JAMB and WAEC Mathematics questions

861 questions with answers and explanations. Tap an option to check yourself.

All JAMB WAEC

Question 501· Permutation and combination· JAMB
In how many ways can a committee of 4 be chosen from 7 people?
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Answer: D. 35
Explanation and discussion
Question 502· Permutation and combination· JAMB
Evaluate ⁶P₂.
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Answer: D. 30
Explanation and discussion
Question 503· Permutation and combination· JAMB
In how many ways can the letters of the word ADMIT be arranged if A must come first?
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Answer: C. 24
Explanation and discussion
Question 504· Permutation and combination· JAMB
The value of 0! is
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Answer: C. 1
Explanation and discussion
Question 505· Permutation and combination· JAMB
In how many ways can first and second places be awarded among 9 runners?
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Answer: A. 72
Explanation and discussion
Question 506· Permutation and combination· JAMB
How many different selections of 10 books can be made from 12 different books?
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Answer: D. 66
Explanation and discussion
Question 507· Permutation and combination· JAMB
Ten people each shake hands once with every other person. How many handshakes take place?
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Answer: A. 45
Explanation and discussion
Question 508· Permutation and combination· JAMB
A committee of 2 boys and 1 girl is to be chosen from 4 boys and 3 girls. In how many ways can this be done?
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Answer: D. 18
Explanation and discussion
Question 509· Permutation and combination· JAMB
In how many ways can 5 people sit in a row of 5 chairs?
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Answer: A. 120
Explanation and discussion
Question 510· Permutation and combination· JAMB
In how many ways can the letters of the word SEED be arranged?
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Answer: C. 12
Explanation and discussion
Question 511· Permutation and combination· JAMB
In how many ways can all the letters of the word LAGOS be arranged?
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Answer: D. 120
Explanation and discussion
Question 512· Permutation and combination· JAMB
In how many ways can first, second and third prizes be awarded among 5 students?
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Answer: B. 60
Explanation and discussion
Question 513· Permutation and combination· JAMB
Evaluate ⁸C₃.
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Answer: A. 56
Explanation and discussion
Question 514· Permutation and combination· JAMB
If ⁿC₂ = 15, find n.
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Answer: D. 6
Explanation and discussion
Question 515· Permutation and combination· JAMB
A team of 3 is to be chosen from 5 boys and 4 girls. In how many ways can this be done if the team must include at least one girl?
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Answer: B. 74
Explanation and discussion
Question 516· Polynomials· WAEC
If (x + 3) is a factor of x² + kx − 6, find k.
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Answer: D. 1
Explanation and discussion
Question 517· Polynomials· JAMB
Solve x/2 − (x − 1)/3 = 1.
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Answer: B. 4
Explanation and discussion
Question 518· Polynomials· JAMB
x = 2 is a root of x³ − 4x² + kx − 2 = 0. Find k.
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Answer: B. 5
Explanation and discussion
Question 519· Polynomials· WAEC
Solve simultaneously: 2x + 3y = 13 and x − y = −1.
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Answer: B. x = 2, y = 3
Explanation and discussion
Question 520· Polynomials· JAMB
Simplify 1/(x − 1) − 1/(x + 1).
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Answer: C. 2/(x² − 1)
Explanation and discussion

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JAMB and WAEC Mathematics questions

In how many ways can a committee of 4 be chosen from 7 people?

Evaluate ⁶P₂.

In how many ways can the letters of the word ADMIT be arranged if A must come first?

The value of 0! is

In how many ways can first and second places be awarded among 9 runners?

How many different selections of 10 books can be made from 12 different books?

Ten people each shake hands once with every other person. How many handshakes take place?

A committee of 2 boys and 1 girl is to be chosen from 4 boys and 3 girls. In how many ways can this be done?

In how many ways can 5 people sit in a row of 5 chairs?

In how many ways can the letters of the word SEED be arranged?

In how many ways can all the letters of the word LAGOS be arranged?

In how many ways can first, second and third prizes be awarded among 5 students?

Evaluate ⁸C₃.

If ⁿC₂ = 15, find n.

A team of 3 is to be chosen from 5 boys and 4 girls. In how many ways can this be done if the team must include at least one girl?

If (x + 3) is a factor of x² + kx − 6, find k.

Solve x/2 − (x − 1)/3 = 1.

x = 2 is a root of x³ − 4x² + kx − 2 = 0. Find k.

Solve simultaneously: 2x + 3y = 13 and x − y = −1.

Simplify 1/(x − 1) − 1/(x + 1).