Let \( x, y, \) and \( z \) be real numbers satisfying:
\( 4(x^2 + y^2 + z^2) = a \),
\( 4(x – y – z) = 3 + a \).
Then \( a \) equals:
Solution
Hit and Trial Table for Solving Using Options
Given equations:
\[ 4(x^2 + y^2 + z^2) = a \quad \text{and} \quad 4(x – y – z) = 3 + a \]
| Option | \( a \) | \( x^2 + y^2 + z^2 = \frac{a}{4} \) | \( x – y – z = \frac{3 + a}{4} \) | Trial Values for \( x, y, z \) | Result |
|---|---|---|---|---|---|
| 1 | \( 1 \) | \( \frac{1}{4} \) | \( 1 \) | \( x = 1, y = 0, z = 0 \) | ❌ Too large for (1) |
| 2 | \( 4 \) | \( 1 \) | \( \frac{7}{4} \) | \( x = 2, y = \frac{1}{8}, z = \frac{1}{8} \) | ❌ Doesn’t satisfy (1) |
| 3 ✅ | \( 3 \) | \( \frac{3}{4} \) | \( \frac{3}{2} \) | Algebra confirms consistency | ✅ Correct |
| 4 | \( \frac{4}{3} \) | \( \frac{1}{3} \) | \( \frac{13}{12} \) | No simple values fit both equations | ❌ Invalid |
✅ Final Answer:
\[ \boxed{3} \quad \text{(Option 3)} \]
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is:
Solution
- \( 1125\pi = 3534.29 \)
- \( 750\pi\sqrt{2} = 3332.16 \)
- \( 1125\pi\sqrt{2} = 4998.24 \)
- \( 750\pi = 2356.19 \)
Volume of Sphere Inscribed with a Rectangular Box
Given:
- Surface area of the box = 846 cm²
- Sum of all edges = 144 cm
- Box is inscribed in a sphere → Diagonal of box = Diameter of sphere
Step 1: Let dimensions be \( l, w, h \)
\[ 4(l + w + h) = 144 \Rightarrow l + w + h = 36 \tag{1} \] \[ 2(lw + wh + hl) = 846 \Rightarrow lw + wh + hl = 423 \tag{2} \]
Step 2: Use identity for cube diagonal
\[ (l + w + h)^2 = l^2 + w^2 + h^2 + 2(lw + wh + hl) \] Substituting (1) and (2): \[ 36^2 = l^2 + w^2 + h^2 + 2(423) \Rightarrow 1296 = l^2 + w^2 + h^2 + 846 \] \[ \Rightarrow l^2 + w^2 + h^2 = 450 \Rightarrow D = \sqrt{450} = 15\sqrt{2} \]
Step 3: Volume of Sphere
\[ V = \frac{4}{3} \pi r^3 = \frac{\pi D^3}{6} \Rightarrow V = \frac{\pi (15\sqrt{2})^3}{6} = \frac{\pi \cdot 3375 \cdot 2\sqrt{2}}{6} = \frac{6750\sqrt{2} \pi}{6} = \boxed{1125\pi\sqrt{2}} \]
Step 4: Test Options
Option 1: \(1125\pi\)
\[ D^3 = \frac{6 \cdot 1125\pi}{\pi} = 6750 \Rightarrow D \approx \sqrt[3]{6750} \approx 18.8 \Rightarrow D^2 \approx 354.2 \neq 450 \Rightarrow \text{❌} \]
Option 2: \(750\pi\sqrt{2}\)
\[ D^3 = 6 \cdot 750\sqrt{2} \approx 6363 \Rightarrow D \approx 18.6 \Rightarrow D^2 \approx 345.6 \neq 450 \Rightarrow \text{❌} \]
Option 3: \(1125\pi\sqrt{2}\)
\[ D^3 = 6 \cdot 1125\sqrt{2} \approx 9544.5 \Rightarrow D \approx \sqrt[3]{9544.5} \approx 21.21 \Rightarrow D^2 = 450 \Rightarrow \text{✅ Matches} \]
Option 4: \(750\pi\)
\[ D^3 = 4500 \Rightarrow D \approx 16.5 \Rightarrow D^2 \approx 272.25 \neq 450 \Rightarrow \text{❌} \]
✅ Final Answer: \( \boxed{1125\pi\sqrt{2}} \) (Option 3)
A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is
1. 36
2. 18
3. 50
4. 42
Solution
Grain Selling Problem – Option Check
Checking Each Option for Grain Selling
Rule: Each customer takes half of the current quantity plus 3 kg.
🔹 Option 1: \( x = 36 \)
First: \( \frac{36}{2} + 3 = 21 \Rightarrow \text{Remaining} = 15 \)
Second: \( \frac{15}{2} + 3 = 10.5 \Rightarrow \text{Remaining} = 4.5 \)
Third: \( \frac{4.5}{2} + 3 = 5.25 \) but only 4.5 left → Invalid
🔹 Option 2: \( x = 18 \)
First: \( \frac{18}{2} + 3 = 12 \Rightarrow \text{Remaining} = 6 \)
Second: \( \frac{6}{2} + 3 = 6 \Rightarrow \text{Remaining} = 0 \)
Third: No grains left to sell → Invalid
🔹 Option 3: \( x = 50 \)
First: \( \frac{50}{2} + 3 = 28 \Rightarrow \text{Remaining} = 22 \)
Second: \( \frac{22}{2} + 3 = 14 \Rightarrow \text{Remaining} = 8 \)
Third: \( \frac{8}{2} + 3 = 7 \Rightarrow \text{Remaining} = 1 \) → Invalid
✅ Option 4: \( x = 42 \)
First: \( \frac{42}{2} + 3 = 24 \Rightarrow \text{Remaining} = 18 \)
Second: \( \frac{18}{2} + 3 = 12 \Rightarrow \text{Remaining} = 6 \)
Third: \( \frac{6}{2} + 3 = 6 \Rightarrow \text{Remaining} = 0 \) → Valid ✅
🎯 Final Answer: \( \boxed{42} \)
Algebra + APGP + Fractions
Find the sum of the series
\[ \frac{1}{5} \left( \frac{1}{5} – \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 – \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 – \left( \frac{1}{7} \right)^3 \right) + \cdots \]
- 5/408
- 7/816
- 7/408
- 5/816
Solution
Solution to the Infinite Series
Given Series:
\[ \frac{1}{5} \left( \frac{1}{5} – \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 – \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 – \left( \frac{1}{7} \right)^3 \right) + \cdots \]
✅ Step 1: Identify the General Term
\[ \sum_{n=1}^{\infty} \left( \frac{1}{5} \right)^n \left[ \left( \frac{1}{5} \right)^n – \left( \frac{1}{7} \right)^n \right] = \sum_{n=1}^{\infty} \left[ \left( \frac{1}{25} \right)^n – \left( \frac{1}{35} \right)^n \right] \] This is a difference of two geometric series.
✅ Step 2: Evaluate Each Series
Using the formula \( \sum_{n=1}^{\infty} r^n = \frac{r}{1 – r} \), we get: \[ \sum_{n=1}^{\infty} \left( \frac{1}{25} \right)^n = \frac{1/25}{1 – 1/25} = \frac{1}{24} \] \[ \sum_{n=1}^{\infty} \left( \frac{1}{35} \right)^n = \frac{1/35}{1 – 1/35} = \frac{1}{34} \]
✅ Step 3: Subtract the Series
\[ \sum = \frac{1}{24} – \frac{1}{34} = \frac{34 – 24}{24 \cdot 34} = \frac{10}{816} = \frac{5}{408} \]
🎯 Final Answer:
\[ \boxed{\frac{5}{408}} \]
