JAMB 1997 · UME · Q21

Two binary operations ∗* and ⊕\oplus are defined by m∗n=mn−n−1m * n = mn - n - 1 and m⊕n=mn+n−2m \oplus n = mn + n - 2 for all real numbers mm, nn. Find the value of 3⊕(4∗5)3 \oplus (4 * 5).

Worked solution (try it first)
  1. Work out the bracket with the ∗* rule: 4∗5=4×5−5−1=144 * 5 = 4 \times 5 - 5 - 1 = 14.
  2. Now use the ⊕\oplus rule with m=3m = 3 and n=14n = 14: 3⊕14=3×14+14−23 \oplus 14 = 3 \times 14 + 14 - 2.
  3. That is 42+14−2=5442 + 14 - 2 = 54, option C.

Report a problem with this question