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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

$\frac{ap}{a+ b + c}$, $\frac{bp}{a+ b + c}$, $\frac{cp}{a+ b + c}$.

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.

4. If the perimeter of a triangle is 45 cm and its sides are in the ratio 2: 3: 4, find the sides of the triangle.

Solution:

Perimeter of the triangle = 45 cm

Ratio of the sides of the triangle = 2 : 3 : 4

Sum of ratio terms = (2 + 3 + 4) = 9

The sides of the triangle $\frac{2}{9}$ × 45 cm, $\frac{3}{9}$ × 45 cm and $\frac{4}{9}$ × 45 cm,

i.е., 10 cm, 15 cm and 20 cm.

Hence, the sides of the triangle are 10 cm, 15 cm and 20 cm.

● Ratio and proportion

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