Tag: ratio, proportion and unitary method

$8 : x :: 16 : 35$

Given, the first three terms of a proportion are $3,5$ and $21$ respectively.

Let the fourth number be $'x'$,

So, Numbers proportion are $3, 5, 21, x$

According to proportionality,

Product of means $=$ Product of extremes

$3 : 5 :: 21 : x$

$\dfrac{3}{5} = \dfrac{21}{x}$

Cross multiply,

$3 \times x = 21 \times 5$

$3x = 105$

$x = 35$

Therefore, The required fourth number is $35$

$12 : 21 = 8 : x$

In $ a:b::c:d;   d $ is the fourth proportional.

In, $ 5: \sqrt {75} :: \sqrt {48} :d $, product of extremes $ = $ product of means

$  5 \times d = \sqrt {75} \times  \sqrt {48} $
$ d = \dfrac{\sqrt {75} \times  \sqrt {48}}{5} = \dfrac{5\sqrt {3} \times 4\sqrt {3}}{5} = 4 \times 3 = 12 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 5:7 :: 8 :d $, product of extremes $ = $ product of means
$ 5 \times d = 7 \times 8 $
$ d = \dfrac {56}{5} = 11.2 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 1.2:3.8 :: 9 :d $, product of extremes $ = $ product of mean
$ 1.2 \times d = 3.8 \times 9 $

$ d = 28.5 $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 2\dfrac{1}{2}:1 \dfrac{1}{4} :: 3\dfrac {1}{3} :d $, product of extremes $ = $ product of means
$ 2\dfrac{1}{2} \times d = 1 \dfrac{1}{4} \times 3\dfrac {1}{3} $ 

$ \dfrac{5}{2} \times d = \dfrac{5}{4} \times \dfrac {10}{3} $

$ d = \dfrac {\dfrac{5}{4} \times \dfrac {10}{3}}{ \dfrac{5}{2}} = \dfrac {5}{3} = 1 \dfrac {2}{3} $

If a : b :: b : c, then we say that a, b, c are in continued proportion, and
c is the third proportional of a and b.
Here, $ {b}^{2} = ac $ or $ c = \dfrac{{b}^{2}}{a} $
So, for $ 12, 16 $, the third proportional is $ c = \dfrac{{b}^{2}}{a} = \dfrac
{{16}^{2}}{16} = \dfrac{16 \times 16}{12} = \dfrac{64}{3} = 21\dfrac{1}{3} $

In $ a:b::c:d;   d $ is the fourth proportional.
For, $ 3:42 :: 7 :d $, product of extremes $ = $ product of means
$ 3 \times d = 42 \times 7 $
$ d = 98 $

If a : b :: b : c, then we say that a, b, c are in continued

proportion, and c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ or $ b = \sqrt {ac} $





So, for $ 0.2, 0.8 $, the third proportional is $ b = \sqrt {ac} = \sqrt {0.2

\times 0.8} = \sqrt {0.16} = 0.4 $