# Divide a Number into Three Parts in a Given Ratio

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