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IBPS Probationary Officer (PO) Β· Topic Collection

Quantitative Aptitude PYQs

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

1305 Total Questions38 Official PapersYears: 2021–2026Search in Quantitative Aptitude
Showing questions 241–260 of 1305Page 13 of 66
IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 241

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

In the question, two quantities I, and II are given. You have to solve both the quantities and establish the correct relation between Quantity - I, and Quantity - II and choose the correct option accordingly.

Quantity I:

A can do a work in 24 days and B can do it in 36 days. They work together for 6 days, then B leaves. Find total time to complete work

Quantity II:

A can do a work in 30 days and B can do it in 20 days. They work together for 5 days, then A leaves. Find total time to complete work

  1. AOption (a)

    Quantity I ≀ Quantity II

  2. BOption (b)

    Quantity I > Quantity II

  3. COption (c)

    Quantity I < Quantity II

  4. DOption (d)

    Quantity I = Quantity II

  5. EOption (e)

    Quantity I β‰₯ Quantity II

View answer & explanation

Correct answer

Answer: (b)B.

Quantity I > Quantity II

Why this answer is correct

Shortcut Trick
Quantity I: Total work (LCM of 24, 36) = 72. Efficiencies: A = 3, B = 2. Together 6 days = 6 Γ— 5 = 30. Remaining = 42. Time by A = 42 Γ· 3 = 14. Total = 6 + 14 = 20 days.
Quantity II: Total work (LCM of 30, 20) = 60. Efficiencies: A = 2, B = 3. Together 5 days = 5 Γ— 5 = 25. Remaining = 35. Time by B = 35 Γ· 3 = 11.67. Total = 5 + 11.67 = 16.67 days.
Comparing both: 20 days > 16.67 days.
∴ The correct answer is Quantity I > Quantity II.
Alternate Method
Quantity I:

Person
Days
Total Work
Efficiency

A
24
72
3

B
36
2

β‡’ Combined efficiency of A and B = 3 + 2 = 5 units/day.
β‡’ Work done in 6 days = 6 Γ— 5 = 30 units.
β‡’ Remaining work = 72 βˆ’ 30 = 42 units.
β‡’ Time taken by A to finish remaining work = 42 Γ· 3 = 14 days.
β‡’ Total time = 6 + 14 = 20 days.
Quantity II:

Person
Days
Total Work
Efficiency

A
30
60
2

B
20
3

β‡’ Combined efficiency of A and B = 2 + 3 = 5 units/day.
β‡’ Work done in 5 days = 5 Γ— 5 = 25 units.
β‡’ Remaining work = 60 βˆ’ 25 = 35 units.
β‡’ Time taken by B to finish remaining work = 35 Γ· 3 = 11.67 days.
β‡’ Total time = 5 + 11.67 = 16.67 days.
Comparison: 20 > 16.67 β‡’ Quantity I > Quantity II.
∴ The correct answer is Quantity I > Quantity II.
Additional Information
LCM Method for Work
Total work is assumed as the LCM of time taken by individuals to simplify efficiency calculations as integers.
Work and Efficiency Relation
Total Work = Efficiency Γ— Time. If total work is constant, efficiency is inversely proportional to time taken.
Leaving and Joining
For such problems, calculate work done during the period of togetherness and subtract from total work to find the remainder.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 242

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

In the question, two quantities I, and II are given. You have to solve both the quantities and establish the correct relation between Quantity - I, and Quantity - II and choose the correct option accordingly.

Quantity I:

Simple Interest on Rs. 3600 for 3 years at 15% per annum?

Quantity II:

Compound Interest on Rs. 3000 for 2 years at 20% per annum?

  1. AOption (a)

    Quantity I > Quantity IIΒ 

  2. BOption (b)

    Β Quantity I < Quantity II

  3. COption (c)

    e) Quantity I β‰₯ Quantity II

  4. DOption (d)

    Quantity I = Quantity IIΒ 

  5. EOption (e)

    Quantity I ≀ Quantity II

View answer & explanation

Correct answer

Answer: (a)A.

Quantity I > Quantity IIΒ 

Why this answer is correct

Shortcut TrickQuantity I: 15% Γ— 3 years = 45% of β‚Ή3600 β†’ 0.45 Γ— 3600 = β‚Ή1620.Quantity II: 20% + 20% + (20 Γ— 20)/100 = 44% effective rate.44% of β‚Ή3000 β†’ 0.44 Γ— 3000 = β‚Ή1320.Comparison: β‚Ή1620 > β‚Ή1320 β‡’ Quantity I > Quantity II.∴ The correct answer is Quantity I > Quantity II.Alternate MethodGiven:For Quantity I: Principal (P1) = β‚Ή3600, Rate (R1) = 15%, Time (T1) = 3 years.For Quantity II: Principal (P2) = β‚Ή3000, Rate (R2) = 20%, Time (T2) = 2 years.Formula Used:Simple Interest (SI) = (P Γ— R Γ— T) Γ· 100Compound Interest (CI) = P[(1 + R/100)T βˆ’ 1]Calculations:Quantity I:β‡’ SI = (3600 Γ— 15 Γ— 3) Γ· 100β‡’ SI = 36 Γ— 45β‡’ SI = β‚Ή1620Quantity II:β‡’ Amount (A) = 3000 Γ— (1 + 20/100)2β‡’ A = 3000 Γ— (1.2)2β‡’ A = 3000 Γ— 1.44 = β‚Ή4320β‡’ CI = Amount βˆ’ Principal = 4320 βˆ’ 3000β‡’ CI = β‚Ή1320

Conclusion:

β‡’ 1620 > 1320∴ The correct answer is Quantity I > Quantity II.Additional InformationEffective Rate of CIFor 2 years, the effective interest rate can be calculated using the formula x + y + (xy/100). For 20%, it is 20 + 20 + 400/100 = 44%.SI vs CI DifferenceThe difference between CI and SI for 2 years is P(R/100)2. Compound interest is always greater than or equal to simple interest for the same rate and principal.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 243

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

In the question, two quantities I, and II are given. You have to solve both the quantities and establish the correct relation between Quantity - I, and Quantity - II and choose the correct option accordingly.

Quantity I:

After giving a discount of 20% on marked price, a shopkeeper still makes a profit of 25%. If cost price is Rs. 800, find the marked price?

Quantity II:

After giving a discount of 25% on marked price, a shopkeeper makes a profit of 20%. If cost price is Rs. 900, find the marked price?

  1. AOption (a)

    Quantity I β‰₯ Quantity II

  2. BOption (b)

    Quantity I = Quantity II

  3. COption (c)

    Quantity I > Quantity II

  4. DOption (d)

    Quantity I < Quantity IIΒ 

  5. EOption (e)

    Quantity I ≀ Quantity II

View answer & explanation

Correct answer

Answer: (d)D.

Quantity I < Quantity IIΒ 

Why this answer is correct

Shortcut Trick
Use the ratio formula: MP / CP = (100 + Profit%) / (100 βˆ’ Discount%)
Quantity I: MP = 800 Γ— (125 / 80) = β‚Ή1250
Quantity II: MP = 900 Γ— (120 / 75) = β‚Ή1440
Comparing the values: 1250 < 1440 β‡’ Quantity I < Quantity II
∴ The correct answer is Quantity I < Quantity II.
Alternate Method

Quantity I:
β‡’ Cost Price (CP) = β‚Ή800, Profit = 25%, Discount = 20%
β‡’ Selling Price (SP) = CP Γ— (100 + P)/100 = 800 Γ— 125/100 = β‚Ή1000
β‡’ Marked Price (MP) = SP Γ— 100 / (100 βˆ’ D) = 1000 Γ— 100/80 = β‚Ή1250
Quantity II:
β‡’ Cost Price (CP) = β‚Ή900, Profit = 20%, Discount = 25%
β‡’ Selling Price (SP) = CP Γ— (100 + P)/100 = 900 Γ— 120/100 = β‚Ή1080
β‡’ Marked Price (MP) = SP Γ— 100 / (100 βˆ’ D) = 1080 Γ— 100/75 = β‚Ή1440
β‡’ Comparison: β‚Ή1250 < β‚Ή1440
∴ The correct answer is Quantity I < Quantity II.
Additional Information
Relationship between CP and MP
The direct relation is given by MP/CP = (100 + P%)/(100 βˆ’ D%). This eliminates the need to calculate SP separately.
Markup Percentage
Markup % = [(MP βˆ’ CP) / CP] Γ— 100. It represents how much the price is raised above the cost price before offering discounts.
Effective Discount
For successive discounts d1 and d2, the single equivalent discount is (d1 + d2 βˆ’ d1Γ—d2/100)%.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 244

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A car cover 5s km distance in 3 hours and 4s km distance in 2 hours. The average speed of the car is 84 km/hr. Find the time taken by the car if the car covers the entire distance at the speed at which the car covers 5s km?

  1. AOption (a)

    2

  2. BOption (b)

    4

  3. COption (c)

    5

  4. DOption (d)

    6

  5. EOption (e)

    3

View answer & explanation

Correct answer

Answer: (c)C.

5

Why this answer is correct

Shortcut Trick

Total distance = 5s + 4s = 9s km

Total time = 3 + 2 = 5 hours

Average speed = Total distance Γ· Total time β‡’ 84 = 9s Γ· 5 β‡’ s = 46.67

Total distance = 9 Γ— 46.67 = 420 km

Speed for the first part = 5s Γ· 3 = 5 Γ— 46.67 Γ· 3 = 77.78 km/hr

Time taken at this speed for entire distance = 420 Γ· 77.78 = 5.4 hours β‰ˆ 5 hours

∴ The correct answer is 5 hours.

Alternate Method

Given:

First segment distance = 5s km, Time = 3 hours

Second segment distance = 4s km, Time = 2 hours

Average speed of the car = 84 km/hr

Formula Used:

Average Speed = Total Distance Γ· Total Time

Speed = Distance Γ· Time

Time = Distance Γ· Speed

Calculations:

β‡’ Total distance = 5s + 4s = 9s km

β‡’ Total time = 3 + 2 = 5 hours

β‡’ Average speed = 9s Γ· 5

β‡’ 84 = 9s Γ· 5

β‡’ 9s = 84 Γ— 5

β‡’ 9s = 420 km (This is the total distance)

β‡’ s = 420 Γ· 9 = 140/3 km

β‡’ First segment distance = 5s = 5 Γ— (140/3) = 700/3 km

β‡’ Speed for the first segment = (700/3) Γ· 3 = 700/9 km/hr

β‡’ Time taken to cover total distance at this speed = Total Distance Γ· First Speed

β‡’ Time = 420 Γ· (700/9) = 420 Γ— 9 Γ· 700

β‡’ Time = (42 Γ— 9) Γ· 70 = 378 Γ· 70 = 5.4 hours β‰ˆ 5 hours

∴ The correct answer is 5 hours.

Additional Information

Average Speed

Average speed is always calculated as the total distance traveled divided by the total time taken, regardless of the individual speeds during the journey.

Equal Distance Formula

If equal distances are covered at speeds u and v, then the average speed is given by the harmonic mean: 2uv Γ· (u + v).

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 245

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A number is first decreased by 40% and then increased by 20%. The number obtained is 140 less than the original number. Find the 60% of the half of the number?

  1. AOption (a)

    120

  2. BOption (b)

    160

  3. COption (c)

    220

  4. DOption (d)

    240

  5. EOption (e)

    150

View answer & explanation

Correct answer

Answer: (e)E.

150

Why this answer is correct

Shortcut Trick

Net percentage change = βˆ’40 + 20 + [(βˆ’40 Γ— 20) Γ· 100] = βˆ’20 βˆ’ 8 = βˆ’28%

Decrease of 28% of original number = 140

β‡’ 100% (Original number) = (140 Γ· 28) Γ— 100 = 500

β‡’ Required value = 500 Γ— (1/2) Γ— 60% = 250 Γ— 0.6 = 150

∴ The correct answer is 150.

Alternate Method

Given: Initial decrease = 40%, Subsequent increase = 20%, Difference = 140.

Formula: Final Value = Initial Value Γ— (1 βˆ’ d1/100) Γ— (1 + i1/100)

β‡’ Let the original number be 100x.

β‡’ After 40% decrease: 100x βˆ’ 40x = 60x

β‡’ After 20% increase: 60x + (20/100 Γ— 60x) = 60x + 12x = 72x

β‡’ Difference = 100x βˆ’ 72x = 28x

β‡’ 28x = 140

β‡’ x = 140 Γ· 28 = 5

β‡’ Original number = 100 Γ— 5 = 500

β‡’ Half of the number = 500 Γ· 2 = 250

β‡’ 60% of 250 = (60 Γ· 100) Γ— 250 = 150

∴ The correct answer is 150.

Additional Information

Successive Percentage Change

For two successive changes of a% and b%, the effective change is [a + b + (ab/100)]%. Use negative sign for decrease.

Base Value Concept

Always ensure the second percentage change is applied to the resultant value of the first change, not the original base.

Percentage to Fraction Conversion

40% = 2/5, 20% = 1/5, 60% = 3/5. Using fractions can often simplify multi-step calculations.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 246

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

The bar graph shows the total number items [ laptop, desktop and tablet] sold in four different shops [A, B, C, and D] and also show the number of laptop sold in each shop.

Note – Ratio of number of laptop and number of tablets sold in each shop is 3:2.

Total laptops sold in Shop A and C together is what percent of total items sold in Shop B?

  1. AOption (a)

    38.85%

  2. BOption (b)

    55.55%

  3. COption (c)

    72.28%

  4. DOption (d)

    68.57%

  5. EOption (e)

    61.25%

View answer & explanation

Correct answer

Answer: (d)D.

68.57%

Why this answer is correct

Common Calculation

Given:

Ratio (Laptop : Tablet) = 3 : 2

Formula Used:

If Laptop = 3x β‡’ Tablet = 2x

Shop A

Calculation:

β‡’ Laptop = 54 = 3x β‡’ x = 18

β‡’ Tablet = 2x = 36

β‡’ Desktop = 180 βˆ’ (54 + 36) = 90

∴ Desktop = 90, Tablet = 36

Shop B

Calculation:

β‡’ Laptop = 69 = 3x β‡’ x = 23

β‡’ Tablet = 2x = 46

β‡’ Desktop = 140 βˆ’ (69 + 46) = 25

∴ Desktop = 25, Tablet = 46

Shop C

Calculation:

β‡’ Laptop = 42 = 3x β‡’ x = 14

β‡’ Tablet = 2x = 28

β‡’ Desktop = 120 βˆ’ (42 + 28) = 50

∴ Desktop = 50, Tablet = 28

Shop D

Calculation:

β‡’ Laptop = 84 = 3x β‡’ x = 28

β‡’ Tablet = 2x = 56

β‡’ Desktop = 240 βˆ’ (84 + 56) = 100

∴ Desktop = 100, Tablet = 56

Given:

Laptop A = 54, Laptop C = 42

Total B = 140

Formula Used:

% = (Part / Total) Γ— 100

Calculation:

β‡’ A + C = 54 + 42 = 96

β‡’ % = (96 / 140) Γ— 100

β‡’ = 68.57%

∴ 68.57%

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 247

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

The bar graph shows the total number items [ laptop, desktop and tablet] sold in four different shops [A, B, C, and D] and also show the number of laptop sold in each shop.

Note – Ratio of number of laptop and number of tablets sold in each shop is 3:2.

Number of desktops in Shop A and Shop C together is what percent more or less than total tablets in Shop B and Shop D together?

  1. AOption (a)

    33.33%

  2. BOption (b)

    48.84%

  3. COption (c)

    25.25%

  4. DOption (d)

    44.44%

  5. EOption (e)

    37.25%

View answer & explanation

Correct answer

Answer: (e)E.

37.25%

Why this answer is correct

Common Calculation

Given:

Ratio (Laptop : Tablet) = 3 : 2

Formula Used:

If Laptop = 3x β‡’ Tablet = 2x

Shop A

Calculation:

β‡’ Laptop = 54 = 3x β‡’ x = 18

β‡’ Tablet = 2x = 36

β‡’ Desktop = 180 βˆ’ (54 + 36) = 90

∴ Desktop = 90, Tablet = 36

Shop B

Calculation:

β‡’ Laptop = 69 = 3x β‡’ x = 23

β‡’ Tablet = 2x = 46

β‡’ Desktop = 140 βˆ’ (69 + 46) = 25

∴ Desktop = 25, Tablet = 46

Shop C

Calculation:

β‡’ Laptop = 42 = 3x β‡’ x = 14

β‡’ Tablet = 2x = 28

β‡’ Desktop = 120 βˆ’ (42 + 28) = 50

∴ Desktop = 50, Tablet = 28

Shop D

Calculation:

β‡’ Laptop = 84 = 3x β‡’ x = 28

β‡’ Tablet = 2x = 56

β‡’ Desktop = 240 βˆ’ (84 + 56) = 100

∴ Desktop = 100, Tablet = 56

Given:

Desktop A = 90, Desktop C = 50

Tablet B = 46, Tablet D = 56

Formula Used:

% Difference = (Diff / Base) Γ— 100

Calculation:

β‡’ A + C = 90 + 50 = 140

β‡’ B + D = 46 + 56 = 102

β‡’ Difference = 140 βˆ’ 102 = 38

β‡’ % = (38 / 102) Γ— 100

β‡’ = 37.25%

∴ 37.25% more

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 248

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Find the missing number of series?

455, 578, 801, 1124, ?

  1. AOption (a)

    1640

  2. BOption (b)

    1950

  3. COption (c)

    1467

  4. DOption (d)

    1547

  5. EOption (e)

    1447

View answer & explanation

Correct answer

Answer: (d)D.

1547

Why this answer is correct

Given series:

455, 578, 801, 1124, ?

Calculation:

β‡’ 455 β†’ 578 = +123

β‡’ 578 β†’ 801 = +223

β‡’ 801 β†’ 1124 = +323

β‡’ Pattern: +123, +223, +323, +423

β‡’ Missing = 1124 + 423 = 1547

∴ Missing number = 1547

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 249

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Find the missing number of series?

10.24,17.64,27.04, 38.44, ?

  1. AOption (a)

    48.84

  2. BOption (b)

    59.64

  3. COption (c)

    58.66

  4. DOption (d)

    54.84

  5. EOption (e)

    51.84

View answer & explanation

Correct answer

Answer: (e)E.

51.84

Why this answer is correct

Given series:

10.24, 17.64, 27.04, 38.44, ?

Calculation:

β‡’ 10.24 = (3.2)Β²

β‡’ 17.64 = (4.2)Β²

β‡’ 27.04 = (5.2)Β²

β‡’ 38.44 = (6.2)Β²

β‡’ Pattern: (3.2)Β², (4.2)Β², (5.2)Β², (6.2)Β²

β‡’ Next = (7.2)Β²

β‡’ 7.2 Γ— 7.2 = 51.84

∴ Missing number = 51.84

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 250

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Find the missing number of series?

456, 467, ?, 555, 632

  1. AOption (a)

    525

  2. BOption (b)

    500

  3. COption (c)

    535

  4. DOption (d)

    545

  5. EOption (e)

    510

View answer & explanation

Correct answer

Answer: (b)B.

500

Why this answer is correct

Given series:

456, 467, ?, 555, 632

Calculation:

β‡’ 456 β†’ 467 = +11

β‡’ Pattern: +11, +33, +55, +77

β‡’ Missing = 467 + 33 = 500

β‡’ 500 + 55 = 555

∴ Missing number = 500

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 251

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Find the missing number of series?

46, 69, ?, 184,276

  1. AOption (a)

    175

  2. BOption (b)

    95

  3. COption (c)

    135

  4. DOption (d)

    115

  5. EOption (e)

    105

View answer & explanation

Correct answer

Answer: (d)D.

115

Why this answer is correct

Given series:

46, 69, ?, 184, 276

Calculation:

β‡’ 46 β†’ 69 = +23

β‡’ Pattern: +23, +46, +69, +92

β‡’ 69 + 46 = 115

β‡’ 115 + 69 = 184

β‡’ 184 + 92 = 276

∴ Missing number = 115

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 252

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Find the missing number of series?

152, 593, ? ,1747,2476

  1. AOption (a)

    1425

  2. BOption (b)

    1224

  3. COption (c)

    1136

  4. DOption (d)

    1226

  5. EOption (e)

    1122

View answer & explanation

Correct answer

Answer: (e)E.

1122

Why this answer is correct

Given series:

152, 593, ?, 1747, 2476

Calculation:

β‡’ 152 β†’ 593 = +441 = 21Β²

β‡’ Pattern: +21Β², +23Β², +25Β², +27Β²

β‡’ Missing = 593 + 23Β² = 593 + 529 = 1122

β‡’ 1122 + 25Β² = 1747

∴ Missing number = 1122

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 253

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A man can do work in 20 days. The efficiency of a man and woman is 5:4. Find approximately how many days 5 men and 4 women complete the double of the work?

  1. AOption (a)

    6

  2. BOption (b)

    3

  3. COption (c)

    10

  4. DOption (d)

    15

  5. EOption (e)

    5

View answer & explanation

Correct answer

Answer: (e)E.

5

Why this answer is correct

Shortcut Trick

Efficiency ratio M : W = 5 : 4

Total Work = 1 Man Γ— 20 days = 5 Γ— 20 = 100 units

Double Work = 100 Γ— 2 = 200 units

Combined Efficiency (5M + 4W) = (5 Γ— 5) + (4 Γ— 4) = 25 + 16 = 41 units/day

Days required = 200 Γ· 41 β‰ˆ 4.87 β‰ˆ 5 days

∴ The correct answer is 5 days.

Alternate Method

EntityEfficiency (e)Total Work (W)New Group
1 Man5100 units5 Men = 25
1 Woman44 Women = 16

Given: Efficiency of 1 Man = 5 units/day, Efficiency of 1 Woman = 4 units/day.

Formula: Total Work = Efficiency Γ— Time; Required Time = Work Γ· Total Efficiency

β‡’ Total Work (Original) = 1 Γ— 5 Γ— 20 = 100 units

β‡’ Target Work (Double) = 100 Γ— 2 = 200 units

β‡’ Total Efficiency of 5 Men and 4 Women = (5 Γ— 5) + (4 Γ— 4)

β‡’ Total Efficiency = 25 + 16 = 41 units/day

β‡’ Required Days = 200 Γ· 41 = 4.878...

β‡’ Approximate Days β‰ˆ 5 days

∴ The correct answer is 5 days.

Additional Information

Work-Efficiency Relationship

Total work remains constant when comparing different groups using the formula M1D1E1/W1 = M2D2E2/W2.

Combined Efficiency

When multiple individuals work together, their individual efficiencies (work done per unit time) are added to find the total output rate.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
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Question 254

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

Speed of the boat upstream is 24 km/hr and the boat cover 72 km in 2 hours downstream. Find the time taken by boat to go 120 km and return?

  1. AOption (a)

    11.25

  2. BOption (b)

    9.33

  3. COption (c)

    15.33

  4. DOption (d)

    8.33

  5. EOption (e)

    5.65

View answer & explanation

Correct answer

Answer: (d)D.

8.33

Why this answer is correct

Shortcut Trick

Upstream speed (u) = 24 km/hr

Downstream speed (v) = 72 km Γ· 2 hr = 36 km/hr

Total time for 120 km round trip = (Distance Γ· u) + (Distance Γ· v)

Total time = (120 Γ· 24) + (120 Γ· 36) = 5 + 3.33 = 8.33 hours

∴ The correct answer is 8.33 hours.

Alternate Method

Given:

Upstream Speed = 24 km/hr

Downstream Distance = 72 km, Downstream Time = 2 hours

Round Trip Distance = 120 km one way

Formula Used:

Speed = Distance Γ· Time

Total Time = (Distance Γ· Upstream Speed) + (Distance Γ· Downstream Speed)

Calculations:

β‡’ Downstream Speed = 72 Γ· 2 = 36 km/hr

β‡’ Time taken to go 120 km (Downstream) = 120 Γ· 36 = 10 Γ· 3 = 3.333... hours

β‡’ Time taken to return 120 km (Upstream) = 120 Γ· 24 = 5 hours

β‡’ Total Time = 5 + 3.333... = 8.333... hours

∴ The correct answer is 8.33 hours.

Additional Information

Speed of Boat and Stream

If speed of boat in still water is x and speed of stream is y, then Downstream Speed (v) = x + y and Upstream Speed (u) = x βˆ’ y.

Calculating x and y from u and v

Speed of boat in still water (x) = (v + u) Γ· 2 and Speed of stream (y) = (v βˆ’ u) Γ· 2.

Average Speed for Round Trip

Average Speed = (2 Γ— u Γ— v) Γ· (u + v). This is the harmonic mean of the upstream and downstream speeds.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
View full paper

Question 255

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A number x is 40% more than y. y is 60% more than z. v is 20% less than y then finds the x is approximately is how much percent of v?

  1. AOption (a)

    325%

  2. BOption (b)

    185%

  3. COption (c)

    195%

  4. DOption (d)

    175%

  5. EOption (e)

    225%

View answer & explanation

Correct answer

Answer: (d)D.

175%

Why this answer is correct

Shortcut Trick

Assume y = 100 units.

x = 140% of y = 140.

v = 80% of y = 80.

Required % = (x Γ· v) Γ— 100 = (140 Γ· 80) Γ— 100 = 1.75 Γ— 100 = 175%.

∴ The correct answer is 175%.

Alternate Method

Given:

x = y + 40% of y = 1.4y

v = y βˆ’ 20% of y = 0.8y

Calculations:

β‡’ Relationship between x and v: x / v = 1.4y / 0.8y

β‡’ x / v = 14 / 8 = 7 / 4

β‡’ Required Percentage = (x / v) Γ— 100

β‡’ Percentage = (7 / 4) Γ— 100

β‡’ Percentage = 7 Γ— 25 = 175%

∴ The correct answer is 175%.

Additional Information

Percentage Change

To increase a value by a%, multiply by (1 + a/100). To decrease by a%, multiply by (1 βˆ’ a/100).

Base Value Selection

When comparing multiple variables to one common variable, assume the common variable (here 'y') as 100 for easier calculation.

Successive Percentage

If a value changes by a% then b%, the net change is [a + b + (aΓ—b)/100]%.

IBPS PO Prelims Memory Based Paper (Held On: 23 August 2025 Shift 4) (Previous Year Paper)Β·Quantitative Aptitude
View full paper

Question 256

IBPS Probationary Officer (PO) 2025β€’ Quantitative Aptitudeβ€’ Quantitative Aptitude

A square inscribed in a circle. The circumference of the circle is 528. The side of the square is equal to the side of a cube. Find the surface are of the cube?Β 

  1. AOption (a)

    80742

  2. BOption (b)

    81654

  3. COption (c)

    89654

  4. DOption (d)

    84672

  5. EOption (e)

    84562

View answer & explanation

Correct answer

Answer: (d)D.

84672

Why this answer is correct

Shortcut Trick

Circumference = 528 β‡’ 2 Γ— (22/7) Γ— Radius = 528 β‡’ Radius = 84.

Diagonal of square = Diameter of circle = 2 Γ— 84 = 168.

Side of square (a) = 168 / √2.

Surface area of cube = 6 Γ— a2 = 6 Γ— (168 / √2)2 = 6 Γ— 168 Γ— 168 / 2 = 3 Γ— 168 Γ— 168.

Surface area = 84672.

∴ The correct answer is 84672.

Alternate Method

Given: Circumference (C) = 528; Side of square = Side of cube (s).

Formula Used: Circumference = 2Ο€r; Diagonal of square = s√2; Total Surface Area (TSA) of cube = 6s2.

β‡’ 2 Γ— (22/7) Γ— r = 528

β‡’ r = (528 Γ— 7) / 44 = 12 Γ— 7 = 84

β‡’ Diameter (d) = 2 Γ— 84 = 168

β‡’ Diagonal of inscribed square = Diameter of circle = 168

β‡’ s√2 = 168 β‡’ s = 168 / √2

β‡’ TSA of cube = 6 Γ— s2 = 6 Γ— (168 / √2)2

β‡’ TSA = 6 Γ— (28224 / 2) = 3 Γ— 28224 = 84672

∴ The correct answer is 84672.

Additional Information

Inscribed Square Property

When a square is inscribed in a circle, the diagonal of the square is always equal to the diameter of the circle (d = s√2).

Cube Surface Area

The Total Surface Area of a cube is given by 6 Γ— (side)2, while the Lateral Surface Area is 4 Γ— (side)2.

Circle Relations

Area of circle = Ο€r2 and Circumference = 2Ο€r. For quick calculations, if radius is 7, circumference is 44 and area is 154.

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How should I prepare for Quantitative Aptitude in IBPS Probationary Officer (PO)?

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.