JSS3: MATHEMATICS – 1ST TERM
- Binary Number System I | Week 15Topics|1 Quiz
- Binary Number System II | Week 26Topics|1 Quiz
- Word Problems I | Week 34Topics|1 Quiz
- Word Problems with Fractions II | Week 41Topic|1 Quiz
- Factorisation I | Week 54Topics|1 Quiz
- Factorisation II | Week 63Topics|1 Quiz
- Factorisation III | Week 73Topics|1 Quiz
- Substitution & Change of Subject of Formulae | Week 82Topics|1 Quiz
- Simple Equations Involving Fractions | Week 93Topics|1 Quiz
- Word Problems | Week 101Topic|1 Quiz
JSS3 – Simple Equations Involving Fractions: 1st term Evaluation questions | Week 9
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- Question 1 of 12
1. Question
Solve the equation \( \frac{ x + 7}{2} + \frac{x + 21}{5} \scriptsize = 0 \)
CorrectIncorrect - Question 2 of 12
2. Question
If \( \frac{ 2y \; – \; 3}{4} + \frac{y \; – \; 1}{8} \scriptsize = 1 \)
find y
CorrectIncorrect - Question 3 of 12
3. Question
What value of x will make \( \frac{ x \: – \: 2}{7} \: – \: \frac{5x \: – \: 3}{3} \scriptsize \: – \: 5 = 0 \)
CorrectIncorrect - Question 4 of 12
4. Question
If \( \frac{ a \: – \: 3}{4} \: – \: \frac{5 \: + \: a}{2} \scriptsize = 0 \)
Find a
CorrectIncorrect - Question 5 of 12
5. Question
Find the value of b if \( \frac{7 + b}{5} \scriptsize = \; – \; 3 \)
CorrectIncorrect - Question 6 of 12
6. Question
If \( \scriptsize p =\normalsize \frac{35 \: – \: 3p}{2} \)
Then p = ?
CorrectIncorrect - Question 7 of 12
7. Question
If \( \frac{14}{1 \: + \: y} \scriptsize = 2 \)
Find y
CorrectIncorrect - Question 8 of 12
8. Question
What value of x makes\( \frac{5}{x \: – \: 3 } = \frac{4}{x} \) true
CorrectIncorrect - Question 9 of 12
9. Question
If \( \frac{2}{a + 3} = \frac{3}{3a – 1} \)
What is the value of a
CorrectIncorrect - Question 10 of 12
10. Question
Solve the equation \( \frac{1}{y \: – \:1} \; – \; \frac{3}{2y\: +\: 1} \scriptsize = 0 \)
CorrectIncorrect - Question 11 of 12
11. Question
Find the value of x in \( \frac{12}{x} \: – \: \frac{10}{x\: + \:1} \scriptsize = 0 \)
CorrectIncorrect - Question 12 of 12
12. Question
Solve \( \frac{3}{x\: + \:2} = \frac{2}{ x \: – \: 3} \)
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