Quantitative Aptitude Quiz For IBPS SO Prelims: 19th December 2018

Dear Aspirants,

Quantitative Aptitude Quiz For IBPS SO Prelims: 18th December 2018


Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.



Directions (1-5): The following line graph shows the number of reams (packets of A4 size paper) in terms of percentage used by three departments of Career Power i.e. Print dept., HR dept. and DTP dept. Provided that total number of reams used per month is 1200 and it remains consistent for all the months. There is no other department using these reams. 



Q1. Find the difference in total number of reams used by print dept. from January to March and that of by DTP dept. from April to May?
700
718
780
468
900
Solution:

Q2. In May, Babu working in Print dept. used 25% of the reams from which he wastedreams. Find the number of reams not wasted by Babu.
68
88
32
44
48
Solution:
Number of reams not wasted by Babu = 132 - 88 = 44

Q3. In July HR dept. demanded 25% more reams than previous month by stationary supervisor Mr. Vinod, which he denied. But he provided them the reams
less than that provided to print dept. in June. Find the ratio of number of reams demanded and actually provided to HR dept.
4 : 3
25 : 24
13 : 12
5 : 3
None of these
Solution:
Reams demanded =1.25 × 0.25 × 1200 = 375
Reams provided = 432 - 72 = 360
∴Required ratio = 375 ∶ 360 = 25:24

Q4. By approximately what percent the total number of reams used by DTP dept. is less than that of by print team throughout the given months?
26%
32%
20%
25%
28%
Solution:

Q5. Total number of reams used by print dept. DTP dept. and HR dept. in February, April and June respectively is what percent of total reams used by HR dept. in given 6 months ?
Solution:

Directions (6-10): What should come in the place of the question mark (?) in following number series problems? 

Q6.12, 12, 18, 45, 180, 1080, ?
9180
9200
11070
12880
12440
Solution:
The pattern is ×1,×1.5,×2.5,×4,×6,×8.5
∴ ? = 1080 ×8.5 = 9180

Q7. 444, 467, 513, 582, 674, 789, ?
950
904
927
881
973
Solution:
The pattern is+23,+46,+69,+92,+115,+138
∴? = 789 + 138 = 927

Q8.1, 16, 81, 256, 625, 1296, ?
4096
2401
1764
3136
6561
Solution:

Q9.23, 25, 53, 163, 657, 3291, ?
16461
13169
9877
23045
19753
Solution:
The pattern is ×1+2,×2+3,×3+4,×4+5,×5+6,×6+7
∴? = 3291 × 6 + 7 = 19753

Q10. 13, 13, 65, 585, 7605, 129285, ?
2456415
2235675
2980565
2714985
2197845
Solution:
The pattern is ×1,×5,×9,×13,×17,×21
∴? = 129285 × 21 = 2714985

Directions (11-15): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer 

Q11. I. 49x² + 84x – 64 = 0 
         II. 25y² - 100y + 64 = 0
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or no relation can be established between x and y.
Solution:

Q12. I. 4x² - 33x + 63 = 0
         II. 10y² - 113y + 318 = 0
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or no relation can be established between x and y.
Solution:

Q13. I. 3x + 4y = 24 
         II. 2x + 3y = 17
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or no relation can be established between x and y.
Solution:
I. 3x + 4y = 24
II. 2x + 3y = 17
Solving (I) & (II)
x = 4, y = 3
x > y

Q14. 
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or no relation can be established between x and y.
Solution:

Q15. I. x² – 15x + 56 = 0 
         II. y² – 13y + 42 = 0
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or no relation can be established between x and y.
Solution:
I. x² – 15x + 56 = 0
x² – 7x – 8x + 56 = 0
x (x – 7) – 8 (x – 7)
x = 7, 8
II. y² – 13y + 42 = 0
y² – 6y – 7y + 42 = 0
y (y – 6) – 7 (y – 6) = 0
y = 7, 6
x ≥ y

               






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