Stepwise
Start practisingPractise

JAMB 1983 · UME · Q9

If 0.0000152×0.00042=A×10B0.0000152 \times 0.00042 = A \times 10^B, where 1≤A<101 \le A < 10, find AA and BB.

Worked solution (try it first)
  1. Write each number in standard form: 0.0000152=1.52×10−50.0000152 = 1.52 \times 10^{-5} and 0.00042=4.2×10−40.00042 = 4.2 \times 10^{-4}.
  2. Multiply the numbers: 1.52×4.2=6.3841.52 \times 4.2 = 6.384.
  3. Add the powers of 10: 10−5×10−4=10−910^{-5} \times 10^{-4} = 10^{-9}.
  4. So the product is 6.384×10−96.384 \times 10^{-9}: A=6.38A = 6.38 and B=−9B = -9, option B.

Report a problem with this question