Question

A 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 found, addition of 8, division by 10 gives the number. Find the starting number?​

Answer 1

Answer:

Let the number be x

Now, going according to word of maiden ,

The number is multiplied by 3 and decreased by 3/4

So, x * 3 + 3/4 ( 3x )

= 21/4x

Then, divided by 7 and diminished by 1/3 of quotient,

21x/4 ÷ 7 = 3x/4

3x/4 - 1/3 ( 3x/4)

= 2/3 ( 3/4 ) x

= x/2

Multiplied by itself,

= x/2 * x/2

= x²/4

Now diminished by 52

= √ ( x²/4 - 52 )

Then added 8 , divided by 10

= [ √ ( x²/4 - 52 ) + 8 ] / 10

According to the question,

[ √ ( x²/4 - 52 ) + 8 ] / 10 = 2

√( x²/4 - 52 ) + 8 = 20

√(x²/4 -52 ) = 12

x²/4 - 52 = 144

x²/4 = 144 + 52

x²/4 = 196

(x/2)² = 14²

x/2 = 14

x = 2 * 14

x = 28 .

Answer 2

Answer:

Step-by-step explanation: