Percentage All Competence Aptitude

Let's denote the two numbers as \(5x\) and \(4x\), where \(x\) is a positive constant.

According to the given information, 40% of the first number is 12:

\[0.40 \times (5x) = 12\]

Solving for \(x\):

\[2x = 12\]

\[x = 6\]

Now that we know \(x = 6\), we can find the second number (\(4x\)):

\[4x = 4 \times 6 = 24\]

Now, 50% of the second number is:

\[0.50 \times 24 = 12\]

So, 50% of the second number is also 12.

Therefore, the answer is \( \boxed{12} \).

Let's denote the quantity of milk as \(M\) and the quantity of water as \(W\).

According to the given information, the milkman gains 25% by selling the mixture at the cost price. This means the selling price of the mixture is 125% of the cost price.

Let's assume the cost price of the mixture is $100.

Selling price of the mixture = $125 (125% of the cost price)

Now, the mixture contains both milk and water. The ratio of milk to the total mixture is given as \(4:1\) (since the milkman gains 25%, the ratio becomes \(5:1\), and if you set up the equation \(4x + x = 5x\), you find that \(x\) corresponds to the quantity of milk).

So, the cost price of milk in the mixture is \(4/5\) of $100, and the cost price of water is \(1/5\) of $100.

\[ \text{Cost price of milk} = \dfrac{4}{5} \times 100 = 80 \]

\[ \text{Cost price of water} = \dfrac{1}{5} \times 100 = 20 \]

Now, let's find the ratio of water to milk:

\[ \text{Ratio of water to milk} = \dfrac{\text{Cost price of water}}{\text{Cost price of milk}} \]

\[ \text{Ratio of water to milk} = \dfrac{20}{80} \]

Simplify the ratio:

\[ \text{Ratio of water to milk} = \dfrac{1}{4} \]

Therefore, the correct ratio of water to milk is \( \boxed{1:4} \).

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

\[ 15\% \text{ of } x = 20\% \text{ of } y \]

Mathematically, this can be expressed as:

\[ 0.15x = 0.20y \]

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

\[ \dfrac{x}{y} = \dfrac{0.20}{0.15} \]

To simplify, we can multiply both the numerator and denominator by 100:

\[ \dfrac{x}{y} = \dfrac{20}{15} \]

Simplify the fraction:

\[ \dfrac{x}{y} = \dfrac{4}{3} \]

Therefore, the correct ratio of \(x\) to \(y\) is \( \boxed{4:3} \).

Let's denote the number of boys as \(3x\) and the number of girls as \(2x\), where \(x\) is a positive constant.

According to the given information:

1. The ratio of the number of boys to girls is \(3:2\).
2. 20% of the boys are scholarship holders, and 25% of the girls are scholarship holders.

Now, let's find the number of students with scholarships:

Number of boys with scholarships = \(0.20 \times 3x = 0.6x\)

Number of girls with scholarships = \(0.25 \times 2x = 0.5x\)

The total number of students with scholarships is \(0.6x + 0.5x = 1.1x\).

Now, the total number of students in the school is \(3x + 2x = 5x\).

The percentage of students with scholarships is \(\dfrac{1.1x}{5x} \times 100\% = 22\%\).

Therefore, the percentage of students who do not get a scholarship is \(100\% - 22\% = 78\%\).

Hence, the correct answer is \( \boxed{78\%} \).

Let the original price of sugar be \(P\) and the original consumption be \(C\). The original expenditure is given by the product \(P \times C\).

After the price is increased by 20%, the new price is \(1.20P\). To keep the expenditure the same, the new consumption (\(C'\)) is given by the equation:

\[ P \times C = 1.20P \times C' \]

Now, let's find the ratio between the reduction in consumption and the original consumption:

\[ \text{Reduction in consumption} = C - C' \]

Substitute the expression for \(C'\) from the equation above:

\[ \text{Reduction in consumption} = C - \left(\dfrac{P \times C}{1.20P}\right) \]

\[ \text{Reduction in consumption} = C - \left(\dfrac{C}{1.20}\right) \]

\[ \text{Reduction in consumption} = C\left(1 - \dfrac{1}{1.20}\right) \]

\[ \text{Reduction in consumption} = C \times \dfrac{0.20}{1.20} \]

\[ \text{Reduction in consumption} = C \times \dfrac{1}{6} \]

Now, find the ratio:

\[ \text{Ratio} = \dfrac{\text{Reduction in consumption}}{\text{Original consumption}} \]

\[ \text{Ratio} = \dfrac{C \times \dfrac{1}{6}}{C} \]

\[ \text{Ratio} = \dfrac{1}{6} \]

Therefore, the correct ratio is \( \boxed{1 : 6} \).

Let's denote the two numbers as \(a\) and \(b\), where \(a\) is the larger number and \(b\) is the smaller number. The given information can be expressed as:

\[ a - b = 0.45(a + b) \]

Now, let's solve this equation to find the relationship between \(a\) and \(b\):

\[ a - b = 0.45a + 0.45b \]

Combine like terms:

\[ 0.55a = 1.45b \]

Now, find the ratio of the larger number to the smaller number:

\[ \dfrac{a}{b} = \dfrac{1.45b}{0.55a} \]

Multiply both sides by \( \dfrac{100}{5} \) to simplify the ratio:

\[ \dfrac{a}{b} = \dfrac{29}{11} \]

Therefore, the correct ratio of the larger number to the smaller number is \( \boxed{29:11} \).

Let's denote the income of Mahesh as \(M\) and the income of Mohan as \(M + 150\% \times M\). The phrase "150% higher than" means the income of Mohan is \(100\% + 150\% = 250\%\) of Mahesh's income.

So, the income of Mohan is \(250\% \times M\) or \(2.5M\).

Now, let's find the percentage by which the income of Mahesh is less than that of Mohan:

\[ \text{Percentage less} = \dfrac{\text{Difference in income}}{\text{Mohan's income}} \times 100\% \]

\[ \text{Percentage less} = \dfrac{(2.5M - M)}{2.5M} \times 100\% \]

Simplify the expression:

\[ \text{Percentage less} = \dfrac{1.5M}{2.5M} \times 100\% \]

\[ \text{Percentage less} = \dfrac{3}{5} \times 100\% \]

\[ \text{Percentage less} = 60\% \]

Therefore, the income of Mahesh is less than that of Mohan by \( \boxed{60\%} \).

Let's denote Christy's income as \(I\).

1. He donated 10% of his income to an orphanage, which means he is left with \(90\%\) of his income after the donation.

   After donation, he has \(0.90I\) left.

2. He deposited 20% of the remainder in his bank, which means he kept \(80\%\) of the remainder.

   The amount left after the deposit is \(0.80 \times 0.90I = 0.72I\).

According to the given information, he now has Rs. 7200 left, so:

\[ 0.72I = 7200 \]

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

\[ I = \dfrac{7200}{0.72} \]

\[ I = 10000 \]

Therefore, Christy's income is Rs. 10000. Hence, the correct answer is \( \boxed{\text{Rs. 10000}} \).

Let's denote the number of boys as \(3x\) and the number of girls as \(2x\), where \(x\) is a positive constant.

According to the given information:

1. The ratio of the number of boys to girls is \(3:2\).
2. 20% of the boys are adults, and 25% of the girls are adults.

Now, let's find the number of adult students:

Number of adult boys = \(0.20 \times 3x = 0.6x\)

Number of adult girls = \(0.25 \times 2x = 0.5x\)

The total number of adult students is \(0.6x + 0.5x = 1.1x\).

Now, the total number of students in the college is \(3x + 2x = 5x\).

The percentage of students who are not adults is:

\[ \dfrac{5x - 1.1x}{5x} \times 100\% = \dfrac{3.9x}{5x} \times 100\% = 78\% \]

Therefore, the correct answer is \( \boxed{78\%} \).

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

\[ 0.10x = 3 \times 0.15y \]

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

\[ \dfrac{x}{y} = \dfrac{3 \times 0.15}{0.10} \]

Simplify the ratio:

\[ \dfrac{x}{y} = \dfrac{0.45}{0.10} \]

\[ \dfrac{x}{y} = \dfrac{9}{2} \]

Therefore, the correct ratio of \(x\) to \(y\) is \( \boxed{9:2} \).