The mean cell count in 2.5 × 10-4 cm3 of a well-mixed suspension is 60.
A student estimates the count in 1.0 cm3 using 60×2.5×10−460\times2.5\times10^{-4}60×2.5×10−4.
Which gives the correct calculation and estimate?
60÷(2.5×10−4)=2.4×10560\div(2.5\times10^{-4})=2.4\times10^{5}60÷(2.5×10−4)=2.4×105
60×(2.5×10−4)=1.5×10−260\times(2.5\times10^{-4})=1.5\times10^{-2}60×(2.5×10−4)=1.5×10−2
60÷(2.5×10−4)=2.4×10−360\div(2.5\times10^{-4})=2.4\times10^{-3}60÷(2.5×10−4)=2.4×10−3
60×(2.5×10−4)=1.5×10260\times(2.5\times10^{-4})=1.5\times10^{2}60×(2.5×10−4)=1.5×102
48 exam-style questions on Edexcel GCSE Biology 1.2 Number, size and scale, covering 1.2.1 Number, size and scale with estimations and 1.2.2 Quantitative units in relation to cells. Each one has a worked solution and a mark scheme showing where the marks go.