Initial Bicycle Price Ratio

Let the initial prices of the two bicycles be $P_1$ and $P_2$.

The ratio between their prices is given as 15:26.

We can represent the initial prices as $P_1 = 15x$ and $P_2 = 26x$, where $x$ is a common multiplier.

Setting Up Prices Algebraically

The price of the first bicycle ($P_1$) is increased by 10%.

New $P_1 = 15x + (10\% \text{ of } 15x) = 15x + 0.10 \times 15x = 15x + 1.5x = 16.5x$.

The price of the second bicycle ($P_2$) is increased by ₹550.

New $P_2 = 26x + 550$.

Ratio Equation Formation

After the price changes, the new ratio between the prices becomes 13:22.

Therefore, we can write the equation:

$ \frac{\text{New } P_1}{\text{New } P_2} = \frac{13}{22} $

$ \frac{16.5x}{26x + 550} = \frac{13}{22} $

Solving for the Price Multiplier

Cross-multiply the equation:

$ 22 \times (16.5x) = 13 \times (26x + 550) $

$ 363x = 338x + (13 \times 550) $

$ 363x = 338x + 7150 $

Subtract $338x$ from both sides:

$ 363x - 338x = 7150 $

$ 25x = 7150 $

Solve for $x$:

$ x = \frac{7150}{25} $

$ x = 286 $

Finding Initial Bicycle Prices

Now, substitute the value of $x$ back into the expressions for the initial prices:

• Initial Price of the first bicycle ($P_1$) = $15x = 15 \times 286 = 4290$.
• Initial Price of the second bicycle ($P_2$) = $26x = 26 \times 286 = 7436$.

The initial prices of the two bicycles were ₹4290 and ₹7436.