A and B have to write 810 and 900 pages respectively in the same time period. But A completes his work 3 days ahead of time and B completes 6 days ahead of time. How many pages did A write per hour if B wrote 21 pages more in each hour?

Options

A. a. 45
B. b. 72
C. c. 54
D. d. 100

Answer

To solve the problem, we need to find out how many pages A wrote per hour, given that B wrote 21 pages more per hour than A. Let's break down the solution step by step.

### Step 1: Define Variables
Let:
- \( x \) = number of pages A writes per hour
- \( y \) = number of pages B writes per hour

From the problem, we know:
- \( y = x + 21 \) (since B writes 21 pages more than A)

### Step 2: Determine Total Time Taken by A and B
Let \( t \) be the total time (in hours) that A and B were supposed to take to complete their work.

- A has to write 810 pages, so the time taken by A can be expressed as:
\[ t_A = \frac{810}{x} \]

- B has to write 900 pages, so the time taken by B can be expressed as:
\[ t_B = \frac{900}{y} \]

### Step 3: Adjust for Early Completion
According to the problem:
- A completes his work 3 days ahead of time, so:
\[ t_A = t - 3 \]

- B completes his work 6 days ahead of time, so:
\[ t_B = t - 6 \]

### Step 4: Set Up Equations
From the expressions for \( t_A \) and \( t_B \):
1. For A:
\[ \frac{810}{x} = t - 3 \]

2. For B:
\[ \frac{900}{y} = t - 6 \]

### Step 5: Substitute \( y \) in B's Equation
Substituting \( y = x + 21 \) into B's equation:
\[ \frac{900}{x + 21} = t - 6 \]

### Step 6: Express \( t \) in Terms of \( x \)
From A's equation:
\[ t = \frac{810}{x} + 3 \]

Now substitute this expression for \( t \) into B's equation:
\[ \frac{900}{x + 21} = \left(\frac{810}{x} + 3\right) - 6 \]
This simplifies to:
\[ \frac{900}{x + 21} = \frac{810}{x} - 3 \]

### Step 7: Clear the Fractions
To eliminate the fractions, multiply through by \( x(x + 21) \):
\[ 900x = 810(x + 21) - 3x(x + 21) \]

### Step 8: Expand and Rearrange
Expanding the right-hand side:
\[ 900x = 810x + 17010 - 3x^2 - 63x \]
Combine like terms:
\[ 900x = 747x + 17010 - 3x^2 \]
Rearranging gives:
\[ 3x^2 + 153x - 17010 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 153, c = -17010 \).
Calculating the discriminant:
\[ b^2 - 4ac = 153^2 - 4 \cdot 3 \cdot (-17010) = 23409 + 204120 = 227529 \]
Now, calculating \( x \):
\[ x = \frac{-153 \pm \sqrt{227529}}{6} \]
Calculating \( \sqrt{227529} \approx 477 \):
\[ x = \frac{-153 + 477}{6} \quad \text{(taking the positive root)} \]
\[ x = \frac{324}{6} = 54 \]

### Step 10: Conclusion
Thus, A writes **54 pages per hour**.