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SBI Clerk Quantitative Aptitude Quiz: 21st June

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SBI-Clerk-Quantitative-Aptitude-Quiz: 21st June

Quantitative Aptitude Quiz For SBI Clerk

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.

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Directions (1-5): study the line graph given below and answer the given questions based on it.


SBI Clerk Quantitative Aptitude Quiz: 21st June |_3.1



Q1. Downstream speed of boat in Ganga is what percent of downstream speed of boat in Tatpi?

60%
80%
90%
85%
75%
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_4.1

Q2. What will be total time taken by boat to go to a distance 28 km from a point A to B and returning from B to A in Yamuna ? (River is flowing from A to B)

4.8 hrs
3.8 hrs
4 hrs
5.6 hrs
8.4 hrs
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_5.1

Q3. What is the average of speed of boat in still water in all the rivers together?

6 kmph
12 kmph
9 kmph
8 kmph
10 kmph
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_6.1

Q4. What is the ratio between downstream speed in Narmada to upstream speed in Yamuna?

3 : 2
2 : 1
1 : 2
1 : 1
2 : 3
Solution:

Downstream speed of boat in Narmada = 6+4 = 10 km/hr
Upstream speed of boat in Yamuna = 12-2 = 10 km/hr
Required average =10/10=1∶1

Q5. In which river swimming will be easiest compare to other against the current>

Ganga
Yamuna
Narmada
Tapti
Narmada and Yamuna
Solution:

Swimming against the current will be easy in that river which rate of current is less
Hence, the required river will be Yamuna

Q6. The radius of a circular field is equal to the side of a square field. If the difference between the perimeter of the circular field and that of the square field is 32m, what is the perimeter of the square field? (in metere)

84
95
56
28
112
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_7.1

Q7. The outer circumference of a 1 cm thick pipe is 44 cm. How much water will 7 cm of the pipe hold?
(take π=22/7)

1078 cm³
1792 cm³
303 cm³
792 cm³
972 cm³
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_8.1

Q8. A bag has twelve strawberry out of which four are rotten. Two strawberry are selected at random. Find the probability that the strawberry are of both types.

2/9
23/40
7/18
4/9
None of these
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_9.1

Q9. Two dice are thrown. The probability that both of them show an even number greater than 4 is:

1/36
1/6
2/3
1/4
5/36
Solution:

Favorable cases = (6, 6)
= 1/36

Q10. In a class, there are 15 girls and 10 boys. Three students are selected at random. The probability that only girls or only boys get selected is:

1/3
2/5
1/4
1/5
3/5
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_10.1

Directions (11-15): In the following questions, two equations in x and y are given. Solve these equations and give answer 


Q11. I. x² + 7x + 12 = 0 
II. 4y ² = 36

if x>y
 if x<y
if x≥y
if x≤y
x=y or relation cannot be established between x and y
Solution:

I. x² + 7x + 12 = 0
⇒ x = –3, –4
II. y² = 9
⇒ y = ± 3
y ≥ x

Q12. I. 2x ² +5x + 3 = 0 
 II. y ² + 3x + 2 = 0

if x>y
 if x<y
if x≥y
if x≤y
x=y or relation cannot be established between x and y
Solution:

I. 2x² + 5x + 3= 0
⇒ 2x² + 2x + 3x + 3 = 0
⇒ (x + 1) (2x + 3) = 0
⇒ x = -1, -3/2
II. y² + 3y + 2 = 0
⇒ (y + 1) (y + 2) = 0
⇒ y = –1, –2
No relation

Q13. I. 2x + 3y = 5 
 II. 3x + 2y = 10

if x>y
 if x<y
if x≥y
if x≤y
x=y or relation cannot be established between x and y
Solution:

SBI Clerk Quantitative Aptitude Quiz: 21st June |_11.1

Q14. I. x² + 19x + 84 = 0 
 II. y² + 14y + 49 = 0

if x>y
 if x<y
if x≥y
if x≤y
x=y or relation cannot be established between x and y
Solution:

I. x² +19x + 84 = 0
⇒ (x + 7) (x+ 12) = 0
⇒ x = -7, –12
II. y² + 14y + 49 = 0
⇒ (y + 7) ² = 0
⇒ y = –7
y ≥ x

Q15. I. 3x² + 4x + 1 = 0 
 II. 2y² + 3y + 1= 0

if x>y
if x<y
if x≥y
if x≤y
x=y or relation cannot be established between x and y
Solution:

I. 3x² + 4x + 1= 0
⇒ 3x² + 3x + x +1 = 0
⇒ (x + 1) (3x +1) = 0
⇒ x = -1, -1/3
II. 2y² + 3y + 1= 0
⇒ 2y² + 2y + y +1 = 0
⇒ (y + 1) (2y+ 1) = 0
⇒ y= -1,-1/2
No relation

               





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