Change Of Subject Of Formula - Mathematics Jamb Past Questions and Answers

Question 1 :

• JAMB 1979

The mean of the numbers 1.2, 1.0, 0.9, 1.4, 0.8, 0.8, 1.2 and 1.1 is

A. 1.5
B. 0.8
C. 1.0
D. 1.02
E. 1.05

Question 2 :

• JAMB 2019

Make q the subject of the formula in the equation \(\frac{mn}{a^2} - \frac{pq}{b^2} = 1\)

A. \(q = \frac{b^2(mn - a^2)}{a^2 p}\)
B. \(q = \frac{m^2 n - a^2}{p^2}\)
C. \(q = \frac{mn - 2b^2}{a^2}\)
D. \(q = \frac{b^2 (n^2 - ma^2)}{n}\)

Question 3 :

• JAMB 2018

If P = [\(\frac{Q(R - T)}{15}\)] \(\frac{1}{3}\) make T the subject of the relation

A. T = \(\frac{R + P^3}{15Q}\)
B. T = \(\frac{R - 15P^3}{Q}\)
C. T = \(\frac{R - 15P^3}{Q}\)
D. T = \(\frac{15R - Q}{P^3}\)

Question 4 :

• JAMB 2018

If \(\frac{x}{a + 1}\) + \(\frac{y}{b}\) = 1. Make y the subject of the relation.

A. \(\frac{b(a + 1 - x)}{a + 1}\)
B. \(\frac{a + 1}{b(a - x + 1)}\)
C. \(\frac{a(b - x + 1)}{b + 1}\)
D. \(\frac{b}{a(b - x + 1)}\)

Question 5 :

• JAMB 2018

Make T the subject of the relation.

A. T = \(\frac{R + P3}{15Q}\)
B. T = \(\frac{R - 15P^3}{Q}\)
C. T =R - \(\frac{15P^3}{Q}\)
D. T = \(\frac{15R + Q}{P^3}\)