The RBI Grade B recruitment exam is a competitive exam conducted by the Reserve Bank of India (RBI) to recruit officers for its Grade B cadre. The exam is conducted in three phases, namely, Phase-I (Preliminary), Phase-II (Mains), and Interview. The exam tests the candidates’ knowledge in various areas such as economics, finance, management, and general awareness.
Quants, short for Quantitative Aptitude, is one of the important sections of the RBI Grade B syllabus,that evaluates the candidates’ ability to solve mathematical problems. This section comprises questions on Series, Arithmetic, algebra, geometry, trigonometry, and data interpretation. The questions in this section test the candidate’s speed, accuracy, and logical reasoning skills. The questions are of varying levels of difficulty and are designed to test the candidate’s understanding of mathematical concepts.
Why Are Series Quantitative Aptitude Questions Important in RBI Grade B exam?
Series questions require finding the pattern and predicting the next term, at different difficulty levels. Hence, dedicating time to practice and master these types of questions is essential for success in the exam. These questions are a popular type of reasoning question that hold significance in exams like the RBI Grade B for several reasons like:
How To prepare for RBI Grade B Quantitative Aptitude Questions
To prepare for the Quants section of the RBI Grade B exam, candidates should have a strong foundation in basic mathematical concepts. They should practice solving different types of problems regularly and aim to improve their speed and accuracy. Candidates can refer to various books and online resources to enhance their preparation for this section. Today, we will be discussing difficult questions on series for the RBI Grade B exam. Let’s look at the questions below
In each of the following questions a number series is given. One term of the series is denoted by X. You have to calculate the value of X and by using the value of X, you have to replace the question mark (?) in the following questions :
1. If 16, 15, 26, 69, 260, x
Find 75% of x + x
a) 2421.25
b) 2521.25
c) 2781.25
d) 2231.25
e) None of these
2. If 2, 4, 6, x, 14, 28, 30
Then, x3 + 2x + 1 = ?
a) 1921
b) 1753
c) 1563
d) 1892
e) None of these
3. If 204, 196, 223, x, 284
Then, what is the average of the numbers of the above series?
a) 213.2
b) 146.2
c) 163.8
d) 292.5
e) None of the above
4. If 3, 12, 108, x, 43200
Then, 47% of (x + 72)= ?
a) 921
b) 846
c) 563
d) 892
e) None of these
5. If 2, 7, 20, x, 110, 235
Then, 7x + 7√x = ?
a) 921
b) 675
c) 563
d) 392
e) None of these
6 In the given series, one number is wrong:
1, 2, 9, 43, 305, 2750, 30256
Now let wrong number in above series is ‘a’ so make another series starting with ‘a’ on same pattern and find 4th term of this new series:
a, , , ?
a) 261
b) 360
c) 363
d) 260
e) None of these
7. There are three series given below which are following with the same pattern.
Series I: 9, 10, 32, 163, 1145 Series II: 4, B, C, D, E Series III: F, G, H, I, J
Also, G = 3C + 5, Find the product of F & H.
a) 7235
b) 6850
c) 9176
d) 8210
e) 9350
In the following number series, two set of series are given. You are expected to find the logic of the series and find the wrong term and answer question accordingly:
Series I: 18, 17, 32, 93, 374, 1835
Series II: 7600, 6498, 5508, 4624, 3844, 3150
8. In series II, if ‘x’ is the right term, then find which of the following statement(s) are true.
x/124 is a factor 6 and 8.
x + 1069 is a perfect cube.
x/961 is a perfect square.
a) Only II follow
b) Only I follow
c) All follows
d) Only II and III follows
e) none follows
9. If X is the wrong number for series I, then find the minimum number must be added to x to make it the perfect cube?
a) 138
b) 128
c) 130
d) 135
e) 140
10. The question consists of three statements numbered “I, II and III” given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. 720 356 174 83 37.5 14.75
If nth term is ‘X’ and (n+1)th term is ‘Y Then, which of the following statements are true
- X = 2Y + 8
- Difference between the first and second term is twice the difference between the second and third term.
- Y = 0.5X – √16
a. Only I follow
b. Only I and II follows
c. I, II and III follows
d. Only I and III follows
e. None of these.
Solutions:
1. d.
16 × 1 – 1², 15 × 2 – 2², 26 × 3 – 3², 69 × 4 – 4², 260 × 5 – 5², 1,275
Hence, x = 1,275
Hence, 75% of x + x = 956.25 + 1,275 = 2,231.25
2. 2(× 2), 4(+2), 6 (× 2), 12 (+2), 14 (× 2), 28 (+2), 30
Hence, x= 12
Hence, x3 + 2 x +1 = (12)3 + 2 × 12 + 1
= 1,728 + 24 + 1 = 1,753
3. a.
204 (-23), 196 (+33), 223 (-4³), 159 (+5³), 284
Hence, x = 159
Hence, average = (204 + 196 + 223 + 159 + 284)/5 = 1066/5 = 213.2
4. b.
3 (× 4), 12 (× 9), 108 (× 16), 1,728 (× 25), 43,200
Hence, x= 49
Hence, 47% of (x + 72) = 47/100 × (1,728 + 72) = 47/100 × 1,800 = 846
5. d.
2 × 2 + 3, 7 × 2 + 6, 20 × 2 + 9, 49 × 2 + 12, 110 × 2 + 15, 235
Hence, x= 49
Hence, 7 x + 7√x = 7 × 49 + 7√49 = 343 + 7 × 7 = 343 + 49 = 392
6. c.
1 × 1 + 1 = 2
2 × 3 + 2 = 8
8 × 5 + 3 = 43
43 × 7 + 4 = 305
305 × 9 + 5 = 2750
2750 × 11 + 6 = 30256
So 8 should come in place of 9, Hence 9 is wrong. So a = 9, new series will be
9 × 1 + 1 = 10
10 × 3 + 2 = 32
32 × 5 + 3 = 363
So required answer = 4th term of new series = 363
7. e.
Series I:
9 × 1 + 1 = 10
10 × 3 + 2 = 32
32 × 5 + 3 = 163
163 × 7 + 4 = 1145
Series II:
4 × 1 + 1 = 5 (B)
5 × 3 + 2 = 17 (C)
17 × 5 + 3 = 88 (D)
88 × 7 + 4 = 620 (E)
Series III:
F × 1 + 1 = G G × 3 + 2 = H
H × 5 + 3 = I I × 7 + 4 = J
And, G = 3C + 5 G = (3 × 17) + 5
G = 56
Now we can find the value of F, F × 1 + 1 = G
F × 1 + 1 = 56
F = 55
Also, we can find the value of H, G × 3 + 2 = H
56 × 3 + 2 = H
H = 170
The product of F & H = (55 × 170) = 9350
8. d.
Series I: 18 17 32 93 374 1835
(18-1) x 1 = 17
(17 – 1) x 2 = 32
(32 – 1) x 3 = 93
(93 – 1) x 4 = 368
(368 – 1) x 4 = 1835
Therefore, 368 will come instead of 374. Series II:The pattern is:
202 × 19 = 7600
192 × 18 = 6498
182 × 17 = 5508
172 × 16 = 4624
162 × 15 = 3840
152 × 14 = 3150
So, 3844 will be replaced by 3840. so, 374 will be replaced by 368.
Alternate Method:
203-20 = 7600
193-19 = 6498
183-18 = 5508
173-17 = 4624
163-16 = 3840
153-15 = 3150
So, 3844 will be replaced by 3840. Now from second series, x = 3844 From statement I,
X/124 is a factor 6 and 8
=> 3840/124 = 31
statement I not follows From statement II:
X + 1069 is a perfect cube.
=> 3844 + 1069 = 4913, is a perfect cube of 17 Hence, statement II follows
From statement III:
x/961 is a perfect square
=> 3844/961 = 4, is a perfect square. Hence, statement II and III follows.
9. a.
For series I, value of x = 374
therefore, 512 – 374 = 138, must be added.
10. c.
From condition,
If X = 720, then Y = 356
If X = 356, then Y = 174, and so on. From statement I:
Now if we take Y = 174 X = 174 × 2 + 8 = 356
Hence, we can see that the value of X is 356. So, statement I is true. From statement II:
Difference between first and second term = 720 – 356 = 364 Difference second and third term = 356 – 174 = 182 Therefore, statement II is also true
From statement III:
=> Y = 0.5X – √16
=> 360 – 4 = 356
=> 356 = 356
Hence, statement III also follows.
Conclusion
The Quantitative Aptitude section, featuring series questions, critically assesses mathematical proficiency, logical reasoning, and problem-solving abilities. Hence, investing time in practicing and mastering these question types is pivotal for achieving a high score and success in the RBI Grade B exam 2024. Understanding these series-based questions thoroughly is imperative to ace the exam. With focused preparation using ixamBee’s resources, you can significantly enhance your success prospects. Using ixamBee’s RBI Grade B Mock Tests ensures reliable assistance, offering valuable guidance and aiding your exam readiness. Try them out today!
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