Question

For a number, greater than one, the difference between itself and its reciprocal is 20% of the sum ofitself and its reciprocal. By how muchpercentage (nearest to an integer) is the square of the number less than its cube?

Answer 1

The square of the number is lesser than its cube by 33% (to the nearest integer).

• Let the number greater than one be x.

Reciprocal of the number = 1/x

• Difference between the number and its reciprocal = x - (1/x)

Sum of the number and its reciprocal = x + (1/x)

• According to the question,

x - (1/x) = 20/100 { x + (1/x) }

Or, (x² - 1 ) / x = 1/5 { ( x² + 1 ) / x}

Or, (x² - 1 ) / x = (x² + 1 ) / 5x

Or, 5 (x² - 1)  = x² + 1

Or, 5x² - 5 = x² + 1

Or, 5x² - x² = 1 + 5

Or, 4x² = 6

Or, x² = 6 / 4

Or, x = √(6 / 4)

Or, x = + 3/2, - 3/2

• Since the number is greater than one, x = - 3/2 gets discarded.

Therefore, x = 3/2

• Square of 3/2 = (3/2)² = 9/4

Cube of 3/2 = (3/2)³ = 27/8

• Difference between the cube and the square of 3/2 = (27/8) - (9/4)

Or, Difference = (27 - 18) / 8

Or, Difference = 9/8

• Percentage by which the square of the number is lesser than its cube = ( Difference between the cube and the square / Cube ) × 100 %

Or, % difference = (9/8) / (27/8) × 100 %

Or, % difference = (100 / 3) %

Or, % difference = 33.33 %

• The difference in percentage to the nearest integer = 33%

Answer 2

Answer:

wrong

ans is 18

Step-by-step explanation:

calculation and understanding mistake