Dear Aspirants,
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.
Q1. The present age of a father is 3 years more than three times the age of his son. Three years hence, father’s age will be 10 years more than twice the age of the son. Find the present age of the father.
32 years
33 years
34 years
35 years
40 years
Solution:
Let the age of father is x and age of son is y.
So, as per question
x – 3y = 3 ……….(i)
x + 3 – 2 (y + 3) = 10
or x – 2y = 13 ……….(ii)
From (i) and (ii) we get
y = 10, x = 33
So, as per question
x – 3y = 3 ……….(i)
x + 3 – 2 (y + 3) = 10
or x – 2y = 13 ……….(ii)
From (i) and (ii) we get
y = 10, x = 33
Q2. How many permutations can be made using the letters of the word ‘BHATNAGAR’ such that ‘N’ always comes in the middle and B and R always come at end places (Repetition of letters is not allowed)?
640
720
360
240
520
Q3. A student has to answer 11 out of 14 questions only in an examination such that he must choose at least 5 from the first 6 questions. The number of ways available to him is. 256
252
156
198
224
Q4. The average age of Megha and Ritu is 18 years. Six years hence, Megha’s age will be two times of Ritu’s age. Find present age of Megha. 24 years
26 years
10 years
32 years
18 years
Solution:
Let Megha’s age = x years
Ritu’s age = y years
∴ x + y = 36 …(i)
And,
x + 6 = 2(y + 6)
⇒ x – 2y = 6 …(ii)
Solving eq. (i) and (ii), we get
x = 26 years, y = 10 years
Ritu’s age = y years
∴ x + y = 36 …(i)
And,
x + 6 = 2(y + 6)
⇒ x – 2y = 6 …(ii)
Solving eq. (i) and (ii), we get
x = 26 years, y = 10 years
Q5. In how many different ways the letters of word ‘DANGER’ can be arranged?
5040
640
720
120
520
Solution:
Required no. of ways = 6! = 720
Directions (6-10): What will come in the place of question (?) mark:
Q6.
850
900
960
1000
1050
Q7. 23% of 600 + 33% of 800 = ? + 53% of 400 170
180
190
210
150
Solution:
? = 23 × 6 + 33 × 8 – 53 × 4
= 138 + 264 – 212
= 190
= 138 + 264 – 212
= 190
Q8. 2341 + 4451 + 6329 – 8431 = ?
4690
4960
4860
4790
4520
Solution:
? = 13121 – 8431 = 4690
Q9. 43% of 500 + 250% of 60 = ? + 150% of 80
145
425
245
345
225
Solution:
? = 43 × 5 + 25 × 6 – 15 × 8
= 365 – 120
= 245
= 365 – 120
= 245
Q10.
14
12
16
18
10
Directions (11-15): The following pie charts show the distribution of students of class 10th and 12th in seven different schools namely P, Q, R, S, T, U and V Q11. Student studying in class 10th in school U are more than the students studying in class 12th of school R by what number?
870
872
857
785
None of these
Q12. What is the difference between total number of students studying in class 10th in school R, S, T, & U together and students studying in class 12th in school P, Q, and V together? 6780
6078
6870
7680
7860
Solution:
Required difference = (0.17 + 0.15 + 0.12 + 0.13) × 24500 – (0.15 + 0.18 + 0.1) × 16500
= 13965 – 7095
= 6870
= 13965 – 7095
= 6870
Q13. 42% of students studying in class 10th in school Q failed in an examination. If the ratio of boys and girls in the failed group is 4 : 3, then find the number of girls who failed in examination.
288
882
828
282
None of these
Q14. Average number of students studying in class 12th of school S & T together are approximately what percent more than the number of students studying in class 10th of school P? 54%
56%
52%
50%
55%
Q15. In a fest, it was decided to select 30% of students studying in class 10th& 20% of students studying in class 12th from every school. What is the average number of students selected for the fest? 5235
5352
5532
3552
5325
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