Question

If A:B = 2:3 and B:C= 4:9, then find the value of A:C.​

Answer 1

Given :-

• A:B = 2:3
• B:C = 4:9

To find :-

• A:C

Solution :-

Here, A:B = 2:3

so,

$\sf \dfrac{A}{B} = \dfrac{2}{3}$

$\sf a = \dfrac{2B}{3}$

Thus, A = 2B/3.

Now, B:C :-

B:C = 4:9

so,

$\sf \dfrac{B}{C} = \dfrac{4}{9}$

$\sf c = \dfrac{4B}{9}$

Hence, C = 4B/9.

At last , A:C

$\sf A:C = \frac{\dfrac{2B}{3}}{\dfrac{4B}{9} }$

B will be cancelled, We get :-

$\sf A:C = \frac{\dfrac{2}{3}}{\dfrac{4}{9} }$

By cross multiplication,

$\sf A:C = \dfrac{2 \times 4}{3 \times 9}$

$\sf A:C = \dfrac{8}{27}$

Therefore, A:C = 8:27.

Answer 2

Given:-

• A:B = 2:3
• B:C= 4:9

To Find:-

• A:C = ?

Solution:-

Here,A:B = 2:3

So,

$\sf\frac{A}{B}$=$\sf\frac{2}{3}$

a=

Thus,A = 2B/3.

Now, B:C:-

B :C = 4:9

So,

$\sf\frac{B}{C}$=$\sf\frac{4}{9}$

C= $\sf\frac{4B}{9}$

Hence,C = 4B/9.

At last, A:C

A :C = $\sf{\frac{{\frac{2}{3}}{{\frac{4}{9}}}$

A:C = $\sf\frac{2×4}{3×9}$

A : C = $\sf\frac{8}{27}$

Therefore,A :C = 8 :27.