Question

Analyze the given statements and answer the following question.
Statement 1: The product of $2\frac{3}{4}$ and $\frac{1}{5}$ is less than $2\frac{3}{4}$ because $2\frac{3}{4}$ is multiplied by a factor less than 1.
Statement 2: The product of $2\frac{3}{4}$ and $\frac{1}{5}$ is greater than $2\frac{3}{4}$ because $2\frac{3}{4}$ is multiplied by a factor less than 1.
Statement 3: The product of $\frac{1}{5}$ and $2\frac{3}{4}$ is greater than $\frac{1}{5}$ because $2\frac{3}{4}$ is multiplied by a factor greater than 1.
Which of the following options is correct?

• A. Both statements $1$ and $3$ are correct.
• B. Only statement $2$ is correct.
  
• C. Both statements $1$ and $2$ are correct.
• D. Only statement $1$ is correct.

Answer 1

The correct option is A Both statements $1$ and $3$ are correct.

We know that $\frac{1}{5}<1$
When a given number is multiplied by a fraction less than one, the product will be less than the given number.
Since $2\frac{3}{4}$ is multiplied by $\frac{1}{5}$ (which is less than $1$), the product will be less than $2\frac{3}{4}$
Therefore, statement $1$ is correct.
This also implies, statement $2$ is incorrect because it states the opposite of statement $1.$
We need to multiply $2\frac{3}{4}$ and $\frac{1}{5}$ and compare the product with $\frac{1}{5}$ to check if the third statemets is correct.
Convert $2\frac{3}{4}$ into an improper fraction.
$2\frac{3}{4}=\frac{11}{4}$, which is greater than $1$
$2\frac{3}{4}\times \frac{1}{5}=\frac{11}{4}\times \frac{1}{5}=\frac{11}{20}$
Let's compare $\frac{11}{20}$ ad $\frac{1}{5}$
Since the denominators of both fractions are not the same, let’s convert them into equivalent, like fractions.
$\frac{1}{5}=\frac{1\times 4}{5\times 4}=\frac{4}{20}$
On comparing the two, we observe that, $\frac{11}{20}>\frac{4}{20}$
$\Rightarrow \frac{11}{20}>\frac{1}{5}$
The product of $\frac{1}{5}$ and $2\frac{3}{4}$ is greater than $\frac{1}{5}$ because $\frac{1}{5}$ is multiplied by a factor greater than $1$.
Therefore, statemet $3$ is correct.
Hence, both statemets $1$ and $3$ are correct.