Question

Mia runs for $2,483$ days in $8$ years. On running, she burns $459$ calories per day. The total calories did Mia burn in $8$ years is .

• A. $1,193,697$
• B. $1,139,679$
• C. $1,139,697$
• D. $1,931,796$

Answer 1

The correct option is C $1,139,697$

On running, in $1$ day, Mia burns $=459$ calories
In $8$ years, Mia runs $=2,483$ days
$\therefore$ Total amount of calories Mia burns in $8$ years $=459\times 2,483$

To solve $459\times 2,483$, we go through the below steps:

$\bullet$ First we write the bigger number at the top and smaller number just below of it with a proper alignment of one's place.


$\bullet$ We first multiply $9$ with $3$, which is $\underset{–}{2}7$, where $2$ is carrying over to the next multiplication, $9\times 8$.

So, the next number, we get $(9\times 8)+2=72+2=\underset{–}{7}4$, where $7$ is carrying over to the next multiplication, $9\times 4$.

Hence, the next number would be $(9\times 4)+7=36+7=\underset{–}{4}3$, where $4$ is carrying over to the next multiplication, $9\times 2$.

And, finally, we get $(9\times 2)+4=18+4=22$.


$\bullet$ For the next multiplication step, we leave one's place (denoted by faded $0$), and start from ten's place.

We first multiply $5$ with $3$, which is $\underset{–}{1}5$, where $1$ is carrying over to the next multiplication, $5\times 8$.

So, the next number, we get $(5\times 8)+1=40+1=\underset{–}{4}1$, where $4$ is carrying over to the next multiplication, $5\times 4$.

Hence, the next number would be $(5\times 4)+4=20+4=\underset{–}{2}4$, where $2$ is carrying over to the next multiplication, $5\times 2$.

And, finally, we get $(5\times 2)+2=10+2=12$.


$\bullet$ For the next multiplication step, we leave one's and ten's places (denoted by faded $0$), and start from hundred's place.

We first multiply $4$ with $3$, which is $\underset{–}{1}2$, where $1$ is carrying over to the next
multiplication, $4\times 8$.

So, the next number, we get $(4\times 8)+1=32+1=\underset{–}{3}3$, where $3$ is carrying over to the next multiplication, $4\times 4$.

Hence, the next number would be $(4\times 4)+3=16+3=\underset{–}{1}9$, where $1$ is carrying over to the next multiplication, $4\times 2$.

And, finally, we get $(4\times 2)+1=8+1=9$.


$\bullet$ Next, we add $22347$, $124150$ and $993200$ to get the final result. Here, $‘‘1+"$ denotes the carry over in addition.



$\therefore \underset{–}{459\times 2,483=1,139,697}$

Hence, the total calories did Mia burn in $8$ years is $1,139,697$.