Question

John needs 20 grams of 54% acid solution for his science project. His school's science lab has bottles of 30% solution and bottles of 60% solution. Hiw much of each should john use?

Answer 1

Answer:

John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

Step-by-step explanation:

Let the number of grams of the 30% solution = x

Let the number of grams of the 60% solution = y

John needs 20 grams of 54% acid solution for his science project.

Hence,

x + y = 20 grams..... Equation 1

x = 20 - y

His school's science lab has bottles of 30% solution and bottles of 60% solution.

30% × x + 60% × y = 54% × 20

0.3x + 0.6y = 10.8......Equation 2

We substitute 20 - y for x in Equation 2

0.3(20 - y) + 0.6y = 10.8

6 - 0.3y + 0.6y = 10.8

- 0.3y + 0.6y = 10.8 - 6

0.3y = 4.8

y = 4.8/3

y = 16 grams

x = 20 - y

x = 20 - 16

x = 4 grams

Therefore, John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

Answer 2

The amount of 30% and 60% bottles he should join is; 4grams and 16grams respectively.

Percentages

The total amount John needs is; 20 grams.

• Let the amount of 30% he needs = x.
• Let the amount of 60% he needs = 20-x.

Therefore, we have;

$(\frac{x}{20} \times 30) + (\frac{20 - x}{20} \times 60) = 54$

Therefore, upon evaluation;

• 1.5x + 60 -3x = 54

• -1.5x = -6.

• x = 4 grams.

• 20-x = 16 grams

Therefore, The amount of 30% and 60% bottles he should join is; 4grams and 16grams respectively.

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