Question

The sum of two numbers is $4.004$. One number has $4$ in the tenths place and $3$ in the thousandths place. The other number has $1$ in the ones place and $8$ in the hundredths place. What are the two numbers?

• A. $3.423$ and $1.581$
• B. $2.423$ and $1.581$
• C. $2.623$ and $1.481$
• D. $2.423$ and $1.681$

Answer 1

The correct option is A $3.423$ and $1.581$

Detailed step-by-step solution:
Sum of the two numbers $=4.004$
From the given condition, the first number $=$_$.4$_$3$
From the given condition, the second number $=1.$_$8$_

	$O$		$T$	$H$	$Th$
		$.$	$4$		$3$
$+$	$1$	$.$		$8$
	$4$	$.$	$0$	$0$	$4$

At the thousandths place, $3+$ number $=4$
So, the number $=4–3=1$
The digit at the thousandths place of the second number is $1$.
At the hundredths place, number$+8=0\left(10\right)$
So, number $=10–8=2$
The digit at the hundredths place of the first number is $2$.
At the tenths place, $4+$ number $=0(10,1$ carry over from hundredths place$)$
So, $4+$ number $=9$
$\Rightarrow$ number $=9–4=5$
Digit at the tenths place of the second number is $5$.
Digit at the ones place of the second number is $1$.
At the ones place, $1+$ number $=4(1$ carried over from the tenths place$)$
So, $1+$ number $=3$
number $=3–1=2$
$\Rightarrow$ The digit at the ones place of the first number is $2$.

	$O$ $1$		$T$ $1$	$H$	$Th$
	$2$	$.$	$4$	$2$	$3$
$+$	$1$	$.$	$5$	$8$	$1$
	$4$	$.$	$0$	$0$	$4$