Quant Test for SBI PO 2018 Prelim Exam Set – 37

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

Click here to know the details of the Examination

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under:

Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam. 

Directions(1-5): What should come at the place of question mark (?) in the following question?

1. (0.76 * 0.76 * 0.76 – 0.008)/(0.76 * 0.76 + 0.76 * 0.2 + 0.04) = ?
   0.33
   0.45
   0.56
   0.81
   0.49
   Option C
   (0.76 * 0.76 * 0.76 – 0.008)/(0.76 * 0.76 + 0.76 * 0.2 + 0.04) = ?
   = (0.438 -0.008)/(0.5776 + 0.152 + 0.04)
   = (0.43)/0.7696
   = 0.56

    

2. 832/24 * 84/26 /252/72 = ?
   27
   40
   35
   22
   32
   Option E
   832/24 * 84/26 /252/72 = ? = 32

    

3. (4096)^2/3/ (65536)^3/4 * (16)^? = 4^4
   3
   7
   5
   4
   2
   Option A
   (4096)^2/3/ (65536)^3/4 * (16)^? = 4^4
   => (16)^3*(2/3) / (16)^4*(3/4) * (16)^? = 16^2
   => (16)^(2-3) * (16)^? = 16^2
   => ? = 3

    

4. 2(17/23) + 1(15/46) – 3(27/92) = ? – 6(13/23)
   3(33/92)
   2(44/90)
   5(11/94)
   7(31/92)
   1(33/90)
   Option D
   2(17/23) + 1(15/46) – 3(27/92) = ? – 6(13/23)
   => (2+1-3+6)(68+30-27+52)/92 = ?
   => 6(123/92) = ?
   => ? = 7(31/92)

    

5. (?)^2 + (65)^2 = (160)^2 – (90)^2 – 7191
   80
   78
   66
   70
   85
   Option B
   (?)^2 + (65)^2 = (160)^2 – (90)^2 – 7191
   => (?)^2 = (160)^2 – (90)^2 – 7191 – (65)^2
   => (?)^2 = 25600 – 8100 – 7191 – 4225
   => ? = 78

    

Directions(6-10): Study the following table carefully and answers the question.


6. How many approximate Bank PO books were given to each distributor by publisher D, if all distributor distributes only bank PO books.
   250
   300
   420
   399
   345
   Option D
   No. of Bank PO books distributed by publisher D
   = 34236 × 63% = 21568.68 ∼ 21569 * 7/18 = 8388
   No. of books given to each distributor by publisher D = 8388/21
   = 399

    

7. The number of books distributed by F is how much percent more or less than the number of books distributed by C?
   14.44%
   18.49%
   11.25%
   20.22%
   19.51%
   Option B
   F = 21590 × 76% = 16408
   C = 25480 × 79% = 20,129
   Required answer = (20129-16408)/20129 x 100 = 18.49%

    

8. If the number of books distributed by publishers B, D & E are increased by 25% & the number of books distributed by publishers A, C & F are decreased by 30%. What will be the new average of books distributed by all the publishers?
   17981
   15480
   18812
   16230
   11745
   Option A
   B = 13200 × 84% = 11088
   D = 34236 × 63% = 21569
   E = 27300 × 82% = 22386
   Books distributed by B, D and E = (11088 + 21569 + 22386 ) × 125% = 68804
   A = 21440 × 90% = 19296
   C = 25480 × 79% = 20129
   F = 21590 × 76% = 16408
   Books distributed by A, C and F = ( 19296 + 20129 + 16408 ) × 70% = 39083
   New average = (68804 + 39083)/6 = 17981

    

9. If the number of distributors by publisher E was increased by 40% and if each publisher gets equal number of books, then the number of books each publisher got before increase in the number of distributor is how much percent more than the number of books each publisher got after increase in number of distributors?
   40%
   20%
   60%
   50%
   30%
   Option A
   No. of distributor = 35+40% of 35 = 35 + 14 = 49
   No. of books each publisher got after 40% increase in number of distributors = = 27300/49 = 557
   No. of books each publisher got before increase = 27300/35 = 780
   Therefore, required percent = (780-557)/557 x 100 = 40%

    

10. If 55.55% of books distributed are of SSC by Publisher A and the number of defected books of SSC by publisher A are 9.11% of total no. of books distributed by Publisher E. Then find the approximate average number of distributed books of SSC which are not defected by publisher A and the number of books which are not distributed by publisher E?
    6797
    4450
    5750
    6670
    8250
    Option A
    Total number of SSC books = 21440 x 9/16 = 12060
    No. of distributed books by publisher A = 21440 x 90% = 19296
    No. of books of SSC distributed by publisher A = 19296 x 55.55 % = 10719
    No. of defected books of SSC by publisher A = 27300 x 82% x 9.11% = 2039
    No. of distributed SSC books which are not defected by publisher A = 10719 – 2039 = 8680
    No. of books which are not distributed by publisher E = 27300 × 18% = 4914
    Average = (8690 + 4914)/2 = 6797