On this page you will get different types of Data Interpretation which is based on different patterns like given below:
- Data Interpretation Based on Time and Distance
- Data Interpretation Based on Profit and Loss
- SI and CI Based Data Interpretation
- Percentage Based Data Interpretation
- Boat and Stream Based Data Interpretation
- Mixed Pattern Based Data Interpretation
Data Interpretation is the most important topic for IBPS PO. It consists of a myriad of graphs, charts and tables form which you have to glean and analyse data. The key to cracking this area is to quickly identify the key pieces of data that you require to work on the questions asked. Here are the DIs Based on differencet maths concept.
- Data Interpretation Based on Time and Distance
The percentage of distance covered by each mode on each day.
1. What is the total distance covered by Bus?
A. 814.9 km
B. 812.3 km
C. 813.3 km
D. 821.23 km
E. None of these
Explanation :
Day | Total distance | By bus |
Day1 | 25% of 3000 = 750 km | 30% of 750 = 225 km |
Day2 | 15% of 3000 = 450 km | 25% of 450 = 112.5 km |
Day3 | 20% of 3000 = 600 km | 45% of 600 = 270 km |
Day4 | 10% of 3000 = 300 km | 20% of 300 = 60 km |
Day5 | 18% of 3000 = 540 km | 15% of 540 = 81 km |
Day6 | 12% of 3000 = 360 km | 18% of 360 = 64.8 km |
Total = 813.3 km |
Hence, option C is correct.
2. If everywhere Ola maintains an average speed of 40 km per hour then what is the total time (In hour) in six days spent on Ola? (approximately)
A. 25 hours
B. 26 hours
C. 27 hours
D. 28 hours
E. 28.5 hours
Explanation :
The total distance travelled by Ola = 1084.2 km
Speed = 40 km per hr
Time = | 1084.2 | = 27.105 ≈ 27 hrs |
40 |
Day | Total distance | By Ola |
Day1 | 25% of 3000 = 750 km | 45% of 750 = 337.5 km |
Day2 | 15% of 3000 = 450 km | 35% of 450 = 157.5 km |
Day3 | 20% of 3000 = 600 km | 15% of 600 = 90 km |
Day4 | 10% of 3000 = 300 km | 20% of 300 = 60 km |
Day5 | 18% of 3000 = 540 km | 60% of 540 = 324 km |
Day6 | 12% of 3000 = 360 km | 32% of 360 = 115.2 km |
Total = 1084.2 km |
Hence, option C is correct.
3. The distance travelled by Ola is approximately how much percentage of the distance travelled by Uber? (round off two decimal)
A. 97.21%
B. 102.32%
C. 98.34%
D. 99.91%
E. 106.29%
Explanation :
The total distance travelled by Ola = 1084.2 km
The total distance travelled by Uber = 3000 – 813.3 – 1084.2 = 1102.5 KM
The reqd. % = | 1084.2 × 100 | = 98.34% |
1102.5 |
Hence, option C is correct.
4. Suppose, Instead of Bus, the person uses Ola and the speed of Ola is 25% more than the speed of bus then approximately how many hours the person would save? (The average speed of bus is 30 km per hour)
A. 6.8 hours
B. 7.2 hours
C. 4.6 hours
D. 5.4 hours
E. 4.8 hours
Explanation :
The average speed of bus is 30 km per hour then the average speed of Ola
= | 30 × 125 | = 37.5 km |
100 |
The total distance travelled by bus = 813.3 km
The reqd. time = | 813.3 | – | 813.3 |
30 | 37.5 |
= 813.3 × | 7.5 | = 5.422 hours |
30 × 37.5 |
= 5.4 hours approximately
Hence, option D is correct.
5. Suppose, on the first day, the person travelled for 18 hours and each day he decreases the traveling time by 2 hours then what can be the average speed of the person during the entire period?
A. 38 | 6 | km per hour |
13 |
B. 37 | 9 | km per hour |
13 |
C. 36 | 2 | km per hour |
13 |
D. 37 | 8 | km per hour |
13 |
E. None of these
Explanation :
Total distance = 3000 km
Total time = 18 + 16+ 14 + 12 + 10 + 8 = 78 hours
The average speed = | total distance |
total time |
= | 3000 | = | 500 | = 38 | 6 | km per hour |
78 | 13 | 13 |
Hence, option A is correct.
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- Data Interpretation Based on Profit and Loss
Articles | Profit % / Loss % | Discount % | Marked up % |
A.C. | Profit 8% | ---- | 28% |
Cooler | Loss 8% | ---- | 16% |
Fridge | ---- | ---- | 15% |
Oven | ---- | 25% | ---- |
Geyser | Profit 10% | 8% | ---- |
Television | ---- | 10% | 12% |
Questions :
1. An owner of the electronics shop sold the cooler to a customer at a loss given in the table. Had he sold it for Rs.475 more, he would have gained a profit of 11%. What is the marked price of the cooler?
B. Rs. 2500
C. Rs. 2700
D. Rs. 2900
E. None of these
Explanation :
Hence, option (D) is correct.
B. Rs. 1575
C. Rs. 1480
D. Rs. 1200
E. None of these
Explanation :
CP of oven = | 100x | = 50x |
2 |
= 50 × 100 = Rs. 5000
MP of oven = 110% of 5000 = Rs. 5500
Hence, option (A) is correct.
B. Rs. 25000
C. Rs. 35000
D. Rs. 20000
E. None of these
Explanation :
B. 2.77%
C. 2.41%
D. 3.24%
E. None of these
Explanation :
Total profit% = | 37000 – 35000 | × 100 ≈ 5.71 |
35000 |
Hence, option (A) is correct.
B. Rs. 23737.5
C. Rs. 22500
D. Rs. 28737.5
E. None of these
Explanation :
CP of Fridge = 11500 × | 100 | = Rs. 10000 |
115 |
SP of Fridge = 108% of 10000 = Rs.10800
MP of oven = 115% of 15000 = Rs.17250
SP of oven = 75% of 17250 = Rs.12937.5
- SI and CI Based Data Interpretation
The table below shows details about invested by different persons:
Person | Rate of interest | Time (Years) | Principal (Rs.) | Amount (Rs.) |
Arun | 6% | 18000 | ||
Sourav | 6% | 30000 | ||
Amit | 5 | 29000 | ||
Piku | 3 | 45000 | ||
Ankit | 8% | 20000 | ||
Anita | 2 | 60000 |
All questions are independent from others.
Questions :
B. 13.45
C. 11.11
D. 15.50
E. 12.54
Explanation :
Rate of interest for Amit = | 6 × 3 | = 9% |
2 |
Let the principal invested by Amit be Rs. x
So,
= x + x × 9 × | 5 | = 29000 |
100 |
= 100x + | 45x | = 29000 |
100 |
= 145x = 29000 × | 100 |
145 |
= x = 20000
Amount invested by Amit = Rs 20000
To double the invested the time required
Let the time be x
= 20000 × | 100 | |
9 × 20000 |
= 11.11 years.
Hence, option C is correct.
B. Rs. 28600
C. Rs. 29600
D. Rs. 35400
E. Rs. 35540
Explanation :
So the time for which Ankit invested = 2x years
So, Interest received by Ankit
= 8 × 2x × | 20000 | = Rs. 3200x |
100 |
Interest received by Sourav
= 6 × x × | 30000 | = Rs. 1800x |
100 |
So, 3200x – 1800x = 4200
1400x = 4200
x = 3 years
So, Interest received by Ankit
= 8 × 6 × | 20000 | = Rs. 9600 |
100 |
Total amount received by Ankit = Rs. (20000 + 9600) = Rs 29600
Hence, option C is correct.
B. Rs. 21992.45
C. Rs. 21292.45
D. Rs. 21929.45
E. Rs. 21229.45
Explanation :
Amount invested by Amit = | 29000 | = 14500 |
2 |
Interest received by Amit = 29000 – 14500 = Rs 14500
Let the rate of interest for Amit be x
According to question,
= x = | 14500 ×100 | |
14500 × 5 |
= x = 20%
Amount he will if compounded half yearly for 2 years
= 14500 | ( | 1 + | 10 | ) | ^{4} |
100 |
= 14500 × | 11 | × | 11 | × | 11 | × | 11 | = Rs. 21229.45 | ||
10 | 10 | 10 | 10 |
Hence, option E is correct.
B. Rs. 130
C. Rs. 200
D. Rs. 1500
E. Rs. 700
Explanation :
Interest received by Anita
= | 20 × 60000 | = Rs. 12000 |
100 |
Rate = 12000 × | 100 | × 60000 |
2 |
Rate = 10%
Interest for 1st year
= 60000 × 10 × | 1 | = Rs. 6000 |
100 |
Principal for 2nd year= 60000 + 6000 = Rs 66000
Interest for second year
= 66000 × 10 × | 1 | = Rs. 6600 |
100 |
Total interest received by Anita = 6600 + 6000 = Rs 12600
So,
Anita will receive (12600 – 12000) = Rs 600 more interest.
Hence, option A is correct.
5. If the interest received by Anita is Rs 7575 more than interest received by Piku and the rate of interest received by Anita 2% more than the rate of interest received by Arun then find the interest rate received by Piku ?
B. 1.3%
C. 2.5%
D. 1.5%
E. 1.4%
Explanation :
Let the rate of interest received by Piku be x
According to question,
60000 × 8 × | 2 | – x × 3 × | 45000 | = 7575 |
100 | 100 |
= 9600 – 1350x = 7575
= 9600 – 7575 = 1350x
= x = | 2025 |
1350 |
= x = 1.5%
Hence, option D is correct.
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- Percentage Based Data Interpretation
Number of students visiting the chemistry lab on different days
Day | Number of students |
Monday | 180 |
Tuesday | 270 |
Wednesday | 150 |
Thursday | 240 |
Friday | 225 |
Saturday | 210 |
Questions :
B. 192.32%
C. 184.09%
D. 186.56%
E. 134.64%
Explanation :
Students handled by Mehandi on Thursday and Friday together = 45% of 240 + 60% of 225 = 243
Students handled by Supriya on Thursday = 55% of 240 = 132
Therefore, % = | 243 | × 100 = 184.09% |
132 |
Hence, option C is correct.
B. Tuesday
C. Friday
D. Saturday
E. Monday
Explanation :
Students handled by Mehandi on Monday = 40% of 180 = 72 Students handled by Mehandi on Tuesday = 70% of 270 = 189
Students handled by Mehandi on Wednesday = 50% of 150 = 75
Students handled by Mehandi on Thursday = 45% of 240 = 108
Students handled by Mehandi on Friday = 60% of 225 = 135
Students handled by Mehandi on Saturday = 30% of 210 = 63
On Saturday Mehandi handled minimum number of students.
Hence, option D is correct.
B. 83 : 26
C. 44 : 31
D. 76 : 25
E. 87 : 55
Explanation :
Number of students handled by Mehandi on Monday and Tuesday together = 72 + 189 = 261
Number of students handled by Supriya on Wednesday and Friday = 50% of 150 + 40% of 225 = 165
Therefore, respected ratio = 87: 55
Hence, option E is correct.
B. 47.37%
C. 50.41%
D. 42.73%
E. 45.56%
Explanation :
Number of students managed by her on Tuesday and Saturday together = 30% of 270 + 70% of 210 = 228
Therefore, % = | 108 | × 100 = 47.37% |
228 |
Hence, option B is correct.
B. 78.56%
C. 87.34%
D. 91.34%
E. 88.30%
Explanation :
Number of students handled by Mehandi from Monday to Friday = 579
Therefore, % = | 486 | × 100 = 83.94% |
579 |
Hence, option A is correct.
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- Boat and Stream Based Data Interpretation
Boat | Distance (Upstream) | Distance (Downstream) |
P | 96 | 288 |
Q | 120 | 240 |
R | 100 | 220 |
S | 150 | 350 |
T | 180 | 540 |
Questions :
1. Find the ratio of the speed of Boats P and Q together in still water to the speed of Boats S and T together in still water.
A. 25 : 32
B. 27 : 19
C. 24 : 39
D. 65 : 48
E. None of these
Explanation :
Ratio = (24 + 24) : (25 + 40)
= 48 : 65
Hence, option E is correct.
Common Explanation:
Speed of Boat P in still water = a km/h
96 | = | 288 |
(a – 12) | (a + 12) |
96 (a + 12) = 288 (a – 12)
96a + 1152 = 288a – 3456
3456 + 1152 = 288a – 96a
a = 24 km/h
Speed of Boat Q in still water = b km/h
120 | = | 240 |
(b – 8) | (b + 8) |
120 (b + 8) = 240 (b – 8)
120b + 960 = 240b – 1920
1920 + 960 = 240b – 120b
b = 24 km/h
Speed of Boat R in still water = c km/h
100 | = | 220 |
(c – 15) | (c + 15) |
100 (c + 15) = 220 (c – 15)
100c + 1500 = 220c – 3300
3300 + 1500 = 220c – 100c
c = 40 km/h
Speed of Boat S in still water = d km/h
150 | = | 350 |
(d – 10) | (d + 10) |
150 (d + 10) = 350 (d – 10)
150d + 1500 = 350d – 3500
3500 + 1500 = 350d – 150d
d = 25 km/h
Speed of Boat T in still water = e km/h
180 | = | 540 |
(e – 20) | (e + 20) |
180 (e + 20) = 540 (e – 20)
180e + 3600 = 540e – 10800
10800 + 3600 = 540e – 180e
e = 40 km/h
2. If the speed of Boat R in still water is increased by 10% and the speed of stream is increased by 20%, Find the time taken by Boat R to cover the distance of 91 km upstream.
A. 3.5 hours
B. 4 hours
D. 2.5 hours
E. None of these
Explanation :
Speed of Boat R in still water = 40 × 110% = 44 km/h
Speed of stream = 15 × 120% = 18 km/h
Time taken by Boat R to cover the distance of 91 km upstream = 91/ (44 – 18)
= 91/ 26 = 3.5 hours
Hence, option A is correct.
Speed of Boat P in still water = a km/h
96 | = | 288 |
(a – 12) | (a + 12) |
96 (a + 12) = 288 (a – 12)
96a + 1152 = 288a – 3456
3456 + 1152 = 288a – 96a
a = 24 km/h
Speed of Boat Q in still water = b km/h
120 | = | 240 |
(b – 8) | (b + 8) |
120 (b + 8) = 240 (b – 8)
120b + 960 = 240b – 1920
1920 + 960 = 240b – 120b
b = 24 km/h
Speed of Boat R in still water = c km/h
100 | = | 220 |
(c – 15) | (c + 15) |
100 (c + 15) = 220 (c – 15)
100c + 1500 = 220c – 3300
3300 + 1500 = 220c – 100c
c = 40 km/h
Speed of Boat S in still water = d km/h
150 | = | 350 |
(d – 10) | (d + 10) |
150 (d + 10) = 350 (d – 10)
150d + 1500 = 350d – 3500
3500 + 1500 = 350d – 150d
d = 25 km/h
Speed of Boat T in still water = e km/h
180 | = | 540 |
(e – 20) | (e + 20) |
180 (e + 20) = 540 (e – 20)
180e + 3600 = 540e – 10800
10800 + 3600 = 540e – 180e
e = 40 km/h
3. The distance between point A and point B is 210 km. Boat T travels from point A to B and comes back. What is the time taken by Boat T to cover the total distance.
A. 10 hours
B. 12.5 hours
C. 14 hours
D. 20 hours
E. 8 hours
Explanation :
Total time = | 210 | + | 210 |
(40 – 20) | (40 + 20) |
= | 210 | + | 210 |
20 | 60 |
Hence, option C is correct.
Speed of Boat P in still water = a km/h
96 | = | 288 |
(a – 12) | (a + 12) |
96 (a + 12) = 288 (a – 12)
96a + 1152 = 288a – 3456
3456 + 1152 = 288a – 96a
a = 24 km/h
Speed of Boat Q in still water = b km/h
120 | = | 240 |
(b – 8) | (b + 8) |
120 (b + 8) = 240 (b – 8)
120b + 960 = 240b – 1920
1920 + 960 = 240b – 120b
b = 24 km/h
Speed of Boat R in still water = c km/h
100 | = | 220 |
(c – 15) | (c + 15) |
100 (c + 15) = 220 (c – 15)
100c + 1500 = 220c – 3300
3300 + 1500 = 220c – 100c
c = 40 km/h
Speed of Boat S in still water = d km/h
150 | = | 350 |
(d – 10) | (d + 10) |
150 (d + 10) = 350 (d – 10)
150d + 1500 = 350d – 3500
3500 + 1500 = 350d – 150d
d = 25 km/h
Speed of Boat T in still water = e km/h
180 | = | 540 |
(e – 20) | (e + 20) |
180 (e + 20) = 540 (e – 20)
180e + 3600 = 540e – 10800
10800 + 3600 = 540e – 180e
e = 40 km/h
4. The ratio of the speeds of the Boat Q to the Boat U in still water is 4 : 5. If the Boat U travels 126 km distance downstream and 81 km distance upstream in 7 hours 30 minutes, What is the speed of stream of Boat U?
A. 15 km/h
B. 10 km/h
C. 20 km/h
D. 12 km/h
E. None of these
Explanation :
Speed of the Boat Q in still water = 24 km/h
Speed of the Boat U in still water
= | 24 | × 5 = 30 km/h |
4 |
Let the speed of stream = x km/h
According to the question,
126 | + | 81 | = | 15 |
(30 + x) | (30 – x) | 2 |
126 (30 – x) + 81 (30 + x) | = | 15 |
(900 – x^{2}) | 2 |
2 (3780 – 126x + 2430 + 81x) = 15 (900 – x^{2})
2 (6210 – 45x) = 13500 – 15x^{2}
12420 – 90x = 13500 – 15x^{2}
15x^{2} – 90x – 1080 = 0
x^{2} – 6x – 72 = 0
x^{2} – 12x + 6x – 72 = 0
x (x – 12) + 6 (x – 12) = 0
(x + 6) (x – 12) = 0
x = –6, 12
Speed of stream = 12 km/h
Hence, option D is correct.
Common Explanation:
Speed of Boat P in still water = a km/h
96 | = | 288 |
(a – 12) | (a + 12) |
96 (a + 12) = 288 (a – 12)
96a + 1152 = 288a – 3456
3456 + 1152 = 288a – 96a
a = 24 km/h
Speed of Boat Q in still water = b km/h
120 | = | 240 |
(b – 8) | (b + 8) |
120 (b + 8) = 240 (b – 8)
120b + 960 = 240b – 1920
1920 + 960 = 240b – 120b
b = 24 km/h
Speed of Boat R in still water = c km/h
100 | = | 220 |
(c – 15) | (c + 15) |
100 (c + 15) = 220 (c – 15)
100c + 1500 = 220c – 3300
3300 + 1500 = 220c – 100c
c = 40 km/h
Speed of Boat S in still water = d km/h
150 | = | 350 |
(d – 10) | (d + 10) |
150 (d + 10) = 350 (d – 10)
150d + 1500 = 350d – 3500
3500 + 1500 = 350d – 150d
d = 25 km/h
Speed of Boat T in still water = e km/h
180 | = | 540 |
(e – 20) | (e + 20) |
180 (e + 20) = 540 (e – 20)
180e + 3600 = 540e – 10800
10800 + 3600 = 540e – 180e
e = 40 km/h
5. The speed of Boat Q and S in still water together is approximately how much percentage more than the speed of stream of the same boats together?
A. 70%
B. 120%
C. 170%
D. 80%
E. 270%
Explanation :
According to the question,
Speed of Boat Q and S in still water together = (25 + 24) = 49 km/h
Speed of stream of Boat Q and S together = 18 km/h
% more = | 49 – 18 | × 100 |
18 |
= | 31 | × 100 = 172.22% ≈ 170% |
18 |
Hence, option C is correct.
Common Explanation:
Speed of Boat P in still water = a km/h
96 | = | 288 |
(a – 12) | (a + 12) |
96 (a + 12) = 288 (a – 12)
96a + 1152 = 288a – 3456
3456 + 1152 = 288a – 96a
a = 24 km/h
Speed of Boat Q in still water = b km/h
120 | = | 240 |
(b – 8) | (b + 8) |
120 (b + 8) = 240 (b – 8)
120b + 960 = 240b – 1920
1920 + 960 = 240b – 120b
b = 24 km/h
Speed of Boat R in still water = c km/h
100 | = | 220 |
(c – 15) | (c + 15) |
100 (c + 15) = 220 (c – 15)
100c + 1500 = 220c – 3300
3300 + 1500 = 220c – 100c
c = 40 km/h
Speed of Boat S in still water = d km/h
150 | = | 350 |
(d – 10) | (d + 10) |
150 (d + 10) = 350 (d – 10)
150d + 1500 = 350d – 3500
3500 + 1500 = 350d – 150d
d = 25 km/h
Speed of Boat T in still water = e km/h
180 | = | 540 |
(e – 20) | (e + 20) |
180 (e + 20) = 540 (e – 20)
180e + 3600 = 540e – 10800
10800 + 3600 = 540e – 180e
e = 40 km/h
Download free PDF Click Here
- Mixed Pattern Based Data Interpretation
Questions :
B. 2300 × 4060 × 904801
C. 2300 × 4060 × 906751
D. 2300 × 4060 × 926801
E. 2300 × 4060 × 925001
Explanation :
Hence, option C is correct.
B. 20160 × 9!
C. 2520 × 8!
D. 6720 × 9!
E. 2520 × 7!
Explanation :
⇒ Easy questions can be set in | 9! | ways |
(9 – 7)! |
∴ Number of ways in which the test can be made if no two easy questions are set next to each other = 8! × 9!/2! = 20160 × 9!
Hence, option B is correct.
B. 30456
C. 27724
D. 32724
E. 28832
Explanation :
Hence, option A is correct.
B. 69000
C. 75000
D. 135000
E. 360000
Explanation :
moderate questions and at most 4 hard questions.
Hence, option D is correct.
B. 17! 16!
C. 18! 17!
D. 33! 16!
E. 18! 16!
Hence, option E is correct.
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Download High Level Data Interpretation PDF for IBPS PO Click Here.
Download Caselet DI PDF for IBPS PO Pre Click Here.
Problems in High Level Data Interpretation for IBPS PO are probably the closest in resemblance to the kind of problems you will be dealing with regular basis. They test your decision-making ability and speed using the minimum possible data. They help you to draw conclusions from collected data.
Folks, you can practice Different chart Based DI for IBPS PO click on the links given below:
Bar Chart Based DI | Table Chart Based DI | Caselet DI | Mixed Chart Based DI | Pie Chart Based DI
Line Chart Based DI | Set Theory Based DI | Web Chart Based DI
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