It's important that all testers should be able to write test cases based on Equivalence Partitioning and Boundary Value Analysis. Taking this into consideration ISTQB is having significant importance for this topic in the **ISTQB Foundation level Certificate exam**. Good practice and logical thinking can make it very easy to solve these questions.

**What is Equivalence Partitioning?**

Equivalence Partitioning is a method for deriving test cases. In this method, equivalence classes (for input values) are identified such that each member of the class causes the same kind of processing and output to occur.

The values at the extremes (start/end values or lower/upper-end values) of such class are known as Boundary values. Analyzing the behavior of a system using such values is called **Boundary Value Analysis** (BVA).

**Here are few sample questions for practice from ISTQB exam papers on Equivalence Partitioning and BVA. (Ordered: Simple to little complex) **

**Question #1)**

One of the fields on a form contains a text box that accepts numeric values in the range of 18 to 25. Identify the invalid Equivalence class.

a) 17

b) 19

c) 24

d) 21

**Solution:**

The text box accepts numeric values in the range 18 to 25 (18 and 25 are also part of the class). So this class becomes our valid class. But the question is to identify invalid equivalence class. The classes will be as follows:

Class I: values < 18 => invalid class

Class II: 18 to 25 => valid class

Class III: values > 25 => invalid class

17 fall under an invalid class. 19, 24 and 21 falls under valid class. **So the answer is ‘A’**

**Question #2)**

In an Examination, a candidate has to score a minimum of 24 marks in order to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalence values if the student clears the exam.

a) 22,23,26

b) 21,39,40

c) 29,30,31

d) 0,15,22

**Solution:**

The classes will be as follows:

Class I: values < 24 => invalid class

Class II: 24 to 40 => valid class

Class III: values > 40 => invalid class

We have to identify Valid Equivalence values. Valid Equivalence values will be there in a Valid Equivalence class. All the values should be in Class II. **So the answer is ‘C’**

**Question #3)**

One of the fields on a form contains a text box that accepts alphanumeric values. Identify the Valid Equivalence class

a) BOOK

b) Book

c) Boo01k

d) Book

**Solution:**

Alphanumeric is a combination of alphabets and numbers. Hence we have to choose an option which has both of these. A valid equivalence class will consist of both alphabets and numbers. Option ‘c’ contains both alphabets and numbers. **So the answer is ‘C’**

**Question #4)**

The Switch is switched off once the temperature falls below 18 and then it is turned on when the temperature is more than 21. When the temperature is more than 21. Identify the Equivalence values which belong to the same class.

a) 12,16,22

b) 24,27,17

c) 22,23,24

d) 14,15,19

**Solution:**

We have to choose values from the same class (it can be a valid or invalid class). The classes will be as follows:

Class I: less than 18 (switch turned off)

Class II: 18 to 21

Class III: above 21 (switch turned on)

Only in Option ‘c’, all values are from one class. Hence the **answer is ‘C’**. (Please note that the question does not talk about valid or invalid classes. It is only about values in the same class)

**Question #5)**

A program validates a numeric field as follows: values less than 10 are rejected, values between 10 and 21 are accepted, values greater than or equal to 22 are rejected. Which of the following input values cover all of the equivalence partitions?

a. 10,11,21

b. 3,20,21

c. 3,10,22

d. 10,21,22

**Solution:**

We have to select values that fall in all the equivalence class (valid and invalid both). The classes will be as follows:

Class I: values <= 9 => invalid class

Class II: 10 to 21 => valid class

Class III: values >= 22 => invalid class

All the values from option ‘c’ fall under all different equivalence classes. **So the answer is ‘C’.**

**Question #6)**

A program validates a numeric field as follows: values less than 10 are rejected, values between 10 and 21 are accepted, values greater than or equal to 22 are rejected. Which of the following covers the MOST boundary values?

a. 9,10,11,22

b. 9,10,21,22

c. 10,11,21,22

d. 10,11,20,21

**Solution:**

We have already come up with the classes as shown in question 5. The boundaries can be identified as 9, 10, 21, and 22. These four values are in option ‘b’. **So the answer is ‘B’**

**Question #7)**

In a system designed to work out the tax to be paid:

An employee has £4000 of salary tax-free.

The next £1500 is taxed at 10%.

The next £28000 after that is taxed at 22%.

Any further amount is taxed at 40%.

To the nearest whole pound, which of these groups of numbers fall into three DIFFERENT equivalence classes?

a) £4000; £5000; £5500

b) £32001; £34000; £36500

c) £28000; £28001; £32001

d) £4000; £4200; £5600

**Solution:**

The classes will be as follows:

Class I : 0 to £4000 => no tax

Class II : £4001 to £5500 => 10 % tax

Class III : £5501 to £33500 => 22 % tax

Class IV : £33501 and above => 40 % tax

Select the values which fall in three different equivalence classes. Option ‘d’ has values from three different equivalence classes. **So the answer is ‘D’.**

**Question #8)**

In a system designed to work out the tax to be paid:

An employee has £4000 of salary tax-free.

The next £1500 is taxed at 10%.

The next £28000 after that is taxed at 22%.

Any further amount is taxed at 40%.

To the nearest whole pound, which of these is a valid Boundary Value Analysis test case?

a) £28000

b) £33501

c) £32001

d) £1500

**Solution:**

The classes are already divided in question # 7. We have to select a value which is a boundary value (start/end value). 33501 is a boundary value. **So the answer is ‘B’**.

**Question #9)**

Given the following specification, which of the following values for age are in the SAME equivalence partition?

If you are less than 18, you are too young to be insured.

Between 18 and 30 inclusive, you will receive a 20% discount.

Anyone over 30 is not eligible for a discount.

a) 17, 18, 19

b) 29, 30, 31

c) 18, 29, 30

d) 17, 29, 31

**Solution:**

The classes will be as follows:

Class I: age < 18 => not insured

Class II: age 18 to 30 => 20 % discount

Class III: age > 30 => no discount

Here we cannot determine if the above classes are valid or invalid, as nothing is mentioned in the question. (But according to our guess we can say I and II are valid and III is invalid. But this is not required here.) We have to select values that are in the SAME equivalence partition. Values from option ‘c’ fall in the same partition. **So the answer is ‘C’.**

**These are a few sample questions for practice from ISTQB papers. We will continue to add more ISTQB question papers with answers in the coming posts. **

**About the Author:**

**This is a guest article by “N. Sandhya Rani”.** She is having around 4 years of experience in Software Testing mostly in Manual Testing. She is helping many aspirant software testers to clear the ISTQB testing certification exam.

**Put your questions related to the ISTQB exam in the comment section below.**

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Hi i’m preparing fot ISTQB foundation level cetification planning to complete next week.can u plz send me latest dumps

My e mail is pudururaja@gmail.com

Hi All,

Although the user “ISTQBPrep” responded with their answers to Mr. Mallikarjun’s 58 thoughtful questions, since Mr Mallikarjun himself has not yet posted his answers, below I am posting my answers also, and hope to see response from Mr. Mallikarjun, for which big thanks in advance!

With the simple explanation given by “softwaretestinghelp” team and by others on the forum, I now feel a bit comfortable on BVA and EP.

However, I need similar help with topics like branch/statement/decision coverage. Any help is appreciated.

Regards

Sai

1c

2c

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==========================

I think 5a

Q. 24: A wholesaler sells printer cartridges. The minimum order quantity is 5. There is

a 20% discount for orders of 100 or more printer cartridges. You have been asked to

prepare test cases using various values for the number of printer cartridges ordered.

Which of the following groups contain three test inputs that would be generated using

Boundary Value Analysis?A. 5, 6, 20

B. 4, 5, 80

C. 4, 5, 99

D. 1, 20, 100

solution to this question

I THINK ANSWER IS D. 1,20,100 …5 IS MINIMUM ORDER QUANTITY..SO 1 IS 1ST TEST CASE..FOR 5-99 NO DISCOUNT…20 WILL BE 2ND TEST CASE…20% DISCOUNT FOR ORDERS OF 100 OR MORE PRINTER CARTRIDGES…100 WILL BE 3RD TEST CASE..

Dear AYUSHI, Question is about Boundary value analysis and not about equivalence partitioning. I am thinking boundary values be : 4, 5, 99, 100, I am thinking the answer would be ‘C’: 4, 5, 99. Please correct me if I am wrong.

There is a temperature controlling system in a Server Room, which takes current server room temperature

(Celsius value) as the input and rotate fans on a specific round per minute (rpm) based on the input values.

The RPM values of the fans based on input temperatures are as follows:

°C < 15 = 500 rpm

15<=°C<20 = 750 rpm

20 <= °C <25 = 1000 rpm

25 <= °C <35 = 2000 rpm

35 <=°C = 5000 rpm

Find the least number of equivalence classes and possible test input values required for this scenario?

i think least number of possible test input values is 5……15<=C>5 ..15,16,17,18,19……20<=C>5…20,21,22,23,24…and least number of equivalence classes is also 5…

1. C<15

2. 15<=C<20

3. 20<=C<25

4. 25<=C=35

i think least number of equivalence classes is 6:

1. C<15

2. 15<=C<20

3. 20<=C<25

4. 25<=C=35

5. 35 5000 (invalid)

Can any one please provide latest ISTQB FL 2016 dumps

Can anyone suggest Imp topics for ISTQB FL exam

Please answer the below question. According to me the answer should be b (x=3 y=2, x3 y2) but it is given as c

17. If the pseudocode below were a programming language ,how many tests are required to achieve

100% statement coverage?

1. If x=3 then

2. Display_messageX;

3. If y=2 then

4. Display_messageY;

5. Else

6. Display_messageZ;

7. Else

8. Display_messageZ;

a. 1

b. 2

c. 3

d. 4

It’s correct, the answer is c

1st: x = 3 and y = 2 -> msg x and msg y.

2nd: x = 3 and y 2 -> msg x and msg z.

3rd: x 3 and y any value -> only msg z.

correct answer B.

1st: x = 3 and y = 2 -> msg x and msg y.

2nd: x =different of 3 and y different of 2 -> msg z and msg z.

Thanks! I couldn’t understand the logic behind Question # 8. Would anyone explain it. Best

Please provide answers of all questions so that we can cross check with our answers and be sure.

design the test case for boundary value analysis of he following consider a program that prompts the user to input 3 number (say x y z) and the data type for input parameters ensures that these will be integers greater than 0 and less or equal to 100.the program should then output the numbers in ascending order

plz i need n answer fst its vry imp ? plz….

the above question is for bscit students. software testing?

What if i have 2 different input in the same program for example

int a, b;

but one of them have to 50<a while the other is 51<b

should i have 2 partitions ?

Can anyone explain below questions with some examples?

18) When testing a grade calculation system, a tester determines that all scores from 90 to 100 will yield a grade of A, but scores below 90 will not. This analysis is known as:

a) Equivalence partitioning

b) Boundary value analysis

c) Decision table

d) Hybrid analysis

Ans – A (not option B)

One technique of Black Box testing is Equivalence Partitioning. In a program

statement that accepts only one choice from among 10 possible choices,

numbered 1 through 10, the middle partition would be from _____ to _____

a) 4 to 6

b) 0 to 10

c) 1 to 10

d) None of the above

Ans -D (Not option C)

Dear,

Any one tell me the correct formula for boundary value analysis

BV = LV-1, LV, LV+1, UV-1, UV, UV+1.

BV = LV-1, LV+1, UV-1,UV+1.

(min-1,min,min+ 1)and (max- 1,max,max+ 1)

Hello,

Plz help me out with following Q. Correct ans is C but I feel it should be D.

An input field takes the year of birth between 1900 and 2004

The boundary values for testing this field are

a. 0,1900,2004,2005

b. 1900, 2004

c. 1899,1900,2004,2005

d. 1899, 1900, 1901,2003,2004,2005?

Dear Pranesh, my answer would be ‘C’.

Input value 1900 and 1901 both belong to the same class (Valid class)

Similarly, Input value 2004 and 2005 both belong to the same class (Valid class)

One value from Valid class be sufficient for boundary value analysis.

Test data planning essentially includes?? BVA or Test Procedure planning ??

Hey Pranesh Joshi , the answer is C becoz only the upper boundary and lower boundary will be taking into consideration becoz the values mentioned in option D 1901,2003, 2004 will be covered cumulatively in answer C btw 1900 to 2004.

Can you explain this

Q28. Minimum Test Required for Statement Coverage :-

Disc = 0

Order-qty = 0

Read Order-qty

If Order-qty >=20 then

Disc = 0.05

If Order-qty >=100 then

Disc =0.1

End if

End if

a) Statement coverage is 4

b) Statement coverage is 1

c) Statement coverage is 3

d) Statement Coverage is 2

answer B

Can you please help and draw flowchart for this task.

1. Pick up and read the newspaper.

2. Look at what is on television.

3. If there is a program that you are interested in watching then switch the television on and watch the program.

4. Otherwise.

5. Continue reading the newspaper.

6. If there is a crossword in the newspaper then try and complete the crossword.

A. SC = 1 and DC = 3.

B. SC = 1 and DC = 2.

C. SC = 2 and DC = 3.

D. SC = 2 and DC = 2.

E. SC = 1 and DC = 1.

Thank you so much. These sums helped me out a lot :)

Can anyone help me put with this?

Below you will find the requirements to identify the Account Diversity Grade of a user. Read the

requirements carefully and identify what test users you need to setup in order to completely

test and make sure all the below requirements are covered. (Note: you should identify the

optimum (minimum) number of users needed to test all of the requirements)

Requirements:

A user can have different types of loan accounts. Now we grade a user’s Account Diversity based

on two factors.

1) loanTypeCount

2) totalAccounts

loanTypeCount = the number of different (distinct) LoanType values for all accounts that the

user has.

However do not include LoanType = Unknown & Collections but include all others

Applicable values for LoanType are ( Home Loan, Heloc, Credit Card, Car Loan, Collections,

Unknown)

totalAccounts = total number of loan accounts user has (do not include LoanType = Unknown &

Collections but include all others)

example-> if user has 3 credit cards and 2 home loans and 1 Collection account, then

totalAccounts = 5 and loanTypeCount = 2)

The logic to determine accountDiversityGrade is the following:

If totalAccounts> 20 or loanTypeCount >= 4, accountDiversityGrade = A

Else if totalAccounts> 10 or loanTypeCount = 3, accountDiversityGrade = B

Else if totalAccounts>= 5 or loanTypeCount= 2, accountDiversityGrade = C

Else if totalAccounts > 0 or loanTypeCount = 1, accountDiversityGrade = D

Else accountDiversityGrade=null (n/a)

An employee has £4000 of salary tax-free. The next £1500 is taxed at 10%. The next £28000 after that is taxed at 22%. Any further amount is taxed at 40%.find equivalence partion

How can be the answer for the above 5th question in the equivalence partitioning be C ?

THANKS

Equivalence partition and bva on date of birth

1. d

2. b

3. d

4. c

5. d

6. a

7. c

8. b

9. a

10. a

11. c

12. a

13. b

14. c

15. b

16. b

17. c

18. c

19. a

20. c

21. b

22. d

23. c

24. a

25. b

26. d

27. a

28. d

29. c

30. b

What is the corect answer of Question 7 ?

A shopping website gives discount based on prices of total purchased items.if customer purchases items of below Rs 2000 then there is no discount.for purchases of Rs 2000 and above it gives 10% discount and above 20000 it gives 15% discount.specify valid equivalence classes for the given input conditions.

Identify minimum number of users to test requirement?

Below you will find the requirements to identify the Account Diversity Grade of a user. Read the requirements carefully and identify what test users you need to setup in order to completely test and make sure all the below requirements are covered. (Note: you should identify the optimum (minimum) number of users needed to test all of the requirements)

Requirements: A user can have different types of loan accounts.Now we grade a user’s account Diversity based on two factors. 1) loanTypeCount 2) totalAccounts

loanTypeCount = the number of different (distinct) LoanType values for all accounts that the user has.

However do not include LoanType = Unknown & Collections but include all others Applicable values for LoanType are ( Home Loan, Heloc, Credit Card, Car Loan, Collections, Unknown)

totalAccounts = total number of loan accounts user has (do not include LoanType = Unknown & Collections but include all others)

example-> if user has 3 credit cards and 2 home loans and 1 Collection account, then totalAccounts = 5 and loanTypeCount = 2)

The logic to determine accountDiversityGrade is the following:

If totalAccounts> 20 or loanTypeCount >= 4, accountDiversityGrade = A

Else if totalAccounts> 10 or loanTypeCount = 3, accountDiversityGrade = B

Else if totalAccounts>= 5 or loanTypeCount= 2, accountDiversityGrade = C

Else if totalAccounts > 0 or loanTypeCount = 1, accountDiversityGrade = D

Else accountDiversityGrade=null (n/a)

You are testing software that controls the amount of water sprayed by an automatic sprinkler

system. The amount to be sprayed in an hour is determined by the weather conditions for the

previous 3 days. The weather conditions can be either sunny, cloudy or rainy. The maximum

amount of water will be sprayed if the previous conditions were sunny, sunny, sunny. No water

will be sprayed if there were two rainy days in the previous three days. Varying amounts will be

sprayed depending on the mix of the previous days. For example, rainy, sunny, sunny will get

more water than sunny, cloudy, rainy.

The software also determines the type of spray to use based on the type of grass being

sprayed. There are five different categories of grasses that are supported.

By applying equivalence partitioning to the weather conditions, how many test cases will be

needed to cover the weather conditions and spray types?

a. 9

b. 15

c. 21

d. 27

Hi Please send me dumps for ISTQB 2011 ……ASAP….

My email address is rodney.mlungisi@gmail.com

Question : Postal rates for ‘light letters’ are 25p up to l0g, 35p up to 50g plus an extra l0p for each additional 25g up to l00g.

Which test inputs (in grams) would be selected using equivalence partitioning?

a. 8,42,82,102

b. 4,15, 65, 92,159

c. 10,50,75,100

d. 5, 20, 40, 60, 80

Can anybody solve the question in detail and send it to my mail.

Since the question does not mention specifically valid or invalid equivalence partitions, the answer should be according to me b as it includes all the valid and invalid EPs.

Ans is C:-

Boundary values for 5 are 4,6

boundary values for 100 are 99 & 101

so the option is C., where 4, 5 are boundary values for 5 and 99, is for 100