Question

The ratio of the prices of two bicycles A and B was \(2:3\). Two years later if the price of A is increased by \(15\%\) and that of cycle B by Rs.475, the ratio of the prices becomes \(3:5\). Then the original price of bicycle A (in Rs.) is

• A. \(1080\)
• B. \(1120\)
• C. \(1140\)
• D. \(1180\)

Hint:

In ratio problems, always assume quantities as multiples of a common variable and build equations from changes.

Answer 1

The Correct Option is C

Concept: Use ratio to assume original prices and form equation after increase.

Step 1: Assume original prices.
Given ratio: \[ A:B=2:3 \] Let: \[ A=2x,\quad B=3x \]

Step 2: Apply the increase.
A increases by \(15\%\): \[ A'=2x+\frac{15}{100}(2x) \] \[ =2x+\frac{3x}{10} \] \[ =\frac{23x}{10} \] B increases by Rs.475: \[ B'=3x+475 \]

Step 3: Use the new ratio.
New ratio: \[ A':B'=3:5 \] So: \[ \frac{\frac{23x}{10}}{3x+475}=\frac35 \] Cross multiply: \[ 5\cdot \frac{23x}{10}=3(3x+475) \] \[ \frac{23x}{2}=9x+1425 \] Multiply by 2: \[ 23x=18x+2850 \] \[ 5x=2850 \] \[ x=570 \]

Step 4: Find original price of A.
\[ A=2x \] \[ =2(570) \] \[ =1140 \] Thus, the original price of bicycle A is: \[ \boxed{1140} \]