷, +

(5 × 6) ÷ 5 + 8

= 6 + 8 = 14 [Accepted]

Option (c): +, ×, ÷

(5 + 6) × 5 ÷ 8

= 55 ÷ 8 ≠ 14 [Rejected]

Option (d): +, ÷, ×

(5 + 6) ÷ 5 × 8

= 11/5 × 8 ≠ 14 [Rejected]

Therefore, Option (b) is correct

Questions 90. If + means ÷ , – means × , ÷ means + , × means – then 36 × 12 + 4 ÷ 6 + 2 – 3 =


a. 28
b. 32
c. 39
d. 42

Answer: (d)

+ → ÷

– → ×

÷ → +

× → –

36 × 12 + 4 ÷ 6 + 2 – 3

= 36 – 12 ÷ 4 + 6 ÷ 2 × 3

= 36 – 3 + 3 × 3

= 36 – 3 + 9

= 42

Direction (91-95): The diagram shows the survey on a sample of 500 persons with respect to their knowledge of Bengali, Hindi and English.

WB NTSE 2021 Stage 1 MAT Paper Question 91 - 95

Question 91. How many persons know the Bengali and English Language but not know the Hindi Language?


a. 53
b. 62
c. 65
d. 37

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 91 solution

Persons who Bangali and English but not know Hindi know

= 62

Therefore, Option (b) is correct

Question 92. How many people know all three Languages?


a. 65
b. 62
c. 53
d. 47

Answer: (c)

WB NTSE 2021 Stage 1 MAT Paper Question 92 solution

Persons who know all languages = 53

Therefore, Option (c) is correct

Question 93. How many people do not know the Hindi Language?


a. 265
b. 200
c. 255
d. 201

Answer: (a)

WB NTSE 2021 Stage 1 MAT Paper Question 93

Persons who do not know the Hindi language

= 120 + 62 + 83

= 265

Therefore, Option (c) is correct

Question 94. What is the ratio of those who know all three Languages to those who do not know the Hindi Language?


a. 3/5
b. 2/5
c. 1/5
d. 4/5

Answer: (c)

Persons who know all three language = 53

Person who do not known Hindi = 265

Ratio = 53/265 = ⅕

Therefore, Option (c) is correct

Question 95. What is the ratio of those who know only the Bengali Language to those who do not know the Bengali Language?


a. 2 : 5
b. 3 : 5
c. 4 : 5
d. 1 : 5

Answer: (b)

Persons who know only bengali = 120

Who do not known bengali = 70 + 47 + 83 = 200

Ratio = 120/200 = ⅗

Direction (96 & 97): Choose the correct one.

Question 96. If M × N means M is the daughter of N; M + N means M is the father of N; M ÷ N means M is the Mother of N and M – N means M is the brother of N then P ÷ Q + R – T × K indicates which relation of P to K?


a. Daughter-in-Law
b. Sister-in-Law
c. Mother-in-Law
d. Aunt

Answer: (c)

M × N → M is the daughter of N

M + N → M is the father of N

M ÷ N → M is the mother of N

M – N → M is the brother of N

P ÷ Q + R – T × K

WB NTSE 2021 Stage 1 MAT Paper Question 97

So, clearly, P is the mother-in-law of T

Therefore, Option (c) is correct

Question 97. Pointing to a person, a man said to a woman, “His mother is the only daughter of your father’. How was the woman related to the person?


a. Wife

b. Daughter

c. Aunt

d. Mother

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 97 solution

So, the woman is the mother of that person.

Direction (98-100): Read the following information and answer the questions given below. There are six children playing football namely A, B, C, D, E and F, A and E are brothers, F is the sister of E; C is the only son of A’s uncle; B and D are the daughters of the brother of C’s father.

Question 98. How many female players are there?


a. Two
b. Three
c. Four
d. Five

Answer: (b)

WB NTSE 2021 Stage 1 MAT Paper Question 98

The total number of females = 3, F, B and D.

Therefore, Option (b) is correct

Question 99. How C is related to F?


a. Brother
b. Uncle
c. Son
d. Cousin

Answer: (d)

WB NTSE 2021 Stage 1 MAT Paper Question 99

So, C and F are cousins

Therefore, Option (d) is correct

Question 100. How D is related to A?


a. Cousin
b. Sister
c. Niece
d. Uncle

Answer: (a, b)

B and D are daughters of A’s uncle’s brother.

So, A’s uncle brother can be A’s father only or may be the third brother of A’s father

So, A & D may be brother-sister or cousin.

Therefore, option (a, b) is correct.

West Bengal NTSE 2021 Stage 1 SAT Paper

PAPER – SAT

Question 1. If the algebraic expression (3x+ px + 3) be always positive, then


a. –6 < p < 6
b. p < – 6
c. p > 6
d. no such p exists

Answer: (a)

3x+ px+3 >0 (given)

So, D must be negative

i.e D < 0

⇒ b– 4ac < 0 …(i)

Compare 3x2+px+3 = 0 and ax2 + bx + c = 0

a = 3, b = p , c = 3

Now, from (1)

⇒ p2– 4(3)(3) < 0

⇒ p2– 36< 0

⇒ (p–6)(p+6) < 0

WB NTSE 2021 Stage 1 SAT Solved Question 1

⇒ p ∈ (–6, 6) or – 6 < p < 6

Question 2. The sum of the roots of ax2+ bx + c = 0 (a≠0) is equal to the sum of the squares of the roots of the equation. Then


a. 2ab = ac + be
b. b2 = ab +2ac
c. 2ac = ab +b2
d. ab = 2ac + b2

Answer: (c)

ax+ bx + c=0 (a ≠ 0)

Given that, the sum of roots = sum of the square of roots.

Let α, β are roots

Then α+β = α22 (given)

⇒ (α+β) = (α+β)2 – 2αβ …(1)

∵ α + β = -b / a, αβ = c / a

Put these values into the equation (1) we get
(-b/a) = (-b/a)2 – 2c/a

⇒ -b/a = b2/a2 – 2c/a

⇒ -b/a = b2 – 2ac/a2

⇒ –ab = b2 – 2ac

⇒ b2 + ab = 2ac

Question 3. The solution set for the equation is 

\(\begin{array}{l}3^{2x^{2}} – 2.3^{x^{2}+x+2} + 3^{2(x-2)} = 0\end{array} \)


a. {–1, 2}
b. { 0, 2}
c. {0, –1}
d. {2, 4}

Answer: (a)

\(\begin{array}{l}3^{2x^{2}} – 2.3^{x^{2}+x+2} + 3^{2(x-2)} = 0\end{array} \)

\(\begin{array}{l}(3x^{2})^{2} – 2.3^{x^{2}}.3^{x+2} + (3^{(x+2)})^{2} = 0\end{array} \)

\(\begin{array}{l}[3x^{2} – 3^{(x+2)}]^{2} = 0\end{array} \)

\(\begin{array}{l}3x^{2} = 3^{(x+2)}\end{array} \)

Comparing both side

x= x+2

⇒ x– x –2 = 0

⇒ x– 2x +x–2 = 0

⇒ x(x–2)+1 (x–2) = 0

⇒ (x–2) (x+1) = 0

⇒ x= –1,2

⇒ x ∈ {–1,2}

Question 4. It is printed on paper that “The length of a diagonal of a rectangle is 10 cm and its area is 62.5 sq. cm”. Then which one of the following statements is true?


a. The perimeter of the rectangle is 30 cm.
b. The sum of the length and breadth is 20 cm.
c. The difference of the length and breadth is 5 cm.
d. No such rectangle can exist.

Answer: (d)

Let the sides of the rectangle be ‘a’ and ‘b’

WB NTSE 2021 Stage 1 SAT Solved Question 4

Given length of diagonal = 10 cm = √(a2+b2) and area = 62.5 cm2

i.e. ab = 62.5 …(1)

a2+b2 = 100 …(2)

Now, (a+b)2 = a2+b2 + 2ab = 100 + 2(62.5)

(a+b)2 = 100 + 125

(a+b)2 = 225

⇒ a+b = 15 …..(3)

From (1) and (3) we get

a(15-a) = 62.5

15a – a= 62.5

a-15a + 62.5 = 0

Solving this,

\(\begin{array}{l}a = \frac{15+\sqrt{15^{2}-4(1)(62.5)}}{2(1)}\end{array} \)
\(\begin{array}{l}a = \frac{15+\sqrt{225-250}}{2}\end{array} \)

Roots are coming imaginary, hence no such rectangle is possible.

Question 5. The ratio of the lengths of the corresponding sides of two similar triangles are in the ratio of 4 : 9. Then the ratio of their areas is


a. 16 : 81
b. 81 : 16
c. 4 : 9
d. 9 : 4

Answer: (a)

WB NTSE 2021 Stage 1 SAT Solved Question 5

Therefore, option (a) is correct.

Question 6. Father says to his son: “On the date of birth of yours, I was twice older than you are now.” Then which one of the following statements is true?


a. The present ages of the father and his son are 30 years and 10 years respectively.
b. The present ages of the father and his son are 48 years and 12 years respectively.
c. Nothing can be said about their ages.
d. The ratio of the present ages of the father to the son is 3 : 1.

Answer: (a) and (d)

Let present age of father = x years

Let present age of son = y years

When son was born then age of father = (x–y) yrs

Now, x–y = 2y (given)

⇒ x = 3y

or x/y = 3/1 = (father’s present age)/(son’s present age)

Therefore, option (d) is correct.

Also since 30/10 = 3/1, (a) is also correct

Question 7. Taking at least one from the numbers 1,2,3, how many different sets can be formed?


a. 8
b. 7
c. 3
d. 33

Answer: (b)

By taking at least one of the number from 1,2,3,

⇒ Total sets = 23–1

⇒ 8–1 = 7

which are

{1} {2} {3}

{1,2} {1,3} {2,3}

{1,2,3}

Question 8. (18)23 is divided by remainder 17 to give the remainder


a. 11
b. 9
c. 5
d. 1

Answer: (d)

=(17+1)23/17

When we open the bracket, all term except last will be divisible by 17

Hence remainder = 1

Question 9. If m2 – 4m + 1 = 0, then the value of (m3 + 1/m3)


a. 52
b. 48
c. 64
d. 68

Answer: (a)

m+ 1 = 4m (given)

Divided both side by , ‘m’ we get

⇒ m + 1/m = 4

Using identity a+ b3 = (a+b)3–3 ab(a+b)

So, m3 + 1/m3 = (m + 1/m)3 – 3m(1/m)(m + 1/m)

= (4)– 3 × 4

= 64 –12

= 52

Question 10. For 3x + y = 81, 81x – y = 3, we get


a. no solution
b. X + 21/2, y = 21/2
c. x = 2, y = 2/3
d. x = 17/8, y = 15/8

Answer: (d)

3x+y = 34 and

34(x–y) =31 (Given)

By comparing the powers , we get

⇒ x + y = 4 …(i)

⇒ x – y = …(ii)

From (i) and (ii), we get

⇒ x = 17/8 and y = 15/8

Question 11. 125 identical cubes are cut from a big cube and all the smaller cubes are arranged in a row to form a long cuboid. What is the percentage of increase in total surface area of the cuboid over the total surface area of the cube?:


a.

\(\begin{array}{l}234\tfrac{2}{3}\end{array} \)
%

b.
\(\begin{array}{l}235\tfrac{1}{3}\end{array} \)
%

c.
\(\begin{array}{l}134\tfrac{2}{3}\end{array} \)
%

d. none of these

Answer: (a)

WB NTSE 2021 Stage 1 SAT Solved Question 11

Clearly volume of larger cube = volume of 125 cubes

⇒ a3 = 125 b3

Now, surface area of larger cube = 6a2 = 6(5b)2 = 150b2

New surface area is equal to 4(125b x 1b) + 2(1bx1b) = 502b2

Increase percentage in surface area = 502b2 – 150b2 / 150b2 × 100

= 352b2 / 150b2 × 100

= 704 / 3 = 234 2/3 %

Question 12. A bag contains 4 Red and 3 Black balls and a second bag contains 2 Red and 4 Black balls. After choosing a bag at random, a ball is also drawn at random. The probability that the ball is Red is;


a. 23/42
b. 17/42
c. 19/42
d. 16/39

Answer: (c)

WB NTSE 2021 Stage 1 SAT Solved Question 12

Now, Probability of red ball = 1/2 × 4/7 + 1/2 × 2/6

= 4/14 + 2/12

= 2/7 + 1/6

= (12+7)/42 = 19/42

Question 13. If [n] denotes the greatest integer ≤ n and (n) denotes the smallest integer ≥ n ; n being a real number, then:

\(\begin{array}{l}(1\tfrac{1}{5}) \times [1\tfrac{1}{5}] – (2 – ⅕) \div [1\tfrac{1}{5}] + (1.5) \end{array} \)
is


a. 1.5
b. 2
c. 2.5
d. 3.5

Answer: (b)

(6/5) x [6/5] – (9/5) ÷ [6/5] + (1.5)

= 2

Question 14. The value of 3 / 12.22 + 5 / 22.32 + ……..+ 19/92.102is


a. 99 / 100
b. 1 / 100
c. 101 / 100
d. 1

Answer: (a)

3 / 1.4 + 5 / 4.9 + 7 / 9.16 +………+ 19 / 81.100

= 4 – 1 / 1.4 + 9 – 4 / 4.9 + 16 – 9 / 9.16 + ……… + 100 – 81 / 81.100

 = (1 – 1 / 4) + (1 / 4 – 1 / 9) + (1 / 9 – 1 / 16) ……….+ (1 / 81 – 1 / 100)

= 1 – 1 / 100 

= 99 / 100

Question 15. In a frequency distribution, Mean = 9.1 and = 132,+ 5k, = 20 , then k is


a. 4
b. 6
c. 10
d. 9

Answer: (c)

Mean = , given that Σfix= 132 + 5k, and =20, mean = 9.1

So, 9.1 =

⇒ 182 = 132 + 5k

⇒ 5k = 50

∴ k = 10

Question 16. ABC is a triangle in which ∠ABC > 90° and AD ⟂ CB produced. Then

WB NTSE 2021 Stage 1 SAT Solved Question 16


a. AB2 = AC2 + BC2+BC × BD
b. AB2 = AC2+BC2+2BC × BD
c. AC2 = AB2 + BC2 + 2BC × BD
d. AC2 = AB2 + BC2 + BC × BD

Answer: (c)

By Pythagoras theorem

AC2 = AD2 +CD….(i)

AB2 = AD2 +BD….(ii)

Also, from (1)

⇒ AC2 = AD2 + (BD+BC)2

⇒ AC2 = AD2 + BD+ BC+ 2BD × BC

⇒ AC2 = AB2 + BC+ 2BC × BD

Therefore, option (c) is correct.

Question 17. Given : 0° < θ < 90°. then, if sinθ + cosθ = x, which one of the following is correct?


a. x < 1
b. x > 1.5
c. 1 ≤ x ≤ √2
d. 1 < x ≤ √2

Answer: (d)

sinθ + cosθ = x (given)

multiply and divide by √2 we get

⇒ √2 (1/2 sin θ + 1 / √2 cos θ = x

⇒ √2 sin (θ + 𝜋/4) = x ( sin (A+B) = sin A cos B + cos A sin B)

Since, 0 < θ, 90°

1/√2 < sin (θ + 𝜋/4) ≤ 1

So, 1 < √2 (θ + 𝜋/4) ≤ √2

Hence , 1 < x ≤ √2 

Question. 18. If sin θ + cos φ = 2 and 0° ≤ = θ, φ ≤ 90°, then 2θ + 2000.φ


a. 180°
b. 90°
c. 2180°
d. Can not be found out

Answer: (a)

sinθ + cosφ =2, (given)

So, sinθ = 1, & cosφ = 1

Hence, θ = 90° and φ = 0°

⇒ 2θ + 2000φ

= (2 × 90°) + (2000 × 0) = 180°

Question 19. secθ + tanθ = a + 

\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)
a, b ∈ Q and 
\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)
∈ Q and sec θ is rational, then


a.

\(\begin{array}{l}\bar{b^{2}} = 1 + a\end{array} \)

b. 
\(\begin{array}{l}{a^{2}} = 1 + \bar{b}\end{array} \)

c. 
\(\begin{array}{l}\bar{b^{2}} = 1 – a\end{array} \)

d. 
\(\begin{array}{l}{a^{2}} = 1 – \bar{b}\end{array} \)

Answer: (b)

secθ + tanθ = a+

\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)
 (given)

By comparing rational and irrational terms, we get

secθ = a (as sec θ is rational)

and tanθ =

\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)
(as
\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)
is irrational)

Since, sec2θ = 1+tan2θ

So, a2 = 1 + 

\(\begin{array}{l}\sqrt{\bar{b}}\end{array} \)

⇒ a2 = 1 + 

\(\begin{array}{l}\bar{b}\end{array} \)

Question 20. The ratio in which 9x – 3y – 14 = 0 divides the join of (2, – 4) and (3, 7) is:


a. 2 : 1
b. 1 : 2
c. 2 : 3
d. 3 : 2

Answer: (a) or (b)

Let the ratio is λ : 1 apply section formula.

WB NTSE 2021 Stage 1 SAT Solved Question 20

x = 3λ +2 / λ + 1; y = (7λ –4 / λ +1)

Since line 9x = 3y + 14 passes through (x,y), it will satisfy (x,y)

So, 9 (3λ +2/ λ +1) = 3 (7λ –4/λ +1) + 14

⇒ 27λ + 18 = 21λ – 12 + 14λ + 14

⇒ -8λ = -16

⇒ λ = 2

∴ 1 = 21

So, the ratio is 2:1

Since any particular order is not mentioned, the ratio can also be 1:2 as well

Hence both (a) and (b) can be the answer

PHYSICS

Question 21. The mass, linear momentum and kinetic energy of a body are m, p and E respectively, then


a. p = √2mE
b. E = √2mp
c. p = √2E
d. E = √2p

Answer: (a)

As we know, linear momentum is given by

P = mv …(1)

And kinetic energy

Ek =E = 1/2 mv2

Or

E = 1/2 mv2 × m/m = 1/2 m2v2/m = 1/2 (mv)2/m From equation (1),

E = 1/2 p2/m 

Or

P2 = 2mE

P = √2mE

Question 22. A stone is allowed to fall freely downwards initially at rest from the top of a tower. The time taken by the stone to reach the bottom of the tower is 4 seconds. What is the height of the tower? Take, acceleration due to gravity = 32 ft/s2


a. 64 ft.
b. 32 ft.
c. 48 ft.
d. 256 ft.

Answer: (d)

Given :

u = 0

t = 4sec.

a = g = 32 ft/sec2

WB NTSE 2021 Stage 1 SAT Solved Question 22

From 2nd equation of motion,

S=ut+ at2

h = 0 + × 32 × (4)2

h = 256 ft

Question 23. The coefficient of linear expansion of a solid is x and the coefficient of volume expansion of the solid is y, then:


a. x = y/3
b. y = x/3
c. x = y/2
d. y = x/2

Answer: (a)

Given :

Coefficient of linear expansion (α) = x

Coefficient of volume expansion (β) = y

As we know, linear expansion is given by

l = lo (1 × ΔT) …(1)

Or

α = x = Δl / loΔT

Now as we know, volume = (length)3

V = l3

From equation (1),

V = [V(1+αΔT)]3 = lo3 (1+αΔT)3

V = V0 (1+αΔT)3

By using the binomial expansion,

V = V0 (1+3αΔT) = V0 + 3VαΔT

Or

= 3α

= 3α

β = 3α

Or

V – V0 / V0ΔT = 3α

ΔV / V0ΔT = 3α 

β = 3α

Or 

y = 3x or x = y/3

Question 24. In case of refraction of light from a medium to air, the critical angle is found to be 45°. What is the refractive index of the medium with respect to air? .


a. √2
b. √3
c. 2
d. 3

Answer: (a)

Given : critical angle (ic) = 45°

As we know, in total internal reflection

μ = 1 / sin sin ic

Or 

μ =1 / sin sin 45° = 1 / 1/√2

μ = √2

Question 25. Which of the following pair have the same unit?


a. Heat and Specific heat
b. Thermal capacity and Water equivalent
c. Specific heat and Thermal capacity
d. Heat and Work

Answer: (d)

unit of heat = Joule

unit of specific heat = Joule/kg × kelvin

unit of thermal capacity = Joule/kelvin

unit of water equivalent = kg

unit of work = joule

∴ unit of heat and work is the same.

Question 26. Which of the following is an electromagnetic wave?


a. α – ray
b. β -ray
c. γ -ray
d. cathode ray

Answer: (c)

When an Electromagnetic wave is passed through the electric or magnetic field, it remains undeflected. Therefore, γ – ray is an electromagnetic wave because it is undeflected.

Question 27. In case of a convex lens, what is the minimum distance between an object and its real image?


a. 2.5 times of focal length
b. 2 times of focal length
c. 4 times of focal length
d. equal to focal length

Answer: (c)

WB NTSE 2021 Stage 1 SAT Solved Question 27

Therefore, the minimum distance of a real image is 2f, when the object is also placed at 2f i.e., the minimum distance between them is 4f.

Question 28. What will be the power consumed by a 50Ω wire if it is kept across a potential difference of 200 V?


a. 0.8 kW
b. 80 kW
c. 400 W
d. 8 kW

Answer: (a)

Given : Resistance R = 50 Ω

Potential difference V = 200 Volt

Now power consumed, P = V2/R = 2002/5 = 800 wall

Or

P = 800 × 10/10 = 8/10 ×103 watt

P = 0.8 Kwatt

Question 29.1 cm of the main scale of a vernier calliper is divided into 10 divisions. The least count of the callipers is 0.005 cm, then what is the number of divisions in the vernier scale?


a. 10
b. 20
c. 25
d. 50

Answer: (b)

Each division on MSR = 10/10 = 0.1 cm

Let N be the number of division that must present on the vernier scale so that each division on MSR corresponds to the maximum value that the vernier scale can measure.

N × 0.005 cm = 0.1 cm

N = 0.1 / 0.005 = 20

N = 20 divisions

Question 30. If an x-ray tube is operated at 20kV, what is the cut-off wavelength? (Take, Planks constant h = 6.62 × 10–34 J.S)


a. 0.89 Å
b. 0.75 Å
c. 0.62 Å
d. 0.31 Å

Answer: (c)

Given :

V = 20 K volt = 20×103 Volt

h = 6.62×10–34 J.S

C = 3×108 m/sec

As we know at the cut-off region,

Emax = eV

hv = eV

hc / λmin = eV

λmin = hc/eV = 6.62 × 10-34 × 3 × 108 / 1.6×10-19 × 20 × 103

λmin = 0.62Å 

Question 31. An ideal gas is found to obey the equation p2V = constant along with the ideal gas equation (here, p = pressure and V= volume). If initial temperature and volume are T0 and V0 respectively and it expands to a volume 3V0, then what is the final temperature?


a. √3 T0 /
b. √2 T0 /
c. T0 / √3
d. T0 / √2

Answer: (a)

Given :

Initial temperature = T0

Initial volume = V0

Final volume = 3V0

P2V = constant …(1)

For an ideal gas PV = nRT,

P = (nRT/V)

From equation (1),

(nRT/V)2 V = constant

T2/V = constant

Now,

T2i / Vi = T2f / Vf 

T2= T2i Vf / Vi

= T2o × 3Vo/Vo 

 Tf = To √3 = √3 To

Question 32. Specific heat (S) of metal at low temperature varies according to S = aT3, where ‘a’ is a constant and T is an absolute temperature. The amount of heat energy needed to raise the temperature of unit mass of the metal from T = 1K to T = 2K is


a. 3a
b. 15a/4
c. 2a/3
d. 13a/4

Answer: (b)

Given :

specific heat (S) = aT3

Initial temperature Ti = 1K.

Final temperature Tf = 2K

Mass m = 1 unit

Heat is given by dQ = msdt

Integrating on the both sides,

\(\begin{array}{l}\int_{O}^{Q} dQ = \int_{T_{i}}^{T_{f}}m(at^{3})dt \: \: \: \: {S = aT^{3}}\end{array} \)
\(\begin{array}{l}[Q]_{o}^{Q} = ma\int_{1}^{2}T^{3}dt\end{array} \)
\(\begin{array}{l}Q = ma \left [ \frac{T^{4}}{4} \right ]_{1}^{2}\end{array} \)

= ma/4 [(2)4–(1)4]

= 15ma/4

Or

Q = 15a/4 { m = 1 unit}

Question 33. An object of weight W and density ρ is submerged wholly in a liquid of density σ, its apparent weight will be


a. (ρ –σ)
b. (ρ –σ)/W
c. W (1 – σ/ρ)
d. (1 – ρ/σ)

Answer: (c)

Given :

weight of object in air/ vacuum = W

Density of object = ρ

Density of liquid = σ

WB NTSE 2021 Stage 1 SAT Solved Question 33

Buoyant force

FB = ρLgVL

Volume of liquid = liquid displaced by object

VL = Vobject = W/ρg

∴ FB = σ g W/ρg

Now apparent weight,

W’ = W –FB

= W – σgW/ρg

W’ = W (1 – σ/ρ)

CHEMISTRY

Question 34. The ratio of σ and π bonds in propyne is


a. 1 : 3
b. 3 : 1
c. 2 : 3
d. 3 : 2

Answer: (b)

The structure of the propyne is as follows:

WB NTSE 2021 Stage 1 SAT Solved Question 34

There are one sigma and two pi bonds in the triple bond between two carbon atoms. Therefore, the total number of sigma bonds is 6 and pi bonds is two in the structure of the propyne.

Question 35. The element having the lowest first ionization energy is


a. He
b. Cl
c. F
d. I

Answer: (d)

Helium has a complete shell and has high ionisation energy. On moving down the halogen group, the number of electron shells increases. With the increase in the distance between the valence electron and the nucleus, less energy is required to remove the electron as they are less tightly held to the nucleus. Therefore, Iodine is having the lowest ionization energy.

Question 36. If the four tubes of a car are filled to the same pressure with N2, O2, H2 and Ne gas separately then which will fill the tube first?


a. N2
b. O2
c. H2
d. Ne

Answer: (c)

Higher the density of a gas, slower will be the tube filling. Hence, H2 will have the highest rate of diffusion as it is least dense.

Question 37. At a given temperature what will be the percentage increase in pressure for a 5% decrease in the volume of the gas


a. 5%
b. 5.26%
c. 6.26%
d. 10.26%

Answer: (b)

We can apply Boyle’s Law ss temperature remains constant, the law can be usefully expressed as P₁V₁ = P₂V₂.

V₂ =0.95 V₁ (as V₂ is 5% less than V₁)

P₁ × V₁ = P₂ × V₂

P₁ × V₁ = P₂ (0.95 × V₁)

P₂ = P₁/ 0.95

P₂ = 1.0526 P₁

% increase in pressure =

\(\begin{array}{l}\frac{Final Pressure – Initial Pressure}{Initial Pressure}\times 100\end{array} \)

% increase in pressure = 

\(\begin{array}{l}\frac{1.0256P_{1} – P_{1}}{P_{1}}\times 100\end{array} \)

Therefore, % increase in pressure is 5.26.

Question 38. O2– is isoelectronic with


a. H2
b. N2
c. F2
d. HF2

Answer: (c)

Two or more molecular entities are described as isoelectronic if they have the same electronic configuration or the same structure.

In O2, the number of electrons is 18 which is similar to the F2 molecule which also contains the same number of electrons.

Question 39. Which of the following forms a homologous series?


a. Ethane, Ethylene, Acetylene
b. Ethane, Propane, Butanone
c. Methanol, Ethanol, Propanoic acid
d. Butane, 2-Methyl Butane, 2, 3, Dimethyl Butane

Answer: (d)

Butane, 2- methyl butane, 2,3-dimethyl butane all are alkanes and belong to the same homologous series with the general formula CnH2n+2 .

Question 40. The gas that gives a black precipitate with aqueous Pb(NO3)2 solution and a white precipitate with aqueous ZnCl2 solution is


a. CO2 /
b. NO2 /
c. NH3 /
d. H2S

Answer: (d)

Lead (II) ions react with hydrogen sulphide to form black ppt of lead sulphide.

Pb(NO3)2 + H2S ➝ PbS + 2HNO3

When H2S gas is passed into an aqueous solution of ZnCl2, Zn2+ ions are completely precipitated forming a white precipitate of zinc sulphide.

ZnCl+ H2S ➝ ZnS +2HCl

Question 41. The organic product that is obtained by absorbing Ethylene into concentrated H2SO4 and subsequently boiling the mixture with water is


a. an aldehyde
b. an amide
c. a ketone
d. an alcohol

Answer: (d)

When ethane gas reacts with conc. H2SO4, it forms ethyl hydrogen sulphate. When this is further reacted with water, it will form ethyl alcohol.

WB NTSE 2021 Stage 1 SAT Solved Question 41

Question 42. Equal volumes of two solutions of pH = 4 and pH = 10 are mixed. pH of the resultant solution will be


a. 6
b. 7
c. 8
d. 9

Answer: (b)

pH= 4; pH=10

For pH= 4, [H+] = 10-4 M

For pH= 10, [H+] = 10-10M or OH = 10-4 M

Thus, they will neutralise each other and form a pH of 7.

Question 43. Find out the position isomers from the following pairs of compounds


a. WB NTSE 2021 Stage 1 SAT Solved Question 43
b. WB NTSE 2021 Stage 1 SAT Solved Question
c. WB NTSE 2021 Stage 1 SAT Solved Question 43
d.WB NTSE 2021 Stage 1 SAT Solved Question 43

Answer: (c)

Position isomers differ in the position of the substituent atom or group or an unsaturated linkage in the same carbon chain. Therefore, in option (c) both the structures are position isomers.

Question 44. Egg albumin in water is a


a. True solution
b. Colloid
c. Suspension
d. Supersaturated Solution

Answer: (b)

A solution of egg albumin in water is a colloidal solution.

Question 45. Lithium is generally used as an electrode in high energy density batteries. This is because:


a. Lithium is the lightest of all metals
b. Lithium has quite a high negative reduction potential
c. Lithium is quite reactive
d. Lithium does not corrode easily

Answer: (b)

Lithium has quite a high negative reduction potential as it is a strong reducing agent. Hence, when a lithium electrode is coupled with any electrode, the cell thus constructed has positive EMF and the cell reaction will be spontaneous. Thus, the cell will be able to generate power.

Question 46. 

\(\begin{array}{l}NH_{3} + O_{2}\xrightarrow[\Delta ]{Pt – Rh}A + H_{2}O\end{array} \)
A + O➝ BB + O2 + H2O ➝ B. A, B and C respectively are


a. N2O, NO2 and HNO3
b. NO, NO2 and HNO3
c. NO2, NO and HNO3
d. N2O, NO and HNO3

Answer: (b)

4NH3 + 5O 4NO + 6H2O

2NO + O 2NO2

4NO2 + O2 + 2H2 4HNO3

BIOLOGY

Question 47. The main plant body of pteridophyte is


a. Sporophyte
b. Prothallus
c. Spore
d. Gametophyte

Answer: (a)

Pteridophytes show alternation of generation. Their main plant body is a sporophyte. It is diploid and differentiated into true root, stem and leaves. The sporophyte produces spores and these spores give rise to gametophytes. The gametophyte is free-living and produces gametes by the mitotic division. The gametophyte is haploid.

Question 48. In human eye, at the blind spot


a. Only rod cells are present.
b. Only cone cells are present.
c. Both rod and cone cells are present.
d. Neither rod nor cone cells are present.

Answer: (d)

The blind spot is where the optic nerve and blood vessels leave the eyeball. It has no light-sensitive cells i.e., rods and cones. So, there is no image deduction in this area and is known as the blind spot.

Question 49. Percentage of O2 present in inhaled air in human beings is approximately


a. 21%
b. 77%
c. 0.04%
d. 3%

Answer: (a)

Atmospheric air contains 21% oxygen. And we inhale this air. So, only 21% of the inhaled air is oxygen.

Question 50. The disease which usually spreads through cuts and wounds is


a. Chickenpox
b. Malaria
c. Tuberculosis
d. Tetanus

Answer: (d)

Tetanus is an infection caused by a bacteria called Clostridium tetani. These bacteria are usually found in soil, dust, and manure and enter the body through cuts or puncture wounds caused by contaminated objects. Chickenpox and tuberculosis are both transmitted from person to person through droplets.

Malaria is a vector-borne disease.

Question 51. Lysosome stores


a. ATP
b. Hydrolytic enzymes
c. Carbohydrate
d. Protein

Answer: (b)

A lysosome is a membrane-bound cell organelle that contains hydrolytic enzymes that have the capability to break down many types of biomolecules. These hydrolytic enzymes help in digesting all the unwanted cell parts, and older cells.

Question 52. Which muscle separates thoracic and abdominal cavities?


a. Abdominal muscle
b. Smooth muscle
c. Diaphragm
d. Cardiac muscle

Answer: (c)

The diaphragm is a dome-shaped, muscular and membranous structure that separates the thoracic cavity, containing the heart and lungs, from the abdominal cavity. It helps in the process of breathing.

Question 53. Which one of the following hormones helps in the contraction of uterine muscles during parturition?


a. Vasopressin
b. Oxytocin
c. Prolactin
d. Relaxin

Answer: (b)

Oxytocin acts on the uterine muscle and causes stronger uterine contractions which result in stronger and stronger contractions. This leads to the expulsion of the baby out of the uterus through the birth canal. This process is called parturition.

Vasopressin plays a key role in maintaining osmolality.

Prolactin is responsible for the secretion of milk.

Relaxin relaxes the ligaments in the pelvis and softens and widens the cervix.

Question 54. Glomerulus and Bowman’s capsule together form


a. Malpighian tubule
b. Malpighian corpuscle
c. Collecting tubule
d. Renal tubule

Answer: (b)

Glomerulus and Bowman’s capsule together form Malpighian corpuscles. They are found in the cortex part of the kidney. They perform the function of filtering blood.

Question 55. Cardiac muscle is:


a. striated and voluntary
b. smooth and voluntary
c. striated and involuntary
d. smooth and involuntary

Answer: (c)

Cardiac muscle is involuntary, striated (have both light and dark bands). It constitutes the main tissue of the walls of the heart and helps in the contraction and relaxation of the heart.

Question 56. In which of the following subphases of meiosis crossing over takes place?


a. Leptotene
b. Pachytene
c. Zygotene
d. Diplotene

Answer: (b)

The Pachytene stage is characterized by the appearance of recombination nodules, the sites at which crossing over occurs. Crossing over is the exchange of genetic material between two homologous chromosomes.

Question 57. The part of the human alimentary canal where no enzyme secretion takes place is


a. Mouth
b. Oesophagus
c. Stomach
d. Ileum

Answer: (b)

The oesophagus is one part of the digestive system where no digestion occurs due to a lack of enzymes. It simply passes the food from mouth to stomach by peristalsis.

Question 58. Marine fish when placed in tap water bursts because of


a. Endosmosis
b. Exosmosis
c. Diffusion
d. Plasmolysis

Answer: (a)

Marine fish when placed in tap water bursts because of endosmosis. The solution inside the fish is hypertonic due to the presence of salt and minerals as compared to tap water. Water enters through the cell membrane of fish. So, it bursts and dies.

Question 59. “Penicillin” obtained from a fungus is an


a. Antiseptic
b. Antiserum
c. Antibody
d. Antibiotic

Answer: (d)

Penicillin is a group of antibiotics originally obtained from the fungus Penicillium notatum. It was discovered by Alexander Flemming. It is used to treat bacterial infections.

Question 60. The immunoglobulin which is transported to the foetus through the placenta from the mother is


a. IgG
b. IgA
c. IgE
d. IgM

Answer: (a)

The mother’s milk colostrum contains a high percentage of IgG. It is the main type of antibody found in blood and extracellular fluid, allowing it to control infection of body tissues. Therefore, in the first six months of life, the newborn can defend itself against all the pathogens encountered. IgA is produced during allergic reactions. IgM is the first antibody produced by the immune system when a virus attacks. IgA is the first line of defence found in tears, saliva etc.

Social

Question 61. France was named as “a museum of economic errors” by


a. Rousseau
b. Adam Smith
c. Montesquieu
d. Quesnay

Answer: (b)

Adam Smith was of the opinion that France was a “museum of economic errors”. He is considered as the “Father of Economics” and known for his magnum opus, The Wealth of Nations.

Question 62. Who was known as the “Tsar the Liberator”?


a. Tsar Nicholas I
b. Tsar Nicholas II
c. Tsar Alexander I
d. Tsar Alexander II

Answer: (d)

Tsar Alexander II or Aleksandr Nikolayevich was the emperor of Russia from 1855-1881. He became the king during the Crimean War and introduced a series of reforms at the end of the war. He also liberated the serfs in Russia in 1861, which earned him the nickname of “Tsar the Liberator”. Serfs belonged to the lowest rung of the Russian feudal society.

Question 63. The day 24th October, 1929 was marked as ‘Black Thursday’ in the U.S.A. because


a. Terrorist Attack
b. Natural Calamity
c. The Great Economic Depression
d. Change of Political Background

Answer: (c)

The day of 24th October, 1929 is known as ‘Black Thursday’ in the United States of America. The day is associated with the collapse in the share prices at the New York Stock Exchange. This heralded the beginning of the Great Depression, which lasted for a decade.

Question 64. The father of British Socialism


a. Louis Blanc
b. Karl Marx
c. Robert Owen
d. Saint Simon

Answer: (c)

Robert Owen is considered the father of British Socialism. He was a Welsh industrialist and social reformer. He was also one of the proponents of utopian socialism and the cooperative movement. In 1825, he built a cooperative community called ‘New Harmony’ in Indiana, United States of America.

Question 65. Jagannath Singh Dhol was the leader of


a. Kol Rebellion
b. Santhal Rebellion
c. Munda Rebellion
d. Chuar Rebellion

Answer: (d)

The Chuar Rebellion was a tribal uprising against the high taxes imposed by the British on zamindars. The rebellion lasted for three decades from 1768-69 to 1799. Jagannath Singh Dhol was one of the first zamindars to take arms against the British East India Company in 1768.

Question 66. The editor of the ‘Samachar Chandrika’ was


a. Rammohan Roy
b. Iswar Gupta
c. Bhabani Charan Bandyopadhyay
d. Gangakishore Bhattacharya

Answer: (c)

Bhabani Charan Bandopadhyay was the editor of the Samachar Chandrika. It was an orthodox Hindu weekly newspaper founded in 1822. The weekly campaigned against the ban on Sati.

Question 67. Madari Pasi was the leader of


a. Santhal Rebellion
b. Munda Uprising
c. Bhil Revolt
d. Eka Movement

Answer: (d)

Madari Pasi was the leader of the Eka Movement. Several peasant rebellions broke out in India after the First World War, Eka Movement was one of them. This movement started in the Hardoi district and later spread to other places like Kanpur, Lucknow etc. It attempted to create a peasant consciousness without the constraints of religion and caste.

Question 68. ‘May Day’ was celebrated for the first time in India at


a. Bombay
b. Calcutta
c. Madras
d. Kanpur

Answer: (c)

1st May is also celebrated as International Workers’ Day or Labour Day or May day in all parts of the world. May Day was celebrated for the first time in Madras on 1st May, 1923. Malayapuram Singaravelu Chettiar, a Communist leader, raised the red flag for the first time in India on this day. At the same event, he explained the ideology of his party and the importance of labour movements in general.

Question 69. The Mahad Satyagraha was organised by


a. Dayananda Saraswati
b. Swami Vivekananda
c. Sree Narayana Guru
d. Dr. B. R. Ambedkar

Answer: (d)

Dr. B.R. Ambedkar organised the Mahad Satyagraha. On 20th March 1927, he led a group of Dalits or untouchables to fetch water from a public tank in Mahad, Maharashtra. This event came to be known as the Mahad Satyagraha. To commemorate this historical event, March 20 is also celebrated as Social Empowerment Day in India.

Question 70. The incident of Chauri Choura took place in


a. 1919 AD
b. 1920 AD
c. 1922 AD
d. 1925 AD

Answer: (c)

On 4th February 1922, a group of protestors participating in the Non-Cooperation Movement clashed with the police who opened fire. The angry protestors in turn set the police station on fire, killing all its occupants. This took place in Chauri Chaura, in Gorakhpur district of erstwhile United Provinces. On seeing the movement turn violent, Mahatma Gandhi put a halt to the Non-Cooperation Movement in the subcontinent.

Question 71. The first language state was formed in Independent India:


a. Andhra Pradesh
b. West Bengal
c. Tamil Nadu
d. Gujarat

Answer: (a)

Andhra Pradesh was the first state to be formed on linguistic grounds in independent India on 1st October, 1953. The state underwent several reorganisations, the latest one being the formation of Telangana on 2nd June, 2014. The linguistic reorganisation of Andhra Pradesh paved the way for the reorganisation and creation of several states in India.

Question 72. The writer of the book named “Chhere Asha Gram” was


a. Manikuntala Sen
b. Dakshina Ranjan Basu
c. Sankha Ghosh
d. Selina Hossain

Answer: (b)

Chere Asa Gram -The Abandoned Village is a collection of after-partition stories of Hindu-Bengalis from various villages. Chhere Asha Gram was written by Dakshina Ranjan Basu.

Question 73. The time difference between Greenwich Mean Time and the Indian Standard Time is


a. 6 hours
b. 5 hours 30 minutes
c. 5 hours 15 minutes
d. 5 hours

Answer: (b)

The latitudinal and longitudinal extent of India’s mainland is about 30°. There is a time lag of 2 hours from Gujarat to Arunachal Pradesh. The standard meridian of India which is at 82°30’E comes in the middle of the country and is therefore considered as the standard time for the whole country. It passes through Mirzapur in Uttar Pradesh creating a time difference of 5 hours 15 minutes between the Greenwich Mean Time and the Indian Standard time.

Question 74. By nature, the Western Ghat is a/an


a. Old fold mountain
b. Young fold mountain
c. Block mountain
d. Igneous mountain

Answer: (a)

The Western Ghats is an old fold mountain that covers the six Indian states of Gujarat, Maharashtra, Goa, Karnataka, Tamil Nadu and Kerala. The mountain range peaks at Anamudi at an altitude of 2695 metres in the borders of Idukki and Ernakulam district in Kerala. The Western Ghats are known as Sahyadri in Maharashtra, Nilgiri Hills in Karnataka and Tamil Nadu and Anaimalai and Cardamom Hills in Kerala. This range is one of the ‘Hottest Hotspot’ of biodiversity in the world.

Question 75. The process by which, the height of the earth surface increases is


a. Aggradation
b. Degradation
c. Weathering
d. Denudation

Answer: (a)

The process by which the height of the earth surface increases is called Aggradation. It is a result of the deposition of solid materials by exogenic forces in basins and depressions. Exogenic forces include actions of wind, water, glaciers, groundwater, waves etc.

Question 76. Deep cracks on the surface of the mountain glacier are called


a. Nunatak
b. Arete
c. Crevasse
d. Cirque

Answer: (c)

A crevasse is a deep, wedge-shaped opening or cracks on the surface of the mountain glacier. It usually forms in the top 50 meters of a glacier, where the ice is fragile. Crevasses range up to 20 meters wide, 45 meters deep, and several hundred metres long.

Question 77. The climate which is found in the Cape Town of South Africa is


a. Equatorial climate
b. Tropical Monsoon climate
c. Mediterranean climate
d. Hot Desert climate

Answer: (c)

Cape Town is a port city on South Africa’s southwest coast. It has a Mediterranean-style climate. This type of climate is characterised by wet and cool winters, dry and warm summers.

Question 78. New South wales current flows along the_________of Australia.


a. northern side
b. southern side
c. eastern side
d. western side

Answer: (c)

The New South Wales current flows along the eastern coast of Australia. It sweeps warm water down the east coast of Australia, carrying nutrient-poor water from the Coral Sea into the cool waters of the Tasman Sea.

Question 79. Which of the following, parallel of latitude, passes through the middle of India?


a. Equator
b. Tropic of Capricorn
c. Prime Meridian
d. Tropic of Cancer

Answer: (d)

The Tropic of Cancer, i.e. 23.5° N latitude passes through the middle of India. It passes through eight Indian states including Gujarat, Rajasthan, Madhya Pradesh, Chhattisgarh, Jharkhand, West Bengal, Tripura and Mizoram.

Question 80. The lake located, in between the deltas of the Godavari river and the Krishna river is


a. Kolleru
b. Pulicat
c. Chilka
d. Vembanad

Answer: (a)

Kolleru Lake, one of the largest freshwater lakes in India located in the state of Andhra Pradesh. Kolleru spans two districts – Krishna and West Godavari and is located between Krishna and Godavari deltas.

Question 81. Salty sea breeze is needed for


a. Sugarcane cultivation
b. Tea cultivation
c. Jute cultivation
d. Coffee cultivation

Answer: (c)

Salty sea breeze is needed for jute cultivation.

Question 82. Which of the following industry is called Footloose Industry?


a. Iron and Steel Industry
b. Engineering Industry
c. Automobile Industry
d. Cotton textile Industry

Answer: (d)

The footloose industry refers to an industry that can be placed and located in various places without much influence from raw material factors. The cotton industry concerns principally with twirling and knitting. Textile manufacturing need not be located close to either the source of the fibre it utilises or its final market. Moreover, the weight of raw material and finished goods is almost the same, this does not affect the location of the industries. Thus, the cotton textile industry has been a “footloose” industry.

Question 83. Diamond Quadrilateral project is related to


a. Air Transport
b. Rail Transport
c. Road Transport
d. Water Transport

Answer: (b)

The Diamond Quadrilateral is a project of the Indian Railways to establish a high-speed rail network, which will connect the four metros, namely Delhi, Mumbai, Kolkata and Chennai. Diamond Quadrilateral project aims to drastically reduce the travelling time by introducing high-speed trains.

Question 84. The colour used for drawing of contour lines in a topographical map is


a. Black
b. Brown
c. Red
d. Blue

Answer: (b)

In a topographical map, a contour line is drawn to show places of equal heights. Contour lines and shapes represent the height and slope or gradient of the landform, which is outlined in brown colour.

Question 85. A candidate for Vidhan Sabha and Lok Sabha Election must not be less than _______years


a. 25
b. 26
c. 27
d. 29

Answer: (a)

The Indian Constitution provides that the minimum age to be eligible to contest in the Lok Sabha election shall be 25 years. A similar provision exists for candidates wishing to contest in the Legislative Assemblies or Vidhan Parishad at the state level.

Question 86. ‘MONEY BILL’ is first introduced in the


a. Lok Sabha
b. Rajya Sabha
c. Supreme Court
d. High Court

Answer: (a)

A Bill is said to be a Money Bill if it only contains provisions related to taxation, borrowing of money by the government, expenditure from or receipt to the Consolidated Fund of India. A Money Bill can be introduced only in Lok Sabha.

Question 87. The number of judges of the International Court of Justice is


a. 9
b. 10
c. 15
d. 16

Answer: (c)

The International Court of Justice (ICJ) is the principal judicial organ of the United Nations (UN). It was established in June 1945. The seat of the Court is at the Peace Palace in The Hague (Netherlands). The Court has 15 judges, each elected for terms of office of nine years.

Question 88. The age of retirement of the judges of the High Court is


a. 65 years
b. 60 years
c. 62 years
d. 70 years

Answer: (c)

Unlike federal judges in the US, judges in India do not hold office for life. In India, the age of retirement of the Supreme Court judges is 65 years and for the judges of the High Court it is at 62 years.

Question 89. The headquarter of the World Health Organization is


a. London
b. Manchester
c. Geneva
d. Paris

Answer: (c)

The World Health Organization(W.H.O.) is a specialized agency of the United Nations. Its primary role is to direct and coordinate international health within the United Nations system. The headquarters of WHO is located in Geneva, Switzerland.

Question 90. The minimum age for the citizen to exercise their right to vote has been reduced to 18 years from 21 years through the


a. 42nd Amendment Act
b. 44th Amendment Act
c. 61st Amendment Act
d. 73rd Amendment Act

Answer: (c)

Before 1988, Indians aged below 21 years were not granted the right to vote. Through the 61st Amendment Act of 1988, the minimum age for the citizen to exercise their right to vote has been reduced to 18 years.

Question 91. The Panch-Sheel Agreement was signed between


a. India and China
b. India and Nepal
c. India and Pakistan
d. Pakistan and China

Answer: (a)

The Five Principles of Peaceful Coexistence, known as the Panch-Sheel Agreement, was signed between India and China. It was signed between the two countries in 1954.

Question 92. The World Trade Organization was founded in __________.


a. 1990
b. 1995
c. 2000
d. 2005

Answer: (b)

The World Trade Organization (WTO) is the only global international organization dealing with trade rules between nations. Its goal is to help producers of goods and services, exporters, and importers conduct their business. The WTO was founded in 1995.

Question 93. Which of the following is not a function of Commercial Banks?


a. Collecting deposits from public
b. Lending loans
c. Issuing Notes
d. Working as an agent of Client

Answer: (c)

A commercial bank is a financial institution. It performs functions like withdrawal and deposit of money for the common public, they also provide loans. Commercial banks are profit-making institutions and conduct business with profit-making motives. They do not have the authority to issue notes. The authority to issue notes lies with the central bank of a country.

Question 94. Stagflation is a situation where


a. production increases and price level increases.
b. production decreases and price level increases.
c. production decreases and price level decreases.
d. production increases and price level decreases

Answer: (b)

Stagflation is a situation in which production decreases and the price level increases. During this time, an economy experiences an increase in inflation and a decrease in economic growth. It was first recognised in the 1970s when several countries saw rapid inflation and high levels of unemployment as a result of an oil crisis.

Question 95. Which of the following is a direct tax?


a. Sales Tax
b. Income Tax
c. Entertainment Tax
d. Service Tax

Answer: (b)

A direct tax is a tax that is paid directly to an imposing entity, usually the government, by an individual or an organisation. This type of tax cannot be transferred to another individual or entity. The collection and levying of direct taxes are regulated by the Central Board of Direct Taxes.

Question 96. In which economy is the policy of Laissez-faire adopted?


a. Capitalist Economy
b. Socialist Economy
c. Mixed Economy
d. Any Economy

Answer: (a)

The policy of Laissez-faire opposes government intervention in business affairs. It was developed by French physiocrats in the 18th century. It forms the basis of a free-market capitalist economy and is criticised for promoting inequality.

Question 97. Exclusion principle is not applicable in the case of __________


a. Capital goods
b. Consumer goods
c. Public goods
d. Private goods

Answer: (c)

In economics, the exclusion principle states “the owner of a private good may exclude others from use unless they pay.” The exclusion principle is not applicable in the case of public goods. This is because public goods are indivisible and have to be paid for by everyone.

Question 98. Railways in India are highlighted by which of the following market forms?


a. Perfect competition
b. Monopolistic competition
c. Monopoly
d. Oligopoly

Answer: (c)

A monopoly is a market structure in which there is one seller or vendor and many buyers. In such a market, the consumer is a price taker. This market system is characterised by profit maximisation.

For example, the Indian Railways, the Calcutta Electric Supply Corporation (CESC).

Question 99. Which of the following taxes follows the ability to pay principle?


a. Wealth Tax
b. Entertainment Tax
c. Goods and Services Tax
d. Excise Duty

Answer: (c)

Good and Services Tax follows the ability to pay principle. Just like progressive taxation, certain goods and services are taxed more on the ability of its buyers.

For example, a car will be taxed more than a bicycle.

Question 100. In underdeveloped countries, most of the labour force is generally engaged in


a. Industrial sector
b. Service sector
c. Agricultural sector
d. Banking sector

Answer: (c)

In underdeveloped and developing countries most of the labour force is generally engaged in the agricultural sector. The agricultural sector also records the highest number of informal employment. This can be attributed to a lack of formal education, sufficient infrastructure and structural deficiencies.

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*