Tag: more on proportion

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 $ 2, 8 $, the third proportional is $ b = \sqrt {ac} = \sqrt {2 \times 8} = \sqrt {16} = 4 $

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 $ 12, 192 $, the third proportional is $ b = \sqrt {12 \times 192} = \sqrt {4

\times 3 \times 64 \times 3} = 2 \times 3 \times 8 = 48 $

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





For, $ 5:6 :: 7 :d $, product of extremes $ = $ product of means





$ 5 \times d = 6 \times 7 $


$ d = \frac {42}{5} = 8.40 $

Let third proportional is x
$\left ( x^{2}-y^{2} \right ):\left ( x-y \right )=\left ( x-y \right ):x$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left ( x^{2}-y^{2} \right )}$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left (x+y  \right )\left (x-y  \right )}$
$\Rightarrow x=\frac{\left ( x-y \right )}{\left (x+y  \right )}$
 

Let the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$ be A.

To convert $\displaystyle \frac{1}{2}:\frac{1}{3}:\frac{1}{4}$ into normal ratio, multiply each fraction by the LCM$(2,3,4) = 12$

$\displaystyle= 12\times \frac{1}{2}:12\times\frac{1}{3}:12\times\frac{1}{4}$
= $\displaystyle 6:4:3 $
Thus, A would have got $\displaystyle \frac{6}{6+4+3} \times 117 = 54$

B would have got $\displaystyle \frac{4}{6+4+3} \times 117 = 36$

And C would have got $\displaystyle \frac{3}{6+4+3} \times 117 = 27$

Instead Rs. $117$ have been divided in the ratio of $2:3:4$

A gets $\displaystyle \frac {2}{2+3+4} \times 117 = 26$

B gets $\displaystyle \frac {3}{2+3+4} \times 117 = 39$

C gets $\displaystyle \frac {4}{2+3+4} \times 117 = 52$

Thus, by changing the ratio,
A gained $26 - 54$ = -$28$
B gained $39 - 36$ = $3$
And C gained $52 - 27 = 25$
Thus, C's gain of $25$ is the most.

Take the parts as x, y & z.

Let the parts be $7x,5x,3x$ and $4x.$ 

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 = \frac{{b}^{2}}{a} $





So, for $ 8, 10 $, the third proportional is $ c = \frac{{b}^{2}}{a} = \frac {{10}^{2}}{8} = 12.50 $

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 $ 





Given  the third proportional of $ 6, x $ is 150 

$ => {x}^{2} = 6 \times 150 = 6 \times 6 \times 25 $

$ => x = \sqrt { 6 \times 6 \times 25 } $

$ x = 6 \times 5 = 30 $