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Quantitative Aptitude For NIACL AO Prelims: 21st January 2019

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Quantitative Aptitude For NIACL AO Prelims: 21st January 2019



Quantitative Aptitude Quiz For NIACL AO

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 bar- graph shows the total number of students in five different coaching institutes 


Table shows the ratio of boys to girls in each institute.
Study the graph carefully to answer the following questions.

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Q1. Girls in institute A and C together is what percent of the boys in institute B and D together?

44.42%
72.22%
54.36%
78.18%
36.24%
Solution:

Quantitative Aptitude For NIACL AO Prelims: 21st January 2019 |_5.1

Q2. Boys in institute E and A together are how much more or less than the total girls in institute D and C together?

32
12
16
21
14
Solution:

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Q3. If one third of the boys in institute C leaves and joins institute D. Then find the ratio of remaining students in institute C to total students in institute D?

None of these
13 : 19
19 : 23
17 : 25
19 : 25
Solution:

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Q4. If in another institute Z total boys is 20% more than number of boys in the institute A and total students in institute Z to institute C is in ratio of 7:8. Then girls in institute Z are what percent of the total students in same institute?

 Quantitative Aptitude For NIACL AO Prelims: 21st January 2019 |_8.1
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Solution:

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Q5. What is the average number of girls in institute B , C and D?

23
21
18
27
15
Solution:

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Q6. How many 5-digit numbers can be formed which contain only one times the digit ‘3’.

25,842
29,889
26,729
26,889
29,349
Solution:

If first digit is ‘3’ then remaining numbers can be formed inQuantitative Aptitude For NIACL AO Prelims: 21st January 2019 |_15.1 ways = 6561 ways
If first digit isn’t ‘3’ then first digit can be formed in 8 ways (excluding 3 and 0). Now out of remaining four one should be ‘3’ and remaining three digits can be formed in 9³ ways = 729 ways
So, total ways to form five digit number in which first digit isn’t ‘3’
= 4 × 8 × 729 = 23,328
‘4’ is multiplied because the digit ‘3’ can take four places.
And, Total ways to form five digit numbers in which first digit is ‘3’
= 6561 ways
Total number of ways = 23,328 + 6561 = 29,889

Q7. Two cards are drawn from a well shuffled pack of cards. If both card are spade, then a dice is thrown and if 2 diamond cards are drawn, then a coin is tossed. Otherwise operation is stopped. In how many ways we can get a tail.

78
52
156
65
102
Solution:

Quantitative Aptitude For NIACL AO Prelims: 21st January 2019 |_16.1

Q8. How many 3 digit number can be formed by 0, 3, 2, 5, 7, 4, 9. Which is divisible by 5 and none of digit is repeated?

55
58
68
54
52
Solution:

When unit digit is ‘0’
Number of ways = 6 × 5 × 1 = 30
When unit digit is ‘1’
Number of ways = 5 × 5 × 1 = 25
Total number of ways = 30 + 25 = 55

Q9. There are 12 points in a plane out of which 8 are collinear. Find the number of triangles formed from the points?

184
None of these
154
164
168
Solution:

Quantitative Aptitude For NIACL AO Prelims: 21st January 2019 |_17.1

Q10. Find total no. of arrangements which can be formed using letters of word ‘MARVELOUS’ so that no vowels come together?

5! × 2!×15
5! × 3!×15
125
4! × 5!×15
(5!/4!)×15
Solution:

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Directions (11-15): What will come at the place of question mark in the following questions? (You are not expected to find exact value) 


Q11. 34.998% of 3499.999 + 24.92% of 2600.01 – 1259.98 = ?

615
635
725
680
585
Solution:

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Q12. 2395.99 + 259.99 × 4.98 – 449.988 – ? = 589.98

2428
2516
2860
2740
2656
Solution:

2396 + 260 × 5 – 450 – ? ≃ 590
? ≃ 2396 + 1300 – 450 – 590
? ≃ 2656

Q13. 54.95% of 1999.99 + ?% of 5000.001 = 1824.999

10
20
22.5
14.5
26
Solution:

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Q14. (14.9 × 19.9) + (4.989)² × 13.010 = (?)²

25
35
15
27
37
Solution:

(15 × 20) + (5² × 13) ≈ (?)²
300 + 325 ≈ (?)²
?² ≈ 625
? ≈ 25

Q15. 2524.001 ÷ √15.9 – 331.01 = (4.998)² × ?

15
16
18
12
21
Solution:

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