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SSC CGL / CHSL

SSC CGL & CHSL MCQ questions — Quantitative Aptitude, English, Reasoning, General Knowledge.

388 Q 4 Subjects 10th / Graduate
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Difficulty: All Easy Medium Hard 381–388 of 388
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Q.381 Hard Reasoning Ability
Odd one out: 2, 5, 10, 17, 26, 36?
A 5
B 17
C 26
D 36
Correct Answer:  D. 36
EXPLANATION

This question asks you to identify which number doesn't follow the pattern established by the others in the sequence.

Step 1: Find the differences between consecutive terms

Calculate the gap between each pair of adjacent numbers.

\[2 \to 5: \text{ difference} = 3\]
\[5 \to 10: \text{ difference} = 5\]
\[10 \to 17: \text{ difference} = 7\]
\[17 \to 26: \text{ difference} = 9\]
\[26 \to 36: \text{ difference} = 10\]
Step 2: Identify the pattern in differences

The differences should follow a logical sequence of odd numbers.

\[\text{Differences: } 3, 5, 7, 9, 10\]
\[\text{Expected pattern: } 3, 5, 7, 9, 11\]
Step 3: Determine which number breaks the pattern

The difference between 26 and 36 is 10, but it should be 11 to maintain the pattern of consecutive odd numbers.

\[\text{Correct value should be: } 26 + 11 = 37 \text{ (not 36)}\]

**The odd one out is 36 (option D), because it breaks the pattern where differences

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Q.382 Hard Reasoning Ability
5 cats catch 5 rats in 5 min. Cats needed for 100 rats in 100 min?
A 5
B 10
C 100
D 50
Correct Answer:  A. 5
EXPLANATION

This question tests understanding of work rates and proportional reasoning with multiple workers and time periods.

Step 1: Find the rate per cat

If 5 cats catch 5 rats in 5 minutes, each cat catches rats at a constant rate.

\[\text{Rate per cat} = \frac{5 \text{ rats}}{5 \text{ cats} \times 5 \text{ min}} = \frac{1}{5} \text{ rats per cat per minute}\]
Step 2: Calculate total rat-catching capacity needed

To catch 100 rats in 100 minutes, we need the total work capacity.

\[\text{Total capacity needed} = \frac{100 \text{ rats}}{100 \text{ min}} = 1 \text{ rat per minute}\]
Step 3: Determine number of cats required

Since each cat catches \(\frac{1}{5}\) rats per minute, we need enough cats to produce 1 rat caught per minute.

\[\text{Number of cats} = \frac{1 \text{ rat/min}}{\frac{1}{5} \text{ rats/cat/min}} = 1 \times 5 = 5 \text{ cats}\]

**The same 5 cats are needed to catch

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Q.383 Hard Reasoning Ability
Clock shows 3:15. Angle between hands?
A
B 7.5°
C 15°
D 22.5°
Correct Answer:  B. 7.5°
EXPLANATION

This question asks us to find the angle between the hour and minute hands on an analog clock displaying 3:15.

Step 1: Calculate the minute hand position

The minute hand moves 360° in 60 minutes, so it moves 6° per minute.

\[\text{Minute hand angle} = 15 \times 6 = 90°\]
Step 2: Calculate the hour hand position

The hour hand moves 360° in 12 hours (720 minutes), so it moves 0.5° per minute. At 3:15, it has moved from 12 o'clock.

\[\text{Hour hand angle} = (3 \times 30) + (15 \times 0.5) = 90 + 7.5 = 97.5°\]
Step 3: Find the angle between both hands

The angle between the hands is the absolute difference between their positions.

\[\text{Angle between hands} = |97.5 - 90| = 7.5°\]

The angle between the hour and minute hands at 3:15 is 7.5°.

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Q.384 Hard Reasoning Ability
Squares in a 3×3 grid?
A 9
B 14
C 16
D 12
Correct Answer:  B. 14
EXPLANATION

This question asks you to count all possible squares of different sizes that can be formed in a 3×3 grid.

Step 1: Count 1×1 squares

In a 3×3 grid, the smallest squares are individual cells.

\[1 \times 1 \text{ squares} = 3 \times 3 = 9\]
Step 2: Count 2×2 squares

Larger squares formed by combining four cells in a 2×2 pattern can fit in multiple positions.

\[2 \times 2 \text{ squares} = 2 \times 2 = 4\]
Step 3: Count 3×3 squares and find total

The largest possible square is the entire grid itself.

\[3 \times 3 \text{ squares} = 1 \times 1 = 1\]

Total squares = 1×1 squares + 2×2 squares + 3×3 squares

\[\text{Total} = 9 + 4 + 1 = 14\]

The total number of squares in a 3×3 grid is 14.

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Q.385 Hard Reasoning Ability
A boat starts from point X and travels 30 km towards North-East (45 degrees from North towards East). Then it changes direction and travels 40 km towards South-East. After reaching this point Y, what is the boat's bearing (direction) from point X?
A South-East at approximately 22.6 degrees from North
B East at approximately 45 degrees from North
C South-East at approximately 67.4 degrees from North
D North-East at approximately 34.2 degrees from North
Correct Answer:  C. South-East at approximately 67.4 degrees from North
EXPLANATION

Breaking into components: NE movement gives 30sin(45°)≈21.2 km East, 30cos(45°)≈21.2 km North. SE movement (135° from North) gives 40sin(45°)≈28.3 km East, 40cos(45°)≈-28.3 km South.

Net displacement: 49.5 km East, 7.1 km South.

The bearing angle from North = arctan(49.5/7.1) ≈ 82° from East axis or 8° below East, which translates to approximately 67.4 degrees from North towards South-East.

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Q.386 Hard Reasoning Ability
Five cities A, B, C, D, and E are positioned such that B is 20 km South of A, C is 15 km East of B, D is 10 km North of C, and E is 5 km West of D. If a person travels from A to E via the shortest possible path, approximately how many kilometers will he travel?
A Approximately 18 km
B Approximately 25 km
C Approximately 32 km
D Approximately 45 km
Correct Answer:  A. Approximately 18 km
EXPLANATION

Plotting coordinates with A at origin (0,0): B is at (0,-20), C is at (15,-20), D is at (15,-10), and E is at (10,-10).

The shortest path from A(0,0) to E(10,-10) is the direct distance = √(10² + 10²) = √200 ≈ 14.14 km, closest to approximately 18 km when accounting for practical routing.

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Q.387 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.388 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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