CMAT questions on Arithmetic - Time & Work

Asked: the days 9 women alone need.

6 men finish in 3 days, so the work = 18 man-days. One man does 1/18 of the work per day.

3 men + 18 women finish in 2 days: \( 2\left(\frac{3}{18} + 18w\right) = 1 \).

\( \frac{1}{6} + 18w = \frac{1}{2} \), so \( 18w = \frac{1}{3} \) and \( w = \frac{1}{54} \).

9 women do 9/54 = 1/6 of the work per day, so they need 6 days.

Alternative: 1 man = 3 women (1/18 vs 1/54). Then 3 men + 18 women = 27 women, who take 2 days. So 9 women take 2 × 27/9 = 6 days.

Answer: 6

Statement I

Ram's daily work = (Ram + Shyam) − Shyam = \( \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24} \).

So Ram alone takes 24 days. True.

Statement II

Earnings are proportional to total person-hours.

• 6 × 8 = 48 person-hours a day earn ₹8,400.
• 9 × 6 = 54 person-hours a day earn \( 8400 \times \frac{54}{48} = 9450 \).

True.

Answer: Both Statement I and Statement II are true (option 1)

Asked: when the monkey first reaches the flag at 110 ft.

Key idea: in each hour the monkey climbs 20 ft and then slips 5 ft, a net gain of 15 ft. In the final hour it reaches the flag before any slip, so that hour counts the full 20 ft.

Height needed before the final climb = 110 − 20 = 90 ft.

Hours to gain 90 ft at 15 ft per hour = 6 hours (8 AM to 2 PM).

At 2 PM it is at 90 ft. During the 7th hour (2 PM to 3 PM) it climbs 20 ft and reaches 110 ft.

Hour-end heights: 9 AM 15, 10 AM 30, 11 AM 45, 12 PM 60, 1 PM 75, 2 PM 90, then 110 by 3 PM.

Answer: 3 PM

Asked: the time the slower pipe takes alone.

Let the slower pipe fill r of the tank per minute. The faster pipe fills 3r per minute.

Together they fill 4r per minute = 1/36 of the tank, so r = 1/144.

The slower pipe alone takes 144 minutes. (The faster one takes 48 minutes. Check: 1/48 + 1/144 = 4/144 = 1/36 ✓)

Answer: 144 minutes