Question

1. sadie wrote a report for school that was $3\frac{1}{4}$ pages long. robert’s report was $4\frac{9}{10}$ pages long. which is the best estimate of how much longer robert’s report was than sadie’s? (7-4)

a $\frac{1}{2}$ page
b 1 page
c 2 pages
d 3 pages

Explanation:

Step1: Estimate the mixed numbers

First, we estimate \( 3\frac{1}{4} \) and \( 4\frac{9}{10} \). For \( 3\frac{1}{4} \), \( \frac{1}{4} \) is close to \( 0 \), so we can approximate \( 3\frac{1}{4} \approx 3 \). For \( 4\frac{9}{10} \), \( \frac{9}{10} \) is close to \( 1 \), so we can approximate \( 4\frac{9}{10} \approx 5 \).

Step2: Calculate the difference

Now we find the difference between the estimated values. We subtract the estimated length of Sadie's report from the estimated length of Robert's report: \( 5 - 3 = 2 \)? Wait, no, wait. Wait, actually, \( 4\frac{9}{10} \) is closer to \( 5 \)? Wait, no, \( 4\frac{9}{10} \) is \( 4.9 \), which is very close to \( 5 \), and \( 3\frac{1}{4} \) is \( 3.25 \), which is close to \( 3 \). But maybe a better estimate is to round \( 3\frac{1}{4} \) to \( 3 \) and \( 4\frac{9}{10} \) to \( 5 \), but wait, maybe another way: \( 3\frac{1}{4} \approx 3 \), \( 4\frac{9}{10} \approx 5 \), but actually, let's do it more accurately. Alternatively, \( 3\frac{1}{4} \) is \( 3.25 \), \( 4\frac{9}{10} \) is \( 4.9 \). The difference is \( 4.9 - 3.25 = 1.65 \), which is close to \( 2 \)? Wait, no, wait, maybe I made a mistake. Wait, let's re - estimate. \( 3\frac{1}{4} \) is \( 3.25 \), we can round it to \( 3 \) or \( 3.25 \). \( 4\frac{9}{10} \) is \( 4.9 \), we can round it to \( 5 \) or \( 4.9 \). The difference between \( 4.9 \) and \( 3.25 \) is \( 1.65 \), which is closest to \( 2 \)? Wait, no, wait, the options are \( \frac{1}{2} \), \( 1 \), \( 2 \), \( 3 \). Wait, maybe my initial estimation was wrong. Let's try another approach. Let's take \( 3\frac{1}{4} \approx 3 \), \( 4\frac{9}{10} \approx 5 \), but \( 5 - 3 = 2 \), but let's check the actual subtraction: \( 4\frac{9}{10}-3\frac{1}{4} \). Let's convert to improper fractions. \( 4\frac{9}{10}=\frac{4\times10 + 9}{10}=\frac{49}{10} \), \( 3\frac{1}{4}=\frac{3\times4+1}{4}=\frac{13}{4} \). Find a common denominator, which is \( 20 \). \( \frac{49}{10}=\frac{98}{20} \), \( \frac{13}{4}=\frac{65}{20} \). Then \( \frac{98}{20}-\frac{65}{20}=\frac{33}{20}=1.65 \), which is approximately \( 2 \)? Wait, no, \( 1.65 \) is closer to \( 2 \) than to \( 1 \) or \( \frac{1}{2} \) or \( 3 \). Wait, but let's check the options again. The options are A. \( \frac{1}{2} \), B. \( 1 \), C. \( 2 \), D. \( 3 \). So \( 1.65 \) is closest to \( 2 \)? Wait, no, \( 1.65 \) is closer to \( 2 \) than to \( 1 \) (since \( 1.65 - 1 = 0.65 \), \( 2 - 1.65 = 0.35 \)), so it's closer to \( 2 \). Wait, but maybe my estimation was wrong. Wait, let's re - estimate the mixed numbers. \( 3\frac{1}{4} \) is \( 3.25 \), we can round it to \( 3 \), and \( 4\frac{9}{10} \) is \( 4.9 \), we can round it to \( 5 \). Then \( 5 - 3 = 2 \). Alternatively, if we round \( 3\frac{1}{4} \) to \( 3.25 \approx 3 \) and \( 4\frac{9}{10}\approx5 \), the difference is \( 2 \). So the best estimate is \( 2 \) pages.

Answer:

C. 2 pages