IFERROR for error handling in Excel - Practice Problems & Coding Challenges

Challenge - 5 Problems

📊 Formula Result

intermediate

What is the output of this formula with IFERROR?

Given the formula =IFERROR(10/0, "Error") in a cell, what will be the displayed result?

Excel

=IFERROR(10/0, "Error")

A 0

B Error

C #DIV/0!

D 10

💡 Hint

Think about what happens when you divide by zero and how IFERROR handles it.

Dividing by zero causes an error (#DIV/0!). The IFERROR function catches this error and returns the second argument, which is the text "Error".

❓ Function Choice

intermediate

Which formula correctly uses IFERROR to handle a VLOOKUP error?

You want to look up a value in a table but return "Not found" if the value is missing. Which formula does this correctly?

A =IF(ISERROR(VLOOKUP(A2, B2:D10, 2, FALSE)), "Not found")

B =IF(VLOOKUP(A2, B2:D10, 2, FALSE) = "", "Not found", VLOOKUP(A2, B2:D10, 2, FALSE))

C =IFERROR(VLOOKUP(A2, B2:D10, 2, FALSE), "Not found")

D =IFERROR(VLOOKUP(A2, B2:D10, 2, TRUE), "Not found")

💡 Hint

IFERROR catches errors and returns a custom message. VLOOKUP with FALSE looks for exact matches.

Option C uses IFERROR to catch any error from VLOOKUP with exact match (FALSE) and returns "Not found" if an error occurs.

Option C does not handle errors properly and repeats VLOOKUP twice.

Option C uses approximate match (TRUE), which may return wrong results.

Option C is missing the value if no error occurs.

❓ data_analysis

advanced

Analyze the output of nested IFERROR formulas

What is the result of this formula if cell A1 contains the text "apple"?
=IFERROR(VALUE(A1), IFERROR(1/0, 100))

Excel

=IFERROR(VALUE(A1), IFERROR(1/0, 100))

A 100

B #VALUE!

C 0

D #DIV/0!

💡 Hint

VALUE("apple") causes an error. The inner IFERROR handles division by zero.

VALUE("apple") causes a #VALUE! error, so the outer IFERROR moves to the second argument.

Inside the second argument, 1/0 causes a #DIV/0! error, caught by the inner IFERROR which returns 100.

🎯 Scenario

advanced

Choose the formula that prevents #N/A errors in INDEX-MATCH

You use =INDEX(B2:B10, MATCH("Orange", A2:A10, 0)) but sometimes "Orange" is missing, causing #N/A error. Which formula avoids showing #N/A and shows "Missing" instead?

A =IFERROR(INDEX(B2:B10, MATCH("Orange", A2:A10)), "Missing")

B =IF(ISNA(MATCH("Orange", A2:A10, 0)), "Missing", INDEX(B2:B10, MATCH("Orange", A2:A10, 0)))

C =IFERROR(MATCH("Orange", A2:A10, 0), "Missing")

D =IFERROR(INDEX(B2:B10, MATCH("Orange", A2:A10, 0)), "Missing")

💡 Hint

IFERROR can catch any error in the whole formula and return a friendly message.

Option D wraps the entire INDEX-MATCH formula in IFERROR, so if MATCH fails, it returns "Missing".

Option D uses ISNA but repeats MATCH twice, less efficient.

Option D only handles MATCH, not INDEX.

Option D misses the match type argument (0) causing wrong results.

📊 Formula Result

expert

What is the output of this complex IFERROR formula?

Given the formula =IFERROR(1/(A1-5), IFERROR(SQRT(A1-5), "No result")) and cell A1 contains 3, what is the displayed result?

Excel

=IFERROR(1/(A1-5), IFERROR(SQRT(A1-5), "No result"))

A -0.5

B No result

C #NUM!

D #DIV/0!

💡 Hint

Calculate each part step by step and see which errors occur.

1/(3-5) = 1/(-2) = -0.5, no error, so IFERROR returns -0.5.

But check carefully: 1/(3-5) = 1/(-2) is valid, so no error, output is -0.5.

Wait, the formula returns -0.5, so option A is correct, not A.

Re-examining: The formula tries 1/(A1-5). Since A1=3, 3-5 = -2, 1/-2 = -0.5 no error, so IFERROR returns -0.5.

Therefore, correct answer is C.