SSC CHSL Tier 1 2025 15 Nov Shift 2nd

Verified by Ajay Kumar

25 questions · Choose an option, then open the answer and solution when you are ready.

Question 51

Mock Test →

Which number is NOT part of the natural numbers?

  1. 0

  2. 1

  3. 2

  4. 9

Answer: A

Solution

Natural numbers are generally 1, 2, 3, 4, ... Therefore, 0 is not a natural number under this convention.

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Question 52

Which number is in both natural and integer sets?

  1. 0

  2. 5

  3. -5

  4. None

Answer: B

Solution

Natural numbers are included in the set of integers. The number 5 belongs to both sets.

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Question 53

A trader buys 200 mixed stationery boxes at a rate of ₹120 each. He spends ₹1,800 on transportation and packaging. He sells 100 boxes at ₹150 each and the remaining at ₹130 each. What is his total profit percentage?

  1. 8%

  2. 8.53%

  3. 8.75%

  4. 9%

Answer: B

Solution

Total cost price = 200 × 120 + 1800 = ₹25,800. Total selling price = 100 × 150 + 100 × 130 = ₹28,000. Profit = 28000 − 25800 = ₹2,200. Profit percentage = (2200/25800) × 100 = 8.53% approximately.

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Question 54

In triangle PQR, O is incentre. ∠QOR = 110°. Find ∠QPR.

  1. 10°

  2. 30°

  3. 40°

  4. 90°

Answer: C

Solution

For the incentre of a triangle, ∠QOR = 90° + 1/2 ∠QPR. Therefore, 110° = 90° + 1/2 ∠QPR. Hence, ∠QPR = 2 × 20° = 40°.

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Question 55

Find the value of (tan 20°)²/(cosec 70°)² + (cot 20°)²/(sec 70°)² + 2 tan 15° tan 45° tan 75°.

  1. 2

  2. 3

  3. 1

  4. 0

Answer: B

Solution

Using cosec 70° = sec 20° and sec 70° = cosec 20°, the first two terms simplify to sin² 20° + cos² 20° = 1. Also, tan 15° × tan 75° = 1 and tan 45° = 1. Therefore, the expression = 1 + 2 × 1 × 1 = 3.

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Question 56

If 2 × sin q × cos(90° − q) = 1/2, where 0° < q < 90°, then find the value of sin(2q) + cos(2q).

  1. (√2 − 1)/2

  2. (√3 − 1)/2

  3. (√3 + 1)/2

  4. (√2 + 1)/2

Answer: C

Solution

Since cos(90° − q) = sin q, 2 sin² q = 1/2, giving sin² q = 1/4. As q is acute, q = 30°. Therefore, sin 60° + cos 60° = √3/2 + 1/2 = (√3 + 1)/2.

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Question 57

An amount invested on simple interest at 'R%' p.a. amounts to Rs. 9,504 and Rs. 12,384 in 4 years and 9 years, respectively. Find the value of 'R'.

  1. 7%

  2. 5%

  3. 11%

  4. 8%

Answer: D

Solution

Interest for 5 years = 12384 − 9504 = ₹2,880. Annual interest = 2880/5 = ₹576. Principal = 9504 − 4 × 576 = ₹7,200. Rate = (576/7200) × 100 = 8%.

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Question 58

A sum of money becomes 240% of itself in 8 years at simple interest. Find the simple interest earned when Rs. 3000 is invested at the same rate for 3 years.

  1. Rs. 1620

  2. Rs. 1430

  3. Rs. 1255

  4. Rs. 1575

Answer: D

Solution

The interest earned in 8 years is 240% − 100% = 140% of the principal. Annual rate = 140/8 = 17.5%. Simple interest for 3 years = (3000 × 17.5 × 3)/100 = ₹1,575.

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Question 59

Rohit distributed 70% of the total money he had between Priya and Meena in the ratio 5:9, respectively. Priya and Meena deposited their respective shares at 6% and 10% simple interest per annum for 4 years. If the difference in the interest earned by Meena and Priya is Rs. 2,304, find the total amount Rohit had.

  1. Rs. 23,400

  2. Rs. 33,400

  3. Rs. 16,900

  4. Rs. 19,200

Answer: D

Solution

Let the total amount be P. Priya receives 70% × 5/14 = P/4, and Meena receives 70% × 9/14 = 9P/20. The difference in their simple interest for 4 years is (9P/20 × 10 × 4/100) − (P/4 × 6 × 4/100) = 0.12P. Thus, 0.12P = 2304, so P = ₹19,200.

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Question 60

A stick of length 48 cm is at an inclination of 60° to a wall with its bottom touching the floor of the wall. Find the length from the floor to the point where the stick touches the wall.

  1. 24√3 cm

  2. 12√3 cm

  3. 12√2 cm

  4. 18√3 cm

Answer: A

Solution

The highlighted answer corresponds to a 60° angle with the ground. Height = 48 × sin 60° = 48 × √3/2 = 24√3 cm. Note that the original wording says the angle is with the wall, so the wording should be verified.

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Question 61

If a 16 metres long rope is stretched from the ground to the top of the house such that the rope makes an angle of 30° with the ground, then find the height of the house.

  1. 8 metres

  2. (16/√3) metres

  3. 3 metres

  4. 16 metres

Answer: A

Solution

The rope is the hypotenuse. Height = 16 × sin 30° = 16 × 1/2 = 8 metres.

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Question 62

A tank has two pipes, one can fill it with water in 14 hours and the other can empty it in 8 hours. In how many hours will the tank be emptied if both the pipes are opened together when 1/4th of the tank is already filled with water?

  1. 6 hours

  2. 5 hours

  3. 3 hour 20 min

  4. 4 hours 40 min

Answer: D

Solution

Net emptying rate = 1/8 − 1/14 = 3/56 tank per hour. The tank initially contains 1/4 water. Time = (1/4)/(3/56) = 14/3 hours = 4 hours 40 minutes.

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Question 63

If 4 women and 6 girls can do a piece of work in 12 days, while 24 women and 46 girls can do the same in 2 days, how much time will 15 women and 20 girls do the same type of work?

  1. 5 days

  2. 3 days 12 hours

  3. 4 days 2 hours

  4. 3 days 4 hours 48 min

Answer: D

Solution

The screenshot highlights option D. However, the given work-rate conditions are inconsistent. If w and g are the daily work rates of one woman and one girl, then 4w + 6g = 1/12 and 24w + 46g = 1/2. Multiplying the first equation by 6 gives 24w + 36g = 1/2. Subtracting yields 10g = 0. Verify the original question before relying on the highlighted answer.

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Question 64

Two vertical poles stand on the ground, 50 meters apart. The height of the smaller pole is 10 meters. From the top of the smaller pole, the angle of elevation of the top of the taller pole is 30°. Find the height of the taller pole.

  1. 35.20 m

  2. 38.87 m

  3. 40.12 m

  4. 42.25 m

Answer: B

Solution

Let the taller pole's height be H. tan 30° = (H − 10)/50. Therefore, H − 10 = 50/√3 and H = 10 + 50/√3 ≈ 38.87 m.

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Question 65

A chord of length 8 cm is 3 cm from the center. Find the radius.

  1. 5 cm

  2. 7 cm

  3. 6 cm

  4. 4 cm

Answer: A

Solution

The perpendicular from the centre bisects the chord. Half the chord = 8/2 = 4 cm. Using Pythagoras' theorem, r² = 3² + 4² = 25. Therefore, r = 5 cm.

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Question 66

The number of chords that can be drawn through a point on a circle is:

  1. 1

  2. 2

  3. Infinite

  4. 0

Answer: C

Solution

A chord can be drawn by joining the given point to any other point on the circle. Since there are infinitely many points on a circle, infinitely many chords can pass through the given point.

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Question 67

The length and breadth of a rectangle are in the ratio 14:11, respectively, and the perimeter of the rectangle is 150 cm. If the area of the rectangle is equal to the area of the top surface of a solid cylinder, then find the curved surface area of the cylinder given that its radius is 105% of its height.

  1. 2640 cm²

  2. 2560 cm²

  3. 2120 cm²

  4. 2250 cm²

Answer: A

Solution

Let the rectangle's length and breadth be 14x and 11x. Its perimeter gives 2(14x + 11x) = 150, so x = 3. The dimensions are 42 cm and 33 cm, giving an area of 1386 cm². Thus, πr² = 1386. Since r = 1.05h, the curved surface area = 2πrh = 2πr²/1.05 = 2 × 1386/1.05 = 2640 cm².

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Question 68

PT is a tangent at point C on a circle with centre O and AB is a diameter. When BA is produced, it meets PT at point P. If ∠BPC = 34°, then what is the measure of ∠ACP?

  1. 26°

  2. 28°

  3. 32°

  4. 29°

Answer: B

Solution

The screenshot highlights 28°. However, the stated geometry appears inconsistent: if AB is a diameter, P lies on the extension of BA, and PC is tangent at C, then the angle ∠BPC would be 90°, not 34°. Verify the original question before using the highlighted answer.

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Question 69

Simplify: (√12 + √27) ÷ √3

  1. 11

  2. -1

  3. 5

  4. 0

Answer: C

Solution

√12 = 2√3 and √27 = 3√3. Therefore, (√12 + √27) ÷ √3 = (2√3 + 3√3)/√3 = 5√3/√3 = 5.

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Question 70

Simplify: (5 + 2√6)(5 − 2√6)

  1. 11

  2. 1

  3. 0

  4. 6

Answer: B

Solution

Using (a + b)(a − b) = a² − b², we get 5² − (2√6)² = 25 − 24 = 1.

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Question 71

Find the value of √(9 + 4√5) + √(9 − 4√5).

  1. 2

  2. 4

  3. 7

  4. 3

Answer: B

Solution

The screenshot highlights option B, 4. However, 9 + 4√5 = (√5 + 2)² and 9 − 4√5 = (√5 − 2)². The expression therefore equals (√5 + 2) + (√5 − 2) = 2√5, approximately 4.472. None of the listed options matches exactly. Verify the original question or options.

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Question 72

Rationalize: 1 / (√3 + √2)

  1. √2 + √3

  2. √2 − √3

  3. √3 + √2

  4. √3 − √2

Answer: D

Solution

Multiply the numerator and denominator by the conjugate (√3 − √2). Then 1/(√3 + √2) = (√3 − √2)/(3 − 2) = √3 − √2.

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Question 73

Simplify: (2√3 + √2)²

  1. 14 − 4√6

  2. 14 + 4√6

  3. 11 + 2√6

  4. 11 − 2√6

Answer: B

Solution

Using (a + b)² = a² + 2ab + b², (2√3 + √2)² = 12 + 4√6 + 2 = 14 + 4√6.

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Question 74

Simplify the expression: (√3 + √5)² + (√3 − √5)² + 2(√3 + √5)(√3 − √5)

  1. 15

  2. 10

  3. 18

  4. 12

Answer: D

Solution

Let a = √3 + √5 and b = √3 − √5. Using a² + b² + 2ab = (a + b)², the expression becomes (2√3)² = 12.

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Question 75

The average weight of 30 students is 60 kg. When the teacher's weight is included, the average becomes 61 kg. What is the weight of the teacher?

  1. 91 kg

  2. 90 kg

  3. 89 kg

  4. 88 kg

Answer: A

Solution

Total weight of 30 students = 30 × 60 = 1800 kg. Total weight of 31 people including the teacher = 31 × 61 = 1891 kg. Teacher's weight = 1891 − 1800 = 91 kg.

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