Quantitative Aptitude Questions (Data Interpretation) for SBI PO Prelims 2018 Day-131

Dear Readers, SBI is conducting Online Examination for the recruitment of probationary officer. To enrich your preparation here we have providing new series of Data Interpretation – Quantitative Aptitude Questions. Candidates those who are appearing in SBI PO Prelims Exams can practice these Quantitative Aptitude average questions daily and make your preparation effective.

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Directions (Q. 1 – 5) Study the following information carefully and answer the given questions:

Following bar graph shows the time taken to travel (in hours) by six vehicles in two different days.

Distance covered (in kilometres) by six vehicles on each day.

Vehicle Day 1 Day 2
P 864 774
Q 516 864
R 550 546
S 703 810
T 636 693
U 931 770

 

  1. Find the ratio between the speed of vehicle P, Q and R on Day 1 to that of the speed of vehicle S, T and U on Day 2?
  1. 152 : 171
  2. 135 : 154
  3. 147 : 163
  4. 123 : 155
  5. None of these
  1. Find the difference between the speed of vehicle P, Q and R on Day 2 to that of the speed of vehicle P, S and T on Day 1?
  1. 12 km/hr
  2. 14 km/hr
  3. 6 km/hr
  4. 8 km/hr
  5. None of these
  1. Total time taken to travel by six vehicles on Day 1 is approximately what percentage of total time taken to travel by six vehicles on Day 2?
  1. 98 %
  2. 85 %
  3. 112 %
  4. 106 %
  5. 123 %
  1. Which of the following vehicles has highest speed on Day 1?
  1. Vehicle T
  2. Vehicle R
  3. Vehicle P
  4. Vehicle U
  5. None of these
  1. Total distance travelled by six vehicles on Day 1 is approximately what percentage more/less than the total distance travelled by six vehicles on Day 2?
  1. 12 % more
  2. 6 % less
  3. 14 % less
  4. 16 % more
  5. 10 % less

 

Directions (Q. 6 – 10) Study the following table carefully and answer the given questions:

Following table shows the population of different cities (In lakhs) and the ratio of males and females among them and the percentage of adult males and adult females.

City Population

(In lakhs)

M : F % Adult males from total males % Adult females from total females
A 8.1 5 : 4 76 % 81 %
B 7.7 6 : 5 69 % 72 %
C 9.8 3 : 4 70 % 60 %
D 7.2 2 : 1 66 % 78 %
E 8.5 2 : 3 82 % 83 %

 

  1. Find the difference between the total adult males in City C to that of total adult females in City A?
  1. 1400
  2. 2400
  3. 3050
  4. 2730
  5. None of these
  1. Find the ratio between the total number of males to that of total number of females of all the given cities together?
  1. 178 : 197
  2. 151 : 176
  3. 256 : 164
  4. 211 : 202
  5. None of these
  1. Total number of minor males in City A and B together is approximately what percentage of total number of minor females in City D and E together?
  1. 94 %
  2. 85 %
  3. 155 %
  4. 117 %
  5. 106 %
  1. Total number of adult females in City B and C together is approximately what percentage more/less than the total number of adult males in City A and E together?
  1. 5 % less
  2. 17 % more
  3. 25 % less
  4. 9 % more
  5. 15 % more
  1. Find the average number of adult females in all the given cities together?
  1. 224840
  2. 289360
  3. 289280
  4. 298020
  5. 275280

Answers:

1). Answer c

The speed of vehicle P, Q and R on Day 1

= > (864/16) + (516/12) + (550/11)

= > 54 + 43 + 50 = 147 km/hr

The speed of vehicle S, T and U on Day 2

= > (810/18) + (693/11) + (770/14)

= > 45 + 63 + 55 = 163 km/hr

Required ratio = 147: 163

2). Answer d

The speed of vehicle P, Q and R on Day 2

= > (774/18) + (864/16) + (546/14)

= > 43 + 54 + 39 = 136 km/hr

The speed of vehicle P, S and T on Day 1

= > (864/16) + (703/19) + (636/12)

= > 54 + 37 + 53 = 144 km/hr

Required difference = 144 – 136 = 8 km/hr

3).Answer a

Total time taken to travel by six vehicles on Day 1

= > 16 + 12 + 11 + 19 + 12 + 19

= > 89 hours

Total time taken to travel by six vehicles on Day 2

= > 18 + 16 + 14 + 18 + 11 + 14

= > 91 hours

Required % = (89/91)*100 = 97.8 % = 98 %

4). Answer c

Speed of P = (864/16) = 54 km/hr

Speed of Q = (516/12) = 43 km/hr

Speed of R = (550/11) = 50 km/hr

Speed of S = (703/19) = 37 km/hr

Speed of T = (636/12) = 53 km/hr

Speed of U = (931/19) = 49 km/hr

Vehicle P has highest speed.

5). Answer b

Total distance travelled by six vehicles on Day 1

= > 864 + 516 + 550 + 703 + 636 + 931 = 4200 km

Total distance travelled by six vehicles on Day 2

= > 774 + 864 + 546 + 810 + 693 + 770 = 4457 km

Required % = [(4457 – 4200)/4457]*100 = 5.766 % = 6 % less

Direction (6-10)

6). Answer b

The total number of adult males in City C

= > 980000 *3/7 *(70/100) =294000

The total number of adult females in City A

= > 810000*4/9 * (81/100) = 291600

Required difference = 294000-291600= 2400

7). Answer d

City Population

(In lakhs)

Males Females
A 810000 450000 360000
B 770000 420000 350000
C 980000 420000 560000
D 720000 480000 240000
E 850000 340000 510000

 

The total number of males of all the given cities together

= > (4.5 + 4.2 + 4.2 + 4.8 + 3.4) lakhs

= > 21.1 lakhs

= > 2110000

The total number of females of all the given cities together

= > (3.6 + 3.5 + 5.6 + 2.4 + 5.1) lakhs

= > 20.2 lakhs

= > 2020000

Required ratio = 2110000: 2020000 = 211: 202

8). Answer c

Total number of minor males in City A and B together

= > 8.1* 5/9 *(24/100)*100000 + 7.7* 6/11*(31/100)*100000

= > (86400+130200) = 216600

Total number of minor females in City D and E together

= > 7.2*1/3 *(22/100)*100000 + 8.5*3/5 *(17/100)*100000

= > 52800+86700=139500

Required % = (216600/139500) *100 = 155%

9). Answer a

Total number of adult females in City B and C together

= > 7.7*5/11*(72/100) * 100000 + 9.8 *4/7 *(60/100) *100000

= > 252000+336000 = 588000

Total number of adult males in City A and E together

= > 8.1* * 5/9 *(76/100) * 100000+ 8.5*2/5 *(82/100) *100000

= > 342000+278800

= > 620800

Required % = [(620800-588000)/620800]*100

= > 5 % less

10. Answer d

Total number of adult females in all the given cities

= > 8.1*4/9 *(81/100) + 7.7*5/11 *(72/100) + 9.8*4/7*(60/100) + 7.2*1/3*(78/100) + 8.5*3/5*(83/100)

=>291600+252000+336000+187200+423300

=>1490100

Required average = 1490100/5 = 298020

Daily Practice Test Schedule | Good Luck

Topic Daily Publishing Time
Daily News Papers & Editorials 8.00 AM
Current Affairs Quiz 9.00 AM
Quantitative Aptitude “20-20” 11.00 AM
Vocabulary (Based on The Hindu) 12.00 PM
General Awareness “20-20” 1.00 PM
English Language “20-20” 2.00 PM
Reasoning Puzzles & Seating 4.00 PM
Daily Current Affairs Updates 5.00 PM
Data Interpretation / Application Sums (Topic Wise) 6.00 PM
Reasoning Ability “20-20” 7.00 PM
English Language (New Pattern Questions) 8.00 PM

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