Algebra and Equations

Let the three numbers be a, b, and c. We know that a² + b² + c² = 138 and ab + bc + ca = 131. Using the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), we get (a + b + c)² = 138 + 2*131 = 138 + 262 = 400. Therefore, a + b + c = √400 = 20.

Let the two numbers be x and y, with x > y. Given x + y = 33 and x - y = 15. Adding both equations: 2x = 48, so x = 24. Substituting back: 24 + y = 33, so y = 9. The smaller number is 9.

The equation $\frac{2x}{25}=\frac{6}{x/3}$ can be rewritten as $\frac{2x}{25}=\frac{6}{x/3}=\frac{18}{x}$ because dividing by $x/3$ is the same as multiplying by $3/x$. Cross-multiplying gives $2x\times x=25\times 18$, which simplifies to $2x^2=450$. Dividing both sides by 2 yields $x^2=225$, so $x=15$ or $x=-15$. Since only positive options are given, the correct answer is 15, which corresponds to option B.

Let the number be x. The problem states that when the number is multiplied by 15, the result increases by 196 compared to the original number. This can be expressed as 15x = x + 196. Simplifying, 15x - x = 196, which gives 14x = 196. Dividing both sides by 14, we get x = 14. Therefore, the correct answer is 14 (option A).

Let the two numbers be x and y. Given x + y = 25 and x - y = 13. Adding the two equations, 2x = 38, so x = 19. Substituting back, y = 6. Their product is 19 × 6 = 114.

The equation given is x + x² = 182, which can be rewritten as x² + x - 182 = 0. Factoring this quadratic gives (x + 14)(x - 13) = 0, so the solutions are x = 13 and x = -14. Among the options provided, 13 is the correct positive solution, which corresponds to option D.

Let the two numbers be 2x and 3x. Their cubes sum to 280: (2x)^3 + (3x)^3 = 8x^3 + 27x^3 = 35x^3 = 280. Solving for x^3 gives x^3 = 280 / 35 = 8, so x = 2. The two numbers are 2x = 4 and 3x = 6. The smallest number is 4, which corresponds to option A.

Starting with 17xy + 7 = 19xy, subtract 17xy from both sides to get 7 = 2xy. Dividing both sides by 2 gives xy = 7/2. Multiplying both sides by 4 gives 4xy = 4 * (7/2) = 14. However, since the options do not include 14 as the correct answer for 4xy, let's re-examine. The original equation is 17xy + 7 = 19xy. Subtract 17xy: 7 = 2xy, so 2xy = 7, then xy = 7/2. Then 4xy = 4 * (7/2) = 14, which matches option D. Therefore, the correct answer is D, not A. The initial correction was incorrect. The original correct_answer D is correct. Explanation updated accordingly.

Let the cost of one mixer be M and one TV be T. From the equations: 2M + T = 7000 and M + 2T = 9800, solving these simultaneously gives T = Rs. 2800.

Al-Khwarizmi, a Muslim mathematician, wrote the first book on algebra titled "Al-Kitab al-Mukhtasar fi Hisab al-Jabr wal-Muqabala," which laid the foundation for the discipline of algebra.

Let the two numbers be x and y. Given x + y = 28 and x - y = 4. Adding these equations gives 2x = 32, so x = 16. Substituting back, 16 + y = 28, so y = 12. Thus, the two numbers are 16 and 12. The pair (16, 12) matches option A (12,16) if order is ignored, since the problem does not specify order. Therefore, option A is the correct answer.

The problem states that the sum of two numbers is 28 and their difference is 4. Let the two numbers be x and y, where x > y. Then we have the system of equations: x + y = 28 and x - y = 4. Adding these two equations gives 2x = 32, so x = 16. Substituting x = 16 into the first equation gives 16 + y = 28, so y = 12. Therefore, the two numbers are 16 and 12. Checking the options, option A (12,16) satisfies both conditions: 12 + 16 = 28 and 16 - 12 = 4. Option D (14,12) does not satisfy the sum condition (14 + 12 = 26). Hence, the correct answer is A.

"Al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wa’l-muqābala" (The Compendious Book on Calculation by Completion and Balancing) was written by Al-Khwarizmi, who is considered the father of algebra. This work laid the foundations for algebra as a discipline.