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solniwko [45]
3 years ago
8

A hamburger bun merchant can ship 8 large boxes or 10 small boxes of hamburger buns into a carton for shipping. In one shipment,

he sent a total of 96 boxes of hamburger buns. If there are more large boxes than small boxes, how many cartoons did he ship?
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
8 0
<h2>Answer:</h2>

56 large boxes and 40 small boxes.

<h2>Step-by-step explanation:</h2>

To solve this problem, we need to use the concept of multiples. By definition, a multiple of a number is that number multiplied by an integer. For instance, 3, 6, 9, 12 are multiples of 3. So let's do a chart and list the multiples of large and small boxes:

\begin{array}{cccccccccc}Number\,of\,boxes & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\Large & 8 & 16 & 24 & 32 & 40 & 48 & 56 & 64 & 72\\Small & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90\end{array}

From this, our goal is to select the number of small and large boxes such that the sum is 96 boxes of hamburger burns and the number of large boxes is greater than the number of small boxes. From the table, the correct solution is:

56 large boxes and 40 small boxes, and this meets our requirement, because 56 + 40 = 96 and 56 > 40

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\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




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