All SSC Combined Higher Secondary Level (CHSL) Topics

SSC Combined Higher Secondary Level (CHSL) · Topic Collection

Quantitative Aptitude PYQs

Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.

13275 Total Questions531 Official PapersYears: 2018–2025Search in Quantitative Aptitude

Showing questions 261–280 of 13275Page 14 of 100

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 261

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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' minutes, then find the distance covered by the bike in (a - 16) hours.

AOption (a)
312 km
BOption (b)
384 km
COption (c)
360 km
DOption (d)
398 km

View answer & explanation

Correct answer

384 km

Why this answer is correct

Given:

Speed of bike = 96 km/h

Speed of car = 60 km/h

Initial distance = 12 km

Formula Used:

Relative speed = Speed of bike - Speed of car

Time = Distance ÷ Relative speed

Distance = Speed × Time

Calculation:

⇒ Relative speed = 96 - 60 = 36 km/h

⇒ Time to catch = 12 ÷ 36 = 1/3 hour

⇒ a = 1/3 × 60 = 20 minutes

Now,

(a - 16) = 20 - 16 = 4 hours

So, distance covered by the bike in (a - 16) hours

⇒ 96 × 4 = 384 km

∴ The correct answer is 384 km

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 262

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

If x = \( \sqrt{10} + \sqrt{3}\) and y = \(\sqrt{10} - \sqrt{3},\) then find x² + y².

AOption (a)
26
BOption (b)
56
COption (c)
25
DOption (d)
15

View answer & explanation

Correct answer

Why this answer is correct

Given:

x = √10 + √3

y = √10 - √3

Formula Used:

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

Calculation:

⇒ x² = (√10 + √3)²

⇒ x² = 10 + 3 + 2√30 = 13 + 2√30

⇒ y² = (√10 - √3)²

⇒ y² = 10 + 3 - 2√30 = 13 - 2√30

⇒ x² + y² = (13 + 2√30) + (13 - 2√30)

⇒ x² + y² = 26

∴ The correct answer is 26.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 263

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

If a = \(\sqrt{8} + \sqrt{3},\) and b = \(\sqrt{8} - \sqrt{3},\) find the value of (a + b) / (a - b).

AOption (a)
\((3\sqrt{6})/5\)
BOption (b)
\((2\sqrt{6})/3\)
COption (c)
\((2\sqrt{3})/7\)
DOption (d)
None of these

View answer & explanation

Correct answer

\((2\sqrt{6})/3\)

Why this answer is correct

Given:

a = √8 + √3

b = √8 - √3

Formula Used:

(a + b)

(a - b)

Calculation:

⇒ a + b = (√8 + √3) + (√8 - √3)

⇒ a + b = 2√8

⇒ a - b = (√8 + √3) - (√8 - √3)

⇒ a - b = 2√3

⇒ (a + b) ÷ (a - b) = (2√8) ÷ (2√3)

⇒ (a + b) ÷ (a - b) = √8 ÷ √3

⇒ (a + b) ÷ (a - b) = √(8 ÷ 3)

⇒ (a + b) ÷ (a - b) = 2√6 ÷ 3

∴ The correct answer is 2√6/3.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 264

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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.

AOption (a)
1200
BOption (b)
1050
COption (c)
1140
DOption (d)
1540

View answer & explanation

Correct answer

1200

Why this answer is correct

Given:

Perimeter of tile = 224 cm

Ratio = 4 : 3

Floor = 25.6 m × 14.4 m

Formula Used:

Perimeter = 2 × (l + b)

Number of tiles = Area of floor ÷ Area of tile

Calculation:

⇒ Let l = 4x, b = 3x

⇒ 2 × (4x + 3x) = 224

⇒ 2 × 7x = 224

⇒ 14x = 224

⇒ x = 16

⇒ l = 64 cm, b = 48 cm

⇒ Area of tile = 64 × 48 = 3072 cm²

⇒ Floor = 25.6 m × 14.4 m = 2560 cm × 1440 cm

⇒ Area of floor = 2560 × 1440 = 3686400 cm²

⇒ Number of tiles = 3686400 ÷ 3072

⇒ Number of tiles = 1200

∴ The correct answer is 1200.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 265

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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

AOption (a)
\(8\sqrt{7} \text{ cm and } 48\sqrt{7} \text{ cm}^2\)
BOption (b)
\(5\sqrt{7} \text{ cm and } 40\sqrt{7} \text{ cm}^2\)
COption (c)
\(2\sqrt{7} \text{ cm and } 33\sqrt{7} \text{ cm}^2\)
DOption (d)
\(3\sqrt{7} \text{ cm and } 44\sqrt{7} \text{ cm}^2\)

View answer & explanation

Correct answer

\(8\sqrt{7} \text{ cm and } 48\sqrt{7} \text{ cm}^2\)

Why this answer is correct

Given:

One side = 6 cm

Diagonal = 22 cm

Formula Used:

Diagonal² = length² + breadth²

Area = length × breadth

Calculation:

⇒ Other side² = 22² - 6²

⇒ Other side² = 484 - 36

⇒ Other side² = 448

⇒ Other side = √448

⇒ Other side = 8√7 cm

⇒ Area = 6 × 8√7

⇒ Area = 48√7 cm²

∴ The correct answer is 8√7 and 48√7.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 266

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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

AOption (a)
48
BOption (b)
49
COption (c)
50
DOption (d)
51

View answer & explanation

Correct answer

Why this answer is correct

Given:

Average of 5 consecutive even numbers = 46

Formula Used:

Average of consecutive numbers = middle number

Average = Sum ÷ Number of terms

Calculation:

⇒ Middle number = 46

⇒ Numbers = 42, 44, 46, 48, 50

⇒ Next numbers = 52, 54, 56

⇒ Total sum = 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56

⇒ Total sum = 392

⇒ New average = 392 ÷ 8

⇒ New average = 49

∴ The correct answer is 49.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 267

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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?

AOption (a)
10.6%
BOption (b)
11.6%
COption (c)
13.6%
DOption (d)
14.6%

View answer & explanation

Correct answer

13.6%

Why this answer is correct

Shortcut Trick

To restore a value after a percentage decrease, use the formula:

Required Increase % = [r ÷ (100 − r)] × 100, where r is the percentage fall.

Required Increase % = [12 ÷ (100 − 12)] × 100

Required Increase % = (12 ÷ 88) × 100 ≈ 13.63%

∴ The correct answer is 13.6%.

Alternate Method

Given: First year population decrease = 12%.

Formula Used: Percentage Increase = (Change ÷ Initial Value) × 100

⇒ Let the original population be 100.

⇒ Population after 1^st year = 100 − (12% of 100) = 100 − 12 = 88.

⇒ To return to 100, the increase needed = 100 − 88 = 12.

⇒ Percentage increase required on the new base (88) = (12 ÷ 88) × 100

⇒ Percentage increase = (3 ÷ 22) × 100 = 300 ÷ 22

⇒ Percentage increase = 13.6363...

∴ The correct answer is 13.6%.

Additional Information

Percentage Reversal Rule

If a value decreases by x%, the percentage increase required to restore the original value is [x ÷ (100 − x)] × 100.

Base Change Principle

Percentage values change when the base value changes. Here, the first change is calculated on 100, while the second must be calculated on 88.

Reciprocal Ratios

Common restoration pairs: a 20% decrease requires a 25% increase; a 25% decrease requires a 33.33% increase; a 50% decrease requires a 100% increase.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 268

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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?

AOption (a)
₹550
BOption (b)
₹540
COption (c)
₹600
DOption (d)
₹450

View answer & explanation

Correct answer

₹540

Why this answer is correct

Shortcut Trick

Net Profit% = Markup% − Discount% − (Markup × Discount / 100)

Net Profit% = 20 − 10 − (20 × 10 / 100) = 20 − 10 − 2 = 8%

Selling Price (SP) = 108% of Cost Price (CP) = 1.08 × 500 = ₹540

∴ The correct answer is ₹540.

Alternate Method

Given:

Cost Price (CP) = ₹500

Markup Percentage = 20%

Discount Percentage = 10%

Formula Used:

Marked Price (MP) = CP × (100 + Markup%) / 100
Selling Price (SP) = MP × (100 − Discount%) / 100

Calculations:

⇒ Marked Price (MP) = 500 × (1 + 20/100)

⇒ Marked Price (MP) = 500 × 1.2 = ₹600

⇒ Selling Price (SP) = 600 × (1 − 10/100)

⇒ Selling Price (SP) = 600 × 0.9 = ₹540

∴ The correct answer is ₹540.

Additional Information

Marked Price and Markup

Marked Price is the price at which a shopkeeper intends to sell a product before any discounts. Markup is the amount added to the Cost Price to arrive at the Marked Price.

Relation between CP and MP

A direct relationship exists between CP and MP: CP/MP = (100 − Discount%) / (100 + Profit%).

Profit and Discount Interaction

Profit is always calculated on Cost Price, whereas Discount is always calculated on the Marked Price. Effective profit change follows the successive percentage rule.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 269

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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

AOption (a)
100%
BOption (b)
200%
COption (c)
300%
DOption (d)
400%

View answer & explanation

Correct answer

300%

Why this answer is correct

Given:

Final amount = 16 × Principal

Time = 2 years

Formula Used:

Amount = Principal × (1 + R ÷ 100)ⁿ

Calculation:

⇒ 16P = P × (1 + R ÷ 100)²

⇒ 16 = (1 + R ÷ 100)²

⇒ 1 + R ÷ 100 = 4

⇒ R ÷ 100 = 3

⇒ R = 300%

∴ The correct answer is 300%.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 270

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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 ?

AOption (a)
70 km/h
BOption (b)
75 km/h
COption (c)
80 km/h
DOption (d)
95 km/h

View answer & explanation

Correct answer

75 km/h

Why this answer is correct

Given:

Distance = 150 km

Time lost = 30 minutes = 1 ÷ 2 hour

Increase in speed = 25 km/h

Formula Used:

Time = Distance ÷ Speed

Calculation:

⇒ Let usual speed = x

⇒ Usual time = 150 ÷ x

⇒ New time = 150 ÷ (x + 25)

⇒ Time difference = 1 ÷ 2

⇒ 150 ÷ x − 150 ÷ (x + 25) = 1 ÷ 2

⇒ 150(x + 25) − 150x = (1 ÷ 2) × x(x + 25)

⇒ 3750 = (1 ÷ 2)(x² + 25x)

⇒ 7500 = x² + 25x

⇒ x² + 25x − 7500 = 0

⇒ x = 75

∴ The correct answer is 75.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 271

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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.

AOption (a)
40
BOption (b)
50
COption (c)
60
DOption (d)
80

View answer & explanation

Correct answer

Why this answer is correct

Given:

Winning rate initially = 70%

Matches = x

Lost matches = 20

Final winning rate = 50%

Formula Used:

Wins = (Winning % ÷ 100) × Total matches

Calculation:

⇒ Initial wins = (70 ÷ 100) × x = 0.7x

⇒ Total matches after losing = x + 20

⇒ Wins remain = 0.7x

⇒ Final winning rate = 50%

⇒ 0.7x = (50 ÷ 100) × (x + 20)

⇒ 0.7x = 0.5(x + 20)

⇒ 0.7x = 0.5x + 10

⇒ 0.2x = 10

⇒ x = 50

∴ The correct answer is 50.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 272

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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?

AOption (a)
16 days
BOption (b)
10 days
COption (c)
12 days
DOption (d)
14 days

View answer & explanation

Correct answer

16 days

Why this answer is correct

Shortcut Trick

Let Total Work be the LCM of 8 and 16 = 16 units.

Efficiency of (A + B) = 16 ÷ 8 = 2 units/day.

Efficiency of A = 16 ÷ 16 = 1 unit/day.

Efficiency of B = 2 − 1 = 1 unit/day.

Time taken by B = 16 ÷ 1 = 16 days.

∴ The correct answer is 16 days.

Alternate Method

Given: Time taken by (A + B) = 8 days | Time taken by A = 16 days.

Formula Used: Total Work = Efficiency × Time | Efficiency of B = Efficiency of (A+B) − Efficiency of A.

Person	Days	Total Work	Efficiency
A + B	8	16 (LCM)	2
A	16	1

⇒ Efficiency of B = Efficiency of (A + B) − Efficiency of A

⇒ Efficiency of B = 2 − 1 = 1 unit/day

⇒ Time taken by B = Total Work ÷ Efficiency of B

⇒ Time taken by B = 16 ÷ 1 = 16 days

∴ The correct answer is 16 days.

Additional Information

Work and Efficiency Relation

Efficiency is inversely proportional to Time taken when Work is constant. If time ratio is x : y, efficiency ratio is y :

x. Total Work Formula

Total Work = Efficiency × Time. For multiple people, total efficiency is the sum of their individual efficiencies.

Pipes and Cisterns

This logic applies to Pipes and Cisterns, where an outlet pipe has negative efficiency (work done in removing water).

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 273

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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

AOption (a)
₹1,500
BOption (b)
₹1,650
COption (c)
₹1,517
DOption (d)
₹1,710

View answer & explanation

Correct answer

₹1,517

Why this answer is correct

Given:

Marked price = 2050

Discount = 26%

Formula Used:

Selling price = Marked price × (100 − discount%) ÷ 100

Calculation:

⇒ Selling price = 2050 × (74 ÷ 100)

⇒ Selling price = 1517

∴ The correct answer is 1517.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 274

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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?

AOption (a)
₹5,292
BOption (b)
₹5,000
COption (c)
₹4,982
DOption (d)
₹5,512

View answer & explanation

Correct answer

₹5,292

Why this answer is correct

Given:

Sides = 60 m, 111 m, 123 m

Cost per meter = 18

Formula Used:

Perimeter = Sum of all sides

Total cost = Perimeter × cost per meter

Calculation:

⇒ Perimeter = 60 + 111 + 123 = 294

⇒ Total cost = 294 × 18 = 5292

∴ The correct answer is 5292.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude

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Question 275

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

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?

AOption (a)
60 m, 100 m, 140 m
BOption (b)
70 m, 110 m, 120 m
COption (c)
50 m, 80 m, 120 m
DOption (d)
65 m, 95 m, 140 m

View answer & explanation

Correct answer

60 m, 100 m, 140 m

Why this answer is correct

Shortcut Trick
Total Ratio of sides = 3 : 5 : 7 ⇒ Sum of parts = 15 units.
Given Total Perimeter = 300 m.
Value of 1 unit = 300 ÷ 15 = 20 m.
Sides = (3 × 20) m, (5 × 20) m, (7 × 20) m = 60 m, 100 m, 140 m.
∴ The correct answer is 60 m, 100 m, 140 m.
Alternate Method
Given: The ratio of the sides of the triangular land is 3 : 5 : 7 and the Perimeter is 300 m.
Formula Used: Perimeter of triangle = Side1 + Side2 + Side3

⇒ Let the actual sides of the triangle be 3x, 5x, and 7x.
⇒ Sum of sides = Perimeter
⇒ 3x + 5x + 7x = 300
⇒ 15x = 300
⇒ x = 300 ÷ 15
⇒ x = 20
⇒ Side1 = 3 × 20 = 60 m
⇒ Side2 = 5 × 20 = 100 m
⇒ Side3 = 7 × 20 = 140 m
∴ The correct answer is 60 m, 100 m, 140 m.
Additional Information
Heron's Formula
Used to find the area of a triangle when all three sides (a, b, c) are known: Area = √[s(s − a)(s − b)(s − c)], where s is the semi-perimeter (a+b+c)/2.
Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle must be greater than the length of the third side (a + b > c).
Semi-Perimeter
It is half of the total perimeter of the triangle. For this land, s = 300 ÷ 2 = 150 m.
Classification of Triangles
Based on sides: Equilateral (all sides equal), Isosceles (two sides equal), and Scalene (no sides equal). This land is a Scalene triangle.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude

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Question 276

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

If x is irrational and y is irrational, then x+y is:

AOption (a)
Always irrational
BOption (b)
Always rational
COption (c)
Rational if x=-y
DOption (d)
Never rational

View answer & explanation

Correct answer

Rational if x=-y

Why this answer is correct

Given:

x is irrational

y is irrational

Formula Used:

Sum of two irrational numbers may be rational or irrational

Calculation:

⇒ Let x = √2

⇒ Let y = -√2

⇒ x + y = √2 + ( -√2 ) = 0

⇒ 0 is rational

⇒ Let x = √2

⇒ Let y = √3

⇒ x + y = √2 + √3

⇒ √2 + √3 is irrational

∴ The correct answer is may be rational or irrational.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude

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Question 277

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

In ΔΑΒC, AB = 7 cm, BC = 24 cm and AC = 25cm Find the value of tanC + sinC + 2cotA.

AOption (a)
2566
BOption (b)
1.233
COption (c)
1.155
DOption (d)
2.35

View answer & explanation

Correct answer

1.155

Why this answer is correct

Given:

AB = 7

BC = 24

AC = 25

Formula Used:

In right triangle: sinθ = Perpendicular ÷ Hypotenuse

tanθ = Perpendicular ÷ Base

cotθ = Base ÷ Perpendicular

Calculation:

⇒ 7² + 24² = 49 + 576 = 625 = 25²

⇒ Triangle is right-angled with hypotenuse = 25

⇒ For angle C: opposite = AB = 7, adjacent = BC = 24

⇒ tanC = 7 ÷ 24

⇒ sinC = 7 ÷ 25

⇒ For angle A: opposite = BC = 24, adjacent = AB = 7

⇒ cotA = 7 ÷ 24

⇒ tanC + sinC + 2cotA = 7 ÷ 24 + 7 ÷ 25 + 2 × 7 ÷ 24

⇒ = 7 ÷ 24 + 7 ÷ 25 + 14 ÷ 24

⇒ = 21 ÷ 24 + 7 ÷ 25

⇒ = 7 ÷ 8 + 7 ÷ 25

⇒ = (175 + 56) ÷ 200

⇒ = 231 ÷ 200

⇒ 1.155

∴ The correct answer is 1.155.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude

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Question 278

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

Sonu and Monu together can do a piece of work in 30 days. If Sonu is 20% less efficient than Monu, in how many days can Sonu, working alone, complete 40% of the work?

AOption (a)
32 days
BOption (b)
25 days
COption (c)
30 days
DOption (d)
27 days

View answer & explanation

Correct answer

27 days

Why this answer is correct

Given:

Sonu + Monu time = 30

Sonu efficiency = 80% of Monu

Required work = 40%

Formula Used:

Efficiency ∝ 1 ÷ Time

Work = Efficiency × Time

Calculation:

⇒ Let Monu efficiency = 5x

⇒ Sonu efficiency = 4x

⇒ Combined efficiency = 9x

⇒ 9x = 1 ÷ 30

⇒ x = 1 ÷ 270

⇒ Sonu efficiency = 4x = 4 ÷ 270 = 2 ÷ 135

⇒ Time taken by Sonu for full work = 1 ÷ (2 ÷ 135) = 135 ÷ 2

⇒ Time for 40% work = 135 ÷ 2 × 40 ÷ 100

⇒ Time = 135 ÷ 2 × 2 ÷ 5

⇒ Time = 135 ÷ 5

⇒ Time = 27

∴ The correct answer is 27.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude

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Question 279

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The angle subtended by a chord at the center is 90°. Its length is 8 cm. What is the radius?

AOption (a)
\(4\sqrt{2} \text{ cm}\)
BOption (b)
\(8\sqrt{2} \text{ cm}\)
COption (c)
\(6\sqrt{2} \text{ cm}\)
DOption (d)
\(10\sqrt{2} \text{ cm}\)

View answer & explanation

Correct answer

\(4\sqrt{2} \text{ cm}\)

Why this answer is correct

Given:

Angle at center = 90

Chord length = 8

Formula Used:

Chord length = 2 × Radius × sin(θ ÷ 2)

Calculation:

⇒ 8 = 2 × r × sin(90 ÷ 2)

⇒ 8 = 2 × r × sin45

⇒ sin45 = 1 ÷ √2

⇒ 8 = 2 × r × 1 ÷ √2

⇒ 8 = √2 × r

⇒ r = 8 ÷ √2

⇒ r = 8√2 ÷ 2

⇒ r = 4√2

∴ The correct answer is 4√2.

SSC CHSL 2025 Tier-1 (Held On: 18 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude

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Question 280

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The length of a chord that is at a distance of 5 cm from the center of a circle of radius 13 cm is:

AOption (a)
12 cm
BOption (b)
18 cm
COption (c)
24 cm
DOption (d)
30 cm

View answer & explanation

Correct answer

24 cm

Why this answer is correct

Given:

Radius = 13

Distance from center = 5

Formula Used:

Chord length = 2 × √(Radius² - Distance²)

Calculation:

⇒ Chord length = 2 × √(13² - 5²)

⇒ = 2 × √(169 - 25)

⇒ = 2 × √144

⇒ = 2 × 12

⇒ = 24

∴ The correct answer is 24.

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How should I prepare for Quantitative Aptitude in SSC Combined Higher Secondary Level (CHSL)?

Solving previous year questions by topic helps identify recurring concepts, question patterns, and frequently asked sub-topics. Attempt each question, review the step-by-step explanation, and note down key formulas or facts for quick revision before the exam.

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