Practice · Write the overall average as a weighted average · Sensim Math

Variant 1 easy

$21$ students took an exam, and $6$ of them failed. The lowest passing score was $3$ points below the average of all $21$ students, $15$ points below the average of the students who passed, and $14$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 21 test takers, 15 passed and 6 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$21$ students; $6$ failed, so $15$ passed.
Cutoff $=$ (overall average) $- 3$ .
Cutoff $=$ (passers' average) $- 15$ .
Cutoff $= 2 \times$ (failers' average) $- 14$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

15

passers scored

15x

in total and the

6

failers

6y

.

\text{overall average} = \frac{15x + 6y}{21} = \frac{5x + 2y}{7}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{5x + 2y}{7} - 3 = x - 15 = 2y - 14

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

7

:

5x + 2y - 21 = 7x - 105

. The second equality rearranges directly.

2x - 2y = 84,\qquad x - 2y = 1

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

15

below the passers' average.

x = 83,\quad 83 - 15 = 68

We only need

x

, because the cutoff is tied to it directly.

Answer:

68

points

4 · Reviewdoes it hold up?

From $x - 2y = 1$ , $y = 41$ . Overall average $= \frac{15 \times 83 + 6 \times 41}{21} = \frac{1245 + 246}{21} = 71$ , and $71 - 3 = 68$ . Also $2 \times 41 - 14 = 68$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 15$ , failers average $\frac{m + 14}{2}$ , overall average $m + 3$ . Totals give $21(m + 3) = 15(m + 15) + 6 \times \frac{m + 14}{2}$ , so $21m + 63 = 18m + 267$ and $m = 68$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 2 easy

$25$ students took an exam, and $10$ of them failed. The lowest passing score was $6$ points below the average of all $25$ students, $25$ points below the average of the students who passed, and $2$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 25 test takers, 15 passed and 10 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$25$ students; $10$ failed, so $15$ passed.
Cutoff $=$ (overall average) $- 6$ .
Cutoff $=$ (passers' average) $- 25$ .
Cutoff $= 2 \times$ (failers' average) $- 2$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

15

passers scored

15x

in total and the

10

failers

10y

.

\text{overall average} = \frac{15x + 10y}{25} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 6 = x - 25 = 2y - 2

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 30 = 5x - 125

. The second equality rearranges directly.

2x - 2y = 95,\qquad x - 2y = 23

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

25

below the passers' average.

x = 72,\quad 72 - 25 = 47

We only need

x

, because the cutoff is tied to it directly.

Answer:

47

points

4 · Reviewdoes it hold up?

From $x - 2y = 23$ , $y = 24.5$ . Overall average $= \frac{15 \times 72 + 10 \times 24.5}{25} = \frac{1080 + 245}{25} = 53$ , and $53 - 6 = 47$ . Also $2 \times 24.5 - 2 = 47$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 25$ , failers average $\frac{m + 2}{2}$ , overall average $m + 6$ . Totals give $25(m + 6) = 15(m + 25) + 10 \times \frac{m + 2}{2}$ , so $25m + 150 = 20m + 385$ and $m = 47$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 3 easy

$27$ students took an exam, and $6$ of them failed. The lowest passing score was $2$ points below the average of all $27$ students, $12$ points below the average of the students who passed, and $20$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 27 test takers, 21 passed and 6 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$27$ students; $6$ failed, so $21$ passed.
Cutoff $=$ (overall average) $- 2$ .
Cutoff $=$ (passers' average) $- 12$ .
Cutoff $= 2 \times$ (failers' average) $- 20$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

21

passers scored

21x

in total and the

6

failers

6y

.

\text{overall average} = \frac{21x + 6y}{27} = \frac{7x + 2y}{9}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{7x + 2y}{9} - 2 = x - 12 = 2y - 20

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

9

:

7x + 2y - 18 = 9x - 108

. The second equality rearranges directly.

2x - 2y = 90,\qquad x - 2y = -8

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

12

below the passers' average.

x = 98,\quad 98 - 12 = 86

We only need

x

, because the cutoff is tied to it directly.

Answer:

86

points

4 · Reviewdoes it hold up?

From $x - 2y = -8$ , $y = 53$ . Overall average $= \frac{21 \times 98 + 6 \times 53}{27} = \frac{2058 + 318}{27} = 88$ , and $88 - 2 = 86$ . Also $2 \times 53 - 20 = 86$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 12$ , failers average $\frac{m + 20}{2}$ , overall average $m + 2$ . Totals give $27(m + 2) = 21(m + 12) + 6 \times \frac{m + 20}{2}$ , so $27m + 54 = 24m + 312$ and $m = 86$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 4 easy

$30$ students took an exam, and $12$ of them failed. The lowest passing score was $4$ points below the average of all $30$ students, $20$ points below the average of the students who passed, and $10$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 30 test takers, 18 passed and 12 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$30$ students; $12$ failed, so $18$ passed.
Cutoff $=$ (overall average) $- 4$ .
Cutoff $=$ (passers' average) $- 20$ .
Cutoff $= 2 \times$ (failers' average) $- 10$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

18

passers scored

18x

in total and the

12

failers

12y

.

\text{overall average} = \frac{18x + 12y}{30} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 4 = x - 20 = 2y - 10

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 20 = 5x - 100

. The second equality rearranges directly.

2x - 2y = 80,\qquad x - 2y = 10

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

20

below the passers' average.

x = 70,\quad 70 - 20 = 50

We only need

x

, because the cutoff is tied to it directly.

Answer:

50

points

4 · Reviewdoes it hold up?

From $x - 2y = 10$ , $y = 30$ . Overall average $= \frac{18 \times 70 + 12 \times 30}{30} = \frac{1260 + 360}{30} = 54$ , and $54 - 4 = 50$ . Also $2 \times 30 - 10 = 50$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 20$ , failers average $\frac{m + 10}{2}$ , overall average $m + 4$ . Totals give $30(m + 4) = 18(m + 20) + 12 \times \frac{m + 10}{2}$ , so $30m + 120 = 24m + 420$ and $m = 50$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 5 medium

$35$ students took an exam, and $10$ of them failed. The lowest passing score was $3$ points below the average of all $35$ students, $18$ points below the average of the students who passed, and $6$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 35 test takers, 25 passed and 10 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$35$ students; $10$ failed, so $25$ passed.
Cutoff $=$ (overall average) $- 3$ .
Cutoff $=$ (passers' average) $- 18$ .
Cutoff $= 2 \times$ (failers' average) $- 6$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

25

passers scored

25x

in total and the

10

failers

10y

.

\text{overall average} = \frac{25x + 10y}{35} = \frac{5x + 2y}{7}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{5x + 2y}{7} - 3 = x - 18 = 2y - 6

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

7

:

5x + 2y - 21 = 7x - 126

. The second equality rearranges directly.

2x - 2y = 105,\qquad x - 2y = 12

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

18

below the passers' average.

x = 93,\quad 93 - 18 = 75

We only need

x

, because the cutoff is tied to it directly.

Answer:

75

points

4 · Reviewdoes it hold up?

From $x - 2y = 12$ , $y = 40.5$ . Overall average $= \frac{25 \times 93 + 10 \times 40.5}{35} = \frac{2325 + 405}{35} = 78$ , and $78 - 3 = 75$ . Also $2 \times 40.5 - 6 = 75$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 18$ , failers average $\frac{m + 6}{2}$ , overall average $m + 3$ . Totals give $35(m + 3) = 25(m + 18) + 10 \times \frac{m + 6}{2}$ , so $35m + 105 = 30m + 480$ and $m = 75$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 6 medium

$40$ students took an exam, and $16$ of them failed. The lowest passing score was $5$ points below the average of all $40$ students, $24$ points below the average of the students who passed, and $4$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 40 test takers, 24 passed and 16 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$40$ students; $16$ failed, so $24$ passed.
Cutoff $=$ (overall average) $- 5$ .
Cutoff $=$ (passers' average) $- 24$ .
Cutoff $= 2 \times$ (failers' average) $- 4$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

24

passers scored

24x

in total and the

16

failers

16y

.

\text{overall average} = \frac{24x + 16y}{40} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 5 = x - 24 = 2y - 4

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 25 = 5x - 120

. The second equality rearranges directly.

2x - 2y = 95,\qquad x - 2y = 20

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

24

below the passers' average.

x = 75,\quad 75 - 24 = 51

We only need

x

, because the cutoff is tied to it directly.

Answer:

51

points

4 · Reviewdoes it hold up?

From $x - 2y = 20$ , $y = 27.5$ . Overall average $= \frac{24 \times 75 + 16 \times 27.5}{40} = \frac{1800 + 440}{40} = 56$ , and $56 - 5 = 51$ . Also $2 \times 27.5 - 4 = 51$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 24$ , failers average $\frac{m + 4}{2}$ , overall average $m + 5$ . Totals give $40(m + 5) = 24(m + 24) + 16 \times \frac{m + 4}{2}$ , so $40m + 200 = 32m + 608$ and $m = 51$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 7 medium

$45$ students took an exam, and $18$ of them failed. The lowest passing score was $6$ points below the average of all $45$ students, $30$ points below the average of the students who passed, and $8$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 45 test takers, 27 passed and 18 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$45$ students; $18$ failed, so $27$ passed.
Cutoff $=$ (overall average) $- 6$ .
Cutoff $=$ (passers' average) $- 30$ .
Cutoff $= 2 \times$ (failers' average) $- 8$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

27

passers scored

27x

in total and the

18

failers

18y

.

\text{overall average} = \frac{27x + 18y}{45} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 6 = x - 30 = 2y - 8

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 30 = 5x - 150

. The second equality rearranges directly.

2x - 2y = 120,\qquad x - 2y = 22

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

30

below the passers' average.

x = 98,\quad 98 - 30 = 68

We only need

x

, because the cutoff is tied to it directly.

Answer:

68

points

4 · Reviewdoes it hold up?

From $x - 2y = 22$ , $y = 38$ . Overall average $= \frac{27 \times 98 + 18 \times 38}{45} = \frac{2646 + 684}{45} = 74$ , and $74 - 6 = 68$ . Also $2 \times 38 - 8 = 68$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 30$ , failers average $\frac{m + 8}{2}$ , overall average $m + 6$ . Totals give $45(m + 6) = 27(m + 30) + 18 \times \frac{m + 8}{2}$ , so $45m + 270 = 36m + 882$ and $m = 68$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 8 medium

$50$ students took an exam, and $20$ of them failed. The lowest passing score was $5$ points below the average of all $50$ students, $30$ points below the average of the students who passed, and $3$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 50 test takers, 30 passed and 20 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$50$ students; $20$ failed, so $30$ passed.
Cutoff $=$ (overall average) $- 5$ .
Cutoff $=$ (passers' average) $- 30$ .
Cutoff $= 2 \times$ (failers' average) $- 3$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

30

passers scored

30x

in total and the

20

failers

20y

.

\text{overall average} = \frac{30x + 20y}{50} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 5 = x - 30 = 2y - 3

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 25 = 5x - 150

. The second equality rearranges directly.

2x - 2y = 125,\qquad x - 2y = 27

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

30

below the passers' average.

x = 98,\quad 98 - 30 = 68

We only need

x

, because the cutoff is tied to it directly.

Answer:

68

points

4 · Reviewdoes it hold up?

From $x - 2y = 27$ , $y = 35.5$ . Overall average $= \frac{30 \times 98 + 20 \times 35.5}{50} = \frac{2940 + 710}{50} = 73$ , and $73 - 5 = 68$ . Also $2 \times 35.5 - 3 = 68$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 30$ , failers average $\frac{m + 3}{2}$ , overall average $m + 5$ . Totals give $50(m + 5) = 30(m + 30) + 20 \times \frac{m + 3}{2}$ , so $50m + 250 = 40m + 930$ and $m = 68$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 9 hard

$55$ students took an exam, and $22$ of them failed. The lowest passing score was $4$ points below the average of all $55$ students, $26$ points below the average of the students who passed, and $8$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 55 test takers, 33 passed and 22 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$55$ students; $22$ failed, so $33$ passed.
Cutoff $=$ (overall average) $- 4$ .
Cutoff $=$ (passers' average) $- 26$ .
Cutoff $= 2 \times$ (failers' average) $- 8$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

33

passers scored

33x

in total and the

22

failers

22y

.

\text{overall average} = \frac{33x + 22y}{55} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 4 = x - 26 = 2y - 8

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 20 = 5x - 130

. The second equality rearranges directly.

2x - 2y = 110,\qquad x - 2y = 18

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

26

below the passers' average.

x = 92,\quad 92 - 26 = 66

We only need

x

, because the cutoff is tied to it directly.

Answer:

66

points

4 · Reviewdoes it hold up?

From $x - 2y = 18$ , $y = 37$ . Overall average $= \frac{33 \times 92 + 22 \times 37}{55} = \frac{3036 + 814}{55} = 70$ , and $70 - 4 = 66$ . Also $2 \times 37 - 8 = 66$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 26$ , failers average $\frac{m + 8}{2}$ , overall average $m + 4$ . Totals give $55(m + 4) = 33(m + 26) + 22 \times \frac{m + 8}{2}$ , so $55m + 220 = 44m + 946$ and $m = 66$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 10 hard

$60$ students took an exam, and $24$ of them failed. The lowest passing score was $5$ points below the average of all $60$ students, $26$ points below the average of the students who passed, and $9$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 60 test takers, 36 passed and 24 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$60$ students; $24$ failed, so $36$ passed.
Cutoff $=$ (overall average) $- 5$ .
Cutoff $=$ (passers' average) $- 26$ .
Cutoff $= 2 \times$ (failers' average) $- 9$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

36

passers scored

36x

in total and the

24

failers

24y

.

\text{overall average} = \frac{36x + 24y}{60} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 5 = x - 26 = 2y - 9

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 25 = 5x - 130

. The second equality rearranges directly.

2x - 2y = 105,\qquad x - 2y = 17

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

26

below the passers' average.

x = 88,\quad 88 - 26 = 62

We only need

x

, because the cutoff is tied to it directly.

Answer:

62

points

4 · Reviewdoes it hold up?

From $x - 2y = 17$ , $y = 35.5$ . Overall average $= \frac{36 \times 88 + 24 \times 35.5}{60} = \frac{3168 + 852}{60} = 67$ , and $67 - 5 = 62$ . Also $2 \times 35.5 - 9 = 62$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 26$ , failers average $\frac{m + 9}{2}$ , overall average $m + 5$ . Totals give $60(m + 5) = 36(m + 26) + 24 \times \frac{m + 9}{2}$ , so $60m + 300 = 48m + 1044$ and $m = 62$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 11 hard

$70$ students took an exam, and $20$ of them failed. The lowest passing score was $4$ points below the average of all $70$ students, $18$ points below the average of the students who passed, and $12$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 70 test takers, 50 passed and 20 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$70$ students; $20$ failed, so $50$ passed.
Cutoff $=$ (overall average) $- 4$ .
Cutoff $=$ (passers' average) $- 18$ .
Cutoff $= 2 \times$ (failers' average) $- 12$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

50

passers scored

50x

in total and the

20

failers

20y

.

\text{overall average} = \frac{50x + 20y}{70} = \frac{5x + 2y}{7}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{5x + 2y}{7} - 4 = x - 18 = 2y - 12

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

7

:

5x + 2y - 28 = 7x - 126

. The second equality rearranges directly.

2x - 2y = 98,\qquad x - 2y = 6

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

18

below the passers' average.

x = 92,\quad 92 - 18 = 74

We only need

x

, because the cutoff is tied to it directly.

Answer:

74

points

4 · Reviewdoes it hold up?

From $x - 2y = 6$ , $y = 43$ . Overall average $= \frac{50 \times 92 + 20 \times 43}{70} = \frac{4600 + 860}{70} = 78$ , and $78 - 4 = 74$ . Also $2 \times 43 - 12 = 74$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 18$ , failers average $\frac{m + 12}{2}$ , overall average $m + 4$ . Totals give $70(m + 4) = 50(m + 18) + 20 \times \frac{m + 12}{2}$ , so $70m + 280 = 60m + 1020$ and $m = 74$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Variant 12 hard

$90$ students took an exam, and $36$ of them failed. The lowest passing score was $4$ points below the average of all $90$ students, $20$ points below the average of the students who passed, and $7$ points below twice the average of the students who failed. Find the lowest passing score.

Your answer

Show solution

1 · Understandwhat's really being asked

Of 90 test takers, 54 passed and 36 failed. The cutoff score is described three ways: relative to the overall average, the passers' average, and the failers' average. Find the cutoff.

Givens

$90$ students; $36$ failed, so $54$ passed.
Cutoff $=$ (overall average) $- 4$ .
Cutoff $=$ (passers' average) $- 20$ .
Cutoff $= 2 \times$ (failers' average) $- 7$ .

Unknowns

The lowest passing score.

Constraints

The overall average is not a plain average of the two group averages, because the groups have different sizes.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#16 Count the Complement

Let the two group averages be unknowns and express the overall average from them, weighting each group by its size. The three descriptions of the cutoff then give two equations.

3 · Execute4 carry out the plan

1Write the overall average

#7 Identify Subproblems 6.SP.B.5

Let the passers' average be

x

and the failers' average be

y

. The

54

passers scored

54x

in total and the

36

failers

36y

.

\text{overall average} = \frac{54x + 36y}{90} = \frac{3x + 2y}{5}

An average times the head count gives the total; totals can be added.

2Set the three descriptions equal

#13 Convert to Algebra 8.EE.C.8

All three expressions name the same cutoff score.

\frac{3x + 2y}{5} - 4 = x - 20 = 2y - 7

Three names for one number give two equations.

3Clear fractions

#13 Convert to Algebra 8.EE.C.8

From the first equality, multiply by

5

:

3x + 2y - 20 = 5x - 100

. The second equality rearranges directly.

2x - 2y = 80,\qquad x - 2y = 13

Collect the letters on one side.

4Solve and read off the cutoff

#13 Convert to Algebra 8.EE.C.8

Subtracting removes

y

at once. The cutoff is

20

below the passers' average.

x = 67,\quad 67 - 20 = 47

We only need

x

, because the cutoff is tied to it directly.

Answer:

47

points

4 · Reviewdoes it hold up?

From $x - 2y = 13$ , $y = 27$ . Overall average $= \frac{54 \times 67 + 36 \times 27}{90} = \frac{3618 + 972}{90} = 51$ , and $51 - 4 = 47$ . Also $2 \times 27 - 7 = 47$ . All three descriptions agree.

Another way: Make the cutoff $m$ itself the unknown: passers average $m + 20$ , failers average $\frac{m + 7}{2}$ , overall average $m + 4$ . Totals give $90(m + 4) = 54(m + 20) + 36 \times \frac{m + 7}{2}$ , so $90m + 360 = 72m + 1206$ and $m = 47$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two equations in the two group averages.
6.SP.B.5 Summarize numerical data sets in relation to context — Combining two group averages into the overall average by head count.

💡Takeaway. To combine averages, go back to totals: average times the number of people.

Write the overall average as a weighted average · 12 practice problems

Generated variants — 12

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