SBI PO Quantitative Aptitude Quiz: 3rd May 2019

Dear Aspirants,

Quantitative Aptitude Quiz For SBI PO/Clerk Prelims

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. Compound interest and simple interest on a certain sum for two years are Rs. 1100 and Rs. 1000 respectively. If rate of interest for both S.I. and C.I is 20% p.a., then find the sum.

Rs. 2500

Rs. 2000

Rs. 3000

Rs. 1800

Rs. 2200

Solution:

Q2. Shikha’s age is one-sixth of her father’s age. Sikha’s father’s age will be twice of Vignesh’s age after 10 years. If Vignesh’s eighth birthday was celebrated 2 years before, then what is Shikha’s present age?

30 years

24 years

6 years

20 years

5 years

Solution:

Q3. Raghav invested

of his total income in scheme x which offers simple interest at the rate of 15% per annum and rest income in bank which offers compound interest at the rate of 10% per annum. If total interest earned by Raghav after two years is Rs. 6480 then find his total income.

Rs. 32, 000

Rs. 30, 000

Rs. 24, 000

Rs. 28, 000

Rs. 20, 000

Solution:

Q4. Rati’s age is three years more than Meera’s age. Sushila’s age is 75% of Rati’s age. If average age of all of them is 12.75 years, then find the age of Rati.

12 yrs.

15 yrs.

14 yrs.

16 yrs.

18 yrs.

Solution:

Q5. The difference between S.I. and C.I. on a certain sum is Rs. 40 for two years. Also, the simple interest on the same sum at same rate of interest for two years is Rs. 200 Find the sum.

Rs. 300

Rs. 250

Rs. 320

Rs. 100

Rs. 200

Solution:

Directions (6 – 10): Read the given bar graph and answer the following questions.

Q6. How many employees taking a loan from UCO in 2009 and PNB in 2010 were defaulters if 23% from UCO in 2009 and 20% from PNB in 2010 have defaulted?

630

650

600

750

840

Solution:

Employee taking loan from UCO in 2009 = 1000
Defaulters (UCO) = 23% of 1000 = 230
Person taking loan from PNB in 2010 = 2000
Defaulters (PNB) = 20% of 2000 = 400
Total desired defaulters = 230 + 400 = 630

Q7. In 2007, no of defaulters in SBI was 5%. However, each year no of defaulters increases by 10% in number. What will be the difference between the number of defaulters of SBI in the year 2009 and 2012?

1500

2000

1325

1456

Cannot be determined

Solution:

Cannot be determined because no. of employee taking a loan from SBI in 2007 is unknown.

Q8. In which of the following years, the difference in no. of employee taking loan from BOB from the previous year is highest?

2008

2009

2010

2012

None of these

Solution:

From graph, it is clear that in 2009, difference between no. of employee taking a loan is highest as compared to previous year.

Q9. If on average, Rs. 175000 per employee loan sanctioned by UCO bank all over the year, what will be total amount sanctioned by UCO in all given years?

1055600000

1055800000

1620000000

1050000000

None of the above

Solution:

No. of employee taking loan from UCO bank all over the year = 500 + 1000 + 1500 + 1000 + 2000
= 6000
Total loan amount sanctioned over the years = 6000 × 1,75,000
= Rs. 1,05,00,00,000

Q10. What is the ratio of number of employee taking Loans from SBI and BOB together in all the years and the total no of employee taking loans in 2010 and 2011 together?

8 : 5

5 : 7

7 : 5

9 : 7

7 : 9

Solution:

Directions (11-15): In the following questions two equations are given. You have to solve both the equations and give answer

Q11. I. x² – 26x + 168 = 0
II. 2y² –19y – 60 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y or relation between x and y cannot be established

if x = y

Solution:

I. x² - 26x + 168 = 0
⇒ x² - 12x – 14x + 168 = 0
⇒ (x - 12) (x- 14) = 0
⇒ x =12, 14
II. 2y² - 19y – 60 = 0
⇒ 2y² - 24y + 5y – 60 =0
⇒ 2y (y - 12) + 5 (y - 12) = 0
⇒ (y - 12) (2y + 5) = 0
⇒ y = 12, -5/2
x ≥ y

Q12. I. 3y² + 13y – 16 = 0
II. 3x² – 13x + 14 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y or relation between x and y cannot be established

if x = y

Solution:

I. 3y² + 13y – 16 = 0
⇒ 3y² + 16y – 3y – 16 = 0
⇒ y (3y + 16) – 1 (3y + 16) = 0
⇒ (3y + 16) (y - 1) = 0
⇒ y=1,-16/3
II. 3x² - 13x + 14 = 0
⇒ 3x² - 6x – 7x + 14 = 0
⇒ 3x (x - 2) -7 (x - 2) = 0
⇒ (x - 2) (3x - 7) = 0
⇒ x = 2, 7/3
x > y

Q13. I. x² – 48x + 576 = 0
II. y² = 576

if x > y

if x < y

if x ≥ y

if x ≤ y or relation between x and y cannot be established

if x = y

Solution:

I. x² - 48x + 576 = 0
⇒ (x - 24)² = 0
⇒ x = 24, 24
II. y² = 576
⇒ y ± 24
x ≥ y

Q14. I. 4x² – 18x – 36 = 0
II. 6y² + 20y + 16 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y or relation between x and y cannot be established

if x = y

Solution:

I. 2x² - 9x – 18 = 0
⇒ 2x² – 12x + 3x – 18 = 0
⇒ 2x (x - 6) + 3 (x - 6) = 0
⇒ (x - 6) (2x + 3) = 0
⇒ x = 6,-3/2
II. 6y² + 20y + 16 = 0
⇒ 3y² + 10y + 8 = 0
⇒ 3y² +6y + 4y + 8 = 0
⇒3y (y + 2) +4 (y + 2) = 0
⇒ (y + 2) (3y + 4) = 0
⇒ y = -2 ,-4/3
No relation

Q15. I. 2x+ 3y = 5
II. 3x + 4y = 9

if x > y

if x < y

if x ≥ y

if x ≤ y or relation between x and y cannot be established

if x = y

Solution: