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Arlecino [84]
3 years ago
7

Jason has 28 books in his library. He bought several books at a yard sale over the weekend. He now has 65 books in his library.

How many books did he buy at the yard sale
A.37
B.93
C.28
Mathematics
2 answers:
melamori03 [73]3 years ago
7 0

Answer:

A.37 books.

Step-by-step explanation:

if you could take 65 books that Jason has currently, then subtract the books he had before he bought the books at the yard sale, you would get 37. And those are the books he bought at the yard sale. (65 - 28 = 37)

Slav-nsk [51]3 years ago
6 0
It’s A. Because you add the answers to 28 or you subtract 28 and 65. GG
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so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

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\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


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\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

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