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How Many Bees in Lilavatiís Swarm

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While solving any word problem, there are four elementary steps – to understand, to translate, to solve and to interpret. Mathematics transforms real life problems into mathematical problems. After obtaining the solution, it needs to be interpreted into real life language (Glaz & Liang, 2009). This essay attempts to solve the problem in the verse by following the four elementary steps.

The first step entails understanding the problem and then trying to solve it by trial and error method. In this problem, 1/5th of the bees sit on the flower of Kadamba, 1/3rd on the flower of Silinda, some on the flower of Krutaja while one bee remains in the air. Now we choose a value for the total number of bees and apply the trial and error method.

Arbitrarily choosing the number 9, we try to solve the problem. 1/3rd of 9 is 3 and 1/5th of 9 is 9/5, which is a fraction. The number of bees in the swarm has to be a whole number and cannot be a fraction. Therefore, 9 is not the correct value. For this reason, we choose the smallest number divisible by both 5 and 3, (that is 15), so that we do not get any fractions.

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Assuming the total number of bees to be 15, 1/5th of 15 is 3, 1/3rd of 15 is 5. The difference between the 1/5th and the 1/3rd,  that is between 3 and 5 is 2 and 3 times 2 is 6. The verse also says that 1 bee is in the air. Therefore, the sum of all these numbers should work out to be 15 if our assumption is correct. The sum of 3, 5, 6 and 1 is 15. Hence, our assumption of the total number of bees is correct.

The second step is to translate the problem into an equation. Let the total number of bees be x. Then, according to the verse, it follows that

1/5 x + 1/3 x + 3(1/3x − 1/5x) + 1= x

Solving the equation, the value of x can be found.

1/5x + 1/3x + 3 (2/15x) + 1 = x

1/5 x + 1/3x + 6/15x + 1 = x

1/5x + 1/3x + 6/15x x = – 1

(14x – 15x) / 15 = − 1

x = −1×15

x = − 15

Therefore, x = 15

To check the solution, let us put all values of x in the original equation.

(1/5)15 + (1/3)15 + 3(15/3 – 15/5) + 1 = 15

3 + 5 + 3(5 – 3) + 1 = 15

3 + 5 + 3(2) + 1 = 15

3 + 5 + 6 + 1 = 15

15 = 15

The left and right side of the equation balance out, hence the calculated value of x is correct. Out of 15 bees, 3 bees settled on Kadamba flower, 5 bees settled on Silinda flower, 6 bees flew over Krutaja flower and 1 bee alone remained in the air.

Bhaskaracharya presents the mathematical problem beautifully in poetic form. Using everyday life and common things like bees and flowers, he conveys the concept of fractions and addition and multiplication of fractions. The trial and error method encourages the student to try out different numbers and study the patterns that emerge. By this method, it quickly becomes clear that the number of bees settling on different flowers cannot be a fraction. Therefore, the total number of bees has to be a number divisible by both 3 and 5. The smallest such number is 15. If the correct answer were not 15, the student would quickly jump to 30 avoiding all other numbers in between as they are not divisible by 5 and 3.

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