Question

Avnish and bhavnesh had chocolates in the ratio 4:5 if each is given 6 more chocolates the ratio becomes 9:11 find the original number each had​

Answer 1

Answer:

Avnish had 60 chocolates and Bhavnesh had 48 chocolates, originally.

Step-by-step explanation:

Let original number of chocolates with Avnish be "a"

Let original number of chocolates with Bhavnesh be "b"

Given that

a:b :: 4:5

=> a/b = 4/5

=> 5a = 4b

=> a = 4b/5      ...Eqn 1

Each is given 6 chocolates more,

=> Number of chocolates with Avnish = a+6

=> Number of chocolates with Bhavnesh = b+6

ATQ,

(a+6):(b+6) :: 9:11

=> (a+6)/(b+6) = 9/11

=> 11(a+6) = 9(b+6)

=> 11a + 66 = 9b + 54

=> 11a - 9b = -12 .......Eqn 2

Substituting for "a" from Eqn1 in Eqn2, we get:

11(4b/5) - 9b = -12

44b/5 - 9b = -12

44b - 45b = -60

-b = -60

b = 60

Substituting for "b" in Eqn 1, we get:

a = 4b/5

a = 4*60/5

a = 48

No. of chocolates with Avnish = 48

No. of chocolates with Bhavnesh = 60

Verify:

48:60

= 48/60

= 4/5         (common factor: 12)

= 4:5   (As stated in the problem)

(48+6):(60+6)

= 54:66

= 54/66

= 9/11        (common factor: 6)

= 9:11 (As stated in the problem)

Thus, verified.

Answer 2

Answer:

48 and 60.

Step-by-step explanation:

Given:- Initial ratio of chocolates is 4:5. After giving 6 more to each it becomes 9:11.

To Find:- The original number of chocolates.

Solution:-

Let the number of chocolates be x .

So, Avnish has 4x chocolates and Bhavesh had 5x.

After giving 6 chocolates to each one of them, ratio becomes

$\frac{4x+6}{5x +6} = \frac{9}{11}$

⇒ $11(4x + 6 ) = 9 ( 5x + 6 )$

⇒ $44x + 66 = 45x + 54$

⇒ $x = 66-54$

⇒ $x = 12$

Therefore, Avnish had 4 × 12 = 48 chocolates.

Bhavesh had 5 × 12 = 60 chocolates.

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