Question

Two whole numbers are such that the cube of first number exceeds the cube of second by 61 and the ratio of the numbers is 5:4. What is the value of larger number?

A) 3 B) 4 C) 5 D) 6

Answer 1

$\large\bf\underline{Given:-}$

• Cube of first number exceeds the cube of second number by 61

• Ratio of the numbers is 5 : 4

$\large\bf\underline {To \: find:-}$

• Value of larger number.

$\huge\bf\underline{Solution:-}$

≫ Ratio of two numbers = 5:4

Let the two numbers be 5x and 4x.

$\large \: \: \: \underbrace{ \dag \bf \: According \: to \: question}$

The cube of first number exceeds the cube of second by 61.

»» (5x)³ - (4x)³ = 61

»» 125x³ - 64x³ = 61

»» 61x³ = 61

»» x³ = 61/61

»» x³ = 1

»» x = 1

• ★ So, we get the Value of x = 1.

Larger number 5x = 5 × 1 = 5

Therefore,

• ≫ the Value of larger number is 5.

Answer 2

Answer:

$\sf{C) \ 5 \ is \ the \ correct \ option. }$

Given:

• Two whole numbers are such that the cube of first number exceeds the cube of second by 61 .

• The ratio of the numbers is 5:4.

To find:

The value of larger number.

Solution:

$\sf{Let \ the \ constant \ be \ x.}$

$\sf{\therefore{The \ numbers \ are \ 5x \ and \ 4x}}$

$\sf{According \ to \ the \ given \ condition. }$

$\sf{(5x)^{3}-4x^{3}=61}$

$\sf{\therefore{125x^{3}-64x^{3}=61}}$

$\sf{\therefore{61x^{3}=61}}$

$\sf{\therefore{x^{3}=\frac{61}{61}}}$

$\sf{\therefore{x^{3}=1}}$

$\sf{On \ taking \ the \ cube \ root \ of \ both \ sides.}$

$\sf{\therefore{x=1}}$

$\sf{The \ larger \ number=5x=5(1)=5}$

$\sf\purple{\tt{\therefore{The \ value \ of \ the \ larger \ number \ is \ 5.}}}$