Question

Two numbers are in the ratio 3: 4. The sum of their cubes is 91. Find the numbers.​

Answer 1

Given :

• Two numbers are in the ratio 3: 4.
• The sum of their cubes is 91.

To Find :

• The numbers.

Solution :

Let's Assume,

• The first number be 3x
• The second number be 4x

According to the Question :

• The sum of their cubes = 91
• (First number)³ + (Second number)³ = 91
• (3x)³ + (4x)³ = 91
• (3 × 3 × 3x) + (4 × 4 × 4x) = 91
• 27x + 64x = 91
• 91x = 91
• x = 91/91
• x = 1

Therefore :

• The first number is 3x
• The first number is 3(1)
• The first number is 3

• The second number is 4x
• The second number is 4(1)
• The second number is 4

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Answer 2

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{Provided\: that:}}}$

➻Two numbers are in the ratio 3:4

➻Sum of their cubes =91

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$\sf \blue {\underline{\underline{To\: be\: found:}}}$

➻The two numbers=?

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$\sf \pink {\underline{\underline{Consideration:}}}$

➻Let the two numbers be,

$\to \tt {1^{st}\:number =3x}$

$\to \tt {2^{nd}\:number=4x}$

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$\sf \orange {\underline{\underline{Condition:}}}$

➻According to the condition given in the question, equation to be framed is,

➻Sum of their cubes =91

$\colon \mapsto \tt {(3x)^{3}+(4x)^{3}=91}$

$\colon \mapsto \tt {27x^{3}+64x^{3}=91}$

$\colon \mapsto \tt {91x^{3}=91}$

$\colon \mapsto \tt {x^{3}=\frac{91}{91}}$

$\colon \mapsto \tt {x^{3}=\frac{\cancel{91}}{\cancel{91}}}$

$\colon \mapsto \tt {x^{3}=1}$

$\colon \mapsto \tt {x=3\sqrt{1}}$

$\colon \implies \tt \green {\underline{\fbox{x=1}}}$

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$\sf \purple {\underline{\underline{Henceforth,}}}$

$\to \tt {1^{st}\:number =3x}$

$\to \tt {1^{st}\:number =3(1)}$

$\implies \tt \green {\underline{\boxed{1^{st}\:number =3}}}$

$\to \tt {2^{nd}\:number=4x}$

$\to \tt {2^{nd}\:number =4(1)}$

$\implies \tt \green {\underline{\boxed{2^{nd}\:number =4}}}$

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➻∴ The two numbers are 3 &4 respectively .

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