Quantitative Aptitude Quiz For IBPS Clerk Mains: 28th December 2018

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. A man invested Rs. 9600 in two equal parts on SI for T years and (T + 2) years at the rates of 12.5% p.a and 16% p.a respectively. If the man got a total interest of Rs. 4272, then find the value of T.

1 years

3 years

1.5 years

2.5 years

2 years

Solution:

Q2. Veer invested Rs. P at the rate of 15% p.a and Rs. (P + 800) at the rate of 8.5% p.a. If veer gets a total simple Interest of Rs. 4836 after two years, then find the value of (P + 800).

9600 Rs.

10400 Rs.

10800 Rs.

11800 Rs.

12600 Rs.

Solution:

Q3. A man invested Rs. 8500 on SI at the rate of R% per annum. If ratio between rate of interest and time period is 4 : 1 and total interest received is Rs. 1360 then find the time for which the man invested his sum?

1.5 years

2.5 years

3 years

4 years

2 years

Solution:

Q4. Mr. Adarsh invested certain principal sum on SI and got 34% more amount after four years. Find the rate for which Adarsh invested his sum?

12.5 %

8.5 %

10 %

8 %

12 %

Solution:

Q5. A man invested his total saving into three different banks namely SBI, UBI and BOB in the ratio of 2 : 1 : 2 for two years. Rate of simple interest offered by SBI is 20%, UBI is 16% and BOB is 12%. If the simple interest received from SBI is Rs. 672 more than interest received from BOB. Find total interest received by the man from UBI after two years.

576 Rs.

484 Rs.

672 Rs.

556 Rs.

772 Rs.

Solution:

Directions (6-10)- Given below is the table which shows the number of registered voters in 5 villages and ratio of male to female in those registered voters in the year 2016
Note - Some values are missing in the table, you have to calculate them if required to answer the question.

Q6. What is the sum of total registered female voters from village A and B together if registered male voters in village A are 200/11% less than registered male voters from village B?

7000

14000

21000

17500

13500

Solution:

Q7. If average of all registered voters from all villages is 18400 and ratio of all registered voters from village C to all registered voters from Village E is 10 : 13 then find the sum of registered male voters from village B and E together.

24000

28000

18000

20000

25000

Solution:

Q8. Registered females from village A and B are equal. If in an election, 20% votes polled by all registered voters of village A are declared invalid in which ratio of male invalid votes to female invalid votes is 5: 3. Then, valid votes casted by females of village A are what percent of total registered female voters of village B.

Solution:

Q9. If total registered voters from village C are 100/9% more than total registered voters from village B then, find the ratio of total registered voters from village C to total registered voters from village D.

4 : 5

3 : 5

5 : 4

5 : 3

5 : 6

Solution:

Q10. If next year, registered voters from village D, are increased by 100/3% while registered male voters are increased by 100/7% and total registered female voters in 2016 in village B are 12.5% less than registered female voters of village D in year 2017 then, find registered male voters of village D in the year 2016.

6000

7000

14000

17500

13500

Solution:

Directions (11-15): Solve the following equations and mark the correct option given below.

Q11. I. x² – 27x + 180 = 0
II. y² – 7y = 60

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x² – 27x + 180 = 0
x² – 12x – 15x + 180 = 0
x (x – 12) –15(x – 12) = 0
(x – 15) (x – 12) = 0
x = 15, 12
II. y² – 7y – 60 = 0
y² – 12y + 5y – 60 = 0
y (y – 12) +5 (y – 12) = 0
(y + 5) (y – 12) = 0
y= -5, 12
⇒ x ≥ y

Q12. I. x² – 59x + 868 = 0
II. y² – 53y + 702 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x² – 59x + 868 = 0
x² – 28x – 31x + 868 = 0
x (x – 28) – 31 (x – 28) = 0
(x – 31) (x – 28) = 0
x = 28, 31
II. y² – 53y + 702 = 0
y² – 27y – 26y + 702 = 0
y (y – 27) – 26(y – 27) = 0
(y– 27) (y – 26) = 0
y = 26, 27
⇒ x > y

Q13. I. 100x² – 120x + 32 = 0
II. 10y² – 17y + 6 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. 100x² – 120x + 32 = 0
100x² – 40x – 80x + 32 = 0
20x (5x – 2) – 16(5x – 2) = 0
(20x – 16) (5x – 2) = 0
x = 4/5, 2/5
II. 10y² – 17y + 6 = 0
10y² – 12y – 5y + 6 = 0
2y (5y – 6) – 1 (5y – 6) = 0
(2y – 1) (5y – 6) = 0
y = 1/2, 6/5
⇒ No relation

Q14. I. 15x² – 22x + 8 = 0
II. 12y² – 5y – 2 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. 15x² – 22x + 8 = 0
15x² – 12x – 10x + 8 = 0
3x (5x – 4) –2 (5x – 4) = 0
(5x – 4) (3x– 2) = 0
x = 4/5, 2/3
II. 12y² – 5y – 2 = 0
12y² – 8y + 3y – 2 = 0
4y (3y – 2) + 1 (3y – 2) = 0
(4y + 1) (3y – 2) = 0
y = -1/4, 2/3
⇒ x ≥ y

Q15. I. x² + 8x + 15 = 0
II. y² – 2y – 8 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x²+ 8x + 15 = 0
x² + 5x + 3x + 15 = 0
x (x + 5) + 3 (x+ 5) = 0
(x + 5) (x + 3) = 0
x = –5, –3
II. y² – 2y – 8 = 0
y² – 4y + 2y – 8 = 0
y (y – 4) + 2(y – 4) = 0
(y – 4) (y + 2) = 0
y= –2, 4
⇒ x < y