Varun and Raju, working together, can complete a job in 40 hours whereas Varun alone can complete the same job in 50 hours. The number of hours required by Raju to complete the job alone is:

Correct — C. Rates add, times do not.

together, rate = 1⁄40 of the job per hour
Varun alone, rate = 1⁄50 of the job per hour

Raju's rate = 1⁄40 − 1⁄50
= 5⁄200 − 4⁄200 = 1⁄200 of the job per hour

Raju's time is the reciprocal of his rate:
200 hours → option (c)

Check in units: take the job as 200 units. The pair does 5 units an hour, Varun 4, so Raju does 1.

Why the others are wrong

• (a)
  170 — At 170 hours Raju's rate is 1⁄170, and 1⁄50 + 1⁄170 = 22⁄850, so the pair would finish in about 38.6 hours, not 40.
• (b)
  150 — At 150 hours, 1⁄50 + 1⁄150 = 4⁄150, a joint time of 37.5 hours. That is too fast for the 40 hours given.
• (d)
  190 — At 190 hours, 1⁄50 + 1⁄190 = 24⁄950, a joint time of about 39.6 hours. Close, but the stem fixes the pair at exactly 40.

Work questions are rate questions. If one person needs a hours and another b hours, their rates 1⁄a and 1⁄b add to the joint rate 1⁄t.

Here t = 40 and a = 50 are given, so the unknown rate is a subtraction: 1⁄40 − 1⁄50 = 1⁄200.

The LCM-units method avoids fractions altogether. Call the job 200 units, the LCM of 40 and 50. The pair completes 5 units an hour and Varun 4, leaving Raju 1 unit an hour, so he needs 200 hours.

A sanity check settles it: adding Raju saves only 10 hours off Varun's 50, so Raju has to be very slow indeed.

Key facts

• 1⁄(time together) = 1⁄(first alone) + 1⁄(second alone).
• 1⁄40 − 1⁄50 = 5⁄200 − 4⁄200 = 1⁄200, so Raju alone needs 200 hours.
• Taking the job as 200 units, the pair does 5 units an hour and Varun 4, leaving Raju 1.

Common traps

• Subtracting the times, 50 − 40 = 10, instead of the rates.
• Expecting Raju's time to be near Varun's when the pair saves only 10 hours.
• Reading 40 hours as Varun's time and 50 as the pair's.

No directly related past PYQ was found.