Question

Albela, Bob and Chulbul have to read a document of seventy eight pages and make a presentation next day. They realize that the article is difficult to understand and they would require teamwork to finish the assignment. Albela can read a page in 2 minutes, Bob in 3 minutes, and Chulbul in 4 minutes. If they divide the article into 3 parts so that all three of them spend the equal amount of time on the article, the number of pages that Bob should read is

• A. 24
• B. 25
• C. 26
• D. 27
• E. 28

Hint:

When several workers (or readers) must spend equal time and have different rates, express each person's work as \(\textrate\times\texttime\) and use the common time to relate quantities. Converting to a common parameter (here \(T\)) often simplifies the algebra.

Answer 1

The Correct Option is A

Step 1: Let the number of pages Albela, Bob and Chulbul read be \(x,\;y,\;z\) respectively. Then \[ x+y+z=78. \] Step 2: Their reading times (in minutes) are \(2x,\;3y,\;4z\). These are equal, so set \[ 2x=3y=4z=T\text{(common time)}. \] Thus \(x=\dfrac{T}{2},\;y=\dfrac{T}{3},\;z=\dfrac{T}{4}.\) Step 3: Sum the pages: \[ \frac{T}{2}+\frac{T}{3}+\frac{T}{4}=78 ⇒ T\Big(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\Big)=78. \] Compute the bracket: \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{6+4+3}{12}=\dfrac{13}{12}.\) Step 4: Solve for \(T\): \[ T\cdot\frac{13}{12}=78⇒ T=78\cdot\frac{12}{13}=6\cdot12=72. \] Hence Bob's pages \(y=\dfrac{T}{3}=\dfrac{72}{3}=24.\)