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SSC Combined Higher Secondary Level (CHSL) Β· 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 161–180 of 13275Page 9 of 100
SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 161

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A circular field is surrounded by a path of uniform width 2.5 m. The total area of the field, including the path, is 1078.125 mΒ². If the area of the circular field (excluding the path) is 803.25 mΒ², find the radius of the circular field and the radius including the path. (Use Ο€ = 3.1416)

  1. AOption (a)

    15.50 m, 18.00 m

  2. BOption (b)

    15.99 m, 18.52 m

  3. COption (c)

    16.20 m, 18.70 m

  4. DOption (d)

    14.00 m, 16.50 m

View answer & explanation

Correct answer

Answer: (b)B.

15.99 m, 18.52 m

Why this answer is correct

Given:

Total area = 1078.125

Inner area = 803.25

Ο€ = 3.1416

Formula Used:

Area of circle = Ο€ Γ— r2

Calculation:

β‡’ Ο€ Γ— r2 = 803.25

β‡’ r2 = 803.25 Γ· 3.1416

β‡’ r2 = 255.75

β‡’ r = √255.75 β‰ˆ 15.99Β 

β‡’ Ο€ Γ— R2 = 1078.125

β‡’ R2 = 1078.125 Γ· 3.1416

β‡’ R2 = 343.177043545

β‡’ R = √343.177043545 β‰ˆ 18.52

∴ The correct answer is 15.99 m and 18.52 m.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 162

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

If (m - n) = 4 and (m3 - n3) = 1216 then find the value of (m X n).

  1. AOption (a)

    89

  2. BOption (b)

    96

  3. COption (c)

    75

  4. DOption (d)

    91

View answer & explanation

Correct answer

Answer: (b)B.

96

Why this answer is correct

Given:

m - n = 4

m3 - n3 = 1216

Formula Used:

m3 - n3 = (m - n)(m2 + mn + n2)

(m - n)2 = m2 + n2 - 2mn

Calculation:

β‡’ 1216 = 4 Γ— (m2 + mn + n2)

β‡’ m2 + mn + n2 = 1216 Γ· 4 = 304

β‡’ (m - n)2 = 16 = m2 + n2 - 2mn

β‡’ m2 + n2 = 16 + 2mn

β‡’ (16 + 2mn) + mn = 304

β‡’ 16 + 3mn = 304

β‡’ 3mn = 288

β‡’ mn = 96

∴ The correct answer is 96.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 163

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

If x + y = 24 and x2 + y2 = 360 then find the value of (x4 + x2y2 + y4).

  1. AOption (a)

    1,17,936

  2. BOption (b)

    1,25,878

  3. COption (c)

    1,24,784

  4. DOption (d)

    2,15,869

View answer & explanation

Correct answer

Answer: (a)A.

1,17,936

Why this answer is correct

Shortcut Trick

Find xy: xy = ((x + y)2 βˆ’ (x2 + y2)) Γ· 2 = (242 βˆ’ 360) Γ· 2 = (576 βˆ’ 360) Γ· 2 = 108.

Use the identity: x4 + x2y2 + y4 = (x2 + y2)2 βˆ’ (xy)2.

Substitute values: 3602 βˆ’ 1082 = (360 + 108)(360 βˆ’ 108) = 468 Γ— 252.

Calculation: 468 Γ— 252 = 1,17,936.

∴ The correct answer is 1,17,936.

Alternate Method

Given:

x + y = 24

x2 + y2 = 360

Formula Used:

  1. (x + y)2 = x2 + y2 + 2xy

  2. x4 + x2y2 + y4 = (x2 + y2 + xy)(x2 + y2 βˆ’ xy)

Calculations:

β‡’ (24)2 = 360 + 2xy

β‡’ 576 = 360 + 2xy

β‡’ 2xy = 576 βˆ’ 360 = 216

β‡’ xy = 108

β‡’ x4 + x2y2 + y4 = (360 + 108) Γ— (360 βˆ’ 108)

β‡’ x4 + x2y2 + y4 = 468 Γ— 252

β‡’ x4 + x2y2 + y4 = 1,17,936

∴ The correct answer is 1,17,936.

Additional Information

Biquadratic Identity

The expression x4 + x2y2 + y4 is also known as the Sophie Germain related identity and can be factored as (x2 + xy + y2)(x2 βˆ’ xy + y2).

Algebraic Expansion

(x2 + y2)2 = x4 + 2x2y2 + y4. Therefore, x4 + x2y2 + y4 = (x2 + y2)2 βˆ’ x2y2.

Difference of Squares

The identity a2 βˆ’ b2 = (a + b)(a βˆ’ b) is frequently used to simplify the product of large numbers like 468 Γ— 252 = (360 + 108)(360 βˆ’ 108).

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 164

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

If sinx + sin2Β x = 1Β  then find the value of cos2Β x + cos4Β x.

  1. AOption (a)

    2

  2. BOption (b)

    0

  3. COption (c)

    1

  4. DOption (d)

    -1

View answer & explanation

Correct answer

Answer: (c)C.

1

Why this answer is correct

Given:

sinx + sin2x = 1

Formula Used:

sin2x + cos2x = 1

Calculation:

β‡’ sinx + sin2x = 1

β‡’ sin2x = 1 - sinx

β‡’ cos2x = 1 - sin2x = 1 - (1 - sinx) = sinx

β‡’ cos4x = (cos2x)2 = sin2x

β‡’ cos2x + cos4x = sinx + sin2x

β‡’ = 1

∴ The correct answer is 1.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 165

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

If, 2 x sin q x cos(90o - q) = (1/2) , such that 0o < q < 90oΒ  then find the value of cot2Β 2q + cos2Β 3q.

  1. AOption (a)

    1/7

  2. BOption (b)

    1/2

  3. COption (c)

    1/5

  4. DOption (d)

    1/3

View answer & explanation

Correct answer

Answer: (d)D.

1/3

Why this answer is correct

Given:

2 Γ— sinq Γ— cos(90Β° - q) = 1 / 2

Formula Used:

cos(90Β° - q) = sinq

sin2q = 2sinq cosq

1 + cot2ΞΈ = cosec2ΞΈ

Calculation:

β‡’ 2 Γ— sinq Γ— sinq = 1 / 2

β‡’ 2sin2q = 1 / 2

β‡’ sin2q = 1 / 4

β‡’ sinq = 1 / 2

β‡’ q = 30Β°

β‡’ cot22q = cot260Β° = (1 / √3)2 = 1/3

β‡’ cos23q = cos290Β° = 0

β‡’ cot22q + cos23q = 1 / 3 + 0 = 1 / 3

∴ The correct answer is 1 / 3.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 166

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A firefighter's ladder of length 41 meters is resting against a wall and makes an angle x with the ground such that sin x = 40/41 Find the horizontal distance of the foot of the ladder from the wall.

  1. AOption (a)

    9 meters

  2. BOption (b)

    5 meters

  3. COption (c)

    6 meters

  4. DOption (d)

    12 meters

View answer & explanation

Correct answer

Answer: (a)A.

9 meters

Why this answer is correct

Given:

Length of ladder = 41

sinx = 40 Γ· 41

Formula Used:

sinx = Perpendicular Γ· Hypotenuse

cosx = Base Γ· Hypotenuse

sin2x + cos2x = 1

Calculation:

β‡’ Perpendicular = 40

β‡’ cos2x = 1 - (40 Γ· 41)2 = 1 - 1600 Γ· 1681 = 81 Γ· 1681

β‡’ cosx = 9 Γ· 41

β‡’ Base = 41 Γ— (9 Γ· 41) = 9

∴ The correct answer is 9 m.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 167

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Rs. 12,000 is invested at the rate of 8% p.a. for 4 years at simple interest in scheme 'A'. The amount received from scheme 'A' is then invested for 2 years at rate of 5% p.a., compounded annually in scheme 'B'. Find the amount received from scheme 'B'.

  1. AOption (a)

    Rs. 17,463.6

  2. BOption (b)

    Rs. 17,677

  3. COption (c)

    Rs. 11,460.5

  4. DOption (d)

    Rs. 17,999

View answer & explanation

Correct answer

Answer: (a)A.

Rs. 17,463.6

Why this answer is correct

Given:

Principal = 12000

Rate1 = 8%

Time1 = 4 years

Rate2 = 5%

Time2 = 2 years

Formula Used:

SI amount = P Γ— (1 + r Γ— t Γ· 100)

CI amount = P Γ— (1 + r Γ· 100)n

Calculation:

β‡’ Amount after scheme A = 12000 Γ— (1 + 8 Γ— 4 Γ· 100) = 12000 Γ— (1 + 32 Γ· 100) = 12000 Γ— (132 Γ· 100) = 15840

β‡’ Amount after scheme B = 15840 Γ— (1 + 5 Γ· 100)2

β‡’ = 15840 Γ— (105 Γ· 100)2

β‡’ = 15840 Γ— (11025 Γ· 10000)

β‡’ = 17463.6

∴ The correct answer is 17463.6.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 168

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A person invested Rs. 18,000 each in two different schemes for 2 years. One scheme offered 15% p.a. simple interest and the other offered 15% p.a. compound interest, compounded annually. Find the difference between the interest earned from both schemes.

  1. AOption (a)

    Rs. 456

  2. BOption (b)

    Rs. 405

  3. COption (c)

    Rs. 489

  4. DOption (d)

    Rs. 411

View answer & explanation

Correct answer

Answer: (b)B.

Rs. 405

Why this answer is correct

Given:

Principal = 18000

Rate = 15%

Time = 2 years

Formula Used:

SI = P Γ— r Γ— t Γ· 100

Difference for 2 years = P Γ— (r Γ· 100)2

Calculation:

β‡’ SI = 18000 Γ— 15 Γ— 2 Γ· 100 = 5400

β‡’ Difference = 18000 Γ— (15 Γ· 100)2

β‡’ = 18000 Γ— (225 Γ· 10000)

β‡’ = 18000 Γ— 0.0225 = 405

∴ The correct answer is 405.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 169

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

If Rs. 60,000 is invested at a compound interest of 20% p.a. for 18 months, then find the interest earned given that the sum is compounded every 6 months.

  1. AOption (a)

    Rs. 21,800

  2. BOption (b)

    Rs. 12,550

  3. COption (c)

    Rs. 14,670

  4. DOption (d)

    Rs. 19,860

View answer & explanation

Correct answer

Answer: (d)D.

Rs. 19,860

Why this answer is correct

Given:

Principal = 60000

Rate = 20% per annum

Time = 18 months

Compounding period = 6 months

Formula Used:

Amount = Principal Γ— (1 + r/n)nt

Interest = Amount βˆ’ Principal

Calculation:

β‡’ Number of periods = 18 Γ· 6 = 3

β‡’ Rate per period = 20 Γ· 2 = 10%

β‡’ Amount = 60000 Γ— (1 + 0.10)3

β‡’ Amount = 60000 Γ— 1.13

β‡’ Amount = 60000 Γ— 1.331 = 79860

β‡’ Interest = 79860 βˆ’ 60000

β‡’ Interest = 19860

∴ The correct answer is 19860.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 170

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A box contains Rs 5, Rs 2 and Rs 1 coins in the ratio 7:3:2. The total amount in the box is Rs 1075. Find the number of Rs 2 coins.

  1. AOption (a)

    70

  2. BOption (b)

    75

  3. COption (c)

    80

  4. DOption (d)

    85

View answer & explanation

Correct answer

Answer: (b)B.

75

Why this answer is correct

Given:

Ratio of coins (β‚Ή5 : β‚Ή2 : β‚Ή1) = 7 : 3 : 2

Total amount = 1075

Formula Used:

Number of coins = Ratio Γ— k

Total value = (number Γ— value)

Calculation:

β‡’ Number of β‚Ή5 coins = 7k, β‚Ή2 coins = 3k, β‚Ή1 coins = 2k

β‡’ Total value = 5Γ—7k + 2Γ—3k + 1Γ—2k

β‡’ Total value = 35k + 6k + 2k = 43k

β‡’ 43k = 1075

β‡’ k = 1075 Γ· 43 = 25

β‡’ Number of β‚Ή2 coins = 3k = 3 Γ— 25 = 75

∴ The correct answer is 75.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 171

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Β A tank contains liquid A and B in the ratio 13:12. 50 litres of the mixture are taken out and filled with 50 litres of B. Now the ratio changes to 7:8. Find the quantity of liquid B initially.

  1. AOption (a)

    222 litres

  2. BOption (b)

    230 litres

  3. COption (c)

    240 litres

  4. DOption (d)

    234 litres

View answer & explanation

Correct answer

Answer: (d)D.

234 litres

Why this answer is correct

Given:

Initial ratio A : B = 13 : 12

50 litres removed and replaced by B

New ratio A : B = 7 : 8

Formula Used:

Quantity = Ratio Γ— Total Γ· Sum of ratios

Calculation:

β‡’ Let total quantity = x

β‡’ Initial A = 13x Γ· 25, B = 12x Γ· 25

β‡’ After removing 50 litres:

β‡’ A removed = (13x Γ· 25) Γ— (50 Γ· x) = 26

β‡’ B removed = (12x Γ· 25) Γ— (50 Γ· x) = 24

β‡’ Remaining A = 13x Γ· 25 βˆ’ 26

β‡’ Remaining B = 12x Γ· 25 βˆ’ 24 + 50

β‡’ Remaining B = 12x Γ· 25 + 26

β‡’ New ratio: (13x Γ· 25 βˆ’ 26) : (12x Γ· 25 + 26) = 7 : 8

β‡’ 8(13x Γ· 25 βˆ’ 26) = 7(12x Γ· 25 + 26)

β‡’ 104x Γ· 25 βˆ’ 208 = 84x Γ· 25 + 182

β‡’ 104x Γ· 25 βˆ’ 84x Γ· 25 = 182 + 208

β‡’ 20x Γ· 25 = 390

β‡’ 4x Γ· 5 = 390

β‡’ x = 390 Γ— 5 Γ· 4 = 487.5

β‡’ Initial B = 12x Γ· 25 = 12 Γ— 487.5 Γ· 25 = 234

∴ The correct answer is 234.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 172

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A mixture containing 18% of methanol is mixed with another mixture containing b% of methanol in the ratio 4: 5. If the percentage of methanol in the final mixture is 16%, what is the percentage of b?

  1. AOption (a)

    12.26%

  2. BOption (b)

    14.4%

  3. COption (c)

    17.2%

  4. DOption (d)

    16.32%

View answer & explanation

Correct answer

Answer: (b)B.

14.4%

Why this answer is correct

Given:

First mixture = 18%

Second mixture = b%

Ratio = 4 : 5

Final mixture = 16%

Formula Used:

Weighted average = (aΓ—x + bΓ—y) Γ· (x + y)

Calculation:

β‡’ (18 Γ— 4 + b Γ— 5) Γ· (4 + 5) = 16

β‡’ (72 + 5b) Γ· 9 = 16

β‡’ 72 + 5b = 16 Γ— 9 = 144

β‡’ 5b = 144 βˆ’ 72 = 72

β‡’ b = 72 Γ· 5 = 14.4

∴ The correct answer is 14.4%.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 173

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A person travels a total distance of 480 km, partly by car at a speed of 80 km/h and the rest by bike at 40 km/h. If 75% of the total distance is covered by car, find the approximate average speed of the entire journey.

  1. AOption (a)

    44 km/h

  2. BOption (b)

    56 km/h

  3. COption (c)

    28 km/h

  4. DOption (d)

    64 km/h

View answer & explanation

Correct answer

Answer: (d)D.

64 km/h

Why this answer is correct

Given:

Total distance = 480

Car speed = 80

Bike speed = 40

Car distance = 75% of 480 = 360

Bike distance = 480 βˆ’ 360 = 120

Formula Used:

Time = Distance Γ· Speed

Average speed = Total distance Γ· Total time

Calculation:

β‡’ Time by car = 360 Γ· 80 = 4.5

β‡’ Time by bike = 120 Γ· 40 = 3

β‡’ Total time = 4.5 + 3 = 7.5

β‡’ Average speed = 480 Γ· 7.5 = 64

∴ The correct answer is 64.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 174

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

How many perfect squares lie between 10,000 and 20,000 (inclusive)?

  1. AOption (a)

    32

  2. BOption (b)

    42

  3. COption (c)

    52

  4. DOption (d)

    62

View answer & explanation

Correct answer

Answer: (b)B.

42

Why this answer is correct

Given:

Lower limit = 10000

Upper limit = 20000

Formula Used:

Perfect squares between a and b = ⌊√bβŒ‹ βˆ’ ⌈√aβŒ‰ + 1

Calculation:

β‡’ √10000 = 100

β‡’ √20000 β‰ˆ 141.42 β‡’ floor = 141

β‡’ Count = 141 βˆ’ 100 + 1 = 42

∴ The correct answer is 42.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 1) (Previous Year Paper)Β·Quantitative Aptitude
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Question 175

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A motorist travels from City P to City Q, a distance of 234 km. For the first 78 km, he drives at 52 km/h. For the next 78 km he drives at 39 km/h. For the remaining distance, he drives at x km/h (constant). His overall average speed for the whole trip is 46.8 km/h.Β What is x?Β 

  1. AOption (a)

    65 km/h

  2. BOption (b)

    72 km/h

  3. COption (c)

    60 km/h

  4. DOption (d)

    52 km/h

View answer & explanation

Correct answer

Answer: (d)D.

52 km/h

Why this answer is correct

Given:

Total distance = 234

First 78 km at 52 km/h

Next 78 km at 39 km/h

Remaining distance = 78 at speed x

Average speed = 46.8

Formula Used:

Time = Distance Γ· Speed

Average speed = Total distance Γ· Total time

Calculation:

β‡’ Time₁ = 78 Γ· 52 = 1.5

β‡’ Timeβ‚‚ = 78 Γ· 39 = 2

β‡’ Time₃ = 78 Γ· x

β‡’ Total time = 234 Γ· 46.8 = 5

β‡’ 1.5 + 2 + 78 Γ· x = 5

β‡’ 3.5 + 78 Γ· x = 5

β‡’ 78 Γ· x = 5 βˆ’ 3.5 = 1.5

β‡’ x = 78 Γ· 1.5 = 52

∴ The correct answer is 52.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 2) (Previous Year Paper)Β·Quantitative Aptitude
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Question 176

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A trader sells two articles at β‚Ή960 each. On one, he gains 20%, and on the other, he loses 20%. What is his overall profit or loss percentage?

  1. AOption (a)

    5% Profit

  2. BOption (b)

    4% Loss

  3. COption (c)

    4% Profit

  4. DOption (d)

    5% Loss

View answer & explanation

Correct answer

Answer: (b)B.

4% Loss

Why this answer is correct

Shortcut Trick

When the Selling Price (SP) of two articles is the same, and one is sold at a gain of x% and the other at a loss of x%, the trader always incurs an overall loss.

Shortcut formula: Overall Loss % = (x/10)2

Given the common percentage change x = 20.

β‡’ Overall Loss % = (20 / 10)2 = 22 = 4%.

∴ The correct answer is 4% Loss.

Alternate Method

Given: Selling Price (SP) of each article = β‚Ή960, Profit% = 20%, Loss% = 20%.

Formula: Cost Price (CP) = SP Γ— [100 / (100 Β± Profit/Loss %)]

β‡’ CP1 = 960 Γ— (100 / 120) = β‚Ή800

β‡’ CP2 = 960 Γ— (100 / 80) = β‚Ή1200

β‡’ Total CP = 800 + 1200 = β‚Ή2000

β‡’ Total SP = 960 + 960 = β‚Ή1920

β‡’ Net Loss = 2000 βˆ’ 1920 = β‚Ή80

β‡’ Loss % = (80 / 2000) Γ— 100 = 4%

&there4$ The correct answer is 4% Loss.

Additional Information

Net Percentage Change

For two successive percentage changes a% and b%, the net change is given by [a + b + (ab/100)]%. When a = +x and b = βˆ’x, it simplifies to βˆ’x2/100.

CP Calculation Shortcut

Profit of 20% means SP = 1.2 Γ— CP, so CP = SP Γ· 1.2. Loss of 20% means SP = 0.8 Γ— CP, so CP = SP Γ· 0.8.

Overall Profit/Loss in Value

If SP is same and profit/loss % is x, Total Loss in Rupees = [2 Γ— SP Γ— x2] / [1002 βˆ’ x2]$.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 2) (Previous Year Paper)Β·Quantitative Aptitude
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Question 177

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A vase has a marked price of Rs. 550. Simran pays Rs. 350 for it because she got 2 successive discounts, one of 20% and the other of:

  1. AOption (a)

    21%

  2. BOption (b)

    22%

  3. COption (c)

    19.25%

  4. DOption (d)

    20.45%

View answer & explanation

Correct answer

Answer: (d)D.

20.45%

Why this answer is correct

Given:

Marked price = 550

Selling price = 350

First discount = 20%

Formula Used:

SP = MP Γ— (1 - d1 Γ· 100) Γ— (1 - d2 Γ· 100)

Calculation:

β‡’ 350 = 550 Γ— (1 - 20 Γ· 100) Γ— (1 - d Γ· 100)

β‡’ 350 = 550 Γ— (4 Γ· 5) Γ— (1 - d Γ· 100)

β‡’ 350 = 440 Γ— (1 - d Γ· 100)

β‡’ 350 Γ· 440 = (1 - d Γ· 100)

β‡’ 35 Γ· 44 = (1 - d Γ· 100)

β‡’ d Γ· 100 = 1 - 35 Γ· 44 = 9 Γ· 44

β‡’ d = (9 Γ· 44) Γ— 100 = 900 Γ· 44 = 20.45

∴ The correct answer is 20.45%.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 2) (Previous Year Paper)Β·Quantitative Aptitude
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Question 178

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A seller marks his goods 32% above their cost price but allows a 16% discount for cash payment. His percentage of profit when sold in cash is:

  1. AOption (a)

    9.12%

  2. BOption (b)

    8.25%

  3. COption (c)

    10.88%

  4. DOption (d)

    11.5%

View answer & explanation

Correct answer

Answer: (c)C.

10.88%

Why this answer is correct

Given:

Marked up = 32%

Discount = 16%

Formula Used:

SP = CP Γ— (1 + markup Γ· 100) Γ— (1 - discount Γ· 100)

Profit % = ((SP - CP) Γ· CP) Γ— 100

Calculation:

β‡’ SP = CP Γ— (1 + 32 Γ· 100) Γ— (1 - 16 Γ· 100)

β‡’ = CP Γ— (132 Γ· 100) Γ— (84 Γ· 100)

β‡’ = CP Γ— (11088 Γ· 10000)

β‡’ SP = CP Γ— 1.1088

β‡’ Profit = 1.1088CP - CP = 0.1088CP

β‡’ Profit % = 0.1088 Γ— 100 = 10.88

∴ The correct answer is 10.88%.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 2) (Previous Year Paper)Β·Quantitative Aptitude
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Question 179

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A company had 4 departments with an average of 45 employees each. Later, a fifth department was added with 90 employees. Due to restructuring, 10 employees from each of the original four departments were transferred to the new department. What is the new average number of employees per department?

  1. AOption (a)

    50

  2. BOption (b)

    52

  3. COption (c)

    54

  4. DOption (d)

    56

View answer & explanation

Correct answer

Answer: (c)C.

54

Why this answer is correct

Given:

Departments = 4

Average = 45

New department employees = 90

Transferred = 10 from each of 4 departments

Formula Used:

Total = Average Γ— Number of departments

Average = Total Γ· Number of departments

Calculation:

β‡’ Initial total = 45 Γ— 4 = 180

β‡’ New department total = 90

β‡’ Total after adding = 180 + 90 = 270

β‡’ Employees transferred = 10 Γ— 4 = 40

β‡’ New total remains = 270

β‡’ New departments = 5

β‡’ New average = 270 Γ· 5 = 54

∴ The correct answer is 54.

SSC CHSL 2025 Tier-1 (Held On: 17 Nov, 2025 Shift 2) (Previous Year Paper)Β·Quantitative Aptitude
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Question 180

SSC Combined Higher Secondary Level (CHSL) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A savings scheme offers compound interest annually. A woman deposits β‚Ή10,000. At the end of the first year, the interest rate is 10%. In the second year, due to inflation, the interest rate rises to 12.5%, and in the third year, it drops to 8%. Calculate the total interest she earned at the end of 3 years.

  1. AOption (a)

    β‚Ή3,265

  2. BOption (b)

    β‚Ή3,365

  3. COption (c)

    β‚Ή3,285

  4. DOption (d)

    β‚Ή3,460

View answer & explanation

Correct answer

Answer: (b)B.

β‚Ή3,365

Why this answer is correct

Shortcut Trick

Apply the Successive Multiplier method for varying interest rates.

Multiplier for Year 1 (10%) = 1.1, Year 2 (12.5%) = 1.125, Year 3 (8%) = 1.08

Total Multiplier = 1.1 Γ— 1.125 Γ— 1.08 = 1.3365

Total Interest = Principal Γ— (Total Multiplier βˆ’ 1) = 10,000 Γ— 0.3365 = β‚Ή3,365

∴ The correct answer is β‚Ή3,365.

Alternate Method

Given:

Principal (P) = β‚Ή10,000

Rate (Year 1) = 10%, Rate (Year 2) = 12.5%, Rate (Year 3) = 8%

Formula Used:

Amount (A) = P Γ— (1 + r1/100) Γ— (1 + r2/100) Γ— (1 + r3/100)

Compound Interest (CI) = Amount βˆ’ Principal

Calculations:

β‡’ Amount at end of Year 1 = 10,000 Γ— (110/100) = β‚Ή11,000

β‡’ Amount at end of Year 2 = 11,000 Γ— (112.5/100) = 11,000 Γ— (9/8) = β‚Ή12,375

β‡’ Amount at end of Year 3 = 12,375 Γ— (108/100) = 12,375 Γ— (27/25) = β‚Ή13,365

β‡’ Total Interest = Final Amount βˆ’ Initial Principal

β‡’ Total Interest = 13,365 βˆ’ 10,000 = β‚Ή3,365

∴ The correct answer is β‚Ή3,365.

Additional Information

Compound Interest with Different Rates

When interest rates vary annually, the amount is calculated as A = P(1 + r1/100)(1 + r2/100)...(1 + rn/100).

Fractional Equivalents for Speed

Using fractions like 12.5% = 1/8 (multiplier 9/8) and 8% = 2/25 (multiplier 27/25) simplifies the calculation of compound growth significantly.

Effective Percentage Rate

The total effective rate for successive percentages a%, b%, and c% is [ (1 + a/100)(1 + b/100)(1 + c/100) βˆ’ 1 ] Γ— 100.

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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.