SSC CHSL Tier 1 2025 18 Nov Shift 2nd

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

Question 51

A shopkeeper sells an article at a discount of 12%. Had he given only a 5% discount, he would have earned ₹504 more as profit. If the cost price of the article is ₹3,000, then find the marked price.

  1. ₹7,200

  2. ₹6,800

  3. ₹6,500

  4. ₹7,500

Answer: A

Solution

The difference between the discounts is 12% - 5% = 7%. Therefore, 7% of the marked price = ₹504. Marked Price = 504 × 100 / 7 = ₹7,200.

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Question 52

The marked prices of a pen and pencil are in the ratio of 5:6. The shopkeeper gives an 18% discount on the pen. If the total discount on both the pen and the pencil is 12%, then the discount offered on the pencil is:

  1. 7%

  2. 10%

  3. 11%

  4. 8%

Answer: A

Solution

Let the discount on the pencil be x%. Using the weighted average, (5 × 18 + 6x) / 11 = 12. Therefore, 90 + 6x = 132, so x = 7%.

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Question 53

In a warehouse, there are 3 stacks of bricks. The average weight of the bricks in Stack A is 112.5 kg, in Stack B is 118.75 kg, and in Stack C is 107.25 kg. If Stack A has 12 bricks, Stack B has 8 bricks, and Stack C has 16 bricks, find the overall average weight of all bricks in the warehouse.

  1. 111.55 kg

  2. 110.25 kg

  3. 113.75 kg

  4. 109.85 kg

Answer: A

Solution

Overall average = (112.5 × 12 + 118.75 × 8 + 107.25 × 16) / (12 + 8 + 16) = 4016 / 36 = 111.555... kg. The listed answer is 111.55 kg.

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Question 54

Which of the following is the largest three-digit number that is a perfect square?

  1. 951

  2. 961

  3. 971

  4. 981

Answer: B

Solution

The largest three-digit number is 999. Since 31² = 961 and 32² = 1024, which is a four-digit number, the largest three-digit perfect square is 961.

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Question 55

A started a business with ₹30,000. B joined after 6 months with ₹40,000. At the end of the year, what is the profit-sharing ratio?

  1. 3:5

  2. 3:4

  3. 4:5

  4. 3:2

Answer: D

Solution

Profit-sharing ratio = (30000 × 12):(40000 × 6) = 360000:240000 = 3:2.

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Question 56

If sin(A + B) = sinAcosB + cosAsinB, then the value of sin 75° is:

  1. (√3 + 1)/2

  2. (√3 + 1)/√2

  3. (√2 + 1)/2√2

  4. (√3 + 1)/2√2

Answer: D

Solution

sin 75° = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2).

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Question 57

If numbers are in a ratio 2 : 5 : 3 : 4, and the numbers are increased by 20%, 30%, 40% and 10% respectively, find the new ratio.

  1. 22:67:47:40

  2. 19:63:41:47

  3. 23:61:41:43

  4. 24:65:42:44

Answer: D

Solution

The new terms are 2 × 1.2 = 2.4, 5 × 1.3 = 6.5, 3 × 1.4 = 4.2 and 4 × 1.1 = 4.4. Multiplying by 10 gives the ratio 24:65:42:44.

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Question 58

The ratio of milk and water in 560 liters of mixture is 4 : 3. 140 liters of mixture are taken out. Find the quantity of milk now.

  1. 300 liters

  2. 320 liters

  3. 240 liters

  4. 280 liters

Answer: C

Solution

Initial milk = 560 × 4/7 = 320 liters. Milk remaining after removing 140 liters of mixture = 320 × (420/560) = 240 liters.

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Question 59

In a mixture of 60 litres, the ratio of wine to water is 1 : 2. If this ratio is to be 1 : 3, then what quantity of water needs to be added?

  1. 24 litres

  2. 18 litres

  3. 22 litres

  4. 20 litres

Answer: D

Solution

Wine = 60 × 1/3 = 20 liters and water = 40 liters. For a 1:3 ratio, water must be 20 × 3 = 60 liters. Water to be added = 60 - 40 = 20 liters.

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Question 60

If sec θ + tan θ = 5, where θ is an acute angle, find the value of sec θ.

  1. 12/5

  2. 13/5

  3. 5/2

  4. 3

Answer: B

Solution

Using (sec θ + tan θ)(sec θ - tan θ) = 1, we get sec θ - tan θ = 1/5. Adding the equations gives 2 sec θ = 5 + 1/5 = 26/5. Therefore, sec θ = 13/5.

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Question 61

A bike is chasing a car, which is 12 km ahead of it. The speeds of the bike and the car are 96 km/hr and 60 km/hr, respectively. If the bike catches the car in 'a' minute, then find the distance covered by the bike in (a - 16) hours.

  1. 312 km

  2. 384 km

  3. 360 km

  4. 398 km

Answer: B

Solution

Relative speed = 96 - 60 = 36 km/hr. Catching time = 12/36 hour = 20 minutes, so a = 20. Therefore, a - 16 = 4 hours. Distance covered by the bike = 96 × 4 = 384 km.

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Question 62

If x = √10 + √3 and y = √10 - √3, then find x² + y².

  1. 26

  2. 56

  3. 25

  4. 15

Answer: A

Solution

Using (a + b)² + (a - b)² = 2(a² + b²), x² + y² = 2(10 + 3) = 26.

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Question 63

If a = √8 + √3 and b = √8 - √3, find the value of (a + b) / (a - b).

  1. (3√6) / 5

  2. (2√6) / 3

  3. (2√3) / 7

  4. None of these

Answer: B

Solution

a + b = 2√8 and a - b = 2√3. Therefore, (a + b)/(a - b) = √8/√3 = √(8/3) = 2√6/3.

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Question 64

A rectangular tile having a perimeter of 224 cm has been used to pave a floor of length and breadth 25.6 m and 14.4 m, respectively. If the ratio of length to breadth of each tile is 4:3, then find the number of tiles required to pave the floor completely.

  1. 1200

  2. 1050

  3. 1140

  4. 1540

Answer: A

Solution

Let the tile dimensions be 4x and 3x. Then 2(4x + 3x) = 224, so x = 16 cm. Tile dimensions are 64 cm × 48 cm. Number of tiles = (2560 × 1440)/(64 × 48) = 1200.

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Question 65

A rectangle has one side of 6 cm, and its diagonal is 22 cm. Find the other side and the area.

  1. 8√7 cm and 48√7 cm²

  2. 5√7 cm and 40√7 cm²

  3. 2√7 cm and 33√7 cm²

  4. 3√7 cm and 44√7 cm²

Answer: A

Solution

By Pythagoras, the other side = √(22² - 6²) = √448 = 8√7 cm. Area = 6 × 8√7 = 48√7 cm².

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Question 66

The average of 5 consecutive even numbers is 46. If the next 3 consecutive even numbers are added, what is the new average?

  1. 48

  2. 49

  3. 50

  4. 51

Answer: B

Solution

The five numbers are 42, 44, 46, 48 and 50. The next three are 52, 54 and 56. New average = (42 + 44 + 46 + 48 + 50 + 52 + 54 + 56)/8 = 49.

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Question 67

A population falls by 12% in the first year. By what percentage must it increase in the second year so that at the end of the second year the population is exactly the same as the original?

  1. 10.6%

  2. 11.6%

  3. 13.6%

  4. 14.6%

Answer: C

Solution

Assume the original population is 100. After a 12% decrease, it becomes 88. Required increase = (100 - 88)/88 × 100 = 13.636...%, approximately 13.6%.

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Question 68

A shopkeeper marks up his goods by 20% and then offers a discount of 10%. If the cost price is ₹500, what is the selling price?

  1. ₹550

  2. ₹540

  3. ₹600

  4. ₹450

Answer: B

Solution

Marked price = ₹500 × 120/100 = ₹600. Selling price after a 10% discount = ₹600 × 90/100 = ₹540.

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Question 69

A sum of money becomes 16 times in 2 years. What is the annual compound interest rate?

  1. 100%

  2. 200%

  3. 300%

  4. 400%

Answer: C

Solution

Let the annual rate be r%. Then (1 + r/100)² = 16. Taking the square root, 1 + r/100 = 4. Therefore, r = 300%.

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Question 70

A bus starts 30 minutes late and to reach its destination 150 km away on time, it increases its speed by 25 km/h. What is the usual speed of the bus?

  1. 70 km/h

  2. 75 km/h

  3. 80 km/h

  4. 95 km/h

Answer: B

Solution

Let the usual speed be x km/h. Then 150/x - 150/(x + 25) = 1/2. Simplifying, x(x + 25) = 7500, so x² + 25x - 7500 = 0. The positive solution is x = 75 km/h.

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Question 71

In a cricket series, Team A had a winning rate of 70% after playing x matches. Then they lost 20 matches consecutively, and their winning rate fell to 50%. Find the total matches played by Team A before the losing streak.

  1. 40

  2. 50

  3. 60

  4. 80

Answer: B

Solution

Let the number of matches before the losing streak be x. Wins = 0.7x. After 20 losses, 0.7x/(x + 20) = 0.5. Therefore, 0.7x = 0.5x + 10, so x = 50.

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Question 72

A and B together can complete a job in 8 days. If A alone can do it in 16 days, how many days will B alone take?

  1. 16 days

  2. 10 days

  3. 12 days

  4. 14 days

Answer: A

Solution

Combined work rate = 1/8 per day and A's work rate = 1/16 per day. B's rate = 1/8 - 1/16 = 1/16. Therefore, B alone takes 16 days.

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Question 73

If the original price of the items is ₹2,050 and the discount is 26%, what will be the final price after the discount?

  1. ₹1,500

  2. ₹1,650

  3. ₹1,517

  4. ₹1,710

Answer: C

Solution

Discount = ₹2050 × 26/100 = ₹533. Final price = ₹2050 - ₹533 = ₹1517.

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Question 74

A farmer wants to fence a triangular field with sides 60 m, 111 m, and 123 m. If the cost of fencing is ₹18 per meter, what will be the total expense?

  1. ₹5,292

  2. ₹5,000

  3. ₹4,982

  4. ₹5,512

Answer: A

Solution

Perimeter = 60 + 111 + 123 = 294 m. Total fencing cost = 294 × ₹18 = ₹5,292.

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Question 75

A triangular piece of land has sides that are in the ratio of 3:5:7, with a total perimeter of 300 meters. What are the sides of this land?

  1. 60 m, 100 m, 140 m

  2. 70 m, 110 m, 120 m

  3. 50 m, 80 m, 120 m

  4. 65 m, 95 m, 140 m

Answer: A

Solution

Sum of ratio parts = 3 + 5 + 7 = 15. One part = 300/15 = 20 m. The sides are 3 × 20 = 60 m, 5 × 20 = 100 m and 7 × 20 = 140 m.

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