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 301–320 of 13275Page 16 of 100
SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 301

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

Which is a mixed recurring decimal?

  1. AOption (a)

    0.25

  2. BOption (b)

    0.6666...

  3. COption (c)

    0.123123...

  4. DOption (d)

    1.436363...

View answer & explanation

Correct answer

Answer: (d)D.

1.436363...

Why this answer is correct

Formula Used:

Mixed recurring decimal: A number with both a non-repeating part and a repeating decimal part.

Calculation:

⇒ Option 1: 0.25 has no repeating decimal part.

⇒ Option 2: 0.6666... is purely recurring, no non-repeating part.

⇒ Option 3: 0.123123... is purely recurring, no non-repeating part.

⇒ Option 4: 1.436363... has 1.43 as the non-repeating part and 63 as the repeating decimal part.

∴ The correct answer is 1.436363...

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 302

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

A microwave is sold at a discount of 25%. If a man buys it for Rs. 7500, what is the marked price?

  1. AOption (a)

    8900

  2. BOption (b)

    10000

  3. COption (c)

    12000

  4. DOption (d)

    9000

View answer & explanation

Correct answer

Answer: (b)B.

10000

Why this answer is correct

Given:

Discount = 25%

Selling Price (SP) = Rs. 7500

Formula Used:

Marked Price (MP) = SP ÷ (1 - Discount/100)

Calculation:

⇒ MP = 7500 ÷ (1 - 25/100)

⇒ MP = 7500 ÷ (1 - 0.25)

⇒ MP = 7500 ÷ 0.75

⇒ MP = Rs. 10000

∴ The correct answer is 10000.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 303

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

In triangle ABC, the orthocenter O and the incenter I. If ∠BIC - ∠BOC = 90o , what is ∠A?

  1. AOption (a)

    100o

  2. BOption (b)

    90o

  3. COption (c)

    110o

  4. DOption (d)

    120o

View answer & explanation

Correct answer

Answer: (d)D.

120o

Why this answer is correct

Given:

∠BIC - ∠BOC = 90°

Formula Used:

∠BIC = 90 + A/2

∠BOC = 180 - A

Calculation:

⇒ (90 + A/2) - (180 - A) = 90

⇒ 90 + A/2 - 180 + A = 90

⇒ -90 + 3A/2 = 90

⇒ 3A/2 = 180

⇒ A = 120°

∴ The correct answer is 120°.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 304

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

If cot \(\theta\) = 3/4, 0° < \(\theta\) < 90°, then find the value of: [(1 + sin \(\theta\)) / cos \(\theta\)] + [(1 - sin \(\theta\)) / cos \(\theta\)].

  1. AOption (a)

    10/3

  2. BOption (b)

    3/17

  3. COption (c)

    -5/7

  4. DOption (d)

    2/9

View answer & explanation

Correct answer

Answer: (a)A.

10/3

Why this answer is correct

Given:

cotθ = 3/4

Formula Used:

cotθ = adjacent ÷ opposite

sinθ = opposite ÷ hypotenuse

cosθ = adjacent ÷ hypotenuse

Calculation:

⇒ adjacent = 3, opposite = 4

⇒ hypotenuse = √(32 + 42) = 5

⇒ sinθ = 4 ÷ 5

⇒ cosθ = 3 ÷ 5

⇒ (1 + sinθ) ÷ cosθ = (1 + 4/5) ÷ (3/5) = (9/5) × (5/3) = 3

⇒ (1 - sinθ) ÷ cosθ = (1 - 4/5) ÷ (3/5) = (1/5) × (5/3) = 1/3

⇒ Total = 3 + 1/3

⇒ Total = 10/3

∴ The correct answer is 10/3.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 305

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

Two tangents are drawn from a point P outside a circle of radius 5 cm. The angle between the tangents is 90°, and the distance from P to the center of the circle is R cm. Find the value of R.

  1. AOption (a)

    6.30 cm

  2. BOption (b)

    5.32 cm

  3. COption (c)

    7.07 cm

  4. DOption (d)

    \(6\sqrt{2} \text{ cm}\)

View answer & explanation

Correct answer

Answer: (c)C.

7.07 cm

Why this answer is correct

Given:

Radius = 5 cm

Angle between tangents = 90°

Formula Used:

Length of tangent = √(R2 - r2)

tan(θ) = opposite ÷ adjacent

Calculation:

⇒ Angle between tangents = 90°

⇒ Each angle = 90 ÷ 2 = 45°

⇒ tan45° = tangent ÷ radius

⇒ 1 = tangent ÷ 5

⇒ tangent = 5

⇒ R2 = 52 + 52

⇒ R2 = 25 + 25 = 50

⇒ R = √50

⇒ R = 7.07 cm

∴ The correct answer is 7.07.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 306

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

In a circle with center O, AC is a chord. OB is perpendicular to AC at point B. If OA = 15 cm and AC = 24cm what is the length of OB?

  1. AOption (a)

    9 cm

  2. BOption (b)

    12 cm

  3. COption (c)

    15 cm

  4. DOption (d)

    10 cm

View answer & explanation

Correct answer

Answer: (a)A.

9 cm

Why this answer is correct

Given:

OA = 15 cm

AC = 24 cm

Formula Used:

Perpendicular from center to chord bisects the chord

OA2 = OB2 + AB2

Calculation:

⇒ AB = 24 ÷ 2 = 12

⇒ OA2 = 152 = 225

⇒ OB2 = 225 - 122

⇒ OB2 = 225 - 144

⇒ OB2 = 81

⇒ OB = 9 cm

∴ The correct answer is 9.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 307

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

The perimeter of a rectangle is \(28\sqrt{2} \text{ cm}\) , while the area of the rectangle is 96 cm². Find the difference between the length and the breadth of the given rectangle.

  1. AOption (a)

    \(3\sqrt{3} \) cm

  2. BOption (b)

    \(\sqrt{2} \) cm

  3. COption (c)

    \(2\sqrt{2} \) cm

  4. DOption (d)

    \(\sqrt{3} \) cm

View answer & explanation

Correct answer

Answer: (c)C.

\(2\sqrt{2} \) cm

Why this answer is correct

Shortcut Trick

Sum of sides (l + b) = Perimeter ÷ 2 = 14√2 cm.

Direct formula: (l − b)2 = (l + b)2 − 4 × Area.

⇒ (l − b)2 = (14√2)2 − 4 × 96 = 392 − 384 = 8.

⇒ (l − b) = √8 = 2√2 cm.

∴ The correct answer is 2√2 cm.

Alternate Method

Given:

Perimeter of the rectangle = 28√2 cm

Area of the rectangle = 96 cm2

Formula Used:

  1. Perimeter = 2 × (l + b)

  2. Area = l × b

  3. (l − b)2 = (l + b)2 − 4lb

Calculations:

⇒ 2(l + b) = 28√2

⇒ l + b = 14√2

⇒ lb = 96

⇒ (l − b)2 = (14√2)2 − 4 × 96

⇒ (l − b)2 = (196 × 2) − 384

⇒ (l − b)2 = 392 − 384 = 8

⇒ l − b = √8 = 2√2 cm

∴ The correct answer is 2√2 cm.

Additional Information

Rectangle Diagonal

The diagonal (d) of a rectangle is calculated using Pythagoras theorem: d = √(l2 + b2).

Algebraic Identity

The relation between sum and difference of two numbers is given by: (a + b)2 − (a − b)2 = 4ab.

Properties of a Rectangle

A rectangle is a quadrilateral where all interior angles are 90° and opposite sides are equal and parallel.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 308

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

The floor of a building consists of 600 rhombus-shaped tiles. If the lengths of the diagonals of each tile are 16 cm and 19 cm, respectively, then find the cost of polishing the floor at a rate of Rs. 10 per m².

  1. AOption (a)

    Rs. 89.5

  2. BOption (b)

    Rs. 98

  3. COption (c)

    Rs. 91.2

  4. DOption (d)

    Rs. 102

View answer & explanation

Correct answer

Answer: (c)C.

Rs. 91.2

Why this answer is correct

Given:

Number of tiles = 600

Diagonal 1 = 16 cm

Diagonal 2 = 19 cm

Rate = 10 per m2

Formula Used:

Area of rhombus = 1/2 × d1 × d2

Total cost = Total area × rate

Calculation:

⇒ Area of one tile = 1/2 × 16 × 19

⇒ Area of one tile = 8 × 19 = 152 cm2

⇒ Total area = 600 × 152 = 91200 cm2

⇒ Total area = 91200 ÷ 10000 = 9.12 m2

⇒ Cost = 9.12 × 10

⇒ Cost = 91.2

∴ The correct answer is 91.2.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 309

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

In a rhombus, the diagonals are in the ratio 8:15. If the side of the rhombus is 17 cm, determine the exact lengths of both diagonals and the area.

  1. AOption (a)

    12 cm and 26 cm, 220 cm² 

  2. BOption (b)

    22 cm and 38 cm, 340 cm²

  3. COption (c)

    16 cm and 30 cm, 240 cm²

  4. DOption (d)

    Can't be determined

View answer & explanation

Correct answer

Answer: (c)C.

16 cm and 30 cm, 240 cm²

Why this answer is correct

Given:

Ratio of diagonals = 8 : 15

Side = 17 cm

Formula Used:

(d1/2)2 + (d2/2)2 = side2

Area of rhombus = 1/2 × d1 × d2

Calculation:

⇒ Let d1 = 8x, d2 = 15x

⇒ (8x/2)2 + (15x/2)2 = 172

⇒ (4x)2 + (7.5x)2 = 289

⇒ 16x2 + 56.25x2 = 289

⇒ 72.25x2 = 289

⇒ x2 = 289 ÷ 72.25 = 4

⇒ x = 2

⇒ d1 = 8 × 2 = 16 cm

⇒ d2 = 15 × 2 = 30 cm

⇒ Area = 1/2 × 16 × 30

⇒ Area = 8 × 30 = 240 cm2

∴ The correct answer is 16 cm, 30 cm and 240 cm2.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 310

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

In a partnership, A and B invest in the ratio 4:5. If the total profit is Rs. 1,10,000, how much profit does A get?

  1. AOption (a)

    Rs. 44,000

  2. BOption (b)

    Rs. 52,678

  3. COption (c)

    Rs. 66,780

  4. DOption (d)

    Rs. 48,889

View answer & explanation

Correct answer

Answer: (d)D.

Rs. 48,889

Why this answer is correct

Given:

Ratio = 4 : 5

Total profit = 110000

Formula Used:

Share = Total profit × individual ratio ÷ sum of ratios

Calculation:

⇒ Total ratio = 4 + 5 = 9

⇒ A's share = 110000 × 4 ÷ 9

⇒ A's share = 440000 ÷ 9

⇒ A's share = 48889

∴ The correct answer is 48889.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 311

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

The greatest four-digit number that is divisible by each of the numbers 12, 18, 21, and 28 is?

  1. AOption (a)

    9288

  2. BOption (b)

    9882

  3. COption (c)

    9828

  4. DOption (d)

    9928

View answer & explanation

Correct answer

Answer: (c)C.

9828

Why this answer is correct

Given:

Numbers = 12, 18, 21, 28

Formula Used:

LCM of given numbers

Required number = Largest 4-digit number ÷ LCM × LCM

Calculation:

⇒ 12 = 22 × 3

⇒ 18 = 2 × 32

⇒ 21 = 3 × 7

⇒ 28 = 22 × 7

⇒ LCM = 22 × 32 × 7 = 252

⇒ 9999 ÷ 252 = 39

⇒ Required number = 252 × 39

⇒ Required number = 9828

∴ The correct answer is 9828.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 312

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

The average of 9 numbers is 63. If the average of the first 4 numbers is 60 and the average of the last 4 numbers is 66, what is the fifth number? 

  1. AOption (a)

    63

  2. BOption (b)

    69

  3. COption (c)

    67

  4. DOption (d)

    65

View answer & explanation

Correct answer

Answer: (a)A.

63

Why this answer is correct

Shortcut Trick

Base average of 9 numbers = 63.

Deviation of first 4 numbers = 4 × (60 − 63) = −12.

Deviation of last 4 numbers = 4 × (66 − 63) = +12.

Net deviation = −12 + 12 = 0.

Fifth number = Base Average + Net Deviation = 63 + 0 = 63.

∴ The correct answer is 63.

Alternate Method

Given:

Average of 9 numbers = 63

Average of first 4 numbers = 60

Average of last 4 numbers = 66

Formula Used:

Sum of observations = Average × Total observations

Calculations:

⇒ Sum of all 9 numbers = 9 × 63 = 567

⇒ Sum of first 4 numbers = 4 × 60 = 240

⇒ Sum of last 4 numbers = 4 × 66 = 264

⇒ Sum of 8 numbers (excluding the 5th) = 240 + 264 = 504

⇒ Fifth number = 567 − 504 = 63

∴ The correct answer is 63.

Additional Information

Deviation Method

The sum of deviations of all observations from their arithmetic mean is always zero. This property is used to find a missing term by balancing positive and negative deviations.

Net Change in Average

If every observation is increased or decreased by a constant value k, the new average also increases or decreases by k.

Weighted Average

When two groups with averages A1 and A2 and sizes n1 and n2 are combined, the combined average = (n1A1 + n2A2) ÷ (n1 + n2).

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 313

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

A batsman typically scores 40 runs in 20 innings. To raise his average by two, how many runs does he need to score in the twenty-first inning?

  1. AOption (a)

    80

  2. BOption (b)

    82

  3. COption (c)

    79

  4. DOption (d)

    86

View answer & explanation

Correct answer

Answer: (b)B.

82

Why this answer is correct

Given:

Initial average = 40 runs

Total innings = 20

New average = 42 runs

21st inning score = ?

Formula Used:

Total runs = Average × Number of innings

21st inning score = New total runs − Previous total runs

Calculation:

⇒ Previous total runs = 40 × 20

⇒ Previous total runs = 800

⇒ New total runs = 42 × 21

⇒ New total runs = 882

⇒ 21st inning score = 882 − 800

⇒ 21st inning score = 82

∴ The correct answer is 82.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 314

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

A cyclist rides at a speed of 30 km/h and covers a distance of 90 km. Then, he increases his speed by 20% and covers another 60 km. What is the total time taken for the journey?

  1. AOption (a)

    4 hour 40 minutes 

  2. BOption (b)

    2 hour 10 minutes

  3. COption (c)

    2 hour 40 minutes

  4. DOption (d)

    3 hour 10 minutes

View answer & explanation

Correct answer

Answer: (a)A.

4 hour 40 minutes 

Why this answer is correct

Given:

Speed 1 = 30 km/h

Distance 1 = 90 km

Increase = 20%

Distance 2 = 60 km

Formula Used:

Time = Distance ÷ Speed

Calculation:

⇒ Time 1 = 90 ÷ 30 = 3 hours

⇒ New speed = 30 × 120 ÷ 100 = 36 km/h

⇒ Time 2 = 60 ÷ 36

⇒ Time 2 = 5/3 hours

⇒ Total time = 3 + 5/3

⇒ Total time = 14/3 hours = 4 hour 40 minutes 

∴ The correct answer is 4 hour 40 minutes.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 315

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

Ravi and Raj are running a 200-meter race. Ravi completes it in 25 seconds, while Raj takes 30 seconds. By what distance will Ravi beat Raj?

  1. AOption (a)

    30.25 m

  2. BOption (b)

    41.27 m

  3. COption (c)

    25.55 m

  4. DOption (d)

    33.33 m

View answer & explanation

Correct answer

Answer: (d)D.

33.33 m

Why this answer is correct

Given:

Distance = 200 m

Time taken by Ravi = 25 s

Time taken by Raj = 30 s

Formula Used:

Speed = Distance ÷ Time

Distance = Speed × Time

Calculation:

⇒ Speed of Ravi = 200 ÷ 25 = 8 m/s

⇒ Speed of Raj = 200 ÷ 30 = 20/3 m/s

⇒ Distance covered by Raj in 25 s = (20/3) × 25

⇒ Distance = 500/3 m

⇒ Difference = 200 - 500/3

⇒ Difference = (600 - 500) ÷ 3

⇒ Difference = 100/3 m

⇒ Difference = 33.33 m

∴ The correct answer is 33.33.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 316

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

A and B can complete a work in 6 days together. If A alone can do it in 12 days, how many days will B alone take?

  1. AOption (a)

    8 days

  2. BOption (b)

    10 days

  3. COption (c)

    12 days

  4. DOption (d)

    14 days

View answer & explanation

Correct answer

Answer: (c)C.

12 days

Why this answer is correct

Given:

A + B = 6 days

A alone = 12 days

Formula Used:

Total work = LCM of given days

Work rate = Work ÷ Days

Calculation:

⇒ LCM of 6 and 12 = 12

⇒ Total work = 12 units

⇒ (A + B)'s 1 day work = 12 ÷ 6 = 2

⇒ A's 1 day work = 12 ÷ 12 = 1

⇒ B's 1 day work = 2 − 1 = 1

⇒ B's time = 12 ÷ 1 = 12 days

∴ The correct answer is 12 days.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 317

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

A watch is sold for ₹840 after a discount of 20% is applied. What was the original marked price of the watch?

  1. AOption (a)

    ₹950

  2. BOption (b)

    ₹1,000

  3. COption (c)

    ₹1,050

  4. DOption (d)

    ₹1,120

View answer & explanation

Correct answer

Answer: (c)C.

₹1,050

Why this answer is correct

Given:

Discount = 20%

Selling Price = ₹840

Formula Used:

Discount = (Discount Percentage ÷ 100) × Marked Price

Selling Price = Marked Price − Discount

Calculation:

Let Marked Price = ₹x

⇒ Discount = (20 ÷ 100) × x = 0.2x

⇒ Selling Price = x − 0.2x

⇒ ₹840 = 0.8x

⇒ x = ₹840 ÷ 0.8

⇒ x = ₹1,050

∴ The correct answer is ₹1,050.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 318

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

Two containers of equal capacity are each filled with a milk-water mixture. In the first container, the ratio of milk to water is 3:2, while in the second container the ratio is 4:

  1. If the contents of both containers are combined thoroughly in a larger vessel, what will be the resulting ratio of milk to water in the final mixture?
  1. AOption (a)

    7:3

  2. BOption (b)

    31:19

  3. COption (c)

    19:31

  4. DOption (d)

    7:1

View answer & explanation

Correct answer

Answer: (a)A.

7:3

Why this answer is correct

Given:

Equal quantities in both containers

Ratio (milk : water) in A = 3 : 2

Ratio (milk : water) in B = 4 : 1

Formula Used:

Quantity = Total × (part ÷ sum of ratio)

Calculation:

⇒ Let each container = 5 litres

⇒ Milk in A = 5 × (3 ÷ 5) = 3

⇒ Water in A = 5 × (2 ÷ 5) = 2

⇒ Milk in B = 5 × (4 ÷ 5) = 4

⇒ Water in B = 5 × (1 ÷ 5) = 1

⇒ Total milk = 3 + 4 = 7

⇒ Total water = 2 + 1 = 3

⇒ Ratio = 7 : 3

∴ The correct answer is 7 : 3.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 319

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

In a circle of radius 10 cm, two parallel chords of lengths 12 cm and 16 cm lie on opposite sides of the centre. Determine the distance separating these two chords.

  1. AOption (a)

    14 cm

  2. BOption (b)

    12 cm

  3. COption (c)

    10 cm

  4. DOption (d)

    11 cm

View answer & explanation

Correct answer

Answer: (a)A.

14 cm

Why this answer is correct

Given:

Radius = 10 cm

Chord1 = 12 cm

Chord2 = 16 cm

Formula Used:

Distance from center = √(r2 − (chord ÷ 2)2)

Calculation:

⇒ Distance1 = √(102 − 62)

⇒ = √(100 − 36) = √64 = 8

⇒ Distance2 = √(102 − 82)

⇒ = √(100 − 64) = √36 = 6

⇒ Total distance = 8 + 6 = 14 cm

∴ The correct answer is 14 cm.

SSC CHSL 2025 Tier-1 (Held On: 19 Nov, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 320

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

Two parallel chords of a circle, with lengths 16 cm and 12 cm, are on opposite sides of the center. If the distance between the two chords is 14 cm, what is the radius of the circle?

  1. AOption (a)

    10 cm

  2. BOption (b)

    8 cm

  3. COption (c)

    12 cm

  4. DOption (d)

    14 cm

View answer & explanation

Correct answer

Answer: (a)A.

10 cm

Why this answer is correct

Given:

Chord 1 = 16 cm

Chord 2 = 12 cm

Distance between chords = 14 cm

Formula Used:

Distance from center to chord = √(r2 - (chord/2)2)

Calculation:

⇒ Let radius = r

⇒ Distance for chord 16 = √(r2 - 82)

⇒ Distance for chord 12 = √(r2 - 62)

⇒ √(r2 - 64) + √(r2 - 36) = 14

⇒ Let √(r2 - 64) = a

⇒ √(r2 - 36) = b

⇒ a + b = 14

⇒ b2 - a2 = 28

⇒ (b - a)(b + a) = 28

⇒ (b - a) × 14 = 28

⇒ b - a = 2

⇒ Solve a + b = 14 and b - a = 2

⇒ b = 8, a = 6

⇒ r2 - 36 = 64

⇒ r2 = 100

⇒ r = 10 cm

∴ The correct answer is 10.

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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.