Averages and charts · applications

Applications: Averages and Charts

10 question types · Model Method and algebra, side by side

PSLE · O-Level · SAT · GCSE Higher

01

The Incomplete / Smudged Bar Chart (Average Deduction)

heuristicAggregate Total Inversion / Value Reconstruction from Mean

A bar chart showed the daily sales of fruit smoothies at a café from Monday to Friday. The bars for Monday, Tuesday, Wednesday, and Thursday recorded sales of 65, 75, 50, and 80 smoothies respectively. The bar for Friday was accidentally smudged and could not be read directly. The café manager noted that the average number of smoothies sold per day from Monday to Friday was 74. (a) How many fruit smoothies were sold on Friday? (b) What was the percentage increase in the number of smoothies sold on Friday compared to Thursday?

Mon65Tue75Wed50Thu80Fri?smudgedAverage7474747474370
Five days at the average of 74 make the total: 370.
5 equal average blocks of 74: Total = 5 × 74 = 370.
step 1 of 7

Represent the 5 days as 5 bars of 74 units each. Sum the 4 known bars and find the difference from the total 370-unit bar.

  1. 5 equal average blocks of 74: Total = 5 × 74 = 370.
  2. Known sales (Mon–Thu): 65 + 75 + 50 + 80 = 270.
  3. Friday's bar = 370 − 270 = 100.
  4. (a) Friday sold 100 smoothies.
  5. Thursday = 80, Friday = 100. Increase = 20.
  6. Compare increase to Thursday's base: 2080 = 14 = 25%.
  7. (b) 25% increase.

answer(a) 100 smoothies; (b) 25%

techniqueThe Mean · Reading Bar Charts

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 270 by 4 (average of 4 days = 67.5) and assuming Friday is 67.5.
  • Calculating percentage increase by dividing by Friday's sales (20 ÷ 100 = 20%) instead of Thursday's base of 80.
02

Unmarked Vertical Axis Scale (Grid Line Intervals)

heuristicGrid Interval Scaling / Difference-to-Unit Conversion

A bar chart displayed the number of visitors to a science museum from Thursday to Sunday. The numerical values on the vertical axis were omitted during printing, but the horizontal grid lines are evenly spaced. The bars for Thursday, Friday, Saturday, and Sunday stood at heights of 3, 5, 9, and 7 grid intervals respectively. Records showed that there were 140 more visitors on Saturday than on Friday. (a) How many visitors does each grid line interval represent? (b) What was the total number of visitors to the museum from Thursday to Sunday?

Thu???Fri?????Sat?????????4 more intervalsSun???????
The axis has no numbers, but Saturday stands 4 intervals above Friday.
Saturday (9 units) has 4 more units than Friday (5 units).
step 1 of 5

Draw unit bars representing each day: Saturday has 9 units and Friday has 5 units. The 4-unit gap equals 140.

  1. Saturday (9 units) has 4 more units than Friday (5 units).
  2. 4 units = 140 ⟹ 1 unit = 140 ÷ 4 = 35 visitors.
  3. (a) 1 grid interval represents 35 visitors.
  4. Total units: 3 + 5 + 9 + 7 = 24 units.
  5. Total visitors: 24 × 35 = 840 visitors.

answer(a) 35 visitors; (b) 840 visitors

techniqueReading Bar Charts · Bar Charts

examsPSLE · O-Level · SAT

Common pitfalls

  • Dividing 140 by 9 or by 5 rather than the difference in intervals (9 − 5 = 4).
  • Forgetting to include Thursday's 3 intervals when calculating total visitors.
03

Change in Average (Member Joining or Leaving)

heuristicSurplus Redistribution / Balancing Around the Mean

A group of students sat for a mathematics assessment. The average score of the group was 76 marks. When another student, who scored 94 marks, joined the group, the new average score of all the students became 78 marks. (a) How many students were in the group at first? (b) If another student who scored 60 marks then joined the 9 students, what would the new average score of the 10 students become?

Each76Newcomer7894
The new average is 78.
New average is 78.
step 1 of 8

The newcomer brings 94 marks. Giving 78 marks to himself leaves a surplus of 16 marks to elevate each of the original students from 76 to 78 (+2 each).

  1. New average is 78.
  2. Newcomer's surplus above the new average: 94 − 78 = 16 marks.
  3. Each of the original students received an increase of: 78 − 76 = 2 marks.
  4. Number of original students: 16 ÷ 2 = 8 students.
  5. (a) 8 students at first.
  6. Total score of 9 students = 9 × 78 = 702.
  7. With 10th student: 702 + 60 = 762.
  8. New average: 762 ÷ 10 = 76.2.

answer(a) 8 students; (b) 76.2 marks

techniqueThe Mean

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 94 by 2 instead of subtracting the new average (78) first to find the surplus.
  • Reporting 9 students as the answer to part (a) instead of the initial count of 8.
04

Two-Scenario Change in Average

heuristicDual-Condition Comparison / Elimination of Invariant Total

A group of children took a coding test. If a child scoring 90 marks was added to the group, the average score of the group would become 82 marks. If a child scoring 66 marks was added to the group instead, the average score would become 78 marks. (a) How many children were in the group originally? (b) What was the original total score of the group?

With 90828282828282With 667878787878784 more marks each
The two newcomers differ by 90 − 66 = 24 marks.
Difference in newcomer's score: 90 − 66 = 24 marks.
step 1 of 7

Compare the difference in hypothetical scores (90 - 66 = 24) to the difference in averages (82 - 78 = 4) across the new group of (n + 1).

  1. Difference in newcomer's score: 90 − 66 = 24 marks.
  2. Difference in resulting average: 82 − 78 = 4 marks per person.
  3. New group size: 24 ÷ 4 = 6 children.
  4. Original group size: 6 − 1 = 5 children.
  5. (a) 5 children originally.
  6. Total score of 6 children in Scenario 1: 6 × 82 = 492.
  7. Subtract newcomer: 492 − 90 = 402 marks.

answer(a) 5 children; (b) 402 marks

techniqueThe Mean · Simultaneous by Elimination

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Forgetting that 6 is the new group size (n + 1) and reporting 6 as the original number of children.
  • Subtracting 66 from 492 instead of matching Scenario 1's score with Scenario 1's average.
05

Balancing Deviations Around an Assumed Mean

heuristicZero-Sum Deviation Property / Assumed Mean Balancing

Seven students took a science assessment. Six of their scores were 74, 82, 68, 91, 79, and 85. The score of the 7th student, Ryan, was withheld, but the overall average score of all 7 students was announced as 80 marks. (a) What was the sum of the deviations of the six known scores from the average of 80? (b) What was Ryan's score on the assessment?

74−682+268−1291+1179−185+5
Below 80: 74, 68 and 79 fall short by 6, 12 and 1: 19 in all.
Deficits below 80: 74 needs 6, 68 needs 12, 79 needs 1. Total deficit = 6 + 12 + 1 = 19.
step 1 of 6

Balance surplus marks against deficit marks relative to an 80-mark baseline.

  1. Deficits below 80: 74 needs 6, 68 needs 12, 79 needs 1. Total deficit = 6 + 12 + 1 = 19.
  2. Surpluses above 80: 82 has 2, 91 has 11, 85 has 5. Total surplus = 2 + 11 + 5 = 18.
  3. Net shortfall: 18 − 19 = −1 mark.
  4. (a) −1.
  5. To bring the entire group's average to 80, Ryan must provide his own 80 marks plus 1 additional mark to cover the net deficit.
  6. Ryan's score: 80 + 1 = 81.

answer(a) −1; (b) 81 marks

techniqueThe Mean

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting 1 from 80 (79) instead of adding 1 to compensate for the negative deviation.
  • Summing large numbers manually (74+82+68+91+79+85 = 479) and making an arithmetic error.
06

Multi-Group Weighted Average

heuristicAggregate Subtotal Fusion / Weighted Mean Proportionality

Class 6A has 40 students with an average score of 75 in an English exam. Class 6B has 30 students with an average score of 82 in the same exam. (a) What was the combined average score of all the students from both classes? (b) If 4 students with an average score of 86 transferred from Class 6B into Class 6A, what was the new average score of Class 6A?

6A757575754 parts6B8282823 parts
40 : 30 is 4 : 3, seven parts.
Student count ratio: 40 : 30 = 4 : 3. Total = 7 parts.
step 1 of 6

Represent Class 6A as 4 units of students and Class 6B as 3 units of students (ratio 4 : 3).

  1. Student count ratio: 40 : 30 = 4 : 3. Total = 7 parts.
  2. Weighted total: (4 × 75) + (3 × 82) = 300 + 246 = 546.
  3. Average: 546 ÷ 7 = 78 marks.
  4. (a) 78 marks.
  5. Part (b): Initial Class 6A marks = 3000. Adding 4 students at 86 marks each adds 4 × 86 = 344 marks.
  6. New total = 3344. Divide by 44 students: 3344 ÷ 44 = 76 marks.

answer(a) 78 marks; (b) 76 marks

techniqueThe Mean · Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Averaging the two averages directly: 75 + 822 = 78.5, ignoring that Class 6A has more students than Class 6B.
  • Dividing 3344 by 40 instead of 44 after the 4 students joined Class 6A.
07

Pie Chart with Angle, Percentage, and Count Hybrids

heuristicAngle-Percentage-Unit Unification / Circular Fraction Allocation

A pie chart represents the enrollment of 360 students across four school activities: Sports, Arts, Robotics, and Uniformed Groups. The sector for Sports has an angle of 120°. The sector for Arts represents 25% of the total students. There are 45 students in Robotics, and the remaining students are in Uniformed Groups. (a) What was the sector angle representing Uniformed Groups on the pie chart? (b) How many students were in Uniformed Groups?

SportsArtsRoboticsUniformed1 student = 1°
360 students share 360°, so one student is one degree.
Total circle = 360° = 360 students. Therefore, 1° = 1 student.
step 1 of 8

Since total students is 360 and total circle degrees is 360°, each 1° corresponds exactly to 1 student.

  1. Total circle = 360° = 360 students. Therefore, 1° = 1 student.
  2. Sports: 120° = 120 students.
  3. Arts: 25% = 14 × 360 = 90 students = 90°.
  4. Robotics: 45 students = 45°.
  5. Uniformed Groups students: 360 − (120 + 90 + 45) = 360 − 255 = 105 students.
  6. (b) 105 students.
  7. Since 1 student = 1°, angle is 105°.
  8. (a) 105°.

answer(a) 105°; (b) 105 students

techniquePie Charts · Finding a Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Adding percentage (25) directly to degrees (120) without converting 25% to 90°.
  • Assuming Uniformed Groups is 100° because 360 minus 255 is miscalculated.
08

Pie Chart with Fractional Remainder & Ratios

heuristicHierarchical Branching in Circular Representation

A pie chart tracks Keith's monthly salary allocation: 13 of his salary was saved. Of the remainder, 25 was spent on food. The remaining money was divided between rent and entertainment in the ratio 3 : 1. Keith spent $450 on entertainment. (a) How much was Keith's total monthly salary? (b) What fraction of his salary was spent on food?

Salarysaved 5remainder 1015 units
Fifteen units, since thirds and fifths both fit.
Let total salary be 15 units (LCM of 3 and 5).
step 1 of 10

Draw a model: 15 base units. 5 units saved (1/3), leaving 10 units. Food takes 2/5 of 10 = 4 units. Remaining 6 units split into rent and entertainment.

  1. Let total salary be 15 units (LCM of 3 and 5).
  2. Savings = 13 × 15 = 5 units. Remainder = 10 units.
  3. Food = 25 × 10 = 4 units.
  4. (b) Food is 4 units out of 15 units = 415.
  5. Remaining for rent & entertainment = 10 − 4 = 6 units.
  6. Subdivide into ratio 3 : 1 (4 parts): each part = 6 ÷ 4 = 1.5 units.
  7. Entertainment = 1.5 units = $450.
  8. 1 unit = 450 ÷ 1.5 = $300.
  9. Total salary (15 units) = 15 × 300 = $4500.
  10. (a) $4500.

answer(a) $4500; (b) 415

techniquePie Charts · A Fraction of an Amount

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Calculating food as 2/5 of the total salary (6/15) instead of 2/5 of the remainder (4/15).
  • Assuming entertainment is 1/4 of total salary.
09

Frequency Distribution Table with Hidden Frequencies

heuristicSimultaneous Equations on Discrete Frequency Distributions

A teacher surveyed 40 students on the number of books they read over the holidays. The table recorded: 4 students read 0 books; 12 students read 1 book; k students read 2 books; 8 students read 3 books; and m students read 4 books. The average number of books read across the 40 students was 2.05. (a) How many students read 2 books? (b) How many students read 4 books?

0 books41 book122 booksk3 books84 booksmk + m = 16
24 students are known; k + m = 16.
Known students = 4 + 12 + 8 = 24. Leftover students for 2 and 4 books = 40 − 24 = 16 students.
step 1 of 9

Use supposition: 16 students read either 2 or 4 books, totaling 46 books.

  1. Known students = 4 + 12 + 8 = 24. Leftover students for 2 and 4 books = 40 − 24 = 16 students.
  2. Total books = 40 × 2.05 = 82. Known books = 12 + 24 = 36. Leftover books = 82 − 36 = 46.
  3. Assume all 16 students read 2 books: Total = 16 × 2 = 32 books.
  4. Shortfall: 46 − 32 = 14 books.
  5. Difference between 4 books and 2 books: 4 − 2 = 2 books per student.
  6. Number of students who read 4 books (m): 14 ÷ 2 = 7 students.
  7. (b) m = 7.
  8. Number of students who read 2 books (k): 16 − 7 = 9 students.
  9. (a) k = 9.

answer(a) 9 students; (b) 7 students

techniqueTally and Frequency Tables · Simultaneous by Elimination

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Forgetting to multiply the books by their frequency (adding 2 + 4 = 6 books instead of 2k + 4m).
  • Including the 0-book students in the books sum as 4 books instead of 0.
10

Supposition Applied to Group Score Boundaries

heuristicExtreme Value Logic / Pigeonhole Score Minimization

Five students, Anna, Brian, Colin, David, and Emma, took a mathematics contest. Each student obtained a distinct integer score between 0 and 100. The average score of the five students was 84 marks. Anna obtained the strictly highest score in the group. Colin scored 82 marks. (a) What was the total score of all five students? (b) What was the lowest possible mark Anna could have scored?

Colin82
(a) Five at an average of 84: 420 in all.
Total = 5 × 84 = 420.
step 1 of 7

Distribute 420 marks into 5 staircase blocks that are as flat as possible, with Colin anchored at 82.

  1. Total = 5 × 84 = 420.
  2. Colin has 82. To keep Anna as low as possible, everyone else must be pushed as high as possible.
  3. Make Colin the lowest score: Colin = 82.
  4. The scores above Colin must be at least 83, 84, 85, 86.
  5. Sum: 82 + 83 + 84 + 85 + 86 = 420.
  6. The sum matches the required 420 perfectly.
  7. (b) Anna's lowest possible mark is 86.

answer(a) 420 marks; (b) 86 marks

techniqueThe Mean · Multiplying and Dividing Undo Each Other

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Assuming Anna's minimum score is simply the average of 84 plus 1 (85), ignoring the requirement that all 5 scores must be distinct integers and sum to 420.
  • Allowing students to have duplicate scores (e.g. 84, 84, 84).
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