Quantitative Aptitude Quiz For IBPS PO Mains: 8th November 2018

Dear Students,


Quantitative Aptitude Quiz For IBPS PO Mains: 7th November 2018
Quantitative Aptitude Quiz For IBPS PO Mains


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.



Directions (1-3): Working efficiency of A is 20% more than that of B. B can alone complete a work ‘X’ in 36 days.
B and C together started to complete the work ‘X’ and after 10 days they both left the work and then remaining work is done by A alone in 15 days.
A and C together started to complete another work ‘Y’ and after working for 12 days they both left the work. Remaining work is done by B alone in 16 days. D first completed work ‘X’ and then completed work ‘Y’ in total 38 days.
It is given that efficiency of all, in completing work ‘X’ and work ‘Y’ is same.

Q1. A, B and C working together completed 1/3rd of work ‘X’, and then A and C are replaced by D. Now remaining of work ‘X’ is completed by B and D together. For how many days B worked?                                                       
(a) 12 days
(b) 10 days
(c) 15 days
(d) 4 days
(e) 7 days

Q2. A, C and D working simultaneously completed work ‘X’ in ‘n’ days and A, B, C and D working simultaneously completed work ‘Y’ in ‘m’ days. Find the value of (m+n).                      (a) 15 days
(b) 10 days
(c) 12 days
(d) 8 days
(e) None of these

Q3. A person E starts the work ‘X’ and leave after 12 days, then B and C together complete the remaining work in 8 days. What is the ratio of number of days taken by A and E together to complete the work ‘X’ to the number of days taken by D, B and C together to complete the both work ‘X’ and ‘Y’ .                                                                                       
(a) 3 : 5
(b) 5 : 3
(c) 8 : 7
(d) 1 : 2
(e) 4: 5

Directions (4-5): Ramesh and Suresh decided to meet at a common point at the same time in the river. Ramesh had to travel 42 km upstream in the river and Suresh had to travel 35 5/7% less distance downstream than that of Ramesh to meet at a common point. They both sets off in their respective boats at the same time and speed of Ramesh’s boat is 20 km/hr more than the speed of Suresh boat. It is given that Suresh covers 280 km upstream in 35 hours.

Q4. Find the speed of stream of river?                                 
(a) 6 km/hr
(b) 8 km/hr
(c) 5 km/hr
(d) 10 km/hr
(e) 4 km/hr

Q5. After meeting, if they decided to return to their original places but Ramesh travelled for 19 km and Suresh travelled for 16 km, then what is the sum of time taken by both in covering these distances?               
(a) 150 min
(b) 120 min
(c) 180 min
(d) 90 min
(e) 60 min

Directions (6-15): What approximate value should come in the place of the question mark (?) in the following questions? (You are not expected to calculate the exact value.) 

Q6. √(?)=(1346.92+53.11)÷99.9-6.98
(a) 121
(b) 441
(c) 1024
(d) 49
(e) 196


Q9. 3/20  of 239.98= ? ÷(1.99×0.99)
(a) 72
(b) 33
(c) 45
(d) 37
(e) 80

Q10. √1296.002÷8.996×9.98+39.99= ?
(a) 80
(b) 8
(c) 4
(d) 120
(e) 40

Q11. 350% of ? ÷50+248=591 
(a) 4900
(b) 4800
(c) 4850
(d) 4950
(e) 5100 


Q14. ?×(523.5+687.5)=24220 
(a) 32 
(b) 22
(c) 28
(d) 20 
(e) 30

Q15. (2198 –1339 – 403)  (159 – 113 – 27) = ?
(a) 15
(b) 24
(c) 37
(d) 49
(e) 53

SOLUTIONS

S1. Ans.(a)
Sol. Working efficiency of A = 120% of working efficiency of B
B can complete ‘X’ work = 36 days
A can complete ‘X’ work = 30 days
Let ‘X’ work = 180 unit
Working efficiency of A = 6 unit/day
Working efficiency of B = 5 unit/day
Now ATQ,
15×6+10×5+C×10=180
So, C’s working efficiency = 4 unit/day
Total units of work ‘Y’= (6+4)×12+16×5=200 units
Total units of both work ‘X’ and work ‘Y’ = 180+200= 380 units
So, Working efficiency of D = 380/38=10 unit/day 
Now: A, B and C work together in ‘x’ work =1/3×180 unit=60 unit
A + B + C = (6 + 4 + 5) unit per day 
So, time required to complete 60 unit by A, B and C =60/15=4 days
Remaining work = 180-60= 120 unit
(B + D) ⇒ (5 + 10) unit per day 
=120/15=8 days.
B worked for = 4 + 8 = 12 days.

S2. Ans.(e)
Sol.
‘X’ work = 180 unit
A + C + D ⇒ 6 + 4 + 10 = 20 unit/day
Days=180/20=9 days
Work = 200 unit
(A + B + C + D) ⇒ (6 + 5 + 4 + 10) = 25 unit/day
=200/25=8 days
Total time = (9 + 8) = 17 days

S3. Ans.(a)
Sol.
Let efficiency of E is Z unit/day 
he works for 12 days
work complete = 12Z unit
B and C work for 8 days = (5 + 4) × 8 unit = 72 unit
Remaining work = 180 - 72 = 108 unit 
Efficiency of E=108/12= 9 unit⁄day
Now,
A and E completed work ‘X’ 
=180/15=12 days
D, B and C completed both work ‘X’ and ‘Y’
=(200+180)/19=20 days
Required Ratio = 12: 20 ⇒ 3: 5



               Sol.8 ans (e)




S15. Ans.(b)
Sol. 456÷19 =?      
 ?=456÷19=24 


You May also like to Read:
  

No comments