Question

Three typists A, B and C working together 8 hours per day can type 900 pages in 20 days. In a day, B types as many pages more than A as C types as many pages more than B. The number of pages typed by A in 4 hours is equal to the number of pages typed by C in 1 hour. How many pages C types in each hour?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Best way to solve is to check options.
Checking for option (c).
3 pages per hour $\to$ C typed $8\times 3=24pagesperday$
C is 4 times faster than A.
So, pages typed by A in a day = $\frac{24}{4}=6pagesperday$
Since, B types as many pages more than A as C types as many pages more than B, this means B's efficiency is arithmetic mean of A and C.
So, number of pages typed by B in a day = $\frac{24+6}{2}=15pagesperday$
Total pages typed by A, B, and C in a day = 15 + 24 + 6 = 45 pages per day.
Number of days needed to type 900 pages = $\frac{900}{45}=20days$, which is true.
Hence, the correct answer is option (c).

Alternative Approach:
Number of pages typed by A, B and C together per day = 45
Now let the number of pages typed by B be x
then the number of pages typed by A = x - d
and the number of pages typed by C = x + d
$\Rightarrow$ (x - d) + (x) + (x + d) = 45 $\Rightarrow x=15$ pages per day.
Again let C types k pages per day then A types $\frac{k}{4}$ pages per day.
Therefore, the ratio of typing of pages per day of A and C = 1 : 4
$\therefore$ Number of pages typed by C in one day
$=\frac{4}{5}\times 30=24$ pages (30 = 45 - 15)
$\therefore NumberofpagestypedbyCperhour=\frac{24}{8}=3pages/hour$