Percentage All Competence Aptitude

Let's denote the numbers as \(x\) and \(y\). According to the given information:

\[ 0.50x = 0.30y \]

Now, let's solve for the ratio \(x:y\):

\[ \dfrac{x}{y} = \dfrac{0.30}{0.50} \]

Simplify the ratio:

\[ \dfrac{x}{y} = \dfrac{3}{5} \]

Therefore, the correct ratio of \(x\) to \(y\) is \( \boxed{3:5} \).

Let the price of a shoe be \(x\). 

Given that the prices of the school bag and the shoe are in the ratio 7:5, the price of the school bag is \(\dfrac{7}{5}x\).

According to the given information, the price of the school bag is 200 more than the price of the shoe:

\[\dfrac{7}{5}x = x + 200\]

Now, let's solve for \(x\):

Multiply both sides of the equation by 5 to get rid of the fraction:

\[7x = 5x + 1000\]

Subtract 5x from both sides:

\[2x = 1000\]

Divide by 2:

\[x = 500\]

Therefore, the price of a shoe is \( \boxed{500} \).

Let's assume Rama's expenditures are \(5x\) and her savings are \(3x\), where \(x\) is a positive constant. Therefore, her income is the sum of expenditures and savings, which is \(5x + 3x = 8x\).

Now, let's consider the changes in income, expenditures, and savings when her income increases by 12% and expenditures increase by 15%.

1. New income after a 12% increase: \(1.12 \times 8x = 8.96x\)
2. New expenditures after a 15% increase: \(1.15 \times 5x = 5.75x\)

Now, let's find the new savings:

\[ \text{New savings} = \text{New income} - \text{New expenditures} \]\[ \text{New savings} = 8.96x - 5.75x \]\[ \text{New savings} = 3.21x \]

Now, let's find the percentage increase in savings:

\[ \text{Percentage increase} = \dfrac{\text{New savings} - \text{Old savings}}{\text{Old savings}} \times 100\% \]

\[ \text{Percentage increase} = \dfrac{3.21x - 3x}{3x} \times 100\% \]

\[ \text{Percentage increase} = \dfrac{0.21x}{3x} \times 100\% \]

\[ \text{Percentage increase} = \dfrac{7}{100} \times 100\% \]

\[ \text{Percentage increase} = 7\% \]

Therefore, her savings increase by \( \boxed{7\%} \).

Let \(x\) be the number of boys in the class. Then, the number of girls is \(1.20x\) (20% more than the number of boys).

The total strength of the class is given by the sum of the number of boys and girls:

\[x + 1.20x = 66\]

Combine like terms:

\[2.20x = 66\]

Now, solve for \(x\):

\[x = \dfrac{66}{2.20} = 30\]

So, the number of boys is 30 and the number of girls is \(1.20 \times 30 = 36\).

Now, if 4 more girls are admitted, the new number of girls becomes \(36 + 4 = 40\).

The ratio of the number of boys to the number of girls is:

\[\dfrac{30}{40} = \dfrac{3}{4}\]

Therefore, the ratio is \( \boxed{3 : 4} \).

Let's assume the number of boys is \(4x\) and the number of girls is \(x\), where \(x\) is a positive constant.

According to the given information:

1. Number of boys = \(4x\)
2. Number of girls = \(x\)

Now, let's calculate the number of boys and girls who are scholarship holders:

Number of boys with scholarship = \(0.75 \times 4x = 3x\)

Number of girls with scholarship = \(0.70 \times x = 0.7x\)

Total number of students with scholarship = \(3x + 0.7x = 3.7x\)

Now, the total number of students in the school is \(4x + x = 5x\).

The percentage of students who do not get a scholarship is:

\[ \dfrac{5x - 3.7x}{5x} \times 100\% = \dfrac{1.3x}{5x} \times 100\% = 26\% \]

Therefore, the correct answer is \( \boxed{26\%} \).

Initially, the solution contains 20% alcohol. That means in the initial solution of 32 liters, there are 20% of 32 liters of alcohol, which is \(0.20 \times 32 = 6.4\) liters of alcohol.

After adding 8 liters of water, the total volume of the solution becomes \(32 + 8 = 40\) liters.

Now, the concentration of alcohol is the amount of alcohol divided by the total volume:

\[\text{Concentration} = \dfrac{\text{Amount of alcohol}}{\text{Total volume}}\]

\[\text{Concentration} = \dfrac{6.4}{40} \times 100\%\]

\[\text{Concentration} = 16\%\]

Therefore, the approximate concentration of alcohol in the solution after adding water is \( \boxed{16\%} \).

Let \(x\) be the amount of sugar to be added.

Initially, the sugar content in the solution is \(40\%\) of 300 grams, which is \(0.40 \times 300 = 120\) grams.

After adding \(x\) grams of sugar, the total amount of sugar becomes \(120 + x\).

The total amount of solution is \(300 + x\) grams.

Now, you want the sugar content to be \(50\%\) of the total solution:

\[\dfrac{120 + x}{300 + x} = 0.50\]

Solve for \(x\):

\[120 + x = 0.50(300 + x)\]

\[120 + x = 150 + 0.50x\]

\[0.50x = 30\]

\[x = 60\]

Therefore, \(60\) grams of sugar should be added to make the solution \(50\%\) sugar.

Let's assume that the first sugar solution is mixed in the ratio \(x\) and the second sugar solution is mixed in the ratio \(y\).

Concentration of the first solution = 15%
Concentration of the second solution = 40%
Target concentration of the mixture = 30%

According to the mixture rule:

\[ \dfrac{x \times 15 + y \times 40}{x + y} = 30 \]

Multiply both sides by \(x + y\) to get rid of the denominator:

\[ 15x + 40y = 30(x + y) \]

Expand and simplify:

\[ 15x + 40y = 30x + 30y \]

\[ 10y = 15x \]

\[ 2y = 3x \]

Now, the ratio \(x : y\) can be represented as \(3 : 2\). However, the question asks for the inverse ratio, so the correct ratio is \( \boxed{2 : 3} \).

Let's denote the amount of pure alcohol to be added as \(x\) (in milliliters).

Initially, in the 400 ml solution containing 15% alcohol, the amount of alcohol is \(0.15 \times 400 = 60\) ml.

After adding \(x\) ml of pure alcohol, the total amount of alcohol becomes \(60 + x\).

The total volume of the mixture is \(400 + x\) ml.

You want the new concentration of alcohol to be 32%, so the equation for the new concentration is:

\[\dfrac{60 + x}{400 + x} = 0.32\]

Now, solve for \(x\):

\[60 + x = 0.32(400 + x)\]

\[60 + x = 128 + 0.32x\]

\[0.68x = 68\]

\[x = 100\]

Therefore, you need to add \( \boxed{100 \text{ ml}} \) of pure alcohol to change the concentration of alcohol in the mixture to 32%.

Let \( x \) be the amount of alcohol (in milliliters) to be added to the mixture.

Initially, in one liter of the mixture, the amount of water is 30% of 1000 ml, which is \( 0.30 \times 1000 = 300 \) ml.

After adding \( x \) ml of alcohol, the total volume of the mixture becomes \( 1000 + x \) ml.

You want the new concentration of water to be 15%, so the equation for the new concentration is:

\[ \dfrac{300}{1000 + x} = 0.15 \]

Now, solve for \( x \):

\[ 300 = 0.15(1000 + x) \]

\[ 300 = 150 + 0.15x \]

\[ 0.15x = 150 \]

\[ x = 1000 \]

Therefore, \( \boxed{1000 \, \text{ml}} \) of alcohol must be added to the mixture to make the part of water in the mixture 15%.