Question

Two numbers are in the ratio 4:5. If difference of their cubes in 61 ,find the numbers.

Please help me to solve this‍♀️‍♀️​

Answer 1

Answer:

5 and 4

Step-by-step explanation:

As their ratio is 4:5, let the numbers are 4x and 5x. Their cubes are: (4x)³ = 64x³ and (5x)³ = 125x³.

Given, difference in their cubes is 61.

=> 125x³ - 64x³ = 61

=> 61x³ = 61

=> x³ = 61/61 = 1

=> x = 1

Therefore, numbers are:

4a = 4(1) = 4

5a = 5(1) = 5

Answer 2

• $\Large\boxed{\sf{\green{Numbers = \bf{4\: and \:5}}}}$

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Explanation :

$\underline{\underline{\sf{\purple{Given\::-}}}}$

• Two numbers are in ratio 4 : 5
• Difference of their cubes = 61

$\underline{\underline{\sf{\purple{To\:Find\::-}}}}$

• Numbers = ?

$\underline{\underline{\sf{\purple{Solution\::-}}}}$

• Here, we have ratio, i.e, 4 : 5 of two numbers and the difference of their cubes, i.e, 61 we have to find out both numbers. We will set up an equation by understanding the given conditions.
• Let the numbers be 4n and 5n

Atq,

$\qquad\longrightarrow\quad\sf (5n)^3 - (4n)^3 = 61$

$\qquad\longrightarrow\quad\sf 125n^3 - 64n^3 = 61$

$\qquad\longrightarrow\quad\sf 61n^3 = 61$

$\qquad\longrightarrow\quad\sf n^3 = \dfrac{61}{61}$

$\qquad\longrightarrow\quad\sf n^3 = \dfrac{\cancel{61}}{\cancel{61}}$

$\qquad\longrightarrow\quad\sf n^3 = 1$

$\qquad\longrightarrow\quad\sf n = \sqrt[3]{1}$

$\qquad\longrightarrow\quad\bf {n = \pink{1}}$

Hence,

• First number = 4n = 4 × 1 = 4
• Second number = 5n = 5 × 1 = 5

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