Question

the present ages of A and B are in the ratio 5 is to 6
3 years ago the ages were in the ratio 4 is to 5 find their present ages​

Answer 1

Given

The present ages of A and B are in the ratio 5:6 . 3 years ago the ages were in the ratio 4:5

Find out

Present ages of A and B

Solution

Let the present age of A be 5x and B be 6x

• 3 years ago
• A's age = (5x - 3)
• B's age = (6x - 3)

$\bigstar\tt{\underline{According\:of\:the\:given\:condition}}\bigstar \\ \\ \implies\tt \dfrac{5x-3}{6x-3}=\dfrac{4}{5} \\ \\ \\ \implies\tt 5(5x-3)=4(6x-3) \\ \\ \\ \implies\tt 25x - 15=24x-12 \\ \\ \\ \implies\tt 25x-24x=15-12 \\ \\ \\ \:\:\:\:\therefore\tt \:x=3\:years \\ \\ \bigstar{\red{\boxed{\tt{Present\:age\:of\:A=5x=15years}}}} \\ \\ \bigstar{\red{\boxed{\tt{Present\:age\:of\:B=6x=18years}}}}$

$\rule{200}3$

Answer 2

$\huge{\bold{\underline{\underline{\blue{\sf{Question:-}}}}}}$

The Present Ages of A and B are in the ratio 5 is to 6 ...3 years ago the ages were in the ratio 4 is to 5..Find their Present ages ..

______________

$\huge{\bold{\underline{\underline{\orange{\sf{Answer:-}}}}}}$

↪Given : -

• Present age of A and B are in ratio = 5:6
• After three years the ages were in ratio = 4:5

↪To Find : -

• Present age of A and B

↪Let : -

• Present age of A = 5x
• Present age of B = 6x

↪After three years : -

• A's age = 5x - 3
• B's age = 6x - 3

↪Now ,

↪According to the Question : -

$\leadsto\sf{\dfrac{5x-3}{6x-3}=\dfrac{4}{5}}$

↪By Cross Multiplication : -

$\implies\sf{5(5x-3)=4(6x-3)}$

$\implies\sf{25x-15=24x-12}$

$\implies\sf{25x-24x=-12+15}$

$\implies\sf{-1x=-3}$

$\leadsto\sf{\large{\boxed{\bold{\bold{\green{\sf{x=3}}}}}}}$

↪On Substituting Values : -

$\implies\sf{Present\:age\:of\:A=5(3)}$

$\implies\sf{15\:years}$

$\implies\sf{Present\:age\:of\:B=6(3)}$

$\implies\sf{18\:years}$