IBPS RRB Office Assistant (Clerk) · Topic Collection

Quantitative Aptitude PYQs

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

1935 Total Questions49 Official PapersYears: 2018–2025Search in Quantitative Aptitude
Showing questions 821–840 of 1935Page 42 of 97
IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 821

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in the place of question mark (?) in the following question?

(124.95 × 5.98) – (110.02 × 7.96) + 669.95 = ? × 8.99

  1. AOption (a)

    40

  2. BOption (b)

    50

  3. COption (c)

    80

  4. DOption (d)

    60

  5. EOption (e)

    30

View answer & explanation

Correct answer

Answer: (d)D.

60

Why this answer is correct

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(124.95 × 5.98) – (110.02 × 7.96) + 669.95 = ? × 8.99

⇒ (125 × 6) – (110 × 8) + 670 = ? × 9

⇒ 750 – 880 + 670 = ? × 9

⇒ 1420 – 880 = ? × 9

⇒ 540 = ? × 9

∴ ? = 60

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 822

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in the place of question mark (?) in the following question?

349.99 – [15.952 – (87.88 + 95.94) ÷ 7.99] = ?

  1. AOption (a)

    123

  2. BOption (b)

    117

  3. COption (c)

    137

  4. DOption (d)

    133

  5. EOption (e)

    113

View answer & explanation

Correct answer

Answer: (b)B.

117

Why this answer is correct

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

349.99 – [15.952 – (87.88 + 95.94) ÷ 7.99] = ?

⇒ 350 – [162 – (88 + 96) ÷ 8] = ?

⇒ 350 – [162 – 184 ÷ 8] = ?

⇒ 350 – [162 – 23] = ?

⇒ 350 – [256 – 23] = ?

⇒ 350 – 233 = ?

∴ ? = 117

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 823

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

An amount of Rs.20000 is invested at the rate of 11% per annum for 2 years at Compound Interest. The interest received is again invested at the same rate for the same time. Find the simple interest received from the second investment.

  1. AOption (a)

    1011.75

  2. BOption (b)

    1021.24

  3. COption (c)

    1031.56

  4. DOption (d)

    1040.45

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

1021.24

Why this answer is correct

Shortcut Trick

Effective Compound Interest rate for 2 years at 11% = 11 + 11 + (11 × 11)/100 = 23.21%.

First Interest amount = 23.21% of ₹20000 = ₹4642.

Simple Interest on second investment = (4642 × 11 × 2) / 100 = 4642 × 0.22 = ₹1021.24.

∴ The correct answer is ₹1021.24.

Alternate Method

Given: Principal (P1) = ₹20000, Rate (R) = 11%, Time (T) = 2 years.

Formula Used: CI = P[(1 + R/100)T − 1]; SI = (P × R × T) / 100.

⇒ 1st Interest (CI) = 20000 × [(1 + 11/100)2 − 1]

⇒ CI = 20000 × [1.2321 − 1] = 20000 × 0.2321 = ₹4642.

⇒ 2nd Principal (P2) = 1st Interest = ₹4642.

⇒ SI received = (4642 × 11 × 2) / 100

⇒ SI = (4642 × 22) / 100 = 102124 / 100 = ₹1021.24.

∴ The correct answer is ₹1021.24.

Additional Information

Effective Rate of Compound Interest

For 2 years, the net percentage change formula (x + y + xy/100) provides the effective CI rate relative to the principal.

Simple Interest vs Compound Interest

While SI is calculated only on the principal, CI is calculated on the principal plus the accumulated interest of previous periods.

CI − SI Difference for 2 years

The difference between CI and SI for 2 years is given by the formula P(R/100)2.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 824

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

In an examination, Ram scored 25% marks and failed by 25 marks. Shyam scored 45% marks and obtained 50 marks more than passing marks. Find the maximum marks.

  1. AOption (a)

    500 

  2. BOption (b)

    375

  3. COption (c)

    250

  4. DOption (d)

    275

  5. EOption (e)

    575

View answer & explanation

Correct answer

Answer: (b)B.

375

Why this answer is correct

Shortcut Trick

Difference in percentage marks = 45% − 25% = 20%

Difference in actual marks = 50 − (−25) = 75 marks

20% of Maximum Marks = 75 marks

100% of Maximum Marks = 75 × 5 = 375 marks

∴ The correct answer is 375.

Alternate Method

Given: Ram score = 25%, Shyam score = 45%, Marks gap = 25 (fail) and 50 (extra).

Let Maximum Marks be M.

Passing Marks = 25% of M + 25

Passing Marks = 45% of M − 50

⇒ 25% of M + 25 = 45% of M − 50

⇒ 45% of M − 25% of M = 50 + 25

⇒ 20% of M = 75

⇒ (20/100) × M = 75

⇒ M = (75 × 100) ÷ 20

⇒ M = 375

∴ The correct answer is 375.

Additional Information

Passing Percentage

To find the passing percentage, substitute M back into the passing marks equation: (25% × 375 + 25) ÷ 375 × 100 = 31.66%.

Percentage Change Base

In examination problems, the Maximum Marks always acts as the base (100%) for calculating scored or passing percentages.

Successive Percentage

For problems involving repeated changes, use the net effect formula: a + b + (ab/100).

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 825

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions:  The given table shows the total number of students (Males + Females) attended the interview, Percentage of girls attended the interview , number of males selected and percentage of selected students in five different schools.

 

SchoolsTotal number of students (Male + Female) attend interviewPercentage of females attend interviewTotal students (Male + Female) selected in interviewNumber of Males selected In interview
A70040%500350
B65060%400250
C75030%450300
D80045%550325
E90055%600300

1/3rd and 1/5 th of the females selected from school A and school  B respectively are from city P and remaining are from city Q and 1/7th and 1/10 th of males selected from school A and school B are from city Q. Find the number of students (male  + female) from school A and school B are from city Q.

  1. AOption (a)

    255

  2. BOption (b)

    265

  3. COption (c)

    275

  4. DOption (d)

    295

  5. EOption (e)

    285

View answer & explanation

Correct answer

Answer: (d)D.

295

Why this answer is correct

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

SchoolsNumber of males attend interviewNumber of females attend InterviewNumber of males selectedNumber of males unselectedNumber of females selectedNumber of females unselected
A42028035070150130
B26039025010150240
C52522530022515075
D440360325115225135
E405495300105300195

 

Solution:

Number of females selected from school A from city P= 150 ×  1/3 = 50

Number of females selected from school A from city Q = (150 - 50) = 100

Number of females selected from school B from city P = 150 ×  1/5 = 30

Number of females selected from school B from city Q = 150 - 30 = 120 

Number of males selected from school A from city Q = 350 ×  1/7 = 50

Number of males selected from school B from city Q = 250 ×  1/10 = 25

Required total = (100 + 120 + 50 + 25) = 295

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 826

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions:  The given table shows the total number of students (Males + Females) attended the interview, Percentage of girls attended the interview , number of males selected and percentage of selected students in five different schools.

 

SchoolsTotal number of students (Male + Female) attend interviewPercentage of females attend interviewTotal students (Male + Female) selected in interviewNumber of Males selected In interview
A70040%500350
B65060%400250
C75030%450300
D80045%550325
E90055%600300

Average number of males selected is what percentage of the total females selected from all the schools.

  1. AOption (a)

    36.78%

  2. BOption (b)

    34.57%

  3. COption (c)

    31.28%

  4. DOption (d)

    30.78%

  5. EOption (e)

    27.89%

View answer & explanation

Correct answer

Answer: (c)C.

31.28%

Why this answer is correct

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

SchoolsNumber of males attend interviewNumber of females attend InterviewNumber of males selectedNumber of males unselectedNumber of females selectedNumber of females unselected
A42028035070150130
B26039025010150240
C52522530022515075
D440360325115225135
E405495300105300195

Solution:

Average number of males selected = (350 + 250 + 300 + 325 + 300)/5 = 305

Total females selected from all the schools = (150 + 150 + 150 + 225 + 300) = 975

Required percentage = 305/975 × 100 = 31.28%

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 827

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions:  The given table shows the total number of students (Males + Females) attended the interview, Percentage of girls attended the interview , number of males selected and percentage of selected students in five different schools.

 

SchoolsTotal number of students (Male + Female) attend interviewPercentage of females attend interviewTotal students (Male + Female) selected in interviewNumber of Males selected In interview
A70040%500350
B65060%400250
C75030%450300
D80045%550325
E90055%600300

If 1/6 th of the males selected from school C for TCS and 1/5 th of the females selected form school C for TCS. Find the number of male and females selected for TCS from school C.

  1. AOption (a)

    130

  2. BOption (b)

    140

  3. COption (c)

    120

  4. DOption (d)

    80

  5. EOption (e)

    110

View answer & explanation

Correct answer

Answer: (d)D.

80

Why this answer is correct

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

SchoolsNumber of males attend interviewNumber of females attend InterviewNumber of males selectedNumber of males unselectedNumber of females selectedNumber of females unselected
A42028035070150130
B26039025010150240
C52522530022515075
D440360325115225135
E405495300105300195

 

Solution:

Number of males selected for TCS from school C = 300 ×  1/6 = 50

Number of females selected for TCS from school C = 150 ×  1/5 = 30

Required total = (50 + 30) = 80

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 828

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions:  The given table shows the total number of students (Males + Females) attended the interview, Percentage of girls attended the interview , number of males selected and percentage of selected students in five different schools.

 

SchoolsTotal number of students (Male + Female) attend interviewPercentage of females attend interviewTotal students (Male + Female) selected in interviewNumber of Males selected In interview
A70040%500350
B65060%400250
C75030%450300
D80045%550325
E90055%600300

Find the difference between average numbers of unselected males and average number of unselected females from all the schools?

  1. AOption (a)

    60

  2. BOption (b)

    45

  3. COption (c)

    30

  4. DOption (d)

    50

  5. EOption (e)

    40

View answer & explanation

Correct answer

Answer: (d)D.

50

Why this answer is correct

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

SchoolsNumber of males attend interviewNumber of females attend InterviewNumber of males selectedNumber of males unselectedNumber of females selectedNumber of females unselected
A42028035070150130
B26039025010150240
C52522530022515075
D440360325115225135
E405495300105300195

 

Solution:

Average number of unselected females = (130 + 240 + 75 + 135 + 195)/5 = 155

Average number of unselected males = (70 + 10 + 225 + 115 + 105)/5 = 105

Required difference = (155 - 105) = 50

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 829

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions:  The given table shows the total number of students (Males + Females) attended the interview, Percentage of girls attended the interview , number of males selected and percentage of selected students in five different schools.

 

SchoolsTotal number of students (Male + Female) attend interviewPercentage of females attend interviewTotal students (Male + Female) selected in interviewNumber of Males selected In interview
A70040%500350
B65060%400250
C75030%450300
D80045%550325
E90055%600300

If 2/7 of the females attending interviews from school A are from section I and 1/5 of the females selected from school A are from section I. Find the number of female students from school A are unselected from section  I in school A.

  1. AOption (a)

    30

  2. BOption (b)

    55

  3. COption (c)

    50

  4. DOption (d)

    40

  5. EOption (e)

    65

View answer & explanation

Correct answer

Answer: (c)C.

50

Why this answer is correct

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

SchoolsNumber of males attend interviewNumber of females attend InterviewNumber of males selectedNumber of males unselectedNumber of females selectedNumber of females unselected
A42028035070150130
B26039025010150240
C52522530022515075
D440360325115225135
E405495300105300195

Solution:

Number of females attending interview from school A from section I = 280 × (2/7) = 80

Number of female selected from school A from section I = 150 × (1/5) = 30

Number of females Unselected from school A from section I = (80 - 30) = 50

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 830

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Average number of books read by Ankur in 3 days (Monday, Tuesday, and Wednesday) is 20. On Monday he read 10 more books than that on Tuesday. On Wednesday he read 10 books less than that on Tuesday. Find the number of books read on each day.

  1. AOption (a)

    30, 20 and 10

  2. BOption (b)

    10, 20 and 40

  3. COption (c)

    20, 20 and 20

  4. DOption (d)

    40, 10 and 10

  5. EOption (e)

    15, 15 and 30

View answer & explanation

Correct answer

Answer: (a)A.

30, 20 and 10

Why this answer is correct

Shortcut Trick

Let the number of books read on Tuesday be x.

According to the given deviations: Monday = x + 10 and Wednesday = x − 10.

The Average of these three days is simply the middle value (Tuesday) because they are in an Arithmetic Progression with a common difference of 10.

∴ Tuesday = Average = 20 books.

Monday = 20 + 10 = 30 and Wednesday = 20 − 10 = 10.

∴ The correct answer is 30, 20 and 10.

Alternate Method

Given: Average = 20 books, Days = 3.

Formula Used: Total Sum = Average × Number of entities.

⇒ Total books = 20 × 3 = 60.

⇒ Let books on Tuesday = x.

⇒ Monday = x + 10, Wednesday = x − 10.

⇒ (x + 10) + x + (x − 10) = 60.

⇒ 3x = 60 ⇒ x = 20.

⇒ Monday = 20 + 10 = 30; Tuesday = 20; Wednesday = 20 − 10 = 10.

∴ The correct answer is 30, 20 and 10.

Additional Information

Average in Arithmetic Progression

If terms are in AP, the Average is always the Middle Term (for odd number of terms) or the mean of the two middle terms.

Average Change Rule

If each observation is increased or decreased by a constant k, the Average also increases or decreases by k.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 831

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

P and Q started a business of garment. P invested Rs 50,000 and Q invested Rs 75,000. After 7 months, P withdraws Rs 20,000 and Q invested additional investment of Rs. 5,000. At the end of the year, business earned a profit of Rs 1,10,000. Find the share of P.(Approx)

  1. AOption (a)

    Rs. 42,000

  2. BOption (b)

    Rs. 38,597

  3. COption (c)

    Rs. 52,000

  4. DOption (d)

    Rs. 48,597

  5. EOption (e)

    Rs. 44,000

View answer & explanation

Correct answer

Answer: (b)B.

Rs. 38,597

Why this answer is correct

Shortcut Trick

Ratio of Profit Share is determined by the ratio of Investment × Time.

Effective investment of P = (50 × 7) + (30 × 5) = 350 + 150 = 500 units.

Effective investment of Q = (75 × 7) + (80 × 5) = 525 + 400 = 925 units.

Profit ratio P : Q = 500 : 925 = 20 : 37.

Share of P = 20 ÷ (20 + 37) × 1,10,000 = 20 ÷ 57 × 1,10,000 ≈ 38,596.49.

∴ The correct answer is Rs. 38,597.

Alternate Method

Given → P initial = Rs. 50,000; Q initial = Rs. 75,000; Time = 12 months; Total Profit = Rs. 1,10,000.
Formulas: Profit Ratio = (I1 × T1) : (I2 × T2) | Profit Share = (Ratio Part ÷ Total Parts) × Total Profit.

⇒ P total investment units = (50,000 × 7) + (30,000 × 5) = 3,50,000 + 1,50,000 = 5,00,000.
⇒ Q total investment units = (75,000 × 7) + (80,000 × 5) = 5,25,000 + 4,00,000 = 9,25,000.
⇒ Ratio P : Q = 5,00,000 : 9,25,000 = 500 : 925 = 20 : 37.
⇒ Total units = 20 + 37 = 57 units.
⇒ Share of P = (20 ÷ 57) × 1,10,000 ≈ Rs. 38,596.49.

∴ The correct answer is Rs. 38,597.

Additional Information

Partnership Basics

In a compound partnership, the profit is shared based on the product of the capital and the duration of investment.

Active vs Sleeping Partner

An active partner manages the business and may receive a salary or extra profit, while a sleeping partner only provides capital.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 832

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Two equations I and II are given below in question. You have to solve these equations and give the answer.

I. x2 - 10x + 25 = 0

II. y2 + 10y + 24 = 0

  1. AOption (a)

    x < y

  2. BOption (b)

    x > y

  3. COption (c)

    x ≤ y

  4. DOption (d)

    x ≥ y

  5. EOption (e)

    x = y or the relation between x and y can't be established.

View answer & explanation

Correct answer

Answer: (b)B.

x > y

Why this answer is correct

Given:

I. x2 – 10x + 25 = 0

II. y2 + 10y + 24 = 0

Calculation:

I. x2 – 10x + 25 = 0

⇒ x2 - 5x - 5x + 25 = 0

⇒ x(x - 5) - 5(x - 5)= 0

⇒ (x - 5)(x - 5) = 0

⇒ x = 5, 5

II. y2 + 10y + 24 = 0

⇒ y2 + 6y + 4y + 24 = 0

⇒ y(y + 6) + 4(y + 6) =0

⇒ (y + 6)(y + 4) = 0

⇒ y = (-6), (-4)

Value of ‘x’RelationValue of ‘y’
5>-6
5>-4

 

When we compared the values of ‘x’ and ‘y’ in the table above, we found that there is only ONE relation between X and Y i.e. >. So, a relation between x and y is “x > y”.

∴ After comparison, all the values of x and y the relation is “x > y”.

Shortcut Trick

I. x2 - 10x + 25 = 0

There are three terms = a, b, and c

Where 'a' = value of x2, 'b' = Value of x, 'c' = 25

If value of 'c' is positive value and value of 'b' is negative value, then both the values of x is POSITIVE value.

II. y2 + 10y + 24 = 0

There are also three terms = a, b, and c

Where 'a' = value of y2, 'b' = Value of y, 'c' = 24

If the value of 'c' is a positive value and the value of 'b' is also a positive value, then both value of y is NEGATIVE value.

∴ After comparison, all the values of x and y, the relation between x and y is "x > y".

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 833

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. 7x2 - 4x - 51 = 0

II. y2 - 24y + 143 = 0

  1. AOption (a)

    x > y

  2. BOption (b)

    x < y

  3. COption (c)

    x ≥ y

  4. DOption (d)

    x ≤ y

  5. EOption (e)

    x = y or relation between x and y can not be established.

View answer & explanation

Correct answer

Answer: (b)B.

x < y

Why this answer is correct

Given:

I. 7x2 - 4x - 51 = 0

II. y2 - 24y + 143 = 0

Calculation:

From I,

7x2 - 4x - 51 = 0

⇒ 7x2 - 21x + 17x - 51 = 0

⇒ 7x(x - 3) + 17(x - 3) = 0

⇒ (x - 3)(7x + 17) = 0

Taking,

⇒ (x - 3) = 0 or (7x + 17) = 0

⇒ x = 3 or (-17/7)

From II,

y2 - 24y + 143 = 0

⇒ y2 - 11y - 13y + 143 = 0

⇒ y(y - 11) - 13(y - 11) = 0

⇒ (y - 11)(y - 13) = 0

Taking,

⇒ (y - 11) = 0 or (y - 13) = 0

⇒ y = 11 or 13

Comparison between x and y (via Tabulation):

Value of xValue of yRelation
311x < y
313x < y
-17/711x < y
-17/713x < y

 

∴ x < y.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 834

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. 2x2 - 17x + 36 = 0

II. y2 + 10y + 24 = 0

  1. AOption (a)

    x > y

  2. BOption (b)

    x < y

  3. COption (c)

    x ≥ y

  4. DOption (d)

    x ≤ y

  5. EOption (e)

    No relation in x and y or x = y

View answer & explanation

Correct answer

Answer: (a)A.

x > y

Why this answer is correct

Given:

I. 2x2 - 17x + 36 = 0

II. y2 + 10y + 24 = 0

Calculation:

From I

2x2 - 17x + 36 = 0

⇒ 2x2 - 8x - 9x + 36 = 0

⇒ 2x(x - 4) - 9(x - 4) = 0

⇒ (x - 4)(2x - 9) = 0

⇒ x = 4, 9/2

From II

y2 + 10y + 24 = 0

⇒ y2 + 6y + 4y + 24 = 0

⇒ y(y + 6) + 4(y + 6) = 0

⇒ (y + 6)(y + 4) = 0

⇒ y = -6, -4

Comparison between x and y (via Tabulation) 

Value of xValue of yRelation between x & y
4-6x > y
4-4x > y
9/2-6x > y
9/2-4x > y

 

∴ x > y

Important Points

Sign Method
EquationSignRemark
ax2 + bx + c = 0-Ve, -VeSame signs
ax2 - bx + c = 0+Ve, +VeSame signs
ax2 + bx - c = 0-Ve, +VeLarger root is negative
smaller root is positive
ax2 - bx - c = 0-Ve, +VeLarger root is positive
smaller root is negative

 

Shortcut Trick

If ax2 - bx + c = 0  then roots will be always +ve

If ax2 + bx + c = 0  then roots will be always -ve

So, without solving the equations we can say the answer.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 835

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. 3x2 + 2x – 8 = 0,

II. 3y2 – 14y + 16 = 0

  1. AOption (a)

    x > y

  2. BOption (b)

    x < y

  3. COption (c)

    x ≥ y

  4. DOption (d)

    x ≤ y

  5. EOption (e)

    x = y or relation cannot be established

View answer & explanation

Correct answer

Answer: (b)B.

x < y

Why this answer is correct

I. 3x2 + 2x – 8 = 0

⇒ 3x2 + 6x – 4x – 8 = 0

⇒ 3x(x + 2) – 4(x + 2) = 0

⇒ (3x – 4) (x + 2) = 0

⇒ x = 4/3 or -2

II. 3y2 – 14y + 16 = 0

⇒ 3y2 – 6y – 8y + 16 = 0

⇒ 3y(y – 2) – 8(y – 2) = 0

⇒ (3y – 8) (y – 2) = 0

⇒ y = 8/3 or 2

Value of xValue of yResult
4/38/3x < y
-28/3x < y
4/32x < y
-22x < y

∴ x < y

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 836

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-

I. x2 - 12x + 11 = 0

II. 3y2 + 12y + 9 = 0

  1. AOption (a)

    x > y

  2. BOption (b)

    x ≥ y

  3. COption (c)

    x < y

  4. DOption (d)

    x ≤ y

  5. EOption (e)

    x = y or the relation cannot be determined

View answer & explanation

Correct answer

Answer: (a)A.

x > y

Why this answer is correct

Equation I:

x2 – 12x + 11 = 0

⇒ x2 – x – 11x + 11 = 0

⇒ (x – 11) (x – 1) = 0

⇒ x = 11 or x = 1

Equation II:

3y2 + 12y + 9 = 0

⇒ 3y2 + 3y + 9y + 9 = 0

⇒ (3y + 9) (y + 1) = 0

⇒ y = - 3 or y = - 1

Comparing the values of x and y we get,

x > y

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 837

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Train passes a platform in 30 seconds and the speed of train is 72 kmph. If the same train crosses another platform which is triple the length of first platform is 45 sec, then find the length of train ?

  1. AOption (a)

     480 

  2. BOption (b)

     450

  3. COption (c)

     510

  4. DOption (d)

     420

  5. EOption (e)

     360

View answer & explanation

Correct answer

Answer: (b)B.

 450

Why this answer is correct

Shortcut Trick

Speed = 72 kmph = 72 × 5/18 = 20 m/s.

Distance 1 (T + P) = 20 × 30 = 600 m.

Distance 2 (T + 3P) = 20 × 45 = 900 m.

Subtracting both: (T + 3P) − (T + P) = 2P = 900 − 600 = 300 m ⇒ P = 150 m.

Length of train (T) = 600 − 150 = 450 m.

∴ The correct answer is 450 m.

Alternate Method

Given:

Speed of train = 72 kmph

Time to cross platform 1 (P1) = 30 sec

Time to cross platform 2 (P2 = 3P1) = 45 sec

Formula Used:

Distance = Speed × Time

Calculations:

⇒ Speed in m/s = 72 × (5/18) = 20 m/s

⇒ Let length of train be L and length of first platform be P

⇒ L + P = 20 × 30 = 600 m ----(Eq 1)

⇒ L + 3P = 20 × 45 = 900 m ----(Eq 2)

⇒ (Eq 2) − (Eq 1) ⇒ 2P = 300

⇒ P = 150 m

⇒ Substitute P in Eq 1: L + 150 = 600

⇒ L = 600 − 150 = 450 m

∴ The correct answer is 450 m.

Additional Information

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Relative Speed

When two objects move in the same direction, relative speed = S1 − S2. In opposite directions, relative speed = S1 + S2.

Crossing Objects

When a train crosses a platform/bridge, total distance = Length of train + Length of platform.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 838

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Pipe A and B can fill the tank in 9 hours and 15 hours respectively and pipe C can empty the tank in 25 hours. If pipe A , B and C are opened simultaneously, pipe A was closed after four hours , pipe B was closed one hour before pipe A was closed and pipe C was closed after 10 hours. Then, Find what part of tank is filled ?

  1. AOption (a)

    \(\frac{11}{45}\)

  2. BOption (b)

    \(\frac{22}{45}\)

  3. COption (c)

    \(\frac{11}{40}\)

  4. DOption (d)

    \(\frac{10}{45}\)

  5. EOption (e)

    \(\frac{14}{45}\)

View answer & explanation

Correct answer

Answer: (a)A.

\(\frac{11}{45}\)

Why this answer is correct

Shortcut Trick

Total Capacity (LCM of 9, 15, 25) = 225 units.

Efficiencies: A = 25, B = 15, C = −9.

Work done: A (4 hrs) + B (3 hrs) + C (10 hrs) = (25 × 4) + (15 × 3) − (9 × 10).

Filled part = 55 ÷ 225 = 11/45.

∴ The correct answer is 11/45.

Alternate Method

PersonTime (hrs)Total CapacityEfficiency
Pipe A922525
Pipe B1515
Pipe C (Outlet)25−9

Given:

Pipe A is opened for 4 hours.

Pipe B is closed 1 hour before Pipe A ⇒ 4 − 1 = 3 hours.

Pipe C is opened for 10 hours.

Calculations:

⇒ Water filled by A = 25 × 4 = 100 units

⇒ Water filled by B = 15 × 3 = 45 units

⇒ Water emptied by C = 9 × 10 = 90 units

⇒ Net water filled = 100 + 45 − 90 = 55 units

⇒ Part of tank filled = 55 ÷ 225

⇒ Simplified fraction = 11/45

∴ The correct answer is 11/45.

Additional Information

Pipes and Cisterns Concept

Efficiency of an inlet pipe is taken as positive (+), while the efficiency of an outlet (leakage) pipe is taken as negative (−).

Work-Time Relationship

Total Work = Efficiency × Time. The part filled is the ratio of actual units filled to the total capacity of the tank.

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 839

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions: The given caselet gives the information about the number of managers (Male + Female)  and number of clerks (Male + Female) working in the three different companies X, Y and Z.

The total  number of managers working in company X is 90 and the number of female clerks working in Company X is 10 less than the total number of managers working in company X. Number of male clerks to female clerks in company X are in the ratio 9: 8. Number of male managers working in company Z is 10 more than the female manager in the same company. Total number of managers in company Z is the same as the number of female clerks in company X. Number of male clerks in company Z is 10 more than the number of male clerks in company X. Total clerks in company Z is 165. Average number of male and female managers in company Z is the same as the number of female clerks in company Y which is 30 less than that of males in the same company. Number of male managers to female managers in company Y are in the ratio 4: 5. Difference between the number of male and female managers in company Y is 15. Number of male managers to female managers in company X is 5: 4.

Number of male clerks in Company X and Y is what percentage of the total number of female clerks in all the companies. 

  1. AOption (a)

    78.90%

  2. BOption (b)

    78.89%

  3. COption (c)

    78.80%

  4. DOption (d)

    83.89%

  5. EOption (e)

    86.48%

View answer & explanation

Correct answer

Answer: (e)E.

86.48%

Why this answer is correct

General Solution:

 

Total number of managers (Male + Female) in company X = 90

Number of male managers in company X= 90/9 ×  5 = 50

Number of female managers in company X = 90 - 50 = 40

Number of female clerks working in Company X is 10 less than the total number of managers working in company X 

Number of female clerks working in Company X = (90 - 10) = 80

Number of male clerks working in company X = 80/8 ×  9 = 90

Number of male managers working in company Z = (x + 10)

Number of female managers working in company Z = x

Total number of managers in company Z is the same as the number of female clerks in company X

Number of managers in company Z = 80

x + x + 10 = 80

2x = 70

x = 35

Number of male managers in company Z = (35 + 10) = 45

Number of female managers in company Z = 35

Number of male clerks in company Z is 10 more than the number of male clerks in company X

Number of male clerks in company Z = 90 + 10 = 100

Number of female clerks in company Z = 165 - 100 = 65

Average number of male and female managers in company Z is the same as the number of female clerks in company Y which is 30 less than that of males in the same company.

Number of female clerks in company Y = (45 + 35)/2 = 80/2 = 40

Number of male clerks in company Y = (40 + 30) = 70

Number of male managers to female managers in company Y are in the ratio 4: 5

Number of male managers in company Y = 4 ×  15 = 60

Number of female managers in company Y = 5 ×  15 = 75

 

CompaniesNumber of male managersNumber of female managersNumber of male clerksNumber of female clerks
X50409080
Y60757040
Z453510065

 

Solution:

Number of male clerks in company X = 90

Number of male clerks in company Y = 70

Total female clerks in all the company = (80 + 40 + 65)

Required percentage = (160/185) ×  100 = 86.48%

IBPS PO Prelims Memory Based Paper (Held On: 30 September 2023 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 840

IBPS RRB Office Assistant (Clerk) 2023• Quantitative Aptitude• Quantitative Aptitude

Directions: The given caselet gives the information about the number of managers (Male + Female)  and number of clerks (Male + Female) working in the three different companies X, Y and Z.

The total  number of managers working in company X is 90 and the number of female clerks working in Company X is 10 less than the total number of managers working in company X. Number of male clerks to female clerks in company X are in the ratio 9: 8. Number of male managers working in company Z is 10 more than the female manager in the same company. Total number of managers in company Z is the same as the number of female clerks in company X. Number of male clerks in company Z is 10 more than the number of male clerks in company X. Total clerks in company Z is 165. Average number of male and female managers in company Z is the same as the number of female clerks in company Y which is 30 less than that of males in the same company. Number of male managers to female managers in company Y are in the ratio 4: 5. Difference between the number of male and female managers in company Y is 15. Number of male managers to female managers in company X is 5: 4.

If the number of male managers and female managers in company Z is increased by 20% and 40% respectively. Find the difference between total male managers in all the given companies and total female managers in all the given companies.

  1. AOption (a)

    20

  2. BOption (b)

    0

  3. COption (c)

    10

  4. DOption (d)

    15

  5. EOption (e)

    5

View answer & explanation

Correct answer

Answer: (b)B.

0

Why this answer is correct

General Solution:

 

Total number of managers (Male + Female) in company X = 90

Number of male managers in company X= 90/9 ×  5 = 50

Number of female managers in company X = 90 - 50 = 40

Number of female clerks working in Company X is 10 less than the total number of managers working in company X 

Number of female clerks working in Company X = (90 - 10) = 80

Number of male clerks working in company X = 80/8 ×  9 = 90

Number of male managers working in company Z = (x + 10)

Number of female managers working in company Z = x

Total number of managers in company Z is the same as the number of female clerks in company X

Number of managers in company Z = 80

x + x + 10 = 80

2x = 70

x = 35

Number of male managers in company Z = (35 + 10) = 45

Number of female managers in company Z = 35

Number of male clerks in company Z is 10 more than the number of male clerks in company X

Number of male clerks in company Z = 90 + 10 = 100

Number of female clerks in company Z = 165 - 100 = 65

Average number of male and female managers in company Z is the same as the number of female clerks in company Y which is 30 less than that of males in the same company.

Number of female clerks in company Y = (45 + 35)/2 = 80/2 = 40

Number of male clerks in company Y = (40 + 30) = 70

Number of male managers to female managers in company Y are in the ratio 4: 5

Number of male managers in company Y = 4 ×  15 = 60

Number of female managers in company Y = 5 ×  15 = 75

 

CompaniesNumber of male managersNumber of female managersNumber of male clerksNumber of female clerks
X50409080
Y60757040
Z453510065

 

Solution:

Male managers in company Z = 45 ×  120/100 = 54

Female managers in company Z = 35 ×  140/100 = 49

Total male managers in all companies = (50 + 60  + 54) = 164

Total female managers in all companies = (40 + 75 + 49) = 164

Required difference = (164 - 164) = 0

How should I prepare for Quantitative Aptitude in IBPS RRB Office Assistant (Clerk)?

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.