Question

The length of a parasite in experiment A is 2.5 × 10–3 inch. The length of a parasite in experiment B is 1.25 × 10–4 inch. How much greater is the length of the parasite in experiment A compared to the length of the parasite in experiment B? A

Answer 1

Answer: 20 times

Step-by-step explanation:

Given : The length of a parasite in experiment A = $2.5\times10^{-3}\text{ inch}$

The length of a parasite in experiment B = $1.25\times10^{-4}\text{ inches}$

Then, the number of times the length of the parasite in experiment A is greater as compared to the length of the parasite in experiment B:-

$n=\dfrac{2.5\times10^{-3}}{1.25\times10^{-4}}\\\\=\dfrac{2.5}{1.25}\times10^{-3-(-4)}\\\\=2\times10^{-3+4}=2\times10^{1}=20$

Hence, the length of the parasite in experiment A is 20 times greater as compared to the length of the parasite in experiment B

Answer 2

We have been given that :-

The length of a parasite in experiment A is $2.5 \times 10^{-3}$

The length of a parasite in experiment B is $1.25 \times 10^{-4}$

Let us write the the  length of the parasite in experiment A in the exponent of -3.

$1.25 \times 10^{-4}= 0.125 \times 10^{-3}$

Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.

The difference in the length is given by

$2.5 \times 10^{-3} -0.125 \times 10^{-4}$

$=2.375 \times 10^{-3}$

Therefore, the length of the parasite in experiment A is $=2.375 \times 10^{-3}$ inches  greater than the length of the parasite in experiment B.