Let $\frac{A}{B}=\frac{B}{C}=\frac{C}{D}=\frac{3}{4}$. If $B$ gets Rs $308$ less than $D$, find the individual amount with $A$, $B$, $C$, and $D$.

Question

Some amount is divided among $A$, $B$, $C$, and $D$ such that $\frac{A}{B}=\frac{B}{C}=\frac{C}{D}=\frac{3}{4}$. If $B$ gets Rs $308$ less than $D$, find the individual amount with $A$, $B$, $C$, and $D$.

What is the logic behind solving these kinds of problems?

I didn't understand the solution $A : B : C : D = 27 : 36 : 48 : 64$. How?

Answer 1

The trick is to solve for one of them and then use $\frac{A}{B}=\frac{B}{C}=\frac{C}{D}=\frac{3}{4}$ to get the other $3$. To be more general, I'll let $\alpha\equiv\frac{3}{4}\in(0,1)$ and $\delta\equiv 308>0$. Then $$ C=\alpha D,\quad B=\alpha C=\alpha^2D,\quad A=\alpha B=\alpha^3D $$ and so $$ \delta=D-B=D(1-\alpha^2)\implies D=\frac{\delta}{1-\alpha^2}\implies A,B,C=\cdots $$

---

Edit: For this particular problem, if you want $A:B:C:D=27:36:48:64$, then note that $$ A=\alpha^3D=\frac{27}{\color{Red}{64}}D,\quad B=\frac{9}{16}D,\quad C=\frac{3}{4}D $$ so if you set $D=64k$ for some $k$ then $$ A=27k,\quad B=36k,\quad C=48k,\quad D=64k $$ which gives $A:B:C:D=27:36:48:64$. Finding $k$ entails noting $D-B=28k=308$.

Answer 2

This is a set of four equations and four unknowns. The four unknowns are $A, B, C, D$ and the four equations are $$ \frac AB = \frac34\\ \frac BC = \frac34\\ \frac CD = \frac34\\ B = D-308 $$

---

To address your last sentence, we see that $A:B = 3:4$. Now, if we want to add $C$ in there to make it $A:B:C$, then we need the $B$ component to be divisible by $3$, since $B:C = 3:4$. Therefore we rewrite $A:B = 9:12$, which lets us insert $C$ as $A:B:C = 9:12:16$. Do the same one more time for $D$, and you're done.

Answer 3

Hint:

$$\frac{A}{B}=\frac{B}{C}=\frac{C}{D}=\frac{3}{4} \quad (1)$$

but $B=D-308 $.

From $(1)$ we get $D=\frac{4C}{3}$ and $B=\frac{3C}{4}$ so backing to $(2)$ we get

$$\frac{3C}{4}=\frac{4C}{3}-308 \Rightarrow C=528$$

Now you can get the others.

Answer 4

Let the amount with $A$ be equal to $x$. Thus, it can be easily checked that the amount with $B,C $ and $D$ are respectively: $\frac{4}{3}x, \frac{16}{9}x, \frac{64}{27}x $.

It is given that the difference of money between $B$ and $D$ is $308$. Thus, it can be mathematically translated to $$\frac{64}{27}x -\frac{4}{3}x = 308$$ $$\Rightarrow \frac{28}{27}x = 308 $$ $$\Rightarrow x=297$$ We can then get the money of $B,C,D$. Hope it helps.

Answer 5

Given,

$A:B = 3:4$

$A=3x, B=4x$

$B:C = 3:4$

$B=3x, C=4x$

Now in above two ratios in first B has 4x value and in second B has 3x value. Take LCM of 3x, 4x. You have 12x.

To make the value of B in first ratio to 12x . Multiply first ratio with 3x and in other ratio with 4x.

You have,

$A:B = 9x:12x$ and $B:C = 12x:16x$

On combining these two not repeat same value,

$A:B:C = 9x:12x:16x$

$C:D = 3:4$

$C = 3x, D = 4x$

Now in above two ratios in first C has 16x value and in second C has 3x value. Take LCM of 16x, 3x. You have 48x.

To make the value of C in first ratio to 48x . Multiply first ratio with 3x and in other ratio with 16x.

$A:B:C = 27x:36x:48x$

$C:D = 48x:64x$

On combining these two,

$A:B:C:D = 27x:36x:48x:64x$

If eliminate x.

We have $A:B:C:D = 27:36:48:64$

To solve question $B = D - 308$

$36x = 64x - 308$

Solve it to get values.

Hope you understand now.