Question

Here is an ancient problem from Bhaskaracharya's Lilavati:
A beautiful maiden, with framing 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? pls give me ans for this with steps i will mark your abswer as brainliest answer
​

Answer 1

Answer:

A beautiful maiden , with beaming eyes, asks of 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 ?

Ahh.. Isn’t it very long sentenced problem? The solution is here:

The method of working out this problem is to reverse the whole process — Multiplying 2 by 10 (20), deducting 8 (12), squaring (144), adding 52 (196), ‘multiplied by itself’ means that 196 was found by multiplying 14 to itself.

Now, Let the number be n.

Then applying initial part of the problem on it.

3n+3n×347–13×3n+3n×347=14

14 is what we already had in first half of solution.

Now as we have:

n2=14

Thus the number is 28 .

Mark me as brainliest

Answer 2

Answer: it's really tricky but I think the square root is 8.

Step-by-step explanation: because 8^2 =64

52 +8 = 64

I've spent over 15minutes trying to help you. I'm sorry I'll be off better help next time!!!