Reading numerical charts — SSC CGL Quant Notes

Concepts and rules

A frequency distribution groups data into class intervals such as 10–20, 20–30. In SSC charts 10–20 means 10 ≤ x < 20: a value of 20 goes into the class 20–30.

Class width (size) = upper limit − lower limit. Class mark (midpoint) = (lower limit + upper limit)/2; for 20–30 it is 25.

Histogram: class intervals on the x-axis, frequencies on the y-axis, bars touching each other with no gaps. The area of a bar is proportional to the frequency of its class.

Unequal class widths: plot adjusted height = frequency × (standard width ÷ class width), also called frequency density. A class twice as wide gets half the height for the same frequency.

Frequency polygon: plot (class mark, frequency) for every class and join the points with straight lines. Close it by adding a class of zero frequency at each end, at the class marks just before the first and after the last class.

With equal class widths, the area under the closed frequency polygon equals the total area of the histogram.

Cumulative frequency (less than type) of a class = sum of frequencies up to and including that class. The last cumulative frequency equals N, the total.

Mean of grouped data = Σ(f × x) ÷ Σf, where x is the class mark. This assumes values in each class are centred at the class mark.

Median = L + ((N/2 − cf) ÷ f) × h, where L = lower limit of the median class (first class whose cumulative frequency reaches N/2), cf = cumulative frequency before it, f = its frequency, h = its width.

Mode = L + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, f₀ is the frequency of the class before it and f₂ of the class after it.

Method and shortcuts

Step 1: Copy the chart into a table: class, frequency, and if needed class mark, f × x and cumulative frequency. Check that the frequencies add up to the stated total.

Step 2: Read the question's range carefully: 'at least 40' means classes from 40 upwards; 'less than 30' means classes ending at 30.

Step 3: For a percentage, divide the count by N, not by the number of classes. For a ratio, simplify only at the end.

Shortcut for the mean: take an assumed mean A (a central class mark) and use mean = A + Σf(x − A)/Σf. The deviations are small multiples of h, so arithmetic is quick.

Shortcut: on a frequency polygon with equal widths, the steepest rising segment is between the 2 adjacent classes with the largest increase in frequency.

Example 1: reading a histogram

Data for Examples 1 to 4 (histogram, marks of 50 students): 0–10 – 4, 10–20 – 6, 20–30 – 12, 30–40 – 15, 40–50 – 8, 50–60 – 5.

Question: What percentage of students scored 40 marks or more?

Solution: Classes 40–50 and 50–60: 8 + 5 = 13 students. Percentage = 13/50 × 100 = 26%.

Answer: 26%

Example 2: mean from class marks

Question: Find the mean marks of the 50 students.

Solution: Class marks 5, 15, 25, 35, 45, 55. Σfx = 4 × 5 + 6 × 15 + 12 × 25 + 15 × 35 + 8 × 45 + 5 × 55 = 20 + 90 + 300 + 525 + 360 + 275 = 1570.

Mean = 1570 ÷ 50 = 31.4 marks.

Answer: 31.4

Example 3: points of the frequency polygon

Question: List the points of the closed frequency polygon and name the segment with the steepest rise.

Solution: Points: (−5, 0), (5, 4), (15, 6), (25, 12), (35, 15), (45, 8), (55, 5), (65, 0). The end points come from imaginary classes −10–0 and 60–70 with frequency 0.

Rises: 4, 2, 6, 3, then falls. The steepest rise (6) is from (15, 6) to (25, 12).

Answer: from (15, 6) to (25, 12)

Example 4: median class and median

Question: Find the median marks. (a) 31.4 (b) 35 (c) 32 (d) 30

Solution: N/2 = 25. Cumulative frequencies: 4, 10, 22, 37, 45, 50. The first to reach 25 is 37, so the median class is 30–40 with cf = 22 before it and f = 15.

Median = 30 + (25 − 22)/15 × 10 = 30 + 2 = 32. Option (a) is the mean, not the median.

Answer: (c) 32

Example 5: histogram with unequal classes

Question: Daily wages (₹) of 60 workers: 100–150 – 10, 150–200 – 15, 200–300 – 20, 300–500 – 10, 500–600 – 5. Taking ₹50 as the standard width, what is the adjusted height of the bar for 300–500? (a) 2.5 (b) 5 (c) 10 (d) 20

Solution: Width of 300–500 is 200 = 4 standard widths. Adjusted height = 10 × 50/200 = 2.5.

All adjusted heights: 10, 15, 10, 2.5, 5. So the tallest bar is 150–200, even though 200–300 has the largest frequency 20. Option (c) is the unadjusted frequency.

Answer: (a) 2.5

Example 6: missing frequency from the mean

Question: The mean of the distribution 0–10 – 5, 10–20 – x, 20–30 – 10, 30–40 – 6, 40–50 – 4 is 25. Find x. (a) 6 (b) 5 (c) 8 (d) 4

Solution: Σf = 25 + x. Σfx = 5 × 5 + 15x + 10 × 25 + 6 × 35 + 4 × 45 = 25 + 15x + 250 + 210 + 180 = 665 + 15x.

(665 + 15x)/(25 + x) = 25 gives 665 + 15x = 625 + 25x, so 10x = 40 and x = 4. Check: 725 ÷ 29 = 25.

Answer: (d) 4

Common mistakes

Reading bar heights as frequencies when class widths differ. With unequal widths, frequency = height × (class width ÷ standard width).

Plotting the frequency polygon at class limits instead of class marks.

Putting a boundary value in the lower class: in 10–20, 20–30, a score of 20 belongs to 20–30.

Using the class with the highest cumulative frequency as the median class. The median class is the first one whose cumulative frequency reaches N/2.

Dividing Σfx by the number of classes instead of by Σf when finding the mean.

Practice questions

Q1. For the marks data of Examples 1 to 4, find the mode. (a) 35 (b) 33 (c) 32 (d) 31.4

Q2. Ages (years) of 40 people: 20–30 – 6, 30–40 – 10, 40–50 – 14, 50–60 – 7, 60–70 – 3. What percentage of them are below 40 years? (a) 40% (b) 25% (c) 35% (d) 60%

Q3. For the data in Q2, find the ratio of the number of people in the modal class to the number aged 50 years or more. (a) 7 : 3 (b) 2 : 1 (c) 5 : 7 (d) 7 : 5

Answers

Q1. (b) Modal class 30–40 with f₁ = 15, f₀ = 12, f₂ = 8. Mode = 30 + (15 − 12)/(30 − 12 − 8) × 10 = 30 + 3 = 33.

Q2. (a) Below 40: 6 + 10 = 16. 16/40 × 100 = 40%.

Q3. (d) Modal class 40–50 has 14; aged 50 or more = 7 + 3 = 10. Ratio 14 : 10 = 7 : 5.

Quick revision

Class mark = (lower + upper)/2; histogram bars touch; area ∝ frequency.

Unequal widths: adjusted height = f × standard width ÷ class width.

Polygon: join (class mark, f); close with zero-frequency classes at both ends.

Mean Σfx/Σf; median L + ((N/2 − cf)/f)h; mode L + ((f₁ − f₀)/(2f₁ − f₀ − f₂))h.

Concepts and rules

A simple bar chart shows 1 quantity across categories or years; a multiple (grouped) bar chart shows 2 or more quantities side by side; a stacked bar shows parts of a total in 1 bar.

Percentage change = (new − old) ÷ old × 100. Always divide by the earlier (base) value. A fall from 50 to 40 is −20%, but a rise from 40 to 50 is +25%.

Average over n years = sum ÷ n. 'X is what percent of Y' = X ÷ Y × 100. 'X is what percent more than Y' = (X − Y) ÷ Y × 100.

Profit = income (sales) − expenditure. Profit percent = profit ÷ expenditure × 100, unless the question defines it on sales.

Pie chart: the whole circle is 360° = 100%. Sector angle = value ÷ total × 360°. Percentage = angle ÷ 3.6. Value = total × angle ÷ 360.

Useful pairs: 10% = 36°, 25% = 90°, 30% = 108°, 15% = 54°, 20% = 72°, 12.5% = 45°.

Ratio of 2 sectors in the same pie = ratio of their angles (or percentages); the total is not needed.

Across 2 pie charts with different totals, a bigger angle does not mean a bigger value. Convert each to its actual value first.

Largest percentage change and largest absolute change can happen in different years, because the base keeps changing.

Method and shortcuts

Step 1: Write the chart values in a small table with units (₹ crore, thousands, degrees). Check the pie angles add to 360°.

Step 2: Identify the base in every comparison word: 'over the year before', 'of sales', 'than expenditure'.

Step 3: Compute only what the options need. If options are far apart, approximate; if close, calculate exactly.

Shortcut: to compare ratios like profit/expenditure across years, compare 1 + profit/expenditure = sales/expenditure instead; the order is the same.

Shortcut: difference between 2 sectors in value = total × (difference of angles) ÷ 360, so you need only 1 multiplication.

Example 1: percentage change from a bar chart

Data for Examples 1, 2, 4 and 6 (multiple bar chart, ₹ crore): Sales: 2019 – 40, 2020 – 56, 2021 – 50, 2022 – 65, 2023 – 85. Expenditure: 2019 – 30, 2020 – 40, 2021 – 45, 2022 – 50, 2023 – 60.

Question: Find the percentage increase in sales from 2021 to 2022.

Solution: (65 − 50) ÷ 50 × 100 = 15 ÷ 50 × 100 = 30%.

Answer: 30%

Example 2: average and years above it

Question: Find the average annual sales and the years in which sales were above this average.

Solution: Total = 40 + 56 + 50 + 65 + 85 = 296. Average = 296 ÷ 5 = ₹59.2 crore. Above average: 2022 (65) and 2023 (85).

Answer: ₹59.2 crore; 2022 and 2023

Example 3: reading a pie chart

Data (pie chart, monthly spending of a family, total ₹60,000): Food – 108°, Rent – 72°, Education – 54°, Transport – 36°, Savings – 90°.

Question: How much is spent on Education, and what percent of the total goes on Food?

Solution: Education = 60000 × 54/360 = ₹9,000. Food = 108 ÷ 3.6 = 30%. Check: 108 + 72 + 54 + 36 + 90 = 360°.

Answer: ₹9,000; 30%

Example 4: highest profit percent

Question: In which year was the profit percent (profit ÷ expenditure × 100) the highest? (a) 2020 (b) 2022 (c) 2019 (d) 2023

Solution: Profits: 10, 16, 5, 15, 25. Profit percents: 2019 – 33.3%, 2020 – 40%, 2021 – 11.1%, 2022 – 30%, 2023 – 41.7%.

2023 is highest, just above 2020. Shortcut: sales/expenditure is 85/60 = 1.417 against 56/40 = 1.4.

Answer: (d) 2023

Example 5: comparing 2 pie charts

Question: Sports chosen by students. School X (1200 students): Cricket – 120°, Football – 90°, Hockey – 60°, Others – 90°. School Y (1500 students): Cricket – 108°, Football – 72°, Hockey – 72°, Others – 108°. By how many does the number of cricket players in Y exceed that in X? (a) 40 (b) 50 (c) 0 (d) 60

Solution: X cricket = 1200 × 120/360 = 400. Y cricket = 1500 × 108/360 = 450. Difference = 50, although Y's angle is smaller.

Similarly football: X = 1200 × 90/360 = 300 and Y = 1500 × 72/360 = 300, equal despite different angles.

Answer: (b) 50

Example 6: percentage change versus absolute change

Question: In which year was the percentage increase in sales over the year before the highest? (a) 2023 (b) 2022 (c) 2020 (d) 2021

Solution: Changes: 2020: +16 on 40 = 40%; 2021: −6 on 56 = −10.7%; 2022: +15 on 50 = 30%; 2023: +20 on 65 = 30.8%.

The largest absolute rise (+20) is in 2023, but the largest percentage rise (40%) is in 2020 because its base 40 is the smallest.

Answer: (c) 2020

Common mistakes

Dividing by the new value in a percentage change. The base is always the earlier value.

Comparing angles across 2 pie charts with different totals. Convert to actual values first.

Treating the largest absolute increase as the largest percentage increase.

Mixing 'what percent of' with 'what percent more than': 60 is 150% of 40 but 50% more than 40.

Ignoring units such as ₹ crore or thousands in the chart title, which changes the final answer by a large factor.

Practice questions

Q1. In the family pie chart of Example 3, how much more is saved than spent on Transport? (a) ₹6,000 (b) ₹12,000 (c) ₹9,000 (d) ₹15,000

Q2. From the bar chart of Example 1, expenditure in 2021 is what percent of sales in 2021? (a) 90% (b) 85% (c) 111.1% (d) 80%

Q3. In the family pie chart of Example 3, what percent of the total is spent on Food and Education together? (a) 40% (b) 50% (c) 54% (d) 45%

Answers

Q1. (c) 60000 × (90 − 36)/360 = 60000 × 54/360 = ₹9,000.

Q2. (a) 45 ÷ 50 × 100 = 90%. (111.1% is sales as a percent of expenditure.)

Q3. (d) (108 + 54) ÷ 3.6 = 162 ÷ 3.6 = 45%.

Quick revision

Percentage change = (new − old)/old × 100; base = earlier value.

Pie: 360° = 100%, 1% = 3.6°, value = total × angle/360.

Same pie: ratio of angles = ratio of values. Different pies: convert to values first.

Profit percent on expenditure; compare sales/expenditure to rank years quickly.