Quantitative Aptitude Prep Plan + Important Question for RBI Grade B 2024

Quantitative Aptitude Preparation Strategy 2024: Preparing for the Quantitative Aptitude section of the RBI Grade B 2024 exam requires a strategic approach to tackle the diverse range of mathematical concepts and problem-solving skills tested. This article aims to provide a comprehensive guide to crafting an effective preparation strategy, along with important questions tailored for the RBI Grade B exam.

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Also Read: RBI Grade B 2024 Notification Details

Quantitative Aptitude Preparation Strategy 2024

Understand the Syllabus: Begin by familiarizing yourself with the Quantitative Aptitude syllabus for RBI Grade B 2024. Identify your strengths and weaknesses in each topic to prioritize your preparation accordingly.

Conceptual Clarity: Focus on building a strong foundation in fundamental mathematical concepts. Understand the underlying principles and formulas associated with each topic to solve problems effectively.

Practice Regularly: Practice is key to mastering Quantitative Aptitude. Solve a variety of questions from different difficulty levels to enhance your problem-solving skills and speed.

Time Management: Time management is crucial during the exam. Practice solving questions within the stipulated time frame to improve your speed and accuracy.

Mock Tests and Previous Year Papers: Take regular mock tests and solve previous year question papers to familiarize yourself with the exam pattern and assess your preparation level. Analyze your performance and identify areas for improvement.

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Quantitative Aptitude Preparation Strategy + Important Question for RBI Grade B

Now, let’s delve into some important questions for the Quantitative Aptitude section of RBI Grade B 2024:

Question 1: One year hence from now, the ratio of ages of ‘J’ and ‘K will be 9:10, respectively. One year ago from now, the ratio of ages of ‘K’ and ‘M’ was 4:5, respectively. The sum of the present ages of ‘K and ‘M’ is 110 years. What was the ratio of ages of ‘K’ and ‘M’, respectively when the youngest among the three was 19 years old?

A) 1:2

B) 3:4

C) 3:5

D) 2:3

E) 4:7

Question 2: ‘A’ alone working at 75% of her full efficiency can finish a work in 24 days. ‘A’ and ‘B’ working together with full efficiency can complete the same work in 10.8 days. If ‘A’ alone starts the work, works on it for ‘t’ days then leaves, then the remaining work is completed by ‘B’ alone in (t – 8) days. What is the value of ‘t’?

A) 14

B) 16

C) 12

D) 15

E) 18

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Question 3: 13 years ago from now, ‘P’ was twice as old as ‘Q’. If __ years hence from now, the ratio of age of ‘P’ and ‘Q’ will be 25:19, respectively, then present age of ‘Q’ is __ years.

The values given in which of the following options will fill the blanks in the same order in which is it given to make the statement true:

I. 15, 28

II. 13, 25

III. 17, 34

A) Only I

B) Only III

C) Only II and III

D) All I, II and III

E) Only II

Question 4: The cost of booking a hotel is partially fixed (irrespective of number of people) and partially variable. When the hotel is booked for 25 people the average cost per person is Rs. 300. However, when the hotel is booked for 80 people the average cost per person is Rs. 162.5. Mr. ‘A’ wants to book the room for 100 people, what is the average cost (in Rs.) per person that he needs to pay?

A) 196

B) 150

C) 200

D) 120

E) 210

Question 5: A bottle in the form of a cylinder whose lid is also in the form of a cylinder. The radius of the base of the bottle is 10.5 cm and the height of the bottle below the lid is 21 cm. If the radius of the lid and the base of the bottle are same and the total curved surface area of the whole bottle (bottle + lid) is 1584 cm², then find the height of the lid of the bottle.

A) 10 cm

B) 7 cm

C) 6 cm

D) 4 cm

E) 3 cm

Question 6: ‘p’ and ‘q’ are roots of the equation x² + Bx + 400 = 0. If ‘p’ and ‘q’ are real and equal, then find the value of ‘B’?

A) 40

B) 30

C) 20

D) 50

E) None of these

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Question 7: ‘p’ and ‘q’ are roots of the equation x² + Bx + 400. If ‘p’ and ‘q’ are real and equal, then find the roots of the equation x² – (B + 5)x + 500 = 0 given that both roots of this equation are also real.

A) 5, 100

B) 30, 35

C) 25, 30

D) 20, 25

E) 50, 10

Question 8: In the given table there are two columns I and II. Column I contains three equations and column II contains the roots of the equations given in column I, not necessarily in the same order. Study the given table carefully and answer the questions accordingly.

Column I	Column II
I. x2 – 3x = 28	a.) (8,13)
II. y2 + 104 = 21y	b.) (-4, 8)
III. z2 = 4z + 32	c.) (7, -4)

Which of the following relation is correct?

A) I – a, II – b, III – c

B) I – a, II – c, III – b

C) I – c, II – b, III – a

D) I – c, II – a, III – b

E) I – b, II – a, III – c

Question 9: Two series I and II following a certain pattern are given below. Determine the value of ‘x’ in series ‘I’ and then find the number that will come in place of ‘?’ in series II.

I. 465, 930, 310, 1240, x, 1488

II. 180, ?, 146, x, 112, 282

A) 196

B) 256

C) 214

D) 228

E) 218

Question 10: Given below are three series: I, II and III. Each of the given series consist an odd one out number. The odd one out number in series I, II and III are represented by ‘P’, ‘Q’ and ‘R’, respectively. You have to find the values of ‘P’, ‘Q’ and ‘R’ and establish the correct relation among them.I. 20, 32, 62, 92, 140, 200II. 14, 21, 42, 110, 315, 1102.5III. 124, 107, 90, 73, 56, 42

A) P > Q < R

B) P < Q > R

C) P > Q > R

D) P = Q > R

E) P > Q = R

प्रश्न 1: None

A) 1:2

B) 3:4

C) 3:5

D) 2:3

E) 4:7

प्रश्न 2: None

A) 14

B) 16

C) 12

D) 15

E) 18

प्रश्न 3: None

A) Only I

B) Only III

C) Only II and III

D) All I, II and III

E) Only II

प्रश्न 4: None

A) 196

B) 150

C) 200

D) 120

E) 210

प्रश्न 5: बेलन के रूप में एक बोतल का ढक्कन भी बेलन के रूप में होता है। बोतल के आधार की त्रिज्या 10.5 cm है और ढक्कन के नीचे की बोतल की ऊंचाई 21 cm है। यदि ढक्कन की त्रिज्या और बोतल का आधार समान है और पूरी बोतल (बोतल + ढक्कन) का कुल वक्र पृष्ठीय क्षेत्रफल 1584 m² है, तो बोतल के ढक्कन की ऊंचाई ज्ञात करें।

A) 10 cm

B) 7 cm

C) 6 cm

D) 4 cm

E) 3 cm

प्रश्न 6: ‘p’ और ‘q’ समीकरण x² + Bx + 400 = 0 के सूत्र हैं। यदि ‘p’ और ‘q’ वास्तविक और समान हैं, तो ‘B’ का मान ज्ञात करें?

A) 40

B) 30

C) 20

D) 50

E) इनमें से कोई नहीं

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प्रश्न 7: ‘p’ और ‘q’ समीकरण x² + Bx + 400 के सित्र हैं। यदि ‘p’ और ‘q’ वास्तविक और समान हैं, तो समीकरण x² – (B + 5)x + 500 = 0 के सूत्र ज्ञात करें। दिया है कि इस समीकरण के दोनों सूत्र भी वास्तविक (real) हैं।

A) 5, 100

B) 30, 35

C) 25, 30

D) 20, 25

E) 50, 10

प्रश्न 8: दी गई तालिका में दो कॉलम I और II हैं। कॉलम I में तीन समीकरण हैं और कॉलम II में कॉलम I में दिए गए समीकरणों के सूत्र हैं, ज़रूरी नहीं कि इसी क्रम में हों। दी गई तालिका का ध्यानपूर्वक अध्ययन करें और उसके अनुसार प्रश्नों के उत्तर दें।

Column I	Column II
I. x2 – 3x = 28	a.) (8,13)
II. y2 + 104 = 21y	b.) (-4, 8)
III. z2 = 4z + 32	c.) (7, -4)

निम्नलिखित में से कौन सा संबंध सही है?

A) I – a, II – b, III – c

B) I – a, II – c, III – b

C) I – c, II – b, III – a

D) I – c, II – a, III – b

E) I – b, II – a, III – c

प्रश्न 9: एक निश्चित पैटर्न का अनुसरण करते हुए दो श्रृंखला I और II नीचे दी गई हैं। श्रृंखला ‘I’ में ‘x’ का मान निर्धारित करें और फिर वह संख्या ज्ञात करें जो श्रृंखला II में ‘?’ के स्थान पर आएगी।

I. 465, 930, 310, 1240, x, 1488

II. 180, ?, 146, x, 112, 282

A) 196

B) 256

C) 214

D) 228

E) 218

प्रश्न 10: नीचे तीन श्रृंखलाएँ : I, II और III दी गई हैं। दी गई श्रृंखला में से प्रत्येक में एक विषम संख्या है। श्रृंखला I, II और III में विषम एक संख्या को क्रमशः ‘P’, ‘Q’ और ‘R’ द्वारा दर्शाया जाता है। आपको ‘P’, ‘Q’ और ‘R’ के मान ज्ञात करने हैं और उनमें सही संबंध स्थापित करना है।

I. 20, 32, 62, 92, 140, 200

II. 14, 21, 42, 110, 315, 1102.5

III. 124, 107, 90, 73, 56, 42

A) P > Q < R

B) P < Q > R

C) P > Q > R

D) P = Q > R

E) P > Q = R

ANSWER KEYS and SOLUTIONS:

1) – D)	2) – A)	3) – E)	4) – B)	5) – E)	6) – E)
7) – D)	8) – D)	9) – C)	10) – B)

Solution 1: D)

1 year hence from now, let the age of ‘J’ will be ‘9x’ years

So, present age of ‘J’ = (9x – 1) years

1 year hence from now, age of ‘K’ = 9x × (10/9) = ‘10x’ years

So, present age of ‘K’ = (10x – 1) years

1 year ago, age of ‘K’ = (10x – 2) years

So, 1 year ago, age of ‘M’ = (10x – 2) × (5/4) = (12.5x – 2.5) years

So, present age of ‘M’ = 12.5x – 2.5 + 1 = (12.5x – 1.5) years

Sum of present ages of ‘K’ and ‘M = (10x – 1) + (12.5x – 1.5) = {22.5x – 2.5} years

ATQ;

22.5x – 2.5 = 110

Or, 22.5x = 110 + 2.5 = 112.5

Or, x = 112.5 ÷ 22.5 = 5

So, present ages of ‘J’, ‘K’ and ‘M’ is 44 (9 × 5 – 1) years, 49 (10 × 5 – 1) years and 61 (12.5 × 5 – 1.5) years, respectively.

So, the youngest person is ‘J’.

Number of years before when ‘J’ was 19 years old = 44 – 19 = 25 years

So, ratio of ages of ‘K’ and ‘M’ 25 years ago = (49 – 25):(61 – 25) = 24:36 = 2:3

Hence, option d.

Solution 2: A)

Time taken by ‘A’ alone complete the work in full efficiency = 24 × 0.75 = 18 days

Let the total work = 54 units

Then, efficiency of ‘A’ alone = (54/18) = 3 units/day

Combined efficiency of ‘A’ and ‘B’ = (54/10.8) = 5 units/day

So, efficiency of ‘B’ alone = 5 – 3 = 2 units/day

According to the question,

3 × t + 2 × (t – 8) = 54

Or, 3t + 2t – 16 = 54

Or, 5t = 54 + 16 = 70

Or, t = (70/5) = 14

Hence, option a.

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Solution 3: E)

Let present ages of ‘P’ and ‘Q’ be ‘a’ years and ‘b’ years, respectively.

Therefore, a – 13 = 2 × (b – 13)

Or, a – 13 = 2b – 26

Or, a = 2b – 13……………. (1)

For I:

{(a + 15)/(b + 15)} = 25/19

Or, 19 × (a + 15) = 25 × (b + 15)

Or, 19a + 285 = 25b + 375

19a – 25b = 90…………… (2)

Putting value of ‘a’ in equation (2)

19 × (2b – 13) – 25b = 90

Or, 38b – 247 – 25b = 90

Or, 13b = 337

Or, b ~ 25.92

Therefore, present age of ‘Q’ = 25.92 years

Hence, I is false.

For II:

{(a + 13)/(b + 13)} = 25/19

Or, 19 × (a + 13) = 25 × (b + 13)

Or, 19a + 247 = 25b + 325

Or, 19a – 25b = 78…………… (3)

Putting value of ‘a’ in equation (3)

19 × (2b – 13) – 25b = 78

Or, 38b – 247 – 25b = 78

Or, 13b = 325

Or, b = 25

Therefore, present age of ‘Q’ = 25 years

Therefore, II is true.

For III:

(a + 17)/(b + 17) = 25/19

Or, 19(a + 17) = 25(b + 17)

Or, 19a + 323 = 25b + 425

19a – 25b = 102………… (4)

Putting value of ‘a’ in equation (4)

19 × (2b – 13) – 25b = 102

Or, 38b – 247 – 25b = 102

Or, 13b = 349

Or, b ~ 26.84

Therefore, present age of ‘Q’ = 26.84 years

Hence, III is false.

Hence, option e.

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Solution 4: B)

Total cost when the hotel is booked for 25 people = 25 × 300 = Rs. 7,500Total cost when the hotel is booked for 80 people = 80 × 162.5 = Rs. 13,000So, variable cost for extra 55 people i.e. (80 – 25) = 13,000 – 7500 = Rs. 5,500Therefore, variable cost per person is = 5500 ÷ 55 = Rs. 100ATQ;25 × 100 + fixed cost = 7,500Or, fixed cost = Rs. 5,000So, total costs when booking the hotel for 100 people = 100 × 100 + 5000 = Rs. 15,000Required average = (15000/100) = Rs. 150

Hence, option b.

Solution 5: E)

Let height of the lid is ‘h’ cm

According to the question,

2 × (22/7) × (21/2) × (21 + h) = 1584

Or, h = 3

So, the height of the lid = 3 cm

Hence, option e.

Solution 6: E)

Since ‘p’ and ‘q’ are both real and equal,

So, √(B² – 4AC) = 0

So, √(B² – 4 × 400) = 0

Or, B² = 1600

So, B = ±40

Hence, option e.

Solution 7: D)

Since ‘p’ and ‘q’ are both real and equal.

So, √(B² – 4AC) = 0

So, √(B² – 4 × 400) = 0

Or, B² = 1600

So, B = ±40

At B = 40, the given equation becomes, x² – 45x + 500 = 0

So, x² – 25x – 20x + 500 = 0

Or, x(x – 25) – 20(x – 25) = 0

Or, (x – 20)(x – 25) = 0

So, x = 20 or x = 25

And, at B = -40, the given equation becomes x² – (-35x) + 500 = 0

So, x² + 35x + 500 = 0

Since, √{35² – (4 × 500)} is irrational, the roots of this equation are not real.

So, required roots = 20 and 25.

Hence, option d.

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Solution 8: D)

From I:

x² – 3x = 28

Or, x² – 3x – 28 = 0

Or, x² – 7x + 4x – 28 = 0

Or, x(x – 7) + 4(x – 7) = 0

Or, (x – 7)(x + 4) = 0

So, x = 7 or x = -4

From II:

y² + 104 = 21y

Or, y² – 21y + 104 = 0

Or, y² – 13y – 8y + 104 = 0

Or, y(y – 13) – 8(y – 13) = 0

Or, (y – 13)(y – 8) = 0

So, y = 13 or y = 8

From III:

z² = 4z + 32

Or, z² – 4z – 32 = 0

Or, z² – 8z + 4z – 32 = 0

Or, z(z – 8) + 4(z – 8) = 0

Or, (z – 8)(z + 4) = 0

So, z = 8 or z = -4

Therefore, correct relation is I – c, II – a, III – b.

Hence, option d.

Solution 9: C)

For series I:

465 × 2 = 930

930 ÷ 3 = 310

310 × 4 = 1240

1240 ÷ 5 = 248

248 × 6 = 1488

Therefore, x = 248

For series II:

180 + (34 × 1) = 214

214 – (34 × 2) = 146

146 + (34 × 3) = 248

248 – (34 × 4) = 112

112 + (34 × 5) = 282

So, 214 is the missing number in series II.

Hence, option c.

Solution 10: B)

For series I:

20 + 12 × 1 = 32

32 + 12 × 2 = 56

56 + 12 × 3 = 92

92 + 12 × 4 = 140

140 + 12 × 5 = 200

So, P = 62

For series II:

14 × 1 = 14

14 × 1.5 = 21

21 × 2 = 42

42 × 2.5 = 105

105 × 3 = 315

315 × 3.5 = 1102.5

So, Q = 110

For series III:

124 – 17 = 107

107 – 17 = 90

90 – 17 = 73

73 – 17 = 56

56 – 17 = 39

So, R = 42

So, P < Q > R

Hence, option b.

Conclusion

Preparing for the Quantitative Aptitude section of the RBI Grade B 2024 exam requires a structured approach, conceptual clarity, regular practice, and time management skills. By understanding the exam pattern, focusing on core concepts, utilizing a variety of resources, and practicing with important questions, aspirants can enhance their preparation and boost their chances of success in the RBI Grade B exam.

Disclaimer: The Quantitative Aptitude strategies, examples, and practice questions shared here are for guidance only. They are illustrative, not official RBI Grade B exam content. Actual exam questions may differ in wording, format, or difficulty. Candidates should always consult official RBI notices for authentic and updated information.