Question

Two numbers are such that the ratio between them is 3:5. If each is increased by 10, the
ratio between the new numbers so formed is 5: 7. Find the original numbers
​

Answer 1

Answer:

Answer:

Given :-

Two numbers are such that the ratio between them is 3:5.

Each is increased by 10, the ratio between the new numbers are 5:7.

To Find :-

What is the original number.

Solution :-

Let, the first number be 3x

And, the second number be 5x

Each number is increased by 10, the ratio between the new numbers are 5:7.

According to the question,

⇒ $\dfrac{3x + 10}{5x + 10}$ = $\dfrac{5}{7}$

By doing cross multiplication we get,

⇒ 7(3x + 10) = 5(5x + 10)

⇒ 21x + 70 = 25x + 50

⇒ 21x - 25x = 50 - 70

⇒ - 4x = - 20

⇒ x = $\dfrac{\cancel{- 20}}{\cancel{- 4}}$

➠ x = 5

Hence, the required number are,

✦ First number = 3x = 3(5) = 15

✦ Second number = 5x = 5(5) = 25

$\therefore$ The number are 15 and 25 .

Let's Verify :-

⇒ 7(3x + 10) = 5(5x + 10)

Put x = 5

⇒ 7(15 + 10) = 5(25 + 10)

⇒ 7(25) = 5(35)

➠ 175 = 175

➥ LHS = RHS

Hence, Verified.

Answer 2

Answer:

Answer:

→ 15 and 25 .

Step-by-step explanation:

Let x and y be the two numbers .

Now,

CASE 1 .

→ Two numbers are such that the ratio between them is 3 : 5.

A/Q,

∵ x : y = 3 : 5

⇒ 5x = 3y .

∵ x = 3y / 5 ....( 1 ).

CASE 2 .

→ If each number in increased by 10, the ratio between the new number so formed is 5 : 7.

A/Q,

∵ ( x + 10 ) : ( y + 10 ) = 5 : 7 .

⇒ 7( x + 10 ) = ( y + 10 ) 5 .

⇒ 7x + 70 = 5y + 50 .

⇒ 7x + 70 - 50 = 5y .

⇒ 7x + 20 = 5y. ....( 2 ).

Put value of 'x' from equation ( 1 ) in ( 2 ) .

⇒ 7× 3y/5 + 20 = 5y .

⇒ ( 21y + 100 ) / 5 = 5y .

⇒ 21y + 100 = 25y .

⇒ 100 = 25y - 21y .

⇒ 100 = 4y .

⇒ 100 / 4 = y .

∴ y = 25 .

Therefore ,

∵ y = 25 ,

Put y = 25 in equation ( 1 ), we get

⇒ x = 3 × 25 / 5

⇒ x = 3 × 5

∴ x = 15

Original numbers are x and y = 15 and 25 .

it is solved .

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