Question

Three numbers are in ratio 2,3,4 the sum of their cubes is 33957 find numbers

Answer 1

Answer:

Let the numbers be = 2x, 3x, 4x

${(2x)}^{3} + {(3x)}^{3} + {(4x)}^{3} = 33957 \\ = > {x}^{3} = 343 \\ = > x = 7$

Numbers = 14, 21, 28

Answer 2

$\huge{\underline{\bigstar{\sf{solution!!}}}}$

The 3 numbers are : 0.3,0.45,0.6

⇛The question says there are three numbers but with a specific ration what that means in that once we pick one of the numbers the other two are know to us through the rations we can therefore replace all 3 of the numbers with a single variable:

2: 3: 4 ⇛ 2x × 3x × 4x

⇒now, no Matter what we chose for c we get the three numbers into the ratios specified we are also told the sum of the cubes of these three numbers which can write :

$({2x})^{3} + ({3x})^{3} + ({4x})^{3} = 0.334125$

⇛Disturbing the powers across the fact is using

$({a×b})^{c} = {a}^{c} {b}^{c} we\: get :$

${8x}^{3}+ {27x}^{3} + {64x}^{3} = {99x}^{3} = 0.334125$

${x}^{3} = \frac{0.334125}{99} = 0.003375$

${x}^{3} = \sqrt[3]{0.003375} = 0.15$

⇛So the 3 numbers are

$2×0.15, 3×0.15, 4×0.15$

$0.3,0.45,0.6$