Problem from Lilavati written by Bhaskaracharya

Algebra Level 3

Here is an ancient problem from Bhaskaracharya's Lilavati. A beautiful maiden, with beaming eyes, asks me which is the number that, multiplied by 3, then increased by three fourths of the product, divided by 7, diminished by one third of the quotient, multiplied by itself, diminished by 52, the square root found, addition of 8, division by 10 gives the number 2? Well, it sounds complicated, doesn't it? No, not if you know how to go about it.


The answer is 28.

Vishwa Tej by Vishwa Tej , 22, India

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2 solutions

Eamon Gupta
Jul 23, 2015

Let the number be x x :

Multiplied by 3: 3 x 3x

Increased by three fourths: 3 x + 3 ⋅ 3 x 4 = 21 x 4 3x+\frac{3\cdot3x}{4} =\frac{21x}{4}

Divided by seven: 3 x 4 \large\frac{3x}{4}

Diminished by a third: 3 x 4 − x 4 = x 2 \large\frac{3x}{4}-\frac{x}{4} = \frac{x}{2}

Multiplied by itself: ( x 2 ) 2 = x 2 4 (\large\frac{x}{2})^{2}=\frac{x^{2}}{4}

Diminished by 52: x 2 4 − 52 \frac{x^{2}}{4}-52

Square root: x 2 4 − 52 \sqrt{\frac{x^{2}}{4}-52}

Addition of 8: x 2 4 − 52 + 8 \sqrt{\frac{x^{2}}{4}-52}+8

Division by 10: x 2 4 − 52 + 8 10 \large\frac{\sqrt{\frac{x^{2}}{4}-52}+8}{10}

Now the calculation:

x 2 4 − 52 + 8 10 = 2 \large\frac{\sqrt{\frac{x^{2}}{4}-52}+8}{10}=2

x 2 4 − 52 + 8 = 20 \sqrt{\frac{x^{2}}{4}-52}+8=20

x 2 4 − 52 = 12 \sqrt{\frac{x^{2}}{4}-52}=12

x 2 4 − 52 = 144 \frac{x^{2}}{4}-52=144

x 2 4 = 196 \frac{x^{2}}{4} = 196

x 2 = ± 14 \frac{x}{2} = \pm14

x = ± 28 \boxed{x=\pm28}

I know this is a longer method but it is easier in some ways when it says for example 'increased by three fourths of the product.'

3 Helpful 0 Interesting 1 Brilliant 1 Confused
Vishwa Tej
Jul 21, 2015

The method of working out this problem is to reverse the whole process - multiplying 2 by 10, deducting 8 , squaring the result and so on

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I'm getting 14 sqrt(2) instead of 28 when I do this. What is everyone else getting?

Jonathan Hocker - 5 years, 10 months ago

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