Jamb Mathematics Questions - Change of subject of formula
Question 6:
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0
- A C = \(\frac{2d - 5 - b}{ad}\)
- B C = \(\frac{2d - 5 - abd}{ad}\)
- C C = \(\frac{2d - 5 - ab}{ad}\)
- D C = \(\frac{5 - 2d - b}{ad}\)
Question 7:
Make c the subject of formula v = 1 - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))
- A [\(\frac{7}{3}\) + \(\frac{5}{a}\)(v - 1)] + b
- B [\(\frac{7b}{3}\) + \(\frac{5}{a}\)(v - 1)]
- C \(\frac{7}{3}\)[b + \(\frac{5}{a}\)(v - 1)]
- D \(\frac{7}{3}\)[b + \(\frac{5v - 1}{a}\)]
Question 8:
Which of the formula below represents the general terms of the following set of numbers(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......) for n = 1, 2, 3, 4.......?
- A \(\frac{2}{n + 1}\)
- B (-1)<sup style='font-size: smaller;'>n + 1</sup> \(\frac{2}{n + 1}\)
- C (-1)<sup style='font-size: smaller;'>n</sup> \(\frac{2}{n + 1}\)
- D \(\frac{n}{2n - 1}\)
- E (-1)<sup style='font-size: smaller;'>n</sup>\(\frac{n}{2n - 1}\)
Question 9:
If b = a + cp and r = ab + \(\frac{1}{2}\)cp2, express b2 in terms of a, c, r.
- A B<sup style='font-size: smaller;'>2</sup> = aV + 2cr
- B B<sup style='font-size: smaller;'>2</sup> = ar + 2c<sup style='font-size: smaller;'>2</sup>r
- C B<sup style='font-size: smaller;'>2</sup> = a<sup style='font-size: smaller;'>2</sup> = \(\frac{1}{2}\) cr<sup style='font-size: smaller;'>2</sup>
- D B<sup style='font-size: smaller;'>2</sup> = \(\frac{1}{2}\)ar<sup style='font-size: smaller;'>2</sup> + c
- E B<sup style='font-size: smaller;'>2</sup> = 2cr - a<sup style='font-size: smaller;'>2</sup>
Question 10:
A variable y is inversely proportional to x2, when y = 10, x = 2. What is y when x = 10?
View Answer & Explanation