Percentage All Competence Aptitude

Let's denote the amount of water to be added as \(x\) (in liters).

Initially, in the 60 liters of solution, there are 80% acid and 20% water. The amount of acid is \(0.80 \times 60 = 48\) liters.

After adding \(x\) liters of water, the total volume becomes \(60 + x\) liters.

Now, you want the new concentration of acid to be 60%, so the equation for the new concentration is:

\[ \dfrac{48}{60 + x} = 0.60 \]

Now, solve for \(x\):

\[ 48 = 0.60(60 + x) \]

\[ 48 = 36 + 0.60x \]

\[ 0.60x = 12 \]

\[ x = 20 \]

Therefore, \( \boxed{20 \, \text{liters}} \) of water must be added to make the solution in which the acid forms 60%.

Initially, in the 3 liters of sugar solution, there are 60% sugar. The amount of sugar is \(0.60 \times 3 = 1.8\) liters.

After adding 1 liter of water, the total volume becomes \(3 + 1 = 4\) liters.

Now, you want to find the percentage of sugar in the new solution:

\[\dfrac{1.8}{4} \times 100\%\]

Simplify the expression:

\[0.45 \times 100\%\]

\[45\%\]

Therefore, the percentage of sugar in the new solution is \( \boxed{45\%} \).

Let's denote the amount of milk to be added as \(x\) (in liters).

Initially, in the 200 liters of mixture, there is 15% water and 85% milk. The amount of milk is \(0.85 \times 200 = 170\) liters.

After adding \(x\) liters of milk, the total volume becomes \(200 + x\) liters.

You want the new concentration of milk to be 87.5%, so the equation for the new concentration is:

\[ \dfrac{170 + x}{200 + x} = 0.875 \]

Now, solve for \(x\):

\[ 170 + x = 0.875(200 + x) \]

\[ 170 + x = 175 + 0.875x \]

\[ 0.125x = 5 \]

\[ x = 40 \]

Therefore, \( \boxed{40 \, \text{liters}} \) of milk must be added to the mixture to make the resulting mixture contain 87.5% milk.

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

Initially, in 100 ml of an 80% solution, the amount of boric acid is \(0.80 \times 100 = 80\) ml.

After adding \(x\) ml of water, the total volume becomes \(100 + x\) ml.

Now, you want the new concentration of boric acid to be 50%, so the equation for the new concentration is:

\[\dfrac{80}{100 + x} = 0.50\]

Now, solve for \(x\):

\[80 = 0.50(100 + x)\]

\[80 = 50 + 0.50x\]

\[0.50x = 30\]

\[x = 60\]

Therefore, \( \boxed{60 \, \text{ml}} \) of water must be added to the 100 ml of an 80% solution of boric acid to reduce it to a 50% solution.

Let's denote the quantity of sugar to be added as \(x\) (in grams).

Initially, in 75 grams of the solution, there is 30% sugar. The amount of sugar is \(0.30 \times 75 = 22.5\) grams.

After adding \(x\) grams of sugar, the total quantity of sugar becomes \(22.5 + x\).

The total quantity of the solution is \(75 + x\) grams.

Now, you want the new concentration of sugar to be 70%, so the equation for the new concentration is:

\[\dfrac{22.5 + x}{75 + x} = 0.70\]

Now, solve for \(x\):

\[22.5 + x = 0.70(75 + x)\]

\[22.5 + x = 52.5 + 0.70x\]

\[0.30x = 30\]

\[x = 100\]

Therefore, \( \boxed{100 \, \text{gm}} \) of sugar should be added to the solution to make the quantity of sugar 70%.

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

Initially, in 50 liters of glycerine, there is 20% impurity. The amount of pure glycerine is \(0.80 \times 50 = 40\) liters.

After adding \(x\) liters of pure glycerine, the total quantity of pure glycerine becomes \(40 + x\).

The total quantity of the mixture is \(50 + x\) liters.

You want the new percentage of impurity to be 5%, so the equation for the new concentration is:

\[\dfrac{10}{50 + x} = 0.05\]

Now, solve for \(x\):

\[10 = 0.05(50 + x)\]

\[10 = 2.5 + 0.05x\]

\[0.05x = 7.5\]

\[x = 150\]

Therefore, \( \boxed{150 \, \text{liters}} \) of pure glycerine should be added to bring down the percentage of impurity to 5%.

Let's assume the original price of the commodity is $100 (or any other currency for consistency). After a 60% increase, the new price becomes $160.

Let \(Q\) be the original quantity of the commodity purchased, and \(P\) be the original price per unit.

The original expenditure (\(E_{\text{original}}\)) is given by the product of the original quantity and the original price per unit:

\[E_{\text{original}} = Q \times P\]

The new expenditure (\(E_{\text{new}}\)) is given by the product of the reduced quantity (let's call it \(Q_{\text{new}}\)) and the increased price per unit:

\[E_{\text{new}} = Q_{\text{new}} \times 160\]

To avoid an increase in expenditure, \(E_{\text{new}}\) should be equal to \(E_{\text{original}}\):

\[Q_{\text{new}} \times 160 = Q \times P\]

Now, let's find the reduced quantity (\(Q_{\text{new}}\)):

\[Q_{\text{new}} = \dfrac{Q \times P}{160}\]

The percentage reduction in consumption is given by:

\[\text{Percentage reduction} = \dfrac{Q - Q_{\text{new}}}{Q} \times 100\%\]

Substitute the expression for \(Q_{\text{new}}\):

\[\text{Percentage reduction} = \dfrac{Q - \dfrac{Q \times P}{160}}{Q} \times 100\%\]

Now, factor out \(Q\) and simplify:

\[\text{Percentage reduction} = \dfrac{Q \times (160 - P)}{Q \times 160} \times 100\%\]

\[ \text{Percentage reduction} = \dfrac{160 - P}{160} \times 100\%\]

\[ \text{Percentage reduction} = \dfrac{160 - 100}{160} \times 100\%\]

\[ \text{Percentage reduction} = \dfrac{60}{160} \times 100\%\]

\[ \text{Percentage reduction} = 37.5\%\]

Therefore, the consumer must reduce the consumption of the commodity by \( \boxed{37.5\%} \) to avoid an increase in expenditure.

Let's assume the original price of the commodity is $100 (or any other currency for consistency), and the original quantity consumed is 1 unit.

1. Original expenditure (\(E_{\text{original}}\)) is given by the product of the original quantity and the original price per unit:

\[E_{\text{original}} = 1 \times 100 = 100\]

2. After a 20% decrease in price, the new price becomes 80% of the original price (\(0.80 \times 100 = 80\)).

3. After a 20% increase in consumption, the new quantity consumed becomes 120% of the original quantity (\(1.20 \times 1 = 1.2\)).

4. New expenditure (\(E_{\text{new}}\)) is given by the product of the new quantity and the new price per unit:

\[E_{\text{new}} = 1.2 \times 80 = 96\]

Now, let's calculate the percentage change in expenditure:

\[\text{Percentage change} = \dfrac{E_{\text{new}} - E_{\text{original}}}{E_{\text{original}}} \times 100\%\]

\[\text{Percentage change} = \dfrac{96 - 100}{100} \times 100\%\]

\[\text{Percentage change} = \dfrac{-4}{100} \times 100\%\]

\[\text{Percentage change} = -4\%\]

Therefore, there is a \( \boxed{4\% \text{ decrease}} \) in expenditure on the commodity.

Let's assume the original quantity purchased is 1 kg, and the original price per kg is $6 (or any other currency for consistency).

1. Original expenditure (\(E_{\text{original}}\)) is given by the product of the original quantity and the original price per kg:

\[E_{\text{original}} = 1 \times 6 = 6\]

2. After the price rises to $7.50 per kg, the new price is $7.50.

3. Let \(x\) be the new quantity that can be purchased with the same expenditure.

The new expenditure (\(E_{\text{new}}\)) is given by the product of the new quantity and the new price per kg:

\[E_{\text{new}} = x \times 7.50\]

Since the expenditure cannot increase, \(E_{\text{new}}\) should be equal to \(E_{\text{original}}\):

\[x \times 7.50 = 6\]

Solve for \(x\):

\[x = \dfrac{6}{7.50} = \dfrac{4}{5}\]

Now, let's find the percentage reduction in consumption:

\[\text{Percentage reduction} = \dfrac{1 - \dfrac{4}{5}}{1} \times 100\%\]

\[\text{Percentage reduction} = \dfrac{1}{5} \times 100\%\]

\[\text{Percentage reduction} = 20\%\]

Therefore, the percentage reduction in consumption is \( \boxed{20\%} \).

Let \( T \) be the total number of marks in the examination.

The examinee has to secure 40% marks to pass, which is \( 0.40 \times T \) marks.

According to the given information, the examinee secures 180 marks and fails by an equal number of marks. So, the passing marks would be \( 180 + 180 \).

The passing marks are equal to 40% of the total marks:

\[ 180 + 180 = 0.40 \times T \]

Solve for \( T \):

\[ 360 = 0.40 \times T \]

\[ T = \dfrac{360}{0.40} \]

\[ T = 900 \]

Therefore, the total number of marks in the examination is \( \boxed{900} \).