LIC AAO Quantitative Aptitude Quiz: 2nd May

Dear Aspirants,

Quantitative Aptitude Quiz For LIC AAO

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 speed of a boat in still water is 8 km/h. If takes 5 hours to go upstream and 3 hours downstream distance between two points. What is the speed of stream?

4 km/h

2 km/h

3 km/h

1 km/h

2.5 km/h

Solution:

Q2. Rishabh can swim 27 km upstream in 9/2 hours while 24 km downstream in 2 hours. Speed of boat in steel water is how much percent more than the speed of current?

150 %

225 %

250 %

175 %

200 %

Solution:

Q3. The distance between two points A and B is 36 km. A boat goes from point A to point B downstream and returns from point B to point A upstream in 18 hours. If speed of boat in steel water is 6 km/h, then what is the speed of current (in kmph)?

3√3

4√3

2√3

1√3

5√3

Solution:

Q4. If the rate of interest is 10% per annum and is compounded half yearly, the principal of Rs. 400 in 3/2 year will amount to

Rs. 436.05

Rs. 463.05

Rs. 563.05

Rs. 363.05

Rs. 263.05

Solution:

Q5. The simple interest on Rs. 4500 at a certain rate of interest for 4 years is Rs. 3600. Find the compound interest on the same sum at the same rate of interest for one year compounded half yearly.

Rs. 945

Rs. 775

Rs. 950

Rs. 1045

Rs. 845

Solution:

Directions (6-10): Study the following pie-chart carefully to answer these questions: Percentage-wise distribution of players who Play Five Different Sports Total players are equal to 4200, out of which Female Players are equal to 2000

Total Players = 4200

Female Players = 2000

Q6. What is the average number of players (both male and female) who play Football and Rugby together?

620

357

230

630

520

Solution:

Q7. What is the difference between the number of female players who play Lawn Tennis and the number of male players who play Rugby?

220

240

194

Solution:

Required difference
=22% of 2000-(13% of 4200-10% of 2000)
=440-[546-200] =94

Q8. What is the ratio of the number of female players who play Cricket to the number of male players who play Hockey?

20 : 7

4 : 21

20 : 3

3 : 20

7 : 20

Solution:

Q9. What is the total number of the male players who play Football, Cricket and Lawn tennis together?

1,724

1,734

1,824

1,964

2,164

Solution:

Required total no. of male players
= (17% of 4200 – 13% of 2000) + (35% of 4200 – 40% of 2000) + (25% of 4200 – 22% of 2000)
= 454 + 670 + 610 = 1734

Q10. The number of male players who play Rugby is approximately what percentage of the total number of players who play Lawn Tennis?

Solution:

Directions (11-15): In the following questions, there are two equations in x and y.
You have to solve both the equations and give answer

Q11. I. x² – 5x – 14 = 0
II. y² – 16y + 64 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y

if x = y or there is no relation between x and y

Solution:

I. x² - 5x – 14 = 0
⇒ x² - 7x + 2x - 14 = 0
⇒ x (x – 7) +2 (x – 7) = 0
⇒ (x - 7) (x + 2) = 0
⇒ x = 7, -2
II. y² - 16y + 64 = 0
⇒ (y -8) ² = 0
⇒ y = 8, 8 = y > x

Q12. I. y² – 7y + 12 = 0
II. x² – 9x + 20 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y

if x = y or there is no relation between x and y

Solution:

I. y² - 7y + 12 = 0
⇒ y² - 4y – 3y + 12 = 0
⇒ (y - 4) (y - 3) = 0 ⇒ y = 4, 3
II. x² - 9x + 20 = 0
⇒ x² - 5x – 4x + 20 = 0
⇒ (x - 5) (x – 4) = 0
⇒ x = 5, 4 x ≥ y

Q13. I. 2x² + 11x + 12 = 0
II. 4y² + 13y + 10 = 0

if x > y

if x < y

if x ≥ y

if x ≤ y

if x = y or there is no relation between x and y

Solution:

I. 2x² + 11x + 12 = 0
⇒ 2x² + 8x + 3x + 12 = 0
⇒ (x + 4) (2x + 3) = 0
⇒x= -4 , -3/2
II. 4y² + 13y + 10 = 0
⇒ 4y² + 8y + 5y + 10 = 0
⇒ (y + 2) (4y + 5) = 0
⇒ y=-2, -5/4 No relation

Q14. I. 2x + 3y = 4
II. 3x + 2y = 6

if x > y

if x < y

if x ≥ y

if x ≤ y

if x = y or there is no relation between x and y

Solution:

I. 2x + 3y = 4
II. 3x + 2y = 6
Multiplying equation (i) by 2 and Equation (ii) by 3 and then substracting,