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The ratio of Val's savings to Dan is 2:3 at first. After Dan spent P3000.00, Val's saving became 11/3 more of that Dan's remaining savings. What is the total amount of saving the two had at first

Sagot :

Answer:

P.7500.00

Step-by-step explanation:

We are told that:

The ratio of Val's savings to Dan is 2:3 at first.

Now, Dan spent P3000.00 and Val's saving became 1⅓ more of that Dan's remaining savings.

Let's say the total amount of savings they had at first was x.

Thus;

Val had: 2x/5

Dan had: 3x/5

Now, Dan spent P3000.00.

So amount Dan has left = (3x/5) - 3000

We are told Val's savings is now 1⅓ = 4/3 times that of Dan's remaining savings

Thus; 2x/5 = (4/3)((3x/5) - 3000)

2x/5 = 12x/15 - 4000

Multiply through by 15 to get;

6x = 12x - (3000 × 15)

6x = 12x - 45000

Rearranging, we have;

12x - 6x = 45000

6x = 45000

x = 45000/6

x = P.7500

The total amount of saving the two had at first is P7500 and this can be determined by using the arithmetic operations and given data.

Given :

  • The ratio of Val's savings to Dan is 2:3 at first.
  • After Dan spent P3000.00, Val's saving became 4/3 more of that Dan's remaining savings.

Given that the ratio of Val's savings to Dan is 2:3 at first. Now, assume that the total amount of savings they had at first was x. So:

Val's had - 2x/5

Dan's had - 3x/5

Now, it is also given that Dan's spent P3000. So, the amount left is:

[tex]\rm =\dfrac{2x}{5}-3000[/tex]

It is given that Val's saving became 4/3 more of that Dan's remaining savings. That is:

[tex]\dfrac{2x}{5} = \dfrac{4}{3} \times\left( \dfrac{3x}{5}-3000 \right)[/tex]

[tex]\dfrac{6x}{20} = \dfrac{3x}{5}-6000[/tex]

[tex]x = \dfrac{45000}{6}[/tex]

x = P.7500

The total amount of saving the two had at first is P7500.

For more information, refer to the link given below:

https://brainly.com/question/13101306

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