Histograms DI – Hard Level – U7 – Data Interpretation – Paper 1

Hard Histogram DI is about 3-step reasoning: overall change, tricky percentages, median class, and “closest value” options. One small reading mistake can change the full answer, so accuracy matters more than speed first. Practice daily one Histogram DI set for 10–15 minutes; once Medium feels easy, shift your focus to Hard.

DI Set 1: Graph DI (Histogram: Daily Online Orders per Store)

Data:

Orders Range (per day)	Number of Stores
0–20	5
21–40	12
41–60	28
61–80	24
81–100	18
101–120	9

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xychart-beta
  title "Histogram: Daily Online Orders per Store"
  x-axis ["0-20","21-40","41-60","61-80","81-100","101-120"]
  y-axis "Stores" 0 --> 30
  bar [5,12,28,24,18,9]

Questions (MCQ):

Q1: What percentage of stores received 61 or more orders (61–80, 81–100, 101–120 combined)?
A) 51.04%;
B) 53.13%;
C) 55.21%;
D) 57.29%;

Q2: What is the estimated average orders per store (using class midpoints)?
A) 63.00;
B) 65.00;
C) 66.33;
D) 68.25;

Q3: By what percent is the number of stores with 61+ orders higher than stores with 60 or less orders?
A) 10.00%;
B) 13.33%;
C) 15.00%;
D) 18.00%;

Q4: The median store (by orders) lies in which orders range?
A) 21–40;
B) 41–60;
C) 61–80;
D) 81–100;

Q5: If 4 stores move from the 41–60 range to the 81–100 range, by how much does the estimated average increase (approx)?
A) 0.83;
B) 1.25;
C) 1.67;
D) 2.00;

Answers and Explanations:

Answer (Q1): B) 53.13%
Explanation:

Stores with 61+ orders = 24 + 18 + 9 = 51.
Total stores = 5+12+28+24+18+9 = 96.
Percentage = (51/96)×100 = 53.125% ≈ 53.13%.

Answer (Q2): C) 66.33
Explanation:

Midpoints: 10, 30.5, 50.5, 70.5, 90.5, 110.5.
Weighted sum = 10×5 + 30.5×12 + 50.5×28 + 70.5×24 + 90.5×18 + 110.5×9 = 6368.
Mean = 6368/96 = 66.33 (approx).

Answer (Q3): B) 13.33%
Explanation:

Stores with 61+ orders = 51; stores with 60 or less = 96 − 51 = 45.
Percent higher = ((51−45)/45)×100.
= (6/45)×100 = 13.33%.

Answer (Q4): C) 61–80
Explanation:

Median position = (96+1)/2 = 48.5th store.
Cumulative stores: 5, 17, 45, 69, 87, 96.
48.5 lies between 45 and 69 → in 61–80.

Answer (Q5): C) 1.67
Explanation:

Midpoint shift: from 50.5 to 90.5 → increase = 40 per store.
Total increase in sum = 4×40 = 160.
Mean increase = 160/96 = 1.666… ≈ 1.67.

DI Set 2: Graph DI (Histogram: Call Handling Time in a Support Center)

Data:

Call Time Range (minutes)	Number of Calls
0–2	14
2–4	26
4–6	30
6–8	22
8–10	10
10–12	6

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xychart-beta
  title "Histogram: Call Handling Time in a Support Center"
  x-axis ["0-2","2-4","4-6","6-8","8-10","10-12"]
  y-axis "Calls" 0 --> 32
  bar [14,26,30,22,10,6]

Questions (MCQ):

Q1: What percentage of calls took more than 6 minutes (6–8, 8–10, 10–12 combined)?
A) 33.33%;
B) 35.19%;
C) 37.04%;
D) 40.00%;

Q2: What is the estimated average call time (in minutes) using class midpoints?
A) 4.91;
B) 5.11;
C) 5.31;
D) 5.51;

Q3: If the team wants to reduce “more than 8 minutes” calls by 40%, how many calls must shift to 8 minutes or less (minimum)?
A) 6;
B) 7;
C) 8;
D) 9;

Q4: What is the ratio of calls in 4–8 minutes to calls in 0–4 minutes?
A) 10:13;
B) 13:10;
C) 5:6;
D) 6:5;

Q5: Estimate the total cost if calls up to 6 minutes are charged at ₹4/min and calls above 6 minutes at ₹6/min (use midpoints).
A) ₹2728;
B) ₹2828;
C) ₹2928;
D) ₹3028;

Answers and Explanations:

Answer (Q1): B) 35.19%
Explanation:

Calls >6 min = 22 + 10 + 6 = 38.
Total calls = 14+26+30+22+10+6 = 108.
Percentage = (38/108)×100 = 35.185…% ≈ 35.19%.

Answer (Q2): B) 5.11
Explanation:

Midpoints: 1, 3, 5, 7, 9, 11.
Weighted sum minutes = 14×1 + 26×3 + 30×5 + 22×7 + 10×9 + 6×11 = 552.
Mean = 552/108 = 5.11 minutes (approx).

Answer (Q3): B) 7
Explanation:

Calls >8 min = 10 + 6 = 16.
40% of 16 = 6.4 calls.
Minimum whole calls to shift = 7.

Answer (Q4): B) 13:10
Explanation:

4–8 min calls = 30 + 22 = 52.
0–4 min calls = 14 + 26 = 40.
Ratio = 52:40 = 13:10.

Answer (Q5): B) ₹2828
Explanation:

Up to 6 min minutes = 14×1 + 26×3 + 30×5 = 242 → cost = 242×4 = 968.
Above 6 min minutes = 22×7 + 10×9 + 6×11 = 310 → cost = 310×6 = 1860.
Total cost = 968 + 1860 = ₹2828.

DI Set 3: Graph DI (Histogram: Daily Steps of Office Employees)

Data:

Steps Range (in thousands)	Number of Employees
0–2	12
2–4	18
4–6	34
6–8	28
8–10	16
10–12	10

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xychart-beta
  title "Histogram: Daily Steps of Office Employees"
  x-axis ["0-2k","2-4k","4-6k","6-8k","8-10k","10-12k"]
  y-axis "Employees" 0 --> 36
  bar [12,18,34,28,16,10]

Questions (MCQ):

Q1: What percentage of employees walked at least 8,000 steps (8–10k and 10–12k combined)?
A) 20.34%;
B) 22.03%;
C) 24.58%;
D) 26.27%;

Q2: What is the estimated average steps (in thousands) using class midpoints?
A) 5.41k;
B) 5.81k;
C) 6.11k;
D) 6.41k;

Q3: The company wants an average of 6.5k steps. How many total extra steps are needed (approx) based on the estimated current mean?
A) 70,948;
B) 80,948;
C) 90,948;
D) 100,948;

Q4: The median employee (by steps) lies in which steps range?
A) 2–4k;
B) 4–6k;
C) 6–8k;
D) 8–10k;

Q5: If 5 employees move from 0–2k to 8–10k, by how many steps does the estimated average increase (approx)?
A) 170;
B) 250;
C) 339;
D) 520;

Answers and Explanations:

Answer (Q1): B) 22.03%
Explanation:

Employees with ≥8k = 16 + 10 = 26.
Total employees = 12+18+34+28+16+10 = 118.
Percentage = (26/118)×100 = 22.033…% ≈ 22.03%.

Answer (Q2): B) 5.81k
Explanation:

Midpoints (k): 1, 3, 5, 7, 9, 11.
Weighted sum (k) = 12×1 + 18×3 + 34×5 + 28×7 + 16×9 + 10×11 = 686k.
Mean = 686/118 = 5.81k (approx).

Answer (Q3): B) 80,948
Explanation:

Target mean = 6.5k; current mean ≈ 5.81k → difference ≈ 0.686k per employee.
Total increase ≈ 0.686k × 118 = 80.948k steps.
≈ 80,948 steps.

Answer (Q4): B) 4–6k
Explanation:

Median position = (118+1)/2 = 59.5th employee.
Cumulative: 12, 30, 64, 92, 108, 118.
59.5 lies between 30 and 64 → in 4–6k.

Answer (Q5): C) 339
Explanation:

Midpoint shift: 1k → 9k, increase = 8k per employee.
Total increase in sum = 5×8k = 40k steps.
Mean increase = 40k/118 = 0.33898k ≈ 339 steps.

DI Set 4: Graph DI (Histogram: Battery Backup Time of Power Banks)

Data:

Backup Time Range (hours)	Number of Power Banks
0–2	6
2–4	14
4–6	26
6–8	30
8–10	18
10–12	10

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xychart-beta
  title "Histogram: Battery Backup Time of Power Banks"
  x-axis ["0-2","2-4","4-6","6-8","8-10","10-12"]
  y-axis "Units" 0 --> 32
  bar [6,14,26,30,18,10]

Questions (MCQ):

Q1: What percentage of power banks last 6 hours or more (6–8, 8–10, 10–12 combined)?
A) 53.85%;
B) 55.77%;
C) 57.69%;
D) 59.62%;

Q2: What is the estimated average backup time (in hours) using class midpoints?
A) 6.05;
B) 6.15;
C) 6.35;
D) 6.55;

Q3: If every unit in 8–12 hours range loses 1 hour due to a defect, what is the new estimated average (approx)?
A) 6.08;
B) 6.18;
C) 6.28;
D) 6.38;

Q4: What is the simplified ratio of units in 4–8 hours range to the total units?
A) 7:13;
B) 13:7;
C) 14:26;
D) 26:14;

Q5: Among units that meet an 8-hour warranty (8–10 and 10–12), what percentage are in the 10–12 range?
A) 31.25%;
B) 35.71%;
C) 40.00%;
D) 44.44%;

Answers and Explanations:

Answer (Q1): B) 55.77%
Explanation:

Units with ≥6 hours = 30 + 18 + 10 = 58.
Total units = 6+14+26+30+18+10 = 104.
Percentage = (58/104)×100 = 55.769…% ≈ 55.77%.

Answer (Q2): C) 6.35
Explanation:

Midpoints: 1, 3, 5, 7, 9, 11.
Weighted sum hours = 6×1 + 14×3 + 26×5 + 30×7 + 18×9 + 10×11 = 660.
Mean = 660/104 = 6.346… ≈ 6.35 hours.

Answer (Q3): A) 6.08
Explanation:

Affected units (8–12) = 18 + 10 = 28; each loses 1 hour.
New sum = 660 − 28 = 632.
New mean = 632/104 = 6.0769… ≈ 6.08 hours.

Answer (Q4): A) 7:13
Explanation:

Units in 4–8 = 26 + 30 = 56.
Ratio to total = 56:104.
Simplify by 8 → 7:13.

Answer (Q5): B) 35.71%
Explanation:

Warranty-meeting units (≥8) = 18 + 10 = 28.
In 10–12 range = 10.
Percentage = (10/28)×100 = 35.714…% ≈ 35.71%.

DI Set 5: Graph DI (Histogram: Scores in UGC NET Mini Test)

Data:

Score Range (out of 200)	Number of Candidates
0–40	9
40–80	21
80–120	38
120–160	26
160–200	14

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xychart-beta
  title "Histogram: Scores in UGC NET Mini Test"
  x-axis ["0-40","40-80","80-120","120-160","160-200"]
  y-axis "Candidates" 0 --> 40
  bar [9,21,38,26,14]

Questions (MCQ):

Q1: What percentage of candidates scored 120 or more (120–160 and 160–200 combined)?
A) 33.33%;
B) 35.19%;
C) 37.04%;
D) 38.89%;

Q2: What is the estimated average score using class midpoints?
A) 101.56;
B) 105.56;
C) 109.56;
D) 113.56;

Q3: By what percent is the number of candidates scoring 80+ higher than those scoring below 80?
A) 140%;
B) 150%;
C) 160%;
D) 170%;

Q4: The median candidate (by score) lies in which score range?
A) 40–80;
B) 80–120;
C) 120–160;
D) 160–200;

Q5: If 6 candidates move from 40–80 to 120–160, by how much does the estimated average score increase (approx)?
A) 3.33;
B) 4.44;
C) 5.56;
D) 6.67;

Answers and Explanations:

Answer (Q1): C) 37.04%
Explanation:

Candidates with ≥120 = 26 + 14 = 40.
Total candidates = 9+21+38+26+14 = 108.
Percentage = (40/108)×100 = 37.037…% ≈ 37.04%.

Answer (Q2): B) 105.56
Explanation:

Midpoints: 20, 60, 100, 140, 180.
Weighted sum = 9×20 + 21×60 + 38×100 + 26×140 + 14×180 = 11400.
Mean = 11400/108 = 105.555… ≈ 105.56.

Answer (Q3): C) 160%
Explanation:

Below 80 = 9 + 21 = 30.
80+ = 38 + 26 + 14 = 78.
Percent higher = ((78−30)/30)×100 = (48/30)×100 = 160%.

Answer (Q4): B) 80–120
Explanation:

Median position = (108+1)/2 = 54.5th candidate.
Cumulative: 9, 30, 68, 94, 108.
54.5 lies between 30 and 68 → in 80–120.

Answer (Q5): B) 4.44
Explanation:

Midpoint shift: 60 → 140, increase = 80 per candidate.
Total increase in sum = 6×80 = 480.
Mean increase = 480/108 = 4.444… ≈ 4.44.

This Tutorial is Related to: Advanced Data Interpretation, Di Multi-Step Questions, Exam Trick Questions, Histogram Di Hard Level, Median In Histogram, Percentage Change Di, Tough Di Practice, Ugc Net Di Hard, Ugc Net Paper 1 Di, Unit 7 Di Preparation

Table of Contents

DI Set 1: Graph DI (Histogram: Daily Online Orders per Store)

DI Set 2: Graph DI (Histogram: Call Handling Time in a Support Center)

DI Set 3: Graph DI (Histogram: Daily Steps of Office Employees)

DI Set 4: Graph DI (Histogram: Battery Backup Time of Power Banks)

DI Set 5: Graph DI (Histogram: Scores in UGC NET Mini Test)