Question

-21. copy and complete each of the diamond problems below. the pattern used in the diamond problems is shown at right. a. diamond with 15 (top), 8 (bottom) b. diamond with 3 (left), 1 (right) c. diamond with 10 (top), 5 (left) d. diamond with \\(\frac{1}{2}\\) (left), \\(\frac{1}{3}\\) (right)

Explanation:

To solve Diamond Problems, we use the pattern: the top cell is the product of the left (\(x\)) and right (\(y\)) cells, and the bottom cell is the sum of \(x\) and \(y\). Let's solve each part:

Part (a)

We know the top cell is \(15\) (product of \(x\) and \(y\)) and the bottom cell is \(8\) (sum of \(x\) and \(y\)). We need two numbers \(x\) and \(y\) such that:
\(x \times y = 15\) and \(x + y = 8\).

Testing factor pairs of \(15\):

• \(1 \times 15 = 15\), but \(1 + 15 = 16

eq 8\)

• \(3 \times 5 = 15\), and \(3 + 5 = 8\)

Thus, \(x = 3\) and \(y = 5\) (or vice versa).

Part (b)

Left cell \(x = 3\), right cell \(y = 1\).

• Top cell (product): \(x \times y = 3 \times 1 = 3\)
• Bottom cell (sum): \(x + y = 3 + 1 = 4\)

Part (c)

Top cell (product) \(= 10\), left cell \(x = 5\).

• Right cell \(y\): \(x \times y = 10 \implies y = \frac{10}{5} = 2\)
• Bottom cell (sum): \(x + y = 5 + 2 = 7\)

Part (d)

Left cell \(x = \frac{1}{2}\), right cell \(y = \frac{1}{3}\).

• Top cell (product): \(x \times y = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)
• Bottom cell (sum): \(x + y = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\)

Final Answers:

a. Left: \(3\), Right: \(5\) (or vice versa)
b. Top: \(3\), Bottom: \(4\)
c. Right: \(2\), Bottom: \(7\)
d. Top: \(\boldsymbol{\frac{1}{6}}\), Bottom: \(\boldsymbol{\frac{5}{6}}\)

Answer:

To solve Diamond Problems, we use the pattern: the top cell is the product of the left (\(x\)) and right (\(y\)) cells, and the bottom cell is the sum of \(x\) and \(y\). Let's solve each part:

Part (a)

We know the top cell is \(15\) (product of \(x\) and \(y\)) and the bottom cell is \(8\) (sum of \(x\) and \(y\)). We need two numbers \(x\) and \(y\) such that:
\(x \times y = 15\) and \(x + y = 8\).

Testing factor pairs of \(15\):

• \(1 \times 15 = 15\), but \(1 + 15 = 16

eq 8\)

• \(3 \times 5 = 15\), and \(3 + 5 = 8\)

Thus, \(x = 3\) and \(y = 5\) (or vice versa).

Part (b)

Left cell \(x = 3\), right cell \(y = 1\).

• Top cell (product): \(x \times y = 3 \times 1 = 3\)
• Bottom cell (sum): \(x + y = 3 + 1 = 4\)

Part (c)

Top cell (product) \(= 10\), left cell \(x = 5\).

• Right cell \(y\): \(x \times y = 10 \implies y = \frac{10}{5} = 2\)
• Bottom cell (sum): \(x + y = 5 + 2 = 7\)

Part (d)

Left cell \(x = \frac{1}{2}\), right cell \(y = \frac{1}{3}\).

• Top cell (product): \(x \times y = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)
• Bottom cell (sum): \(x + y = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\)

Final Answers:

a. Left: \(3\), Right: \(5\) (or vice versa)
b. Top: \(3\), Bottom: \(4\)
c. Right: \(2\), Bottom: \(7\)
d. Top: \(\boldsymbol{\frac{1}{6}}\), Bottom: \(\boldsymbol{\frac{5}{6}}\)