Averages and charts · applications

Applications: Averages and Charts

12 problems

PSLE · O-Level · SAT · GCSE Higher

01

Friday's Smudged Bar on a Café's Smoothie Chart

heuristicThe average times the number of days is the total; the missing day is what the known days leave of it

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.
Five days at an average of 74 smoothies make 5 × 74 = 370 smoothies in all: draw five equal bars of 74 beside the chart.
step 1 of 7

An average of 74 over five days means the five days together sold as much as five days of 74 each. Draw that total beside the chart: Friday is the part of it that the other four days do not fill.

All steps
  1. Five days at an average of 74 smoothies make 5 × 74 = 370 smoothies in all: draw five equal bars of 74 beside the chart.
  2. Monday to Thursday sold 65 + 75 + 50 + 80 = 270 smoothies.
  3. Friday's bar is what is left of the 370: 370 − 270 = 100 smoothies.
  4. (a) 100 smoothies were sold on Friday. Check: 270 + 100 = 370, and 370 ÷ 5 = 74.
  5. Friday sold 100 − 80 = 20 more smoothies than Thursday.
  6. A percentage increase compares the increase with the earlier day, so measure the 20 against Thursday's 80: 2080 = 14.
  7. (b) That is 14 × 100% = 25% more smoothies on Friday than on Thursday.

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

techniqueThe Mean · Reading Bar Charts

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Averaging the four days you can read, 270 ÷ 4 = 67.5, and giving that as Friday. Friday is whatever makes the five-day total 370.
  • Dividing the increase by Friday's sales, 20 ÷ 100 = 20%. A percentage increase is measured against the starting value, Thursday's 80.
02

A Museum's Visitor Chart with the Axis Numbers Missing

heuristicA difference in grid intervals matches a difference in visitors, which gives the value of one interval

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.
The grid lines are evenly spaced, so the bars can be compared in intervals. Saturday's bar reaches 9 intervals and Friday's 5: Saturday is 4 intervals higher.
step 1 of 5

Without numbers on the axis, the bars can still be measured in grid intervals. The one fact given in visitors, Saturday's 140 more than Friday, is a difference, so match it to the difference in intervals.

All steps
  1. The grid lines are evenly spaced, so the bars can be compared in intervals. Saturday's bar reaches 9 intervals and Friday's 5: Saturday is 4 intervals higher.
  2. Saturday had 140 more visitors than Friday, so those 4 intervals stand for 140 visitors.
  3. (a) One interval is 140 ÷ 4 = 35 visitors.
  4. Together the four bars reach 3 + 5 + 9 + 7 = 24 intervals.
  5. (b) The museum had 24 × 35 = 840 visitors from Thursday to Sunday.

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

techniqueReading Bar Charts · Bar Charts

examsPSLE · O-Level · SAT

Common pitfalls

  • Dividing 140 by Saturday's 9 intervals, or by Friday's 5. The 140 is the difference between the two days, so it matches the difference in intervals, 9 − 5 = 4.
  • Leaving out Thursday's 3 intervals from the total. All four bars count: 24 intervals.
03

A 94 Joins a Group with an Average of 76

heuristicThe newcomer's marks above the new average are shared out, lifting everyone else to it

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.
After the newcomer joins, every student, the newcomer included, is counted at the new average of 78.
step 1 of 8

Measure the newcomer's score against the NEW average. The marks above it are exactly what is needed to raise each of the original students from the old average to the new one.

All steps
  1. After the newcomer joins, every student, the newcomer included, is counted at the new average of 78.
  2. The newcomer scored 94, which is 94 − 78 = 16 marks above that average: 16 marks to share among the students who were there before.
  3. Those 16 marks lift each of the original students from the old average of 76 to the new one of 78: 2 marks each.
  4. So the 16 marks are shared out 2 at a time: 16 ÷ 2 = 8 students.
  5. (a) There were 8 students at first. Check: 8 × 76 + 94 = 702, and 702 ÷ 9 = 78.
  6. Now 9 students have an average of 78: 9 × 78 = 702 marks.
  7. A tenth student who scored 60 brings the total to 702 + 60 = 762 marks.
  8. (b) The new average is 762 ÷ 10 = 76.2 marks.

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

techniqueThe Mean

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the newcomer's whole 94 by 2. Only the 16 marks above the new average are shared out; the other 78 are the newcomer's own share.
  • Answering (a) with 9, the number of students after the newcomer joined. The question asks how many there were at first.
04

Two Ways a Newcomer Could Change a Group's Average

heuristicThe original group's total is the same in both cases, so the newcomers' difference is the whole difference between the new totals

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.
The two cases differ only in the newcomer: one scores 90, the other 66, which is 90 − 66 = 24 marks more.
step 1 of 7

In both cases the original children and their total are the same; only the newcomer changes. So the 24 marks between the two newcomers account for the whole difference between the two new totals.

All steps
  1. The two cases differ only in the newcomer: one scores 90, the other 66, which is 90 − 66 = 24 marks more.
  2. The averages differ by 82 − 78 = 4 marks, and that difference is shared by everyone in the new group.
  3. So the 24 extra marks raise each person's share by 4: the new group has 24 ÷ 4 = 6 children.
  4. The new group includes the newcomer, so the original group had 6 − 1 = 5 children.
  5. (a) There were 5 children in the group originally.
  6. With the 90, the 6 children have 6 × 82 = 492 marks.
  7. (b) Taking away the newcomer's 90 leaves the original total: 492 − 90 = 402 marks. Check with the other case: 6 × 78 − 66 = 468 − 66 = 402.

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

techniqueThe Mean · Simultaneous by Elimination

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Answering (a) with 6. Dividing 24 by 4 counts everyone sharing the new average, newcomer included; the original group is one fewer.
  • Taking the 66 away from 492. The total of 492 belongs to the case with the 90, so it is the 90 that comes off.
05

Ryan's Missing Score, Found from the Class Average

heuristicDeviations from the average always add to zero, so the missing score makes up what the others lack

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.
Measure each known score against the average of 80. Three are below it: 74, 68 and 79 fall short by 6, 12 and 1, which is 19 marks in all.
step 1 of 6

Instead of adding seven large numbers, measure each score against the average of 80. The amounts above it and below it must balance exactly, so the missing score is whatever restores the balance.

All steps
  1. Measure each known score against the average of 80. Three are below it: 74, 68 and 79 fall short by 6, 12 and 1, which is 19 marks in all.
  2. Three are above it: 82, 91 and 85 are over by 2, 11 and 5, which is 18 marks in all.
  3. The deviations add up to 18 − 19 = −1: the six known scores are 1 mark short of six scores of 80.
  4. (a) The sum of the deviations of the six known scores is −1.
  5. For the average of all seven to be 80, the deviations of all seven must add up to 0, so Ryan's score must be 1 above 80.
  6. (b) Ryan scored 80 + 1 = 81 marks. Check: 74 + 82 + 68 + 91 + 79 + 85 + 81 = 560 = 7 × 80.

answer(a) −1; (b) 81 marks

techniqueThe Mean

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking 1 away from 80 and answering 79. The known scores are 1 mark short, so Ryan's score has to be 1 more than 80 to make up for it.
  • Adding the deviations without their signs, 6 + 2 + 12 + 11 + 1 + 5 = 37. Scores below the average count as negative and scores above it as positive; only then do they cancel.
06

Two Classes Combined, Then Four Students Move

heuristicWeight each class's average by its size; the ratio of the sizes is enough

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.
Class 6A has 40 students and 6B has 30, in the ratio 40 : 30 = 4 : 3. Draw 6A as 4 parts and 6B as 3 parts, 7 parts in all.
step 1 of 6

A combined average counts every student once, so the bigger class pulls harder. Writing the class sizes as a ratio, 4 : 3, keeps the weights and keeps the numbers small.

All steps
  1. Class 6A has 40 students and 6B has 30, in the ratio 40 : 30 = 4 : 3. Draw 6A as 4 parts and 6B as 3 parts, 7 parts in all.
  2. Each part of 6A averages 75 and each part of 6B averages 82, so the 7 parts together hold 4 × 75 + 3 × 82 = 300 + 246 = 546.
  3. Shared equally over the 7 parts, that is 546 ÷ 7 = 78.
  4. (a) The combined average is 78 marks. Check with the whole classes: 40 × 75 + 30 × 82 = 3000 + 2460 = 5460, and 5460 ÷ 70 = 78.
  5. Class 6A had 40 × 75 = 3000 marks. The 4 students who moved in bring 4 × 86 = 344 marks, so 6A now has 3344 marks.
  6. (b) Class 6A now has 40 + 4 = 44 students, so its new average is 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, (75 + 82) ÷ 2 = 78.5. Class 6A has more students, so its average counts for more: the combined average is 78.
  • Dividing 6A's new total of 3344 by 40. The four students who moved in are now in 6A, so there are 44 of them.
07

A School's Activity Pie Chart Given in Degrees, Percent and Students

heuristicWith 360 students in a 360° circle, one student is one degree

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.
The pie chart shows 360 students, and a full circle is 360°, so each degree of the circle stands for exactly one student.
step 1 of 8

The pie chart describes its parts three ways: by an angle, by a percentage and by a number of students. Turn each into the same thing before adding. Here that is easy, because the 360 students fill a 360° circle.

All steps
  1. The pie chart shows 360 students, and a full circle is 360°, so each degree of the circle stands for exactly one student.
  2. The Sports sector is 120°, so 120 students do Sports.
  3. Arts is 25%, a quarter of the students: 14 × 360 = 90 students, a sector of 90°.
  4. Robotics has 45 students, so its sector is 45°.
  5. Uniformed Groups has everyone else: 360 − (120 + 90 + 45) = 360 − 255 = 105 students.
  6. (b) 105 students are in Uniformed Groups.
  7. One student is one degree, so the Uniformed Groups sector is 105°.
  8. (a) The sector for Uniformed Groups is 105°. Check: 120° + 90° + 45° + 105° = 360°.

answer(a) 105°; (b) 105 students

techniquePie Charts · Finding a Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the 25 from Arts to the 120° of Sports. The 25% is a percentage of the students, 90 of them, so it has to be converted before it is added to anything.
  • Expecting the angle and the number of students always to match. They match here only because there are exactly 360 students; with 180 students, each student would take 2°.
08

Keith's Salary Split into Savings, Food, Rent and Entertainment

heuristicEach fraction is of what is left at that stage, so give the salary enough units for every fraction

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.
Thirds and fifths of the salary are both needed, so draw the salary as 15 equal units: 15 is the smallest number that both 3 and 5 divide.
step 1 of 10

The fractions are taken one after another, each from what is left, so a single bar of units has to hold all of them. Choose 15 units, because the salary is split into thirds and the remainder into fifths.

All steps
  1. Thirds and fifths of the salary are both needed, so draw the salary as 15 equal units: 15 is the smallest number that both 3 and 5 divide.
  2. Keith saved a third: 13 × 15 = 5 units, leaving 10 units.
  3. He spent 25 of the remaining 10 units on food: 25 × 10 = 4 units.
  4. (b) Food took 4 of the 15 units, which is 415 of his salary.
  5. Rent and entertainment share the last 10 − 4 = 6 units.
  6. In the ratio 3 : 1 there are 4 parts, so one part is 6 ÷ 4 = 1.5 units: rent is 4.5 units and entertainment is 1.5 units.
  7. Entertainment, 1.5 units, cost $450.
  8. So one unit is $450 ÷ 1.5 = $300.
  9. The salary is 15 units: 15 × $300 = $4500.
  10. (a) Keith's monthly salary was $4500. Check: he saved $1500 and spent $1200 on food, $1350 on rent and $450 on entertainment, which add up to $4500.

answer(a) $4500; (b) 415

techniquePie Charts · A Fraction of an Amount

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking food as 25 of the whole salary, 615. The 25 is of what is left after saving, 10 units, so food is 4 units: 415.
  • Taking entertainment as 14 of the whole salary. The ratio 3 : 1 splits only the 6 units left after savings and food.
09

Holiday Reading: Two Missing Counts in a Frequency Table

heuristicSuppose every unknown student read the smaller number of books, then let the shortfall count the others

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.
The students whose counts are given number 4 + 12 + 8 = 24, so the other 40 − 24 = 16 students read 2 books or 4 books: k + m = 16.
step 1 of 9

Two counts are unknown, but two facts pin them down: how many students they are, and how many books they read. Suppose all of them read 2 books, and let the shortfall say how many really read 4.

All steps
  1. The students whose counts are given number 4 + 12 + 8 = 24, so the other 40 − 24 = 16 students read 2 books or 4 books: k + m = 16.
  2. The 40 students read 40 × 2.05 = 82 books in all. The known groups read 4 × 0 + 12 × 1 + 8 × 3 = 36, so the 16 students read 82 − 36 = 46 books.
  3. Suppose all 16 of them read 2 books. That would be 16 × 2 = 32 books.
  4. That is 46 − 32 = 14 books too few.
  5. Each student who read 4 books instead of 2 adds 2 more books.
  6. So 14 ÷ 2 = 7 of the 16 students read 4 books.
  7. (b) 7 students read 4 books, so m = 7.
  8. The rest of the 16 read 2 books: 16 − 7 = 9.
  9. (a) 9 students read 2 books, so k = 9. Check: 12 + 2 × 9 + 24 + 4 × 7 = 82, and 82 ÷ 40 = 2.05.

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

techniqueTally and Frequency Tables · Simultaneous by Elimination

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the numbers of books, 2 + 4, instead of multiplying each by its number of students. The k students who read 2 books read 2k books between them.
  • Counting the 4 students who read nothing as reading 4 books. They are among the 40 students but add nothing to the 82 books.
10

The Lowest Top Score Among Five Different Marks

heuristicTo make the top score small, push every other score as close under it as the conditions allow

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?

score 184score 284score 384score 484score 584
(a) Five scores at an average of 84: 420 in all.
(a) Five scores with an average of 84 make 5 × 84 = 420 marks in all.
step 1 of 7

Anna's score is smallest when the other scores crowd up just beneath it. Try a value for Anna, give everyone else the highest scores still allowed, and see whether the five can reach the total of 420.

All steps
  1. (a) Five scores with an average of 84 make 5 × 84 = 420 marks in all.
  2. Anna has the highest score. To make her score as low as it can be, the scores of the other three, besides Colin's 82, must be as high as they can be: just below Anna's, each a different whole number.
  3. Try 85 for Anna. The highest the others can be is 84 and 83, then Colin's 82, then 81: 85 + 84 + 83 + 82 + 81 = 415, which is less than 420.
  4. Try 86 for Anna. The others can be 85, 84 and 83, with Colin's 82: 86 + 85 + 84 + 83 + 82 = 420.
  5. That is exactly 420, so Anna can score 86. Any lower score for Anna leaves the total short, as 85 did.
  6. The five scores 86, 85, 84, 83 and 82 are different whole numbers, Anna's is the highest and Colin's is 82, so every condition holds.
  7. (b) The lowest possible mark for Anna 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

  • Taking Anna's lowest score as one more than the average, 85. With Anna on 85 the other four can add up to at most 84 + 83 + 82 + 81 = 330, so the five fall short of 420.
  • Letting two students share a score, such as 84, 84 and 84. The question says the five scores are all different.
11

Six Different Scores Around a Fixed Third Place: the Least Top Score and the Greatest Bottom Score

heuristicPush Every Other Score to Its Limit / Let the Order Decide, Then Let the Total Check

Six students, Ahmad, Bala, Cheryl, Daniel, Emily and Farhan, sat a mathematics contest marked out of 100. Their scores were six different whole numbers, and every student scored at least 65 marks. The average score of the six students was 82.5 marks. Ahmad had the highest score and Farhan the lowest. Cheryl scored 88 marks, the third highest score. (a) What was the total score of the six students? (b) What is the lowest score Ahmad could have obtained? (c) What is the highest score Farhan could have obtained?

Ahmad?highest2nd?Cheryl88third4th?5th?Farhan?lowesttotal: 6 × 82.5 = 495
(a) Six scores at an average of 82.5: 6 × 82.5 = 495 marks.
(a) Six scores at an average of 82.5 make 6 × 82.5 = 495 marks.
step 1 of 6

Six places in order, with Cheryl's 88 fixed in the third. To make one score as low or as high as it can be, push every other score as far the other way as the order allows, then check that the total still works.

All steps
  1. (a) Six scores at an average of 82.5 make 6 × 82.5 = 495 marks.
  2. Cheryl is third, so exactly two scores are above 88, Ahmad's and one other, and three are below it, Farhan's the lowest of them.
  3. (b) Two different whole numbers above 88 are at least 89 and 90, and the larger is Ahmad's, so Ahmad's lowest possible score is 90. Check: the bottom three then share 495 − (90 + 89 + 88) = 228, for example 77 + 76 + 75, all below 88 and at least 65.
  4. To make Farhan's score high, push the top three as low as they go: 90 + 89 + 88 = 267. That leaves the largest possible share for the bottom three, 495 − 267 = 228.
  5. The bottom three are different, so Farhan's score is highest when they are one apart: Farhan's score, 1 more and 2 more. Three of Farhan's score and 3 make 228, so three of his score make 225, and his score is 225 ÷ 3 = 75.
  6. (c) Farhan's highest possible score is 75 marks. Check: 90 + 89 + 88 + 77 + 76 + 75 = 495.

answer(a) 495 marks; (b) 90 marks; (c) 75 marks

techniqueThe Mean · Ordering Numbers

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Pushing the bottom three up to 87, 86 and 85 to find Ahmad's least score. That leaves 495 − 88 − 258 = 149 for the top two, less than 88 each, which breaks the order. The order already forces the top two to be at least 89 and 90; the total only has to be checked.
  • Dividing the bottom three's 228 by 3 and giving Farhan 76. The three scores are different, so 76 is the middle one of them; Farhan, the lowest, has one less: 75.
12

The Size of a Third Class Found from the Average of All Three

heuristicMeasure Every Class Against the Overall Average / The Shortfall of Two Classes Equals the Surplus of the Third

Three classes, 6A, 6B and 6C, sat the same mathematics test. The 35 pupils of 6A had an average score of 76 marks and the 45 pupils of 6B had an average score of 82 marks. The pupils of 6C had an average score of 86 marks, and the average score of all the pupils in the three classes was 81 marks. (a) What was the total score of 6A and 6B together? (b) How many pupils are there in 6C? (c) Five pupils from 6C, whose average score was 92 marks, then moved into 6A. What was the new average score of 6A?

6A35 × 76 = 26606B45 × 82 = 3690
6A: 35 × 76 = 2660 marks. 6B: 45 × 82 = 3690 marks.
6A: 35 × 76 = 2660 marks. 6B: 45 × 82 = 3690 marks.
step 1 of 6

Measure every class against the overall average of 81. The marks that 6A and 6B together fall short of it must be made up exactly by the marks that 6C has above it.

All steps
  1. 6A: 35 × 76 = 2660 marks. 6B: 45 × 82 = 3690 marks.
  2. (a) 6A and 6B together: 2660 + 3690 = 6350 marks.
  3. At an average of 81, their 80 pupils would have 80 × 81 = 6480 marks, so they are 6480 − 6350 = 130 marks short.
  4. Each 6C pupil has 86 − 81 = 5 marks above the overall average, and these make up the 130 marks: 130 ÷ 5 = 26 pupils.
  5. (b) There are 26 pupils in 6C. Check: 6350 + 26 × 86 = 8586, and 8586 ÷ 106 = 81.
  6. (c) The five pupils bring 5 × 92 = 460 marks. 6A now has 2660 + 460 = 3120 marks for 35 + 5 = 40 pupils, an average of 3120 ÷ 40 = 78 marks.

answer(a) 6350 marks; (b) 26 pupils; (c) 78 marks

techniqueThe Mean · Solving Linear Equations

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Averaging the three class averages, (76 + 82 + 86) ÷ 3 ≈ 81.3, and treating the classes as the same size. The overall average weights each class by its number of pupils, and the number in 6C is what decides it.
  • Dividing 6A's new total of 3120 by 35. The five pupils who moved in are now 6A pupils, so there are 40 of them: 3120 ÷ 40 = 78.