ISTQB CTFL-Foundation Exam - Topic 3 Question 47 Discussion

Actual exam question for ISTQB's CTFL-Foundation exam

Question #: 47
Topic #: 3

A garden irrigation system allows the user to specify 2 inputs:

1. Frequency - The number of times the system should be automatically switched on per day; minimum once per day, maximum 5 times

2. Duration - The duration of operation, in whole minutes, each time it is switched on; ranging from 1 to 60

Applying 2-value boundary value analysis Which option best options has the correct test set of valid and invalid boundary values?

CFrequency 0, 1, 5, 6; Duration 0, 1, 30, 60, 61

DFrequency 0, 1, 2, 5, 6; Duration 0, 1, 30, 60, 61

BFrequency 0, 1, 5, 6; Duration 59 seconds, 1 minute, 60 minutes, 60 minutes 1 second

AFrequency 1, 5; Duration 1, 60

Show Suggested Answer

Suggested Answer: C

Boundary value analysis involves testing at the boundaries between partitions. For the Frequency input, valid boundaries are 1 and 5, while invalid boundaries are 0 and 6. For the Duration input, valid boundaries are 1 and 60 minutes, while invalid boundaries are 0 and 61 minutes. Therefore, Option C includes the correct set of valid and invalid boundary values for both inputs.

by Jeffrey at Jan 26, 2026, 02:38 AM

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Tonja

5 days ago

I think the valid boundaries would be 1 and 5 for frequency, and 1 and 60 for duration, but I might be mixing it up with another question we did.

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Rosendo

10 days ago

I remember we practiced boundary value analysis, but I'm not sure if I should include the minimum and maximum values for both inputs.

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Teri

15 days ago

This seems straightforward enough. I'll just need to be careful to double-check my work and make sure I don't miss any of the key boundary conditions.

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Minna

20 days ago

I think I've got a good handle on the boundary values now. I'll make sure to thoroughly test each of those cases to ensure I cover all the bases.

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Winfred

25 days ago

The duration input is a bit trickier. The valid range is 1 to 60 minutes, so I'll need to test 1, 30, and 60. For the invalid values, I'll try 0 and 61.

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Julieta

1 month ago

Okay, let's see. For frequency, the valid range is 1 to 5, so I'll need to test the values 1, 2, 4, and 5. For the invalid values, I'll try 0 and 6.

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Oliva

1 month ago

Hmm, this looks like a classic boundary value analysis problem. I'll need to identify the valid and invalid boundaries for both the frequency and duration inputs.

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