SBI PO Quantitative Aptitude Quiz: 30th April

| Updated On December 6th, 2019 at 06:21 pm

Dear Aspirants,

 

SBI PO Quantitative Aptitude Quiz: 30th April
 
 

Quantitative Aptitude Quiz For SBI PO/Clerk Prelims

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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 train covers 3584 km in 2 days 8 hours. If it covers 1440 km on the first day and 1608 km on the second day, by how much does the average speed of the train for the remaining part of the journey differ from that for the entire journey?

3 km/hour more
3 km/hour less
4 km/hour more
5 km/hour less
4 km/hour less
Solution: 

Total time required to cover 3584 km is 2 days 8 hrs
Avg. speed for entire journey = 3584/56 = 64 km/hr
ATQ
Remaining distance (3584 – 1440 – 1608) = 536 km has to be covered in 8 hrs.
∴ speed of remaining journey = 536/8 = 67 kmph
Difference = 3 kmph more

Q2. The average age of A and B is 22 years. If C were to replace A, the average would be 18 and if C were to replace B, the average would be 23. What are the ages of A, B and C?

27, 17, 19
18, 22, 20
22, 20, 18
18, 20, 22
17, 23, 29
Solution: 

A + B = 22 × 2
⇒ A + B = 44
and B + C = 36
and C + A = 46
∴ A + B + C = 126/2 = 63
∴ Age of A = 63 – 36 = 27 years
Age of B = 63 – 46 = 17 years
Age of C = 63 – 44 = 19 years

Q3. If 1 is added to the age of the elder sister, then the ratio of the ages of two sisters becomes 0.5 : 1, but if 2 is subtracted from the age of the younger one, the ratio becomes 1 : 3. Find the age of the two sisters.

8 and 5 years
11 and 6 years
9 and 5 years
8 and 6 years
7 and 4 years
Solution: 

Q4. If the ages of P and R are added to twice the age of Q, the total becomes 59. If the ages of Q and R are added to thrice the age of P, the total becomes 68 and if the age of P is added to thrice the age of Q and thrice the age of R, the total becomes 108. What is the age of P?

19 years
15 years
17 years
12 years
8 years
Solution: 

P + R + 2Q = 59 ……….(i)
Q + R + 3P = 68 …………(ii)
P + 3Q + 3R = 108 ………….(iii)
From 3 × (ii) – (iii)
P = 12 years

Q5. The average age of a class is 40 years. 12 new students with an average age of 32 years join the class, thereby decreasing the average age by 4 years. The original strength of the class was:

10
11
12
15
17
Solution: 

 
 

Q6. Sneh’s age is one-sixth of her father’s age. Sneh’s father’s age will be twice of Vimal’s age after 10 years. If Vimal’s eighth birthday was celebrated 2 years before, then what is Sneh’s present age?

30 years
24 years
6 years
20 years
5 years
Solution: 

let sneh, father and vimal’s age are S, F and V respectively
Given,
S = F/6
F + 10 = 2(V + 10)
And
V – 2 = 8
∴ V = 10 years
F = 2 (10 + 10) – 10 = 30 years.
So, Sneh’s present age = 5 years

Q7. The average income of a person for the first 6 days is Rs. 29, for the next 6 days it is Rs. 24, for the next 10 days it is Rs. 32 and for the remaining days of the month it is Rs. 30. Find the average income per day (Assume month is of 30 days).

Rs. 31.64
Rs. 30.64
Rs. 29.27
Rs. 34.27
Rs. 32.27
Solution: 

Q8. A man covers half of his journey by train at 60 km/hr, half of the remainder by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey.

36 kmph
30 kmph
18 kmph
24 kmph
28 kmph
Solution: 

Q9. There are 5 consecutive odd numbers. If the difference between square of the average of first two odd number and the square of the average of last two odd numbers is 492, what is the smallest odd number?

37
42
41
35
39
Solution: 

Let 5 consecutive odd numbers are x–4, x–2, x, x+2, x+4
ATQ,
(x + 3) ² – (x – 3) ² = 492
⇒ 12x = 492
⇒ x = 41
∴ smallest odd no. = 41 – 4 = 37

Q10. With an average speed of 40 km/hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The length of the total journey is

40 km
70 km
30 km
80 km
50 km
Solution: 

Directions (11-15): In each of the following questions two equations are given. Solve the equations and give answer— 


Q11. I. y² + 2y – 3 = 0 
II. 2x² – 7x + 6 = 0

If x less than y
If x≤y
If x>y
If x≥y
If x=y or relationship between x and y cannot be established
Solution: 

I. y² + 2y – 3 = 0
y² +3y – y – 3 = 0
(y + 3) (y- 1) = 0
y = -3 or 1
II. 2x² -7x + 6 = 0
2x² – 4x -3x + 6 = 0
2x(x – 2) – 3 (x- 2) = 0
(x – 2) (2x – 3) = 0
x = 2 or 3/2
x > y

Q12. I. (x)² = 961 
II. y =√961

If x less than y
If x≤y
If x>y
If x≥y
If x=y or relationship between x and y cannot be established
Solution: 

I. x² = 961
x = ± 31
II. y = √961 =31
y ≥ x

Q13. I. 15x² – 19x + 6 = 0 
II. 45y² – 47y + 12 = 0

If x less than y
If x≤y
If x>y
If x≥y
If x=y or relationship between x and y cannot be established
Solution: 

I. 15x² – 19x +6 = 0
15x² – 9x – 10x + 6 = 0
3x (5x – 3) -2 (5x – 3) = 0
(5x- 3) (3x- 2) = 0
x = 3/5 or 2/3
II. 45y² – 47y + 12 = 0
45y² – 20y – 27y + 12= 0
5y (9y – 4) – 3 (9y – 4) = 0
(9y – 4) (5y – 3) = 0
y = 4/9 or 3/5
x ≥ y

Q14. I. 2x² + 5x + 2 = 0 
 II. 12y² + 7y + 1 = 0

If x less than y
If x≤y
If x>y
If x≥y
If x=y or relationship between x and y cannot be established
Solution: 

I. 2x² + 5x + 2 = 0
2x² + 4x + x + 2 = 0
2x (x + 2) +1 (x+ 2) = 0
x = -2 or -1/2
II. 12y² + 7y + 1 = 0
12y² + 4y+ 3y+ 1 = 0
4y(3y + 1) + 1 (3y+ 1) = 0
(3y + 1) (4y+ 1) = 0
y = -1/3 or -1/4
y > x

Q15. I. 2x² – 13x + 21 = 0 
 II. 5y² – 22y + 21 = 0

If x less than y
If x≤y
If x>y
If x≥y
If x=y or relationship between x and y cannot be established
Solution: 

I. 2x² –13x + 21 = 0
2x² – 7x – 6x + 21 = 0
x(2x – 7) – 3 (2x – 7) = 0
(2x – 7) (x – 3) = 0
x=7/2 or 3
II. 5y² – 22y + 21 = 0
5y² – 15y – 7y + 21 = 0
5y(y- 3) – 7 (y- 3) = 0
(y – 3) (5y – 7) = 0
y=3 or 7/5
x≥y

           

 

 
 
          
 


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