To divide a number into three parts in a given ratio
Let the number be p. It is to be divided into three parts in the ratio a : b : c.
Let the parts be x, y and z. Then, x + y + z = p .................... (i)
and x = ak, y =bk, z = ck.................... (ii)
Substituting in (i), ak + bk + ck = p
⟹ k(a + b + c) = p
Therefore, k = \(\frac{p}{a + b + c}\)
Therefore, x = ak = \(\frac{ap}{a+ b + c}\), y = bk = \(\frac{bp}{a+ b + c}\), z = ck = \(\frac{cp}{a+ b + c}\).
`The three parts of p in the ratio a : b : c are
\(\frac{ap}{a+ b + c}\), \(\frac{bp}{a+ b + c}\), \(\frac{cp}{a+ b + c}\).
Solved examples on dividing a number into three parts in a given ratio:
1. Divide 297 into three parts that are in the ratio 5 : 13 : 15
Solution:
The three parts are \(\frac{5}{5 + 13 + 15}\) ∙ 297, \(\frac{13}{5 + 13 + 15}\) ∙ 297 and \(\frac{15}{5 + 13 + 15}\) ∙ 297
i.e., \(\frac{5}{33}\) ∙ 297, \(\frac{13}{33}\) ∙ 297 and \(\frac{15}{33}\) ∙ 297 i.e., 45, 117 and 135.
2. Divide 432 into three parts that are in the ratio 1 : 2 : 3
Solution:
The three parts are \(\frac{1}{1 + 2 + 3}\) ∙ 432, \(\frac{2}{1 + 2 + 3}\) ∙ 432 and \(\frac{3}{1 + 2 + 3}\) ∙ 432
i.e., \(\frac{1}{6}\) ∙ 432, \(\frac{2}{6}\) ∙ 432 and \(\frac{3}{6}\) ∙ 432
i.e., 72, 144 and 216.
3. Divide 80 into three parts that are in the ratio 1 : 3 : 4.
Solution:
The three parts are \(\frac{1}{1 + 3 + 4}\) ∙ 80, \(\frac{3}{1 + 3 + 4}\) ∙ 80 and \(\frac{4}{1 + 3 + 4}\) ∙ 80
i.e., \(\frac{1}{8}\) ∙ 80, \(\frac{3}{8}\) ∙ 80 and \(\frac{4}{8}\) ∙ 80
i.e., 10, 30 and 40.
● Ratio and proportion
10th Grade Math
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