Set Calculations

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Write each formula used for solving the question.
If the question includes a requirement to draw a Venn Diagram, draw the Venn Diagram.
Show all work.

Corequisites: Probability

(1.) ACT Fifty shoppers at a pet store were asked if they owned at least 1 cat or at least 1 dog.
Data from their answers were recorded below.

Ownership	Number of shoppers
Cat(s) only	13
Dog(s) only	24
Both cat(s) and dog(s)	7

How many of these shoppers said that they owned NEITHER a cat NOR a dog?

$ A.\;\; 0 \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 7 \\[3ex] D.\;\; 13 \\[3ex] E.\;\; 43 \\[3ex] $

$ Let \\[3ex] Cat = C \\[3ex] Dog = D \\[3ex] Cat(s)\;\;only = n(C \cap D') = 13 \\[3ex] Dog(s)\;\;only = n(D \cap C') = 24 \\[3ex] Both\;\;cat(s)\;\;and\;\;dogs = n(C \cap D) = 7 \\[3ex] Neither\;\;a\;\;cat\;\;nor\;\;a\;\;dog = n(C' \cap D') = ? \\[3ex] Universal\;\;set = n(\mu) = 50 \\[3ex] n(C \cap D') + n(D \cap C') + n(C \cap D) + n(C' \cap D') = n(\mu) \\[3ex] 13 + 24 + 7 + n(C' \cap D') = 50 \\[3ex] 44 + n(C' \cap D') = 50 \\[3ex] n(C' \cap D') = 50 - 44 \\[3ex] n(C' \cap D') = 6 $

(2.) WASSCE Out of 30 candidates applying for a post, 17 have degrees, 15 diplomas and 4 neither degree nor diploma.
How many of them have both?

$ n(\xi) = 30 \\[3ex] Let: \\[3ex] degree = DE \\[3ex] diploma = DI \\[3ex] n(DE) = 17 \\[3ex] n(DI) = 15 \\[3ex] n(neither) = n(DE \cup DI)' = 4 \\[3ex] n(both) = n(DE \cap DI) = d \\[3ex] n(DE\;\;only) = n(DE \cap DI') = 17 - d \\[3ex] n(DI\;\;only) = n(DI \cap DE') = 15 - d \\[3ex] \implies \\[3ex] (17 - d) + (15 - d) + d + 4 = 30 \\[3ex] 17 - d + 15 - d + d + 4 = 30 \\[3ex] 36 - d = 30 \\[3ex] 36 - 30 = d \\[3ex] d = 6 \\[3ex] $ 6 candidates have both degrees and diplomas

(4.) ACT At Brookfield High School, 55 seniors are enrolled in the sociology class and 40 seniors are enrolled in the drawing class.
Of these seniors, 20 are enrolled in both the sociology class and the drawing class.
How many of the 120 seniors enrolled at Brookfield High School are NOT enrolled in either the sociology class or the drawing class?

$ F.\;\; 5 \\[3ex] G.\;\; 15 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 35 \\[3ex] K.\;\; 45 \\[3ex] $

$ n(\mu) = 120 \\[3ex] Let: \\[3ex] sociology\;\;class = C \\[3ex] drawing\;\;class = D \\[3ex] n(C) = 55 \\[3ex] n(D) = 40 \\[3ex] n(C \cap D) = 20 \\[3ex] n(C\;\;only) = n(C \cap D') = 55 - 20 = 35 \\[3ex] n(D\;\;only) = n(D \cap C') = 40 - 20 = 20 \\[3ex] n(C \cup D)' = k \\[3ex] \implies \\[3ex] 35 + 20 + 20 + k = 120 \\[3ex] 75 + k = 120 \\[3ex] k = 120 - 75 \\[3ex] k = 45 \\[3ex] $ 45 students are NOT enrolled in either the sociology class or the drawing class

(5.) BeGood Tutors offer tutorials on Physics, Chemistry, and Biology classes.
They divided their student population into two groups: Group $A$ and Group $B$

(a.) Group $A$ has a hundred and eighty students.
Each student is enrolled in at least one of the three classes.
Sixty students are taking Physics
Seventy five students are taking Chemistry.
Seventy students are taking Biology.
Twelve students are taking Physics and Chemistry.
Twenty one students are taking Physics and Biology.
Ten students are taking Chemistry and Biology.
How many students are taking all three classes?

(b.) Group $B$ has a certain number of students.
Thirty students are taking Physics.
Twenty students are taking Chemistry.
Sixteen students are taking Biology.
Ten students are taking Physics and Chemistry
Seven students are taking Physics and Biology.
Twelve students are taking Chemistry and Biology.
Five students are taking all three classes.
Nine students are not taking any class.
How many students are in Group $B$

$ Let: \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] Biology = B \\[3ex] (a.) \\[3ex] n(P \cap C \cap B) = x \\[3ex] n(\mu) = 180 \\[3ex] n(P) = 60 \\[3ex] n(C) = 75 \\[3ex] n(B) = 70 \\[3ex] n(P \cap C) = 12 \\[3ex] n(P \cap B) = 21 \\[3ex] n(C \cap B) = 10 \\[3ex] \underline{Physics\;\;and\;\;Chemistry\;\;only} \\[3ex] n(P \cap C \cap B') = n(P \cap C) - n(P \cap C \cap B) \\[3ex] = 12 - x \\[3ex] \underline{Physics\;\;and\;\;Biology\;\;only} \\[3ex] n(P \cap B \cap C') = n(P \cap B) - n(P \cap C \cap B) \\[3ex] = 21 - x \\[3ex] \underline{Chemistry\;\;and\;\;Biology\;\;only} \\[3ex] n(C \cap B \cap P') = n(C \cap B) - n(P \cap C \cap B) \\[3ex] = 10 - x \\[3ex] \underline{Physics\;\;only} \\[3ex] = n(P) - [n(P \cap C \cap B') + n(P \cap B \cap C') + n(P \cap C \cap B)] \\[3ex] = 60 - [(12 - x) + (21 - x) + x] \\[3ex] = 60 - (12 - x + 21 - x + x) \\[3ex] = 60 - (33 - x) \\[3ex] = 60 - 33 + x \\[3ex] = 27 + x \\[3ex] \underline{Chemistry\;\;only} \\[3ex] = n(C) - [n(P \cap C \cap B') + n(C \cap B \cap P') + n(P \cap C \cap B)] \\[3ex] = 60 - [(12 - x) + (10 - x) + x] \\[3ex] = 60 - (12 - x + 10 - x + x) \\[3ex] = 60 - (22 - x) \\[3ex] = 60 - 22 + x \\[3ex] = 38 + x \\[3ex] \underline{Biology\;\;only} \\[3ex] = n(B) - [n(P \cap B \cap C') + n(C \cap B \cap P') + n(P \cap C \cap B)] \\[3ex] = 60 - [(21 - x) + (10 - x) + x] \\[3ex] = 60 - (21 - x + 10 - x + x) \\[3ex] = 60 - (31 - x) \\[3ex] = 60 - 31 + x \\[3ex] = 29 + x \\[3ex] Based\;\;on\;\;the\;\;information\;\;given \\[3ex] This\;\;implies\;\;that: \\[3ex] (27 + x) + (38 + x) + (29 + x) + (12 - x) + (21 - x) + (10 - x) + x = 180 \\[3ex] 27 + x + 38 + x + 29 + x + 12 - x + 21 - x + 10 - x + x = 180 \\[3ex] 137 + x = 180 \\[3ex] x = 180 - 137 \\[3ex] x = 43\;\;students \\[3ex] $ Forty three students are taking all three classes.

$ (b.) \\[3ex] n(P \cap C \cap B) = 5 \\[3ex] n(\mu) = ? \\[3ex] n(P) = 30 \\[3ex] n(C) = 20 \\[3ex] n(B) = 16 \\[3ex] n(P \cap C) = 10 \\[3ex] n(P \cap B) = 7 \\[3ex] n(C \cap B) = 12 \\[3ex] n(P' \cap C' \cap B') = 9 \\[3ex] \underline{Physics\;\;and\;\;Chemistry\;\;only} \\[3ex] n(P \cap C \cap B') = n(P \cap C) - n(P \cap C \cap B) \\[3ex] = 10 - 5 \\[3ex] = 5 \\[3ex] \underline{Physics\;\;and\;\;Biology\;\;only} \\[3ex] n(P \cap B \cap C') = n(P \cap B) - n(P \cap C \cap B) \\[3ex] = 7 - 5 \\[3ex] = 2 \\[3ex] \underline{Chemistry\;\;and\;\;Biology\;\;only} \\[3ex] n(C \cap B \cap P') = n(C \cap B) - n(P \cap C \cap B) \\[3ex] = 12 - 5 \\[3ex] = 7 \\[3ex] \underline{Physics\;\;only} \\[3ex] = n(P) - [n(P \cap C \cap B') + n(P \cap B \cap C') + n(P \cap C \cap B)] \\[3ex] = 30 - (5 + 2 + 5) \\[3ex] = 30 - 12 \\[3ex] = 18 \\[3ex] \underline{Chemistry\;\;only} \\[3ex] = n(C) - [n(P \cap C \cap B') + n(C \cap B \cap P') + n(P \cap C \cap B)] \\[3ex] = 20 - (5 + 7 + 5) \\[3ex] = 20 - 17 \\[3ex] = 3 \\[3ex] \underline{Biology\;\;only} \\[3ex] = n(B) - [n(P \cap B \cap C') + n(C \cap B \cap P') + n(P \cap C \cap B)] \\[3ex] = 16 - (2 + 7 + 5) \\[3ex] = 16 - 14 \\[3ex] = 2 \\[3ex] Based\;\;on\;\;the\;\;information\;\;given \\[3ex] This\;\;implies\;\;that: \\[3ex] n(\mu) = \\[3ex] Physics\;\;only + Chemistry\;\;only + Biology\;\;only + \\[3ex] Physics\;\;and\;\;Chemistry\;\;only + Physics\;\;and\;\;Biology\;\;only + Chemistry\;\;and\;\;Biology\;\;only + \\[3ex] Physics\;\;and\;\;Chemistry\;\;and\;\;Biology + \\[3ex] Neither\;\;Physics\;\;nor\;\;Chemistry\;\;nor\;\;Biology \\[3ex] n(\mu) = 18 + 3 + 2 + 5 + 2 + 7 + 5 + 9 \\[3ex] n(\mu) = 51 \\[3ex] $ $51$ students are in Group $B$

(6.) Every student at Divine Mercy Bible School is required to study at least one of the book of Genesis from the Old Testament or the Book of Revelation from the New Testament.
During the Fall Semester of a certain year, seventy students studied the Book of Genesis.
One hundred students studied the book of Revelation.
One hundred and sixty students studied the book of Genesis OR the book of Revelation.

How many students:
(I.) studied both books?
(II.) studied the book of Genesis but not the book of Revelation?
(III.) studied the book of Revelation but not the book of Genesis?
(IV.) did not study any of the books?
(V.) are enrolled in the school?
(VI.) studied at least one book?
(VII.) studied at least two books?
(VIII.) studied exactly one book?
(IX.) studied exactly two books?
(X.) studied at most one book?
(XI.) studied at most two books?
(XII.) more than one book?
(XIII.) more than two books?
(XIV.) less than one book?
(XV.) less than two books?

$ Let: \\[3ex] Genesis = G \\[3ex] Revelation = R \\[3ex] n(G) = 70 \\[3ex] n(R) = 100 \\[3ex] (I.) \\[3ex] n(G \cup R) = 160 \\[3ex] n(G \cap R) = n(G) + n(R) - n(G \cup R) \\[3ex] = 70 + 100 - 160 \\[3ex] = 10 \\[3ex] $ $10$ students studied BOTH books.

$ (II.) \\[3ex] n(G \cap R') = n(G) - n(G \cap R) \\[3ex] = 70 - 10 \\[3ex] = 60 \\[3ex] $ $60$ students studied ONLY the book of Genesis.

$ (III.) \\[3ex] n(G' \cap R) = n(R) - n(G \cap R) \\[3ex] = 100 - 10 \\[3ex] = 90 \\[3ex] $ $90$ students studied ONLY the book of Revelation.

(IV.)
$n(G \cup R)' = 0$
Based on the Question: Every student at Divine Mercy Bible School is required to study the book of Genesis from the Old Testament or the Book of Revelation from the New Testament.

No student did not study any book.

$ (V.) \\[3ex] n(\mu) = n(G \cup R) + n(G \cup R)' \\[3ex] = 160 + 0 \\[3ex] = 160 \\[3ex] $ $160$ students are enrolled in the school.

(VI.)
"At least one" book means "one book or more"
"At least one" means $\ge 1$
They are the students who studied ONLY the book of Genesis and the students who studied ONLY the book of Revelation and the students who studied BOTH books.

$ n(\gt 1) = n(G \cap R') + n(G' \cap R) + n(G \cap R) \\[3ex] = 60 + 90 + 10 \\[3ex] = 160 \\[3ex] $ $160$ students studied AT LEAST one book.

(VII.)
"At least two" books means "two books or more"
"At least two" means $\ge 2$
Because we have only two books in this case, we shall find the students who studied BOTH books.

$ n(\gt 2) = n(G \cap R) \\[3ex] = 10 \\[3ex] $ $10$ students studied AT LEAST two books.

(VIII.)
"Exactly one book" means "only one book"
"Exactly one" means $=1$
They are the students who studied ONLY the book of Genesis and the students who studied ONLY the book of Revelation.

$ n(= 1) = n(G \cap R') + n(G' \cap R) \\[3ex] = 60 + 90 \\[3ex] = 150 \\[3ex] $ $150$ students studied EXACTLY one books.

(IX.)
"Exactly two" books means "only two books"
"Exactly two" means $= 2$
Because we have only two books in this case, we shall find the students who studied BOTH books.

$ n(= 2) = n(G \cap R) \\[3ex] = 10 \\[3ex] $ $10$ students studied EXACTLY two books.

(X.)
"At most one" book means "one book or less"
"At most one" means $\le 1$
They are the students who studied ONLY the book of Genesis and the students who studied ONLY the book of Revelation and the students who did not study any book.

$ n(\le 1) = n(G \cap R') + n(G' \cap R) + n(G \cup R)' \\[3ex] = 60 + 90 + 0 \\[3ex] = 150 \\[3ex] $ $150$ students studied AT MOST one book.

(XI.)
"At most two" books means "two books or less"
"At most two" means $\le 2$
They are the students who studied BOTH books and the students who studied ONLY the book of Genesis and the students who studied ONLY the book of Revelation and the students who did not study any book.

$ n(\le 2) = n(G \cap R) + n(G \cap R') + n(G' \cap R) + n(G \cup R)' \\[3ex] = 10 + 60 + 90 + 0 \\[3ex] = 160 \\[3ex] $ $160$ students studied AT MOST two books.

(XII.)
"More than one" means $\gt 1$
They are the students who studied BOTH books

$ n(\gt 1) = n(G \cap R) \\[3ex] = 10 \\[3ex] $ $10$ students studied MORE THAN one book.

(XIII.)
"More than two" means $\gt 2$
We have only two books. So, there are no students who studied more than two books.
$0$ student studied MORE THAN two books.

(XIV.)
"Less than one" means $\lt 1$
They are the students who did not study any book.

$ n(\lt 1) = n(G \cup R)' \\[3ex] = 0 \\[3ex] $ $0$ student studied less than one book.

(XV.)
"Less than one" means $\lt 1$
They are the students studied ONLY the book of Genesis and the students who studied ONLY the book of Revelation and the students who did not study any book.

$ n(\lt 2) = n(G \cap R') + n(G' \cap R) + n(G \cup R)' \\[3ex] = 60 + 90 + 0 \\[3ex] = 150 \\[3ex] $ $150$ students studied less than two books.

(8.) ACT Among a group of 20 students, 13 students are members of the Math Club, 11 students are members of the Drama Club, and 9 students are members of both clubs.
How many of the 20 students are NOT members of either club?

$ F.\;\; 4 \\[3ex] G.\;\; 5 \\[3ex] H.\;\; 6 \\[3ex] J.\;\; 11 \\[3ex] K.\;\; 13 \\[3ex] $

$ n(\mu) = 20 \\[3ex] Let: \\[3ex] Math\;\;Club = M \\[3ex] Drama\;\;Club = D \\[3ex] n(M) = 13 \\[3ex] n(D) = 11 \\[3ex] n(M \cap D) = 9 \\[3ex] n(M \cap D') = 13 - 9 = 4 \\[3ex] n(M' \cap D) = 11 - 9 = 2 \\[3ex] n(M' \cap D') = ? \\[3ex] \implies \\[3ex] n(M' \cap D') + n(M \cap D') + n(M' \cap D) + n(M \cap D) = n(\mu) \\[3ex] n(M' \cap D') + 4 + 2 + 9 = 20 \\[3ex] n(M' \cap D') + 15 = 20 \\[3ex] n(M' \cap D') = 20 - 15 \\[3ex] n(M' \cap D') = 5 \\[3ex] $ 5 students are NOT members of either club.

(10.) ACT In a small high school with 20 seniors, 8 of the seniors are in soccer, 9 of the seniors are in band, and 5 of the seniors are in both.
How many seniors are in neither soccer nor band?

$ A.\;\; 15 \\[3ex] B.\;\; 12 \\[3ex] C.\;\; 11 \\[3ex] D.\;\; 8 \\[3ex] E.\;\; 3 \\[3ex] $

$ n(\mu) = 20 \\[3ex] Let: \\[3ex] Soccer = C \\[3ex] Band = B \\[3ex] n(C) = 8 \\[3ex] n(B) = 9 \\[3ex] n(C \cap B) = 5 \\[3ex] n(C \cap B') = 8 - 5 = 3 \\[3ex] n(C' \cap B) = 9 - 5 = 4 \\[3ex] n(C' \cap B') = ? \\[3ex] \implies \\[3ex] n(C' \cap B') + n(C \cap B') + n(C' \cap B) + n(C \cap B) = n(\mu) \\[3ex] n(C' \cap B') + 3 + 4 + 5 = 20 \\[3ex] n(C' \cap B') + 12 = 20 \\[3ex] n(C' \cap B') = 20 - 12 \\[3ex] n(C' \cap B') = 8 \\[3ex] $ 8 seniors are in neither soccer nor band.

(14.) ACT A health club surveyed 175 members about which types of equipment they had used in the past month.
Of the 175 members, 117 had used treadmills, 89 had used stationary bikes, and 53 had used both types of equipment.
Some members had used neither type of equipment.
Of the 175 members, how many had used treadmills, stationary bikes, or both?

$ A.\;\; 53 \\[3ex] B.\;\; 81 \\[3ex] C.\;\; 122 \\[3ex] D.\;\; 134 \\[3ex] E.\;\; 153 \\[3ex] $

The members that had used treadmills, stationary bikes, or both are those that:
used treadmills ONLY
used stationary bikes ONLY
used both treadmills and stationary bikes

$ Let: \\[3ex] treadmills = T \\[3ex] stationary\;\;bikes = B \\[3ex] n(\xi) = 175 \\[3ex] n(T) = 175 \\[3ex] n(B) = 89 \\[3ex] n(T \cap B) = 53 \\[3ex] n(T\;only) = n(T \cap B') = 117 - 53 = 64 \\[3ex] n(B\;only) = n(B \cap T') = 89 - 53 = 36 \\[3ex] n(T \cup B) \\[3ex] = n(T \cap B') + n(B \cap T') + n(T \cap B) \\[3ex] = 64 + 36 + 53 \\[3ex] = 153 \\[3ex] $ 153 members had used treadmills, stationary bikes, or both

(20.) ACT Kelly asked 120 students questions about skiing.
The results of the poll are shown in the table below.

Question	Yes	No
1. Have you skied either cross-country or downhill?	65	55
2. If you answered Yes to Question 1, did you ski downhill?	28	37
3. If you answered Yes to Question 1, did you ski cross-country?	45	20

After completing the poll, Kelly wondered how many of the students polled had skied both cross-country and downhill.
How many of the students polled indicated that they had skied both cross-country and downhill?

$ A.\;\; 73 \\[3ex] B.\;\; 65 \\[3ex] C.\;\; 47 \\[3ex] D.\;\; 18 \\[3ex] E.\;\; 8 \\[3ex] $

$ Let: \\[3ex] C = cross-country \\[3ex] D = downhill \\[3ex] n(C \cup D) = 65 \\[3ex] n(D) = 28 \\[3ex] n(C) = 45 \\[3ex] n(C \cap D) = e \\[3ex] n(C \cap D') = 45 - e \\[3ex] n(C' \cap D) = 28 - e \\[3ex] n(C \cap D) + n(C \cap D') + n(C' \cap D) = n(C \cup D) \\[3ex] e + (45 - e) + (28 - e) = 65 \\[3ex] e + 45 - e + 28 - e = 65 \\[3ex] 73 - e = 65 \\[3ex] 73 - 65 = e \\[3ex] e = 8 \\[3ex] $ 8 students skied both cross-country and downhill

(24.) ACT After polling a class of 20 music students by a show of hands, you find that 8 students play the guitar and 9 students play the piano.
Given that information, what is the minimum number of students in this music class who play both the guitar and the piano?

$ F.\;\; 0 \\[3ex] G.\;\; 1 \\[3ex] H.\;\; 8 \\[3ex] J.\;\; 9 \\[3ex] K.\;\; 17 \\[3ex] $

$ Let: \\[3ex] guitar = G \\[3ex] piano = P \\[3ex] n(G) = 8 \\[3ex] n(P) = 9 \\[3ex] n(G \cap P) = k \\[3ex] n(\mu) = 20 \\[3ex] n(G \cap P') = 8 - k \\[3ex] n(G \cap P') = 9 - k \\[3ex] \implies \\[3ex] n(G \cap P') + n(G' \cap P) + n(G \cap P) = n(\mu) \\[3ex] 8 - k + 9 - k + k = 20 \\[3ex] 17 - k = 20 \\[3ex] 17 - 20 = k \\[3ex] k = -3 \\[3ex] $ This is not possible because the number that play both the guitar and the piano cannot be negative
That number must be non-negative (zero and positive integers only)
So, let us analyze the options.
The number of students that play ONLY the guitar cannot be negative. This eliminates Options J. and K.
Any of Options F., G., and H. can be the number of students that play both the guitar and the piano.
However, the question is asking for the minimum number
That leaves us with zero...Option F. as the correct answer

(32.) ACT All 30 students in a history class took 2 tests.
Of these students, 28 passed the first test and 25 passed the second test.
What is the maximum possible number of students who passed both tests?

$ F.\;\; 23 \\[3ex] G.\;\; 25 \\[3ex] H.\;\; 27 \\[3ex] J.\;\; 28 \\[3ex] K.\;\; 30 \\[3ex] $

Let: \\[3ex] first\;\;test = F \\[3ex] second\;\;test = E \\[3ex] n(F) = 28 \\[3ex] n(E) = 25 \\[3ex] n(both\;\;tests) = n(F \cap E) = p \\[3ex] n(F\;\;ONLY) = n(F \cap E') = 28 - p \\[3ex] n(E\;\;ONLY) = n(E \cap F') = 25 - p \\[3ex] n(\xi) = 30 \\[3ex] \implies \\[3ex] n(F \cap E) + n(F \cap E') + n(F' \cap E) = n(\xi) \\[3ex] p + 28 - p + 25 - p = 30 \\[3ex] 53 - p = 30 \\[3ex] 53 - 30 = p \\[3ex] p = 23 \\[3ex] $ The number of students who passed both tests is 23
However, the maximum number of students who passed both tests is 25

Student: How is it 25?
Teacher: Because the number of students who passed either test ONLY cannot be negative
The number of students who passed the either test ONLY should be at least the number of people who passed either test
For example: the number of people who passed the second test ONLY should be at least the number of people who passed the second test.
In other words, the number of people who passed the second test ONLY cannot be negative.
Student: You used the second test as the example because 28 is greater than 25?
Teacher: That is correct.