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Another problem from Mah?vira: One-third of a herd of elephants and three times the square root of the remaining part of the herd were seen on a mountain slope; and in a lake was seen a male elephant along with three female elephants constituting the ultimate remainder. How many were the elephants here?

Short Answer

Expert verified
Answer: There are 6 elephants in total in the herd.

Step by step solution

01

Assign variables and create equations based on the given information

Let the total number of elephants in the herd be x. According to the problem, one-third of the herd and three times the square root of the remaining part of the herd were seen on the mountain slope. We can represent the mountain slope group as \(\frac{1}{3}x + 3\sqrt{x - \frac{1}{3}x}\), which simplifies to \(\frac{1}{3}x + 3\sqrt{\frac{2}{3}x}\). There are 4 elephants (1 male and 3 females) seen in the lake. This means that the remaining part of the herd is 4 elephants less than the total number of elephants in the herd: \(x - 4\). Now, we can equate these two expressions as follows: $$ \frac{1}{3}x + 3\sqrt{\frac{2}{3}x} = x - 4 $$
02

Solve the equation for x

To solve the equation, let's first isolate the square root term: $$ 3\sqrt{\frac{2}{3}x} = \frac{2}{3}x - 4 $$ Next, divide both sides by 3: $$ \sqrt{\frac{2}{3}x} = \frac{2}{9}x - \frac{4}{3} $$ Now square both sides of the equation to get rid of the square root: $$ \frac{2}{3}x = \left(\frac{2}{9}x - \frac{4}{3}\right)^2 $$ Expand the right-hand side of the equation: $$ \frac{2}{3}x = \frac{4}{81}x^2 - \frac{16}{27}x + \frac{16}{9} $$ To solve this quadratic equation, let's first multiply through by 81 to eliminate the fractions: $$ 54x = 4x^2 - 48x + 144 $$ Rearrange the equation into the standard quadratic form: $$ 4x^2 - 102x + 144 = 0 $$ Now, we can either use the quadratic formula or factor to solve for x. In this case, we can factor the equation with a factorization of \((2x - 12)(2x - 12)\) . $$ 0 = (2x - 12)(2x - 12) $$
03

Find the value of x

Now, we can solve for x by setting each factor equal to zero: $$ 2x - 12 = 0 $$ Add 12 to both sides and then divide by 2: $$ x = 6 $$ Since x represents the total number of elephants in the herd, there are 6 elephants in total.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
A quadratic equation is a fundamental concept in mathematics representing equations of the form \( ax^2 + bx + c = 0 \). In this format, \( a \), \( b \), and \( c \) are coefficients, and \( a \) is not zero. Quadratic equations are characterized by the presence of the \( x^2 \) term, which is their defining feature.

To solve quadratic equations, several methods can be employed:
  • Factoring: This involves expressing the equation as a product of two binomials. It's effective when these factors are easily identifiable.
  • Completing the Square: By transforming the equation, it becomes possible to convert it into the form \((x + p)^2 = q\).
  • Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides a solution for any quadratic equation.
The quadratic nature of Mahāvīra's problem requires understanding and manipulating quadratic equations to find the total number of elephants in the herd through algebraic transformations and solving techniques.
Ancient Indian Mathematics
Ancient Indian Mathematics is a rich field with contributions that significantly advanced the understanding of mathematical concepts well ahead of the Western world. One major component was the development and application of arithmetic and algebraic techniques, including the concept of zero as a number, and sophisticated use of fractions and square roots.

Mathematicians like Aryabhata and Brahmagupta made monumental contributions. They produced techniques and formulae that laid the groundwork for calculus and trigonometry. Their work included solutions to quadratic equations, work on finite series, and geometric principles.

In Mahāvīra's time, mathematics was used to solve practical problems, often in beautifully creative ways that included elements of abstract thought that are detailed even compared to today’s standards. They used language ingeniously to depict complex mathematical problems, as seen in Mahāvīra's problem where elephants' grouping was presented as a textual puzzle to be solved using algebra.
Mahāvīra's Problems
Mahāvīra, a notable Indian mathematician from the 9th century, is renowned for his work on algebra and his attempt to link it with day-to-day problems. His mathematical work, called the "Gaṇita-sāra-saṅgraha," encapsulates a series of mathematical exercises expressed in a story-like form.

Mahāvīra's problems, like the elephant problem described, are essential because they reflect the practical application of algebra. These problems present a narrative that involves finding unknown quantities, which encourage logical reasoning and the application of mathematical principles such as quadratic equations.
  • His storytelling approach makes complex ideas more relatable and easier to understand.
  • By blending everyday scenarios with mathematical reasoning, Mahāvīra engaged learners of his time in problem-solving that was both educational and entertaining.
  • His methods showcased advanced problem modeling skills, introducing multiple variables and equation-balancing to arrive at a solution.
Understanding Mahāvīra's problems can offer insights into ancient algorithmic thinking, shedding light on early practical applications of algebra.

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Most popular questions from this chapter

Show that if \((u, v)\) is a solution to \(D x^{2}+2=y^{2}\), then \(\left(u_{1}, v_{1}\right)=\left(u v, v^{2}-1\right)\) is a solution to \(D x^{2}+1=y^{2} .\) Deduce a similar rule if \((u, v)\) is a solution to \(D x^{2}-2=y^{2}\).

Solve the following problem from Mah?vira: There are 4 pipes leading into a well. Among these, each fills the well (in order) in \(1 / 2,1 / 3,1 / 4\), and \(1 / 5\) of a day. In how much of a day will all of them together fill the well and each of them to what extent?

Brahmagupta asserts that if \(A B C D\) is a quadrilateral inscribed in a circle, with side lengths \(a, b, c, d\) (in cyclic order) (see Fig. 8.8), then the lengths of the diagonals \(A C\) and \(B D\) are given by $$ A C=\sqrt{\frac{(a c+b d)(a d+b c)}{a b+c d}} $$ and similarly $$ B D=\sqrt{\frac{(a c+b d)(a b+c d)}{a d+b c}} $$ Prove this result as follows: a. Let \(\angle A B C=\theta\). Then \(\angle A D C=\pi-\theta\). Let \(x=A C\). Use the law of cosines on each of triangles \(A B C\) and \(A D C\) to express \(x^{2}\) two different ways. Then, since \(\cos (\pi-\theta)=-\cos \theta\), use these two formulas for \(x^{2}\) to determine \(\cos \theta\) as a function of \(a, b, c\), and \(d\). b. Replace \(\cos \theta\) in your expression for \(x^{2}\) in terms of \(a\) and \(b\) by the value for the cosine determined in part \(\mathrm{a}\). c. Show that \(c d\left(a^{2}+b^{2}\right)+a b\left(c^{2}+d^{2}\right)=(a c+b d)(a d+b c)\). d. Simplify the expression for \(x^{2}\) found in part \(\mathrm{b}\) by using the algebraic identity found in part c. By then taking square roots, you should get the desired expression for \(x=A C\). (Of course, a similar argument will then give you the expression for \(y=B D\).)

Solve \(61 x^{2}+1=y^{2}\) by Bh?skara's process. The solution is \(x=226,153,980, y=1,766,319,049\).

Solve the problem \(N \equiv 5(\bmod 6) \equiv 4(\bmod 5) \equiv\) \(3(\bmod 4)) \equiv 2(\bmod 3))\) by the Indian procedure and by the Chinese procedure. Compare the methods.

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