Time and Work: Aptitude Questions with Solutions

Ans: B

To find the true time when the watch indicates 4:15 p.m., we first must determine how much the watch gains by this time, starting from 7:00 a.m.

Given:

– The watch gains 5 seconds every 3 minutes.

Let’s calculate the duration from 7:00 a.m. to 4:15 p.m.:

– Time from 7:00 a.m. to 12:00 p.m. is 5 hours.

– Time from 12:00 p.m. to 4:15 p.m. is 4 hours and 15 minutes.

– Total time from 7:00 a.m. to 4:15 p.m. is 5 hours + 4 hours 15 minutes = 9 hours 15 minutes.

Convert 9 hours 15 minutes to minutes:

$ 9 \text{ hours} \times 60 \text{ minutes/hour} = 540 \text{ minutes} $

$ 540 \text{ minutes} + 15 \text{ minutes} = 555 \text{ minutes} $

Now calculate how much the watch gains in 555 minutes. Since it gains 5 seconds every 3 minutes, let’s use a proportion:

$ \text{Gain in seconds} = \frac{555 \text{ minutes}}{3 \text{ minutes}} \times 5 \text{ seconds} $

$ = 185 \times 5 \text{ seconds} = 925 \text{ seconds} $

Convert 925 seconds to minutes:

$ \frac{925 \text{ seconds}}{60 \text{ seconds/minute}} \approx 15.4167 \text{ minutes} $

The watch gains an additional 15.4167 minutes over the period from 7:00 a.m. to 4:15 p.m. Therefore, the true time, when the watch shows 4:15 p.m., can be found by subtracting this gain from 4:15 p.m.:

First, convert 4:15 p.m. into a decimal:

$ 4:15 = 4 + \frac{15}{60} = 4.25 \text{ hours} $

Subtract the gain:

$ 4.25 \text{ hours} – \frac{15.4167}{60} \text{ hours} = 4.25 – 0.25695 \approx 3.99305 \text{ hours} $

$ \approx 3 \text{ hours }+ 0.99305 \times 60 \text{ minutes} $

$ = 3 \text{ hours } + 59.583 \text{ minutes} \approx 4:00 p.m. $

Thus, although the inaccurate watch indicates 4:15 p.m., the true time is approximately 4:00 p.m.

Ans: A) 24 days

Total work = $( \text{LCM}(40, 30) = 120 )$ units. A and B’s rate = $( \frac{120}{40} + \frac{120}{30} = 3 + 4 = 7 )$ units/day. In 10 days, A and B do $( 10 \times 7 = 70 )$ units. Remaining work = $( 120 – 70 = 50 )$ units in 5 days. C’s rate = $( \frac{50}{5} = 10 )$ units/day, so C alone takes $( \frac{120}{10} = 24 )$ days.

Have a look at common Time and Work Scenarios.

Ans: C) 30

Total work = $( 10 \times 15 = 150 )$ woman-days. Required number of women = $( \frac{150}{5} = 30 )$.

Ans: A) 30 days

Let total work be ( W ). A’s rate = $( \frac{W}{20} )$, and (A+B)’s rate = $( \frac{W}{12} )$. B’s rate = $( \frac{W}{12} – \frac{W}{20} = \frac{5W – 3W}{60} = \frac{2W}{60} = \frac{W}{30} )$. Thus, B alone can complete it in 30 days.

Ans: C) 80 days

Let 1 man’s work = ( x ) and 1 woman’s work = ( y ). $( 8x + 6y = \frac{1}{10} )$ and $( 10x + 8y = \frac{1}{8} )$. Solving, $( x + y = \frac{1}{80} )$, so they take 80 days together.

Ans: B) 25 days

A’s 1-day work = $( \frac{75\%}{15} = 5\% )$. Remaining work = 25%, done by A and B in 5 days, so (A + B)’s daily work = $( \frac{25\%}{5} = 5\% )$. B’s work rate = $( 5\% – 5\% = 3\% )$, so B alone needs $( \frac{100}{5} = 25 )$ days.

Ans: B) 10 hours

Total work = ( 1 ) tank (100%). A + B’s combined rate = $( \frac{1}{20} + \frac{1}{30} = \frac{1}{12} )$. In 10 hours, $( \frac{10}{12} = \frac{5}{6} )$ of the tank is filled. Remaining work = $( \frac{1}{6} )$, with net rate $( \frac{1}{12} – \frac{1}{15} = \frac{1}{60} )$. Time = $( \frac{1}{6} \div \frac{1}{60} = 10 )$ hours.

Ans: B) 8 days

One day of A = $( \frac{1}{15} )$, B = $( \frac{1}{20} )$, C = $( \frac{1}{30} )$. Combined work of (A + B + C) in 2 days = $( \frac{1}{15} + \frac{1}{20} + \frac{1}{30} = \frac{3}{20} )$. In 8 days (4 cycles), work done = $( \frac{12}{15} = 1 )$.

Common Time and Work Problems Guide

Ans: C) 12 days

Let total work be ( W = 48 ) units (LCM of 16 and 12). Work rate of A = 3 units/day, and B = 4 units/day. If they work together for ( x ) days, then A works alone for the last 4 days. Work done by A in last 4 days = $( 3 \times 4 = 12 )$ units, so work done together = 48 – 12 = 36. $( 7x = 36 \Rightarrow x = 8 )$. Total time = 8 + 4 = 12 days.

Ans: C) 12 days

A’s work rate = $( \frac{1}{18} )$, and B’s work rate = $( \frac{1}{36} )$. Combined rate = $( \frac{1}{18} + \frac{1}{36} = \frac{1}{12} )$. Thus, they will finish in 12 days.

Ans: D

Rate of work = 1/ 10
Total work done = (1/10) * 4 * 100 %

Ans: D

Rate of work = 1/ 10
Total work done = (1/10) * 5 * 100 %

Ans: C

Rate of work = 1/ (25*36). So, total time it will take to do the work = 25*36/6 = 150

Ans: A

Rate of work together will be = 1/ 14 + 1/13). So, total time it will take to do the work will be inverse of their combined rate of work = 14*13/27 = 6.7

Ans: A

Rates of three robots respectively are $\frac{1}{6}$,

$\frac{1}{7}$, and $\frac{1}{8}$.

To maximize the greatest part of the work that gets done, we need the two fastest robots, i.e., those which take less time to do more work, and we need their combined rate.

$\frac{1}{6}$ + $\frac{1}{7}$ = $\frac{13}{42}$.

Greatest part of work that can be done by two fastest robots together in 1 hour?

$r*t = W$ —> $\frac{13}{42}$ * 1 = $\frac{13}{42}$.

Ans: D

To solve this problem, we need to find where Alex and Ben meet and the time it takes for them to meet after Alex starts driving.

Let’s denote:

– the time after Alex starts driving when they meet as $ t $ hours,

– the distance Alex travels as $ 35t $ miles (since she travels at 35 mph),

– the time Ben drives when they meet as $ t – 1 $ hours (since he starts one hour later),

– and the distance Ben travels as $ 45(t – 1) $ miles.

Since they are traveling towards each other, the sum of the distances they travel must equal the total distance between towns A and B, which is 355 miles. Therefore, the equation connecting their distances is:

$ 35t + 45(t – 1) = 355 $

Expanding and combining like terms:

$ 35t + 45t – 45 = 355 $

$ 80t – 45 = 355 $

$ 80t = 400 $

$ t = \frac{400}{80} = 5 \text{ hours} $

Thus, Alex has been driving for 5 hours when they meet. To find the distance she has traveled, we simply multiply her driving time by her speed:

$ \text{Distance Alex has traveled} = 35 \times 5 = 175 \text{ miles} $

Therefore, Alex has traveled 175 miles when she and Ben pass each other.

Ans: C

There are 60 seconds in a minute. So, in x minutes, there are $x \times 60 = 60x$ seconds.

At the rate of c calories per s seconds, the number of calories burned in x minutes is $c \times 60x = 60cx$ calories.

Ans: D

We will start by assuming the last time:

(E) 4:42 pm

If she starts typing 18 mins before the deadline, she has 18 mins to type 1950 words. Her speed at this time will be 20 + (the mins passed after 3)/2 words i.e. her speed at 4:42 will be

20 + (120 – 18)/2 = 71 words/min

She can type a total of 18 * 71 = 1278 words.

Not enough since she has to type 1950 words. She must start earlier.

(D) 4:30 pm

If she starts 30 mins before deadline, she has 30 mins to type 1950 words. Her speed at this time will be 20 + (120 – 30)/2 = 65 words/min

She can type a total of 30*65 = 1950 words

This is enough.

So she can start at 4:30

Let’s look at the algebraic solution too though backsolving is the best way way here.

Let’s say she starts her paper T mins before 5 pm.

Speed = 20 + (120 – T)/2

Time to type = T min

Words typed = 80 – T/2

We need this expression to be not less than 1950

(80 – T/2)*T = 1950

160T – T^2 = 2*1950

T^2 – 160T + 2*1950 = 0

(T – 30)(T – 130) = 0

T can be 30 or 130. The only possible option when two hrs are left is 30 mins.