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rodikova [14]
3 years ago
11

A buyer owes a supplier $820. The terms were 3/10, n/60. Within 10 days, the buyer sent in a payment of $140. How much is the ne

w balance? (Round your answers to the nearest hundredth)
Mathematics
1 answer:
Margaret [11]3 years ago
7 0

Answer:

The new balance is $675.8

Step-by-step explanation:

Solution

Given that

The total amount of loan = $820.

The terms were = 3/10, n/60

Within 10 days, the buyer sent in a payment of  =$140

What is the new balance =?

Now,

3/10 =  if the amount is paid in 10 days, a discount of is included

n/60 = This means that all amount should be paid within 60 days

Thus,

$140 for 3/10 as this is paid within 10 days

140 *3% = 140 * 3/100

we get

=$4.2 of discount

Then,

The balance amount becomes $ 820 - 140 -4.2

=$675.8

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Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

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\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

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