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.
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.
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