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Defence NDA / CDS

NDA & CDS MCQ questions — Mathematics, English, GK, Reasoning for defence exams.

1,216 Q 4 Subjects 12th / Graduate
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Difficulty: All Easy Medium Hard 1181–1190 of 1,216
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Q.1181 Easy Reasoning Ability
Amit starts walking from point A towards North. After walking 10 km, he turns right and walks 5 km. Then he turns left and walks 8 km. Finally, he turns right and walks 3 km to reach point B. In which direction is point B with respect to point A?
A North-East
B East
C South-East
D North-West
Correct Answer:  A. North-East
EXPLANATION

Tracking Amit's movements: Starting North, he goes 10 km North; turns right (East) and walks 5 km; turns left (North) and walks 8 km; turns right (East) and walks 3 km.

His net displacement is 8 km North and 8 km East (5+3), placing point B in the North-East direction from point A.

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Q.1182 Hard Reasoning Ability
A ship starts from port X and sails 60 km towards North-East (exactly 45° from North). It then turns and sails 80 km towards South-East (exactly 45° from South). How far is the ship from port X?
A 100 km
B 140 km
C 70 km
D 20√13 km
Correct Answer:  D. 20√13 km
EXPLANATION
Step 1: Establish coordinate system and resolve first displacement

Set port X at origin with North as positive y-axis and East as positive x-axis. The ship sails 60 km at 45° from North towards North-East.

\[x_1 = 60 \sin(45°) = 60 \times \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}} = 30\sqrt{2} \text{ km}\]
\[y_1 = 60 \cos(45°) = 60 \times \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}} = 30\sqrt{2} \text{ km}\]
Step 2: Resolve second displacement from new position

The ship then sails 80 km at 45° from South towards South-East. This means 45° East of South direction, or equivalently, the angle is -45° from East (or 315° from North).

\[x_2 = 80 \sin(45°) = 80 \times \frac{1}{\sqrt{2}} = \frac{80}{\sqrt{2}} = 40\sqrt{2} \text{ km}\]
\[y_2 = -80 \cos(45°) = -80 \times \frac{1}{\sqrt{2}} = \frac{-80}{\sqrt{2}} = -40\sqrt{2} \text{ km}\]
Step 3: Calculate total displacement and distance from port X

Total displacement components from port X:

$$x_{total} = x_1 + x_2 = 30\sqrt{2} + 40

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Q.1183 Hard Reasoning Ability
In a city grid, Apartment A is 4 km directly East of Apartment B. Apartment C is 3 km directly North of Apartment B. Apartment D is 2 km directly West of Apartment C. From Apartment D, if you travel in a straight line to Apartment A, what is the total distance covered and the general direction?
A 5 km in South-East direction
B √37 km in South-East direction
C √29 km in East-South-East direction
D 6 km in South-East direction
Correct Answer:  C. √29 km in East-South-East direction
EXPLANATION
Step 1: Establish Coordinate System

Let Apartment B be at the origin (0, 0). Apartment A is 4 km East, so A is at (4, 0). Apartment C is 3 km North of B, so C is at (0, 3). Apartment D is 2 km West of C, so D is at (-2, 3).

\[\text{Coordinates: } B(0,0), A(4,0), C(0,3), D(-2,3)\]
Step 2: Calculate Distance from D to A

Using the distance formula between points D(-2, 3) and A(4, 0):

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
\[d = \sqrt{(4 - (-2))^2 + (0 - 3)^2} = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}\]
\[\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \approx 6.71 \text{ km}\]
Step 3: Determine Direction from D to A

The displacement vector from D to A is (6, -3), meaning 6 km East and 3 km South. The angle from East toward South is calculated as:

\[\tan(\theta) = \frac{3}{6} = \frac{1}{2}\]
\[\theta \approx 26.57° \text{ South of East, which is East-South-East direction}\]

**The

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Q.1184 Medium Reasoning Ability
Priya faces North. She turns 90° clockwise, then 45° counter-clockwise, then 135° clockwise. Finally, she turns 90° counter-clockwise. Which direction is she facing now?
A North-West
B East
C South
D South-West
Correct Answer:  C. South
EXPLANATION
Step 1: Initial Position and First Turn

Priya starts facing North. She turns 90° clockwise.

\[\text{North} + 90° \text{ clockwise} = \text{East}\]
Step 2: Second and Third Turns Combined

From East, she turns 45° counter-clockwise, then 135° clockwise.

\[45° \text{ counter-clockwise} - 135° \text{ clockwise} = -90° \text{ (net clockwise)}\]
\[\text{East} + 90° \text{ clockwise} = \text{South}\]
Step 3: Final Turn

From South, she turns 90° counter-clockwise.

\[\text{South} + 90° \text{ counter-clockwise} = \text{East}\]

Priya is now facing East, not South. However, if we verify the total rotation: North → 90° CW → 45° CCW → 135° CW → 90° CCW gives a net rotation of 90° clockwise from North, which is East.

Note: The given answer of South appears to be incorrect based on the step-by-step calculation. The correct answer should be East.

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Q.1185 Easy Reasoning Ability
A person is standing at the origin of a coordinate system. He walks 5 km South, then 12 km West, then 9 km North, and finally 7 km East. At which direction is he now with respect to his starting point?
A South-West
B North-West
C South-East
D North-East
Correct Answer:  A. South-West
EXPLANATION
Step 1: Track the North-South displacement

Starting at origin (0, 0), walking 5 km South gives position (0, -5), then 9 km North gives displacement of -5 + 9 = 4 km North.

\[\text{Net North-South} = -5 + 9 = 4 \text{ km North}\]
Step 2: Track the East-West displacement

Walking 12 km West gives position (-12, 4), then 7 km East gives displacement of -12 + 7 = -5 km (or 5 km West).

\[\text{Net East-West} = -12 + 7 = -5 \text{ km (5 km West)}\]
Step 3: Determine final position and direction

The final position is 5 km West and 4 km North of the starting point. Since the person is West and North of the origin, the direction is North-West.

\[\text{Final position} = (-5, 4) \text{ km}\]

The person is now in the North-West direction with respect to the starting point, not South-West as given in option A. However, if the answer key states South-West, there may be an error in either the problem statement or the provided answer.

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Q.1186 Medium Reasoning Ability
Rahul starts from point A and walks 10 km towards the North. He then turns right and walks 15 km. After this, he turns left and walks 8 km. Finally, he turns right and walks 12 km to reach point B. What is his net displacement from point A in terms of direction?
A 27 km towards North-East
B 25 km towards East-North-East
C 23 km towards North-East
D 20 km towards East
Correct Answer:  A. 27 km towards North-East
EXPLANATION
Step 1: Establish Coordinate System and Track North-South Movement

Let point A be at origin (0, 0). North is positive y-direction, East is positive x-direction.

Starting position: (0, 0)

- Walks 10 km North: position becomes (0, 10)

- Turns right (now facing East) and walks 15 km: position becomes (15, 10)

- Turns left (now facing North) and walks 8 km: position becomes (15, 18)

- Turns right (now facing East) and walks 12 km: position becomes (27, 18)

\[\text{Final position B} = (27, 18)\]
Step 2: Calculate Net Displacement

The net displacement is the straight-line distance from point A to point B using the Pythagorean theorem.

\[\text{Displacement} = \sqrt{(27-0)^2 + (18-0)^2}\]
\[= \sqrt{27^2 + 18^2} = \sqrt{729 + 324} = \sqrt{1053}\]
\[= \sqrt{729 + 324} = \sqrt{1053} \approx 32.45 \text{ km}\]
Step 3: Determine Direction of Displacement

Find the angle from North towards East using trigonometry.

\[\tan(\theta) = \frac{\text{East component}}{\text{North component}} = \frac{27}{18} = \frac{3}{2} = 1.5\]
\[\theta = \arctan(1.5) \approx 56.3°\]

This means the displacement is approximately 56.3° East of

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Q.1187 Medium Mathematics
In GP: 2,6,18,54... 5th term?
A 108
B 162
C 216
D 180
Correct Answer:  B. 162
EXPLANATION

This question asks you to find the 5th term of a geometric progression using the formula for the nth term.

Step 1: Identify the first term and common ratio.

First term: a = 2

Common ratio: r = 6/2 = 3 (or 18/6 = 3, or 54/18 = 3)

Step 2: Write the formula for the nth term of a GP.

The nth term of a GP is given by:

\[a_n = a \cdot r^{n-1}\]
Step 3: Substitute values for the 5th term (n = 5).
\[a_5 = 2 \cdot 3^{5-1}\]
\[a_5 = 2 \cdot 3^4\]
Step 4: Calculate the result.
\[a_5 = 2 \cdot 81 = 162\]

Therefore, the 5th term = 162

The answer is (B) 162

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Q.1188 Medium Mathematics
Determinant of [[1,2],[3,4]]?
A -2
B 2
C -4
D 4
Correct Answer:  A. -2
EXPLANATION

For a 2×2 matrix, the determinant is calculated using the cross-product formula: (top-left × bottom-right) minus (top-right × bottom-left).

Step 1: Identify the matrix elements.

Matrix = [[1, 2], [3, 4]]

where a = 1, b = 2, c = 3, d = 4

Step 2: Apply the determinant formula for 2×2 matrix.
\[\text{Det} = ad - bc\]
Step 3: Substitute the values.
\[\text{Det} = (1)(4) - (2)(3)\]
Step 4: Calculate.
\[\text{Det} = 4 - 6 = -2\]

Therefore, the determinant of [[1,2],[3,4]] = -2

The answer is (A) -2

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Q.1189 Easy Mathematics
cos 60° = ?
A 1/2
B √3/2
C 1
D 0
Correct Answer:  A. 1/2
EXPLANATION

This question tests your knowledge of standard trigonometric values for common angles.

Step 1: Recall the standard angle values

The trigonometric ratios for 0°, 30°, 45°, 60°, and 90° are fixed values that must be memorized for competitive exams.

Step 2: Identify cos 60° from the standard table

For 60°, the cosine value is:

\[\cos 60° = \frac{1}{2}\]
Step 3: Verify using the unit circle concept

At 60°, if you place a point on the unit circle, the x-coordinate (which represents cosine) equals \(\frac{1}{2}\).

Step 4: Eliminate other options

- \(\sin 60° = \frac{\sqrt{3}}{2}\) (option B is sine, not cosine)

- \(\cos 0° = 1\) (option C)

- \(\cos 90° = 0\) (option D)

Therefore, \(\cos 60° = \frac{1}{2}\)

The correct answer is A) 1/2

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Q.1190 Medium Mathematics
10th term of AP: 2,5,8,11...?
A 29
B 32
C 28
D 30
Correct Answer:  A. 29
EXPLANATION

To find the nth term of an arithmetic progression, we use the formula \(a_n = a + (n-1)d\) where a is the first term and d is the common difference.

Step 1: Identify the first term and common difference.

From the AP: 2, 5, 8, 11...

First term a = 2

Common difference d = 5 - 2 = 3

Step 2: Write the formula for the 10th term.

We need to find \(a_{10}\) using:

\[a_n = a + (n-1)d\]
Step 3: Substitute n = 10, a = 2, and d = 3.
\[a_{10} = 2 + (10-1) \times 3\]
\[a_{10} = 2 + 9 \times 3\]
\[a_{10} = 2 + 27\]
\[a_{10} = 29\]

Therefore, the 10th term of the AP is 29.

The answer is (A) 29.

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