Question

If A:B = 3:4, B:C = 5:6 and C:D = 11:9, then A:D is :
Options are
A) 50:60
B) 55:72
C) 60:70
D) 65:75
Please provide the steps for the solution and the answer. Thank you.

Answer 1

Answer

Option B) 55 : 72

$\rule{200}1$

Explanation

Since A : B = 3 : 4, let A be 3k and B be 4k.

Now, B : C = 5 : 6

→ B/C = 5/6

→ C = 6B/5 (by cross multiplication)

We know that B = 4k

→ C = 6 × 4k/5

→ C = 24k/5

Again, C : D = 11 : 9

→ C/D = 11/9

→ D = 9C/11 (by cross multiplication)

We know that C = 24k/5

→ D = 9/11 × 24k/5

→ D = 216k/55

Now, A = 3k and D = 216k/55

→ A : D = 3k : 216k/55

→ A/D = $\sf\frac{3k}{\frac{216k}{55}}$

→ A/D = $\sf\frac{3k\times 55}{216k}$

→ A/D = $\sf\frac{55}{72}$

Thus, A : D = 55 : 72

Answer 2

Answer:

$\large{\underline{\boxed{\mathfrak\green{Answer:- A : D = B) 55 : 72 }}}}$

Given:-

• A : B = 3 : 4.............(i)
• B : C = 5 : 6.............(ii)
• C : D = 11 : 9............(iii)

To find:-

• A : D = ?

From (i) :-

$\dashrightarrow\sf \: \: \: \: \: \: A : B = 3 : 4 \\\\\dashrightarrow\sf \: \: \: \: \: \: \dfrac{A}{B} = \dfrac{3}{4} \\\\\dashrightarrow\sf \: \: \: \: \: \:4A = 3B \\\\\dashrightarrow\sf\: \: \: \: \: \: A = \dfrac{3B}{4} \: \: \: \: \: \: \: \: -(eq.1)\\\\\dashrightarrow\sf \: \: \: \: \: \:B = \dfrac{4A}{3} \: \: \: \: \: \: \: \: -(eq.2)$

Now from (ii):-

$\dashrightarrow\sf \: \: \: \: \: \:\dfrac{B}{C} = \dfrac{5}{6}\\\\\sf From \: (eq.2):- \\\\\dashrightarrow\sf \: \: \: \: \: \:\dfrac{\dfrac{4A}{3}}{C} = \dfrac{5}{6} \\\\\dashrightarrow\sf \: \: \: \: \: \:5C = 6 * \dfrac{4A}{3} \\\\\dashrightarrow\sf \: \: \: \: \: \:5C = 8A \\\\\dashrightarrow\sf \: \: \: \: \: \:C = \dfrac{8A}{5} \: \: \: \: \: \: \: \: -(eq.3)$

Now from (iii) :-

$\dashrightarrow\sf \: \: \: \: \: \:\dfrac{C}{D} = \dfrac{11}{9} \\\\\sf From \: (eq.3) \::- \\\\\dashrightarrow\sf \: \: \: \: \: \:\dfrac{\dfrac{8A}{5}}{D} = \dfrac{11}{9} \\\\\dashrightarrow\sf \: \: \: \: \: \:11D = 9 * \dfrac{8A}{5} \\\\\dashrightarrow\sf \: \: \: \: \: \:11D = \dfrac{72A}{5}\\\\\dashrightarrow\sf \: \: \: \: \: \:D = \dfrac{72A}{55}\: \: \: \: \: \: \: \: -(eq.4)$

Now solving from (eq.1) & (eq.4) :-

$\dashrightarrow\sf \: \: \: \: A : D = \dfrac{3B}{4} : \dfrac{72A}{55} \\\\\dashrightarrow\sf \: \: \: \:A : D = \dfrac{3 * \dfrac{4A}{3}}{4} : \dfrac{72A}{55} \: \: \: \:\: \:\: \: [\because B = \dfrac{4A}{3}] \\\\\dashrightarrow\sf \: \: \: \:A : D = A : \dfrac{72A}{55} \\\\\dashrightarrow\sf \: \: \: \:A : D = \dfrac{A}{\dfrac{72A}{55}} \\\\\dashrightarrow\sf \: \: \: \:A : D = A \times \dfrac{55}{72A} \\\\\dashrightarrow\sf \: \: \: \:A : D = \dfrac{55}{72}\\\\\dashrightarrow \: \: \: \: \: \:\large{\underline{\boxed{\sf\blue{A : D = 55 : 72 }}}}$

$\green{\boxed{\therefore{\sf\purple{A : D = 55 : 72 }}}}$