Two numbers are in the ratio 3 : 5. If 4 is subtracted from each of these two numbers, the new numbers then are in the ratio 4 : 7. Find the smaller of the two original numbers.
36
Let the two original numbers be represented in terms of a variable based on their given ratio.
The problem states that the two numbers are in the ratio 3 : 5. This means we can represent the numbers as \(3x\) and \(5x\), where \(x\) is a common multiplier.
Since \(3x < 5x\) for any positive \(x\), the smaller of the two original numbers is \(3x\).
According to the question, 4 is subtracted from each of these two numbers. The new numbers become:
The problem also states that these new numbers are then in the ratio 4 : 7. We can write this relationship as an equation:
\[ \frac{3x - 4}{5x - 4} = \frac{4}{7} \]
To solve for \(x\), we can cross-multiply:
\[ 7(3x - 4) = 4(5x - 4) \] Distribute the numbers on both sides of the equation:
\[ 21x - 28 = 20x - 16 \] Now, collect the \(x\) terms on one side and the constant terms on the other side:
\[ 21x - 20x = -16 + 28 \] Simplify both sides:
\[ x = 12 \]
Now that we have the value of \(x\), we can find the original numbers:
The two original numbers are 36 and 60. The smaller of these two numbers is 36.
Let's check if these numbers satisfy the conditions given in the problem:
Both conditions are met, so our calculated numbers are correct.
The smaller of the two original numbers is 36.
| Step | Description | Calculation/Representation |
|---|---|---|
| 1 | Represent original numbers | \(3x, 5x\) |
| 2 | Represent numbers after subtraction | \(3x-4, 5x-4\) |
| 3 | Set up equation from new ratio | \(\frac{3x-4}{5x-4} = \frac{4}{7}\) |
| 4 | Solve for \(x\) | \(x=12\) |
| 5 | Calculate original numbers | \(3 \times 12 = 36, 5 \times 12 = 60\) |
| 6 | Identify smaller number | 36 |
A ratio is a comparison of two or more quantities. It shows how much of one quantity there is compared to another. Ratios can be written in several ways, such as \(3:5\), 3 to 5, or \(\frac{3}{5}\).
When solving problems involving ratios of unknown numbers, it's common practice to represent the numbers using a common multiplier, like \(x\). If the ratio is \(a:b\), the numbers can be represented as \(ax\) and \(bx\). This allows you to set up algebraic equations when the relationship between the numbers changes.
Changing the numbers (like adding or subtracting) will change the actual values, but not necessarily the ratio, unless the change is proportional to the original numbers. In this problem, subtracting a constant value (4) from each number changes their relationship, resulting in a new ratio.
Solving ratio problems often involves setting up a proportion (an equation stating that two ratios are equal) and then solving for the unknown variable using cross-multiplication.
The train fare, bus fare and air fare between 2 places are in the ratio 5 : 8 : 12, the number of passenger travelled by them is in the ratio 3 : 4 : 5 and the total fare collected on a particular day for these modes of transportation for a single trip is Rs. 1,07,000. Find the fare collected from the air passengers.
The compounded ratio of (1 ∶ 3), (6 ∶ 5) and (7 ∶10) is:
Rs. 750 are divided among A, B and C in such a manner that A : B = 5 : 2 and B : C = 7 : 13, What is A’s share?
Rs. 750 are divided among A, B and C in such a manner that A : B = 5 : 2 and B : C = 7 : 13, What is B’s share?
A person has some coins of Rs. 10, Rs. 5, and Rs. 2 denominations. The ratio of the products of the numbers of Rs. 10 and Rs. 5 coins, the numbers of Rs. 5 and Rs. 2 coins, and the numbers of Rs. 2 and Rs. 10 coins is 3 ∶ 4 ∶ 2 respectively. What could be the minimum amount of money this person has?