The average of the two numbers is 48. If 3 is added to a smaller number, then the ratio between them is 11 ∶ 20. Find the larger number. (Round up to two decimal places).
63.87
Solving the Average and Ratio Math Problem
The problem asks us to find the larger of two numbers given their average and a condition involving their ratio after a modification.
Understanding the Problem Conditions
Let the two numbers be \(x\) and \(y\). We are given two pieces of information:
- The average of the two numbers is 48.
- If 3 is added to the smaller number, the ratio between the modified smaller number and the larger number becomes 11:20.
Let's assume \(x\) is the smaller number and \(y\) is the larger number (\(x < y\)).
Setting up the Equations
From the first condition, the average of \(x\) and \(y\) is 48. The formula for the average of two numbers is the sum of the numbers divided by 2.
$$ \frac{x + y}{2} = 48 $$
Multiplying both sides by 2 gives us the first equation:
$$ x + y = 96 \quad \cdots(1) $$
From the second condition, if 3 is added to the smaller number (\(x\)), the new number is \(x+3\). The ratio of this new number to the larger number (\(y\)) is 11:20.
$$ \frac{x + 3}{y} = \frac{11}{20} $$
We can cross-multiply to get the second equation:
$$ 20(x + 3) = 11y $$
$$ 20x + 60 = 11y \quad \cdots(2) $$
Solving the System of Equations
Now we have a system of two linear equations with two variables:
1. \(x + y = 96\)
2. \(20x + 60 = 11y\)
We need to find the value of \(y\), the larger number. We can use the substitution method. From equation (1), we can express \(x\) in terms of \(y\):
$$ x = 96 - y $$
Substitute this expression for \(x\) into equation (2):
$$ 20(96 - y) + 60 = 11y $$
Distribute the 20:
$$ 1920 - 20y + 60 = 11y $$
Combine the constant terms on the left side:
$$ 1980 - 20y = 11y $$
Add \(20y\) to both sides to isolate the \(y\) terms:
$$ 1980 = 11y + 20y $$
$$ 1980 = 31y $$
Now, divide by 31 to find the value of \(y\):
$$ y = \frac{1980}{31} $$
Let's perform the division:
$$ \frac{1980}{31} \approx 63.870967... $$
Rounding the Result
The question asks us to round the larger number to two decimal places. Looking at the third decimal place (0), since it is less than 5, we round down (keep the second decimal place as it is).
$$ y \approx 63.87 $$
The larger number is approximately 63.87.
| Step | Description | Equation/Calculation |
|---|---|---|
| 1 | Define variables | Let numbers be \(x\) (smaller) and \(y\) (larger) |
| 2 | Set up average equation | \(\frac{x+y}{2} = 48 \implies x+y=96\) |
| 3 | Set up ratio equation | \(\frac{x+3}{y} = \frac{11}{20} \implies 20(x+3)=11y\) |
| 4 | Express x from Eq. 1 | \(x = 96-y\) |
| 5 | Substitute x into Eq. 2 | \(20(96-y+3) = 11y \implies 20(99-y)=11y\) |
| 6 | Solve for y | \(1980 - 20y = 11y \implies 1980 = 31y \implies y = \frac{1980}{31}\) |
| 7 | Calculate and round y | \(y \approx 63.8709... \implies 63.87\) |
Revision Table: Average and Ratio Concepts
| Concept | Definition | Formula/Example |
|---|---|---|
| Average (Arithmetic Mean) | The sum of a set of numbers divided by the count of the numbers. | Average of \(a\) and \(b\) is \(\frac{a+b}{2}\). |
| Ratio | A comparison of two quantities. It can be written as \(a:b\) or \(\frac{a}{b}\). | Ratio of 11 to 20 is 11:20 or \(\frac{11}{20}\). |
| System of Linear Equations | A set of two or more linear equations involving the same variables. | \(x+y=96\) and \(20x+60=11y\) is a system. |
| Substitution Method | A method to solve a system of equations by solving one equation for one variable and substituting that expression into the other equation. | Used here by substituting \(x=96-y\) into the second equation. |
| Rounding Decimals | Reducing the number of decimal places. Look at the digit to the right of the desired place. If ≥ 5, round up; if < 5, round down. | 63.8709 rounded to two decimal places is 63.87. |
Additional Information: Applying Average and Ratio Concepts
Problems involving averages and ratios are common in quantitative aptitude. Understanding how to translate word problems into algebraic equations is key. In this problem, the average gave us a simple linear equation (sum of numbers). The ratio condition, after a slight modification to one number, provided a second linear equation. Solving the system allowed us to find the values of the unknown numbers.
When working with ratios, it's often helpful to express the ratio as a fraction. For example, a ratio of \(a:b\) means \(\frac{a}{b}\). If the ratio is given as \(m:n\), it implies that the first quantity is proportional to \(m\) and the second is proportional to \(n\), or that the fraction \(\frac{\text{first quantity}}{\text{second quantity}}\) equals \(\frac{m}{n}\).
Solving systems of equations like the one in this problem can be done using substitution (as shown) or elimination. Both methods aim to reduce the system to a single equation with one variable, which can then be solved.
Always double-check the question to ensure you are solving for the correct variable (smaller number, larger number, sum, difference, etc.) and paying attention to any specific formatting or rounding requirements.
Important Questions from Simple Ratios
The ratio of two numbers A and B is 5 : 8. If 5 is added to each of A and B, then the ratio of A and B becomes 2 : 3. The sum of A and B is:
The ratio of two numbers A and B is 5: 8. If 5 is added to each of A and B, then the ratio becomes 2 : 3. The difference between A and B is:
A sum of Rs. 6342 is divided amongst A, B, C and D in the ratio 3 : 4 : 8 : 6. What is the difference between the shares of B and D?
The ratio of monthly incomes of A and B is 4 ∶ 5 and that of their monthly expenditures is 3 ∶ 8. If the income of A is equal to the expenditure of B, then what is the ratio of savings of A and B?
The ratio of the monthly incomes of A and B is 11 : 13 and the ratio of their expenditures is 9 : 11. If both of them manage to save Rs. 4,000 per month, then find the difference in their incomes (in Rs.)
