{"id":17628,"date":"2023-11-24T14:11:22","date_gmt":"2023-11-24T08:41:22","guid":{"rendered":"https:\/\/www.ixambee.com\/blog\/?p=17628"},"modified":"2023-11-24T16:15:02","modified_gmt":"2023-11-24T10:45:02","slug":"new-pattern-question-on-series-for-rbi-grade-b-exam","status":"publish","type":"post","link":"https:\/\/blog.ixambee.com\/new-pattern-question-on-series-for-rbi-grade-b-exam","title":{"rendered":"New pattern Question on Series For RBI Grade B exam\u00a0"},"content":{"rendered":"\n

The RBI Grade B recruitment exam<\/a> 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\u2019 knowledge in various areas such as economics, finance, management, and general awareness.<\/p>\n\n\n\n

Quants, short for Quantitative Aptitude<\/strong>, is one of the important sections of the RBI Grade B syllabus<\/a>,that evaluates the candidates\u2019 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\u2019s speed, accuracy, and logical reasoning skills. The questions are of varying levels of difficulty and are designed to test the candidate\u2019s understanding of mathematical concepts.<\/p>\n\n\n\n

Why Are Series Quantitative Aptitude Questions Important in RBI Grade B exam? <\/strong><\/h2>\n\n\n\n

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:<\/p>\n\n\n\n

<\/strong><\/p>\n\n\n\n

How To prepare for RBI Grade B Quantitative Aptitude Questions <\/strong><\/h2>\n\n\n\n

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\u2019s look at the questions below<\/p>\n\n\n\n

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 :<\/strong><\/p>\n\n\n\n

1. If   16,  15, 26,  69, 260, x<\/p>\n\n\n\n

Find 75% of x + x<\/p>\n\n\n\n

a)   2421.25<\/p>\n\n\n\n

b)   2521.25<\/p>\n\n\n\n

c)   2781.25<\/p>\n\n\n\n

d)   2231.25<\/p>\n\n\n\n

e) None of these<\/p>\n\n\n\n

2. If  2, 4, 6, x, 14, 28,   30<\/strong><\/p>\n\n\n\n

Then, x3 <\/sup>+ 2x + 1 = ?<\/p>\n\n\n\n

a)   1921<\/p>\n\n\n\n

b)   1753<\/p>\n\n\n\n

c)    1563<\/p>\n\n\n\n

d)   1892<\/p>\n\n\n\n

e) None of these<\/p>\n\n\n\n

3. If   204, 196, 223, x, 284<\/strong><\/p>\n\n\n\n

Then, what is the average of the numbers of the above series?<\/p>\n\n\n\n

a)   213.2<\/p>\n\n\n\n

b)   146.2<\/p>\n\n\n\n

c)   163.8<\/p>\n\n\n\n

d)   292.5<\/p>\n\n\n\n

e) None of the above<\/p>\n\n\n\n

4. If 3, 12, 108, x,  43200<\/strong><\/p>\n\n\n\n

Then, 47% of (x + 72)= ?<\/p>\n\n\n\n

a)   921<\/p>\n\n\n\n

b)   846<\/p>\n\n\n\n

c)   563<\/p>\n\n\n\n

d)   892<\/p>\n\n\n\n

e) None of these<\/p>\n\n\n\n

5. If 2,   7, 20,   x, 110, 235<\/strong><\/p>\n\n\n\n

Then, 7x + 7\u221ax = ?<\/p>\n\n\n\n

a)   921<\/p>\n\n\n\n

b)   675<\/p>\n\n\n\n

c)   563<\/p>\n\n\n\n

d)   392<\/p>\n\n\n\n

e) None of these<\/p>\n\n\n\n

6 In the given series, one number is wrong:<\/strong><\/p>\n\n\n\n

1, 2, 9, 43, 305, 2750, 30256<\/p>\n\n\n\n

Now let wrong number in above series is \u2018a\u2019 so make another series starting with \u2018a\u2019 on same pattern and find 4th<\/sup> term of this new series:<\/p>\n\n\n\n

a,  ,     , ?<\/p>\n\n\n\n

a)   261<\/p>\n\n\n\n

b)   360<\/p>\n\n\n\n

c)   363<\/p>\n\n\n\n

d)   260<\/p>\n\n\n\n

e) None of these<\/p>\n\n\n\n

7. There are three series given below which are following with the same pattern.<\/strong><\/p>\n\n\n\n

Series I: 9, 10, 32, 163, 1145 Series II: 4, B, C, D, E Series III: F, G, H, I, J<\/p>\n\n\n\n

Also, G = 3C + 5, Find the product of F & H.<\/p>\n\n\n\n

a)   7235<\/p>\n\n\n\n

b)   6850<\/p>\n\n\n\n

c)   9176<\/p>\n\n\n\n

d)   8210<\/p>\n\n\n\n

e)   9350<\/p>\n\n\n\n

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:<\/p>\n\n\n\n

Series I: 18,  17, 32, 93, 374, 1835<\/p>\n\n\n\n

Series II: 7600, 6498, 5508, 4624, 3844, 3150<\/p>\n\n\n\n

8. In series II, if \u2018x\u2019 is the right term, then find which of the following statement(s) are true.<\/strong><\/p>\n\n\n\n

x\/124 is a factor 6 and 8.<\/p>\n\n\n\n

x + 1069 is a perfect cube.<\/p>\n\n\n\n

x\/961 is a perfect square.<\/p>\n\n\n\n

a) Only II follow<\/p>\n\n\n\n

b) Only I follow<\/p>\n\n\n\n

c) All follows<\/p>\n\n\n\n

d) Only II and III follows<\/p>\n\n\n\n

e) none follows<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

a)   138<\/p>\n\n\n\n

b)   128<\/p>\n\n\n\n

c)   130<\/p>\n\n\n\n

d)   135<\/p>\n\n\n\n

e)   140<\/p>\n\n\n\n

10. The question consists of three statements numbered \u201cI, II and III\u201d 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<\/strong><\/p>\n\n\n\n

If nth term is \u2018X\u2019 and (n+1)th term is \u2018Y Then, which of the following statements are true<\/p>\n\n\n\n

    \n
  1. X = 2Y + 8<\/li>\n<\/ol>\n\n\n\n
      \n
    1. Difference between the first and second term is twice the difference between the second and third term.<\/li>\n<\/ol>\n\n\n\n
        \n
      1. Y = 0.5X – \u221a16<\/li>\n<\/ol>\n\n\n\n

        a. Only I follow<\/p>\n\n\n\n

        b. Only I and II follows<\/p>\n\n\n\n

        c. I, II and III follows<\/p>\n\n\n\n

        d. Only I and III follows<\/p>\n\n\n\n

        e. None of these. <\/p>\n\n\n\n

        Solutions:<\/strong><\/p>\n\n\n\n

        1. d.<\/strong><\/p>\n\n\n\n

        16 \u00d7 1 – 1\u00b2, 15 \u00d7 2 – 2\u00b2, 26 \u00d7 3 – 3\u00b2, 69 \u00d7 4 – 4\u00b2, 260 \u00d7 5 – 5\u00b2, 1,275<\/p>\n\n\n\n

        Hence, x = 1,275<\/p>\n\n\n\n

        Hence, 75% of x + x = 956.25 + 1,275 = 2,231.25<\/p>\n\n\n\n

        2. 2(\u00d7 2), 4(+2), 6 (\u00d7 2), 12 (+2), 14 (\u00d7 2), 28 (+2), 30<\/p>\n\n\n\n

        Hence, x= 12<\/p>\n\n\n\n

        Hence, x3 <\/sup>+ 2 x +1 = (12)3 <\/sup>+ 2 \u00d7 12 + 1<\/p>\n\n\n\n

        = 1,728 + 24 + 1 = 1,753<\/p>\n\n\n\n

        3. a.<\/strong><\/p>\n\n\n\n

        204 (-23<\/sup>), 196 (+33<\/sup>), 223 (-4\u00b3), 159 (+5\u00b3), 284<\/p>\n\n\n\n

        Hence, x = 159<\/p>\n\n\n\n

        Hence, average = (204 + 196 + 223 + 159 + 284)\/5 = 1066\/5 = 213.2<\/p>\n\n\n\n

        4. b.<\/strong><\/p>\n\n\n\n

        3 (\u00d7 4), 12 (\u00d7 9),  108 (\u00d7 16), 1,728 (\u00d7 25), 43,200<\/p>\n\n\n\n

        Hence, x= 49<\/p>\n\n\n\n

        Hence, 47% of (x + 72) = 47\/100 \u00d7 (1,728 + 72) = 47\/100 \u00d7 1,800 = 846<\/p>\n\n\n\n

        5. d.<\/strong><\/p>\n\n\n\n

        2 \u00d7 2 + 3, 7 \u00d7 2 + 6,   20 \u00d7 2 + 9,   49 \u00d7 2 + 12, 110 \u00d7 2 + 15, 235<\/p>\n\n\n\n

        Hence, x= 49<\/p>\n\n\n\n

        Hence, 7 x + 7\u221ax = 7 \u00d7 49 + 7\u221a49 = 343 + 7 \u00d7 7 = 343 + 49 = 392<\/p>\n\n\n\n

        6. c.<\/strong><\/p>\n\n\n\n

        1 \u00d7 1 + 1 = 2<\/p>\n\n\n\n

        2 \u00d7 3 + 2 = 8<\/strong><\/strong><\/p>\n\n\n\n

        8 \u00d7 5 + 3 = 43<\/p>\n\n\n\n

        43 \u00d7 7 + 4 = 305<\/p>\n\n\n\n

        305 \u00d7 9 + 5 = 2750<\/p>\n\n\n\n

        2750 \u00d7 11 + 6 = 30256<\/p>\n\n\n\n

        So 8 should come in place of 9, Hence 9 is wrong. So a = 9, new series will be<\/p>\n\n\n\n

        9 \u00d7 1 + 1 = 10<\/p>\n\n\n\n

        10 \u00d7 3 + 2 = 32<\/p>\n\n\n\n

        32 \u00d7 5 + 3 = 363<\/p>\n\n\n\n

        So required answer = 4th <\/sup>term of new series = 363<\/p>\n\n\n\n

        7. e.<\/strong><\/p>\n\n\n\n

        Series I:<\/strong><\/p>\n\n\n\n

        9 \u00d7 1 + 1 = 10<\/p>\n\n\n\n

        10 \u00d7 3 + 2 = 32<\/p>\n\n\n\n

        32 \u00d7 5 + 3 = 163<\/p>\n\n\n\n

        163 \u00d7 7 + 4 = 1145<\/p>\n\n\n\n

        Series II:<\/strong><\/p>\n\n\n\n

        4 \u00d7 1 + 1 = 5 (B)<\/p>\n\n\n\n

        5 \u00d7 3 + 2 = 17 (C)<\/p>\n\n\n\n

        17 \u00d7 5 + 3 = 88 (D)<\/p>\n\n\n\n

        88 \u00d7 7 + 4 = 620 (E)<\/p>\n\n\n\n

        Series III:<\/strong><\/p>\n\n\n\n

        F \u00d7 1 + 1 = G G \u00d7 3 + 2 = H<\/p>\n\n\n\n

        H \u00d7 5 + 3 = I I \u00d7 7 + 4 = J<\/p>\n\n\n\n

        And, G = 3C + 5 G = (3 \u00d7 17) + 5<\/p>\n\n\n\n

        G = 56<\/p>\n\n\n\n

        Now we can find the value of F, F \u00d7 1 + 1 = G<\/p>\n\n\n\n

        F \u00d7 1 + 1 = 56<\/p>\n\n\n\n

        F = 55<\/strong><\/p>\n\n\n\n

        Also, we can find the value of H, G \u00d7 3 + 2 = H<\/p>\n\n\n\n

        56 \u00d7 3 + 2 = H<\/p>\n\n\n\n

        H = 170<\/strong><\/p>\n\n\n\n

        The product of F & H = (55 \u00d7 170) = 9350<\/p>\n\n\n\n

        8. d.<\/strong><\/p>\n\n\n\n

        Series I: 18 17 32 93 374 1835<\/p>\n\n\n\n

        (18-1) x 1 = 17<\/p>\n\n\n\n

        (17 – 1) x 2 = 32<\/p>\n\n\n\n

        (32 – 1) x 3 = 93<\/p>\n\n\n\n

        (93 – 1) x 4 = 368<\/p>\n\n\n\n

        (368 – 1) x 4 = 1835<\/p>\n\n\n\n

        Therefore, 368 will come instead of 374. Series II:The pattern is:<\/p>\n\n\n\n

        202 \u00d7 19 = 7600<\/p>\n\n\n\n

        \"\"<\/figure>\n\n\n\n

        192 \u00d7 18 = 6498<\/p>\n\n\n\n

        182 \u00d7 17 = 5508<\/p>\n\n\n\n

        172 \u00d7 16 = 4624<\/p>\n\n\n\n

        162 \u00d7 15 = 3840<\/p>\n\n\n\n

        152 \u00d7 14 = 3150<\/p>\n\n\n\n

        So, 3844 will be replaced by 3840. so, 374 will be replaced by 368.<\/p>\n\n\n\n

        Alternate Method:<\/p>\n\n\n\n

        203-20 = 7600<\/p>\n\n\n\n

        193-19 = 6498<\/p>\n\n\n\n

        183-18 = 5508<\/p>\n\n\n\n

        173-17 = 4624<\/p>\n\n\n\n

        163-16 = 3840<\/p>\n\n\n\n

        153-15 = 3150<\/p>\n\n\n\n

        So, 3844 will be replaced by 3840. Now from second series, x = 3844 From statement I,<\/p>\n\n\n\n

        X\/124 is a factor 6 and 8<\/p>\n\n\n\n

        => 3840\/124 = 31<\/p>\n\n\n\n

        statement I not follows From statement II:<\/p>\n\n\n\n

        X + 1069 is a perfect cube.<\/p>\n\n\n\n

        => 3844 + 1069 = 4913, is a perfect cube of 17 Hence, statement II follows<\/p>\n\n\n\n

        \"\"<\/figure>\n\n\n\n

        From statement III:<\/p>\n\n\n\n

        x\/961 is a perfect square<\/p>\n\n\n\n

        => 3844\/961 = 4, is a perfect square. Hence, statement II and III follows.<\/p>\n\n\n\n

        9. a.<\/strong><\/p>\n\n\n\n

        For series I, value of x = 374<\/p>\n\n\n\n

        therefore, 512 \u2013 374 = 138, must be added.<\/p>\n\n\n\n

        10. c.<\/strong><\/p>\n\n\n\n

        From condition,<\/p>\n\n\n\n

        If X = 720, then Y = 356<\/p>\n\n\n\n

        If X = 356, then Y = 174, and so on. From statement I:<\/p>\n\n\n\n

        Now if we take Y = 174 X = 174 \u00d7 2 + 8 = 356<\/p>\n\n\n\n

        Hence, we can see that the value of X is 356. So, statement I is true. From statement II:<\/p>\n\n\n\n

        Difference between first and second term = 720 \u2013 356 = 364 Difference second and third term = 356 \u2013 174 = 182 Therefore, statement II is also true<\/p>\n\n\n\n

        From statement III:<\/p>\n\n\n\n

        => Y = 0.5X – \u221a16<\/p>\n\n\n\n

        => 360 \u2013 4 = 356<\/p>\n\n\n\n

        => 356 = 356<\/p>\n\n\n\n

        Hence, statement III also follows.<\/p>\n\n\n\n

        Conclusion<\/strong><\/h2>\n\n\n\n

        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<\/strong> ensures reliable assistance, offering valuable guidance and aiding your exam readiness. Try them out today!<\/p>\n\n\n\n

        Phase 1 Mock Tests – https:\/\/www.ixambee.com\/free-mock-tests\/rbi-grade-b<\/a><\/p>\n\n\n\n

        Phase 2 Mock Tests – https:\/\/www.ixambee.com\/free-mock-tests\/rbi-grade-b-phase-2<\/a><\/p>\n\n\n\n

        To help you prepare 50% faster for competitive exams, ixamBee provides free Mock Test Series and all the Current Affairs in English and Current Affairs in Hindi in the BeePedia capsules for GA Preparation. You can also get the latest updates for Bank PO, Bank Clerk, SSC, RBI NABARD and Other Government Jobs<\/strong>.<\/p>\n\n\n\n

        Also Read<\/h3>\n\n\n\n

        How to start RBI Grade B Exam preparation from scratch?<\/a><\/p>\n\n\n\n

        All About Puzzle Reasoning Questions in the RBI Grade B Exam<\/a><\/p>\n\n\n\n

        Seating Arrangement Questions For RBI Grade B Exam<\/a><\/p>\n\n\n\n

        <\/p>\n","protected":false},"excerpt":{"rendered":"

        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\u2019 knowledge in various areas such as economics, finance, management, and […]<\/p>\n","protected":false},"author":46,"featured_media":17642,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[328],"tags":[450,85,459,1085],"class_list":["post-17628","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-rbi-grade-b-exam","tag-quantitative-aptitude","tag-rbi-grade-b","tag-reasoning","tag-reserve-bank-of-india"],"yoast_head":"\nNew pattern Question on Series For RBI Grade B exam\u00a0<\/title>\n<meta name=\"description\" content=\"New pattern Question on Series For RBI Grade B exam: Informative and practice Question series by ixamBee..\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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