Time and Work is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Time and Work questions in the<\/span> CAT Previous year’s papers<\/a>. <\/strong><\/span>If you want to learn the basics, you can watch these videos on Time and Work basics<\/strong><\/span>. This article will look into some important Time and Work Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Time and Work Most Important Questions PDF below, which is completely Free.<\/p>\n Download Time and Work Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?<\/p>\n a)\u00a0$\\frac{7}{3}$<\/p>\n b)\u00a0$\\frac{4}{3}$<\/p>\n c)\u00a0$\\frac{5}{3}$<\/p>\n d)\u00a0$\\frac{8}{3}$<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $12\/(R – S) = T$ Question 2:\u00a0<\/b>Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom\u2019s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?<\/p>\n a)\u00a040<\/p>\n b)\u00a050<\/p>\n c)\u00a060<\/p>\n d)\u00a080<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of steps on the escalator be x.<\/p>\n So, by the time Shyama covered 25 steps, the escalator moved ‘x-25’ steps.<\/p>\n Hence, the ratio of speeds of Shyama and escalator = 25:(x-25)<\/p>\n Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20)<\/p>\n But the ratio is 3:2<\/p>\n Ratio of speeds of Shyama and Vyom = 25(x-20)\/20*(x-25) = 3\/2<\/p>\n => 10(x-20) = 12(x-25)<\/p>\n => 2x = 100 => x = 50<\/p>\n Question 3:\u00a0<\/b>There\u2019s a lot of work in preparing a birthday dinner. Even after the turkey is in the oven, there\u2019s still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. a)\u00a020 min<\/p>\n b)\u00a030 min<\/p>\n c)\u00a040 min<\/p>\n d)\u00a050 min<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the time taken working together be t. Alternatively, Question 4:\u00a0<\/b>On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km\/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend to the patient at the hospital. a)\u00a04 min<\/p>\n b)\u00a02.5 min<\/p>\n c)\u00a01.5 min<\/p>\n d)\u00a0The patient died before reaching the hospital<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.<\/p>\n Hence, x + y + 2y + x = 20 => 2x+3y=20<\/p>\n Also x = 30kmph * 5\/60 = 2.5km<\/p>\n Hence, y = 5km<\/p>\n After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5<\/p>\n It takes 1 min to load and unload the patient.<\/p>\n Hence, total time = 5 + 32.5 + 1 = 38.5 mins<\/p>\n So, the doctor would get 1.5 min to attend to the patient.<\/p>\n Question 5:\u00a0<\/b>It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed? a)\u00a06.40 pm<\/p>\n b)\u00a07 pm<\/p>\n c)\u00a07.20 pm<\/p>\n d)\u00a08 pm<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the work done by each technician in one hour be 1 unit. Checkout: CAT Free Practice Questions and Videos<\/p>\n Question 6:\u00a0<\/b>Three small pumps and a large pump are filling a tank. Each of the three small pump works at 2\/3 the rate of the large pump. If all four pumps work at the same time, they should fill the tank in what fraction of the time that it would have taken the large pump alone? a)\u00a04\/7<\/p>\n b)\u00a01\/3<\/p>\n c)\u00a02\/3<\/p>\n d)\u00a03\/4<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the work done by the big pump in one hour be 3 units. Question 7:\u00a0<\/b>Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?<\/p>\n a)\u00a012 hours<\/p>\n b)\u00a010 hours<\/p>\n c)\u00a08 hours<\/p>\n d)\u00a06 hour<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n As machine B’s efficiency is twice as of A’s, Hence, it will complete its work in 30 hours. $\\frac{1}{x} = \\frac{1}{20} + \\frac{1}{30} + \\frac{1}{60}$<\/p>\n or x = 10 hours<\/p>\n Question 8:\u00a0<\/b>Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?<\/p>\n a)\u00a01 hour<\/p>\n b)\u00a050 minutes<\/p>\n c)\u00a01\/2 hour<\/p>\n d)\u00a055 minutes<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Since we know that Neera’s husband drives at a uniform speed to and from his residence. Question 9:\u00a0<\/b>A man travels from A to B at a speed x km\/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km\/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be<\/p>\n a)\u00a03 hr<\/p>\n b)\u00a06 hr<\/p>\n c)\u00a02 hr<\/p>\n d)\u00a04 hr<\/p>\n
\n$12\/(R + S) = T – 6$
\n$12\/(2R – S) = t$
\n$12\/(2R + S) = t – 1$
\n=> $12\/(R – S) – 12\/(R + S) = 6$ and $12\/(2R – S) – 12\/(2R + S) = 1$
\n=> $12R + 12S – 12R + 12S = 6R^2 – 6S^2$ and $24R + 12S – 24R + 12S = 4R^2 – S^2$
\n=> $24S = 6R^2 – 6S^2 and 24S = 4R^2 – S^2$
\n=> $6R^2 – 6S^2 = 4R^2 – S^2$
\n=> $2R^2 = 5S^2$
\n=> $24S = 10S^2 – S^2 = 9S^2$
\n=> $S = 24\/9 = 8\/3$<\/p>\n
\nThree friends \u2014 Asit, Arnold and Afzal \u2014 work together to get all of these chores done. The time it takes them to do the work together is 6 hr less than Asit would have taken working alone, 1 hr less than Arnold would have taken alone, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?<\/p>\n
\nTime taken by Arnold = t+1
\nTime taken by Asit = t+6
\nTime taken by Afzal = 2t
\nWork done by each person in one day =\u00a0$\\frac{1}{(t+1)}+\\frac{1}{(t+6)}+\\frac{1}{2t}$
\nTotal portion of workdone in one day $=\\frac{1}{t}$
\n$\\frac{1}{(t+1)}+\\frac{1}{(t+6)}+\\frac{1}{2t}=\\frac{1}{t}$
\n$\\frac{1}{(t+1)}+\\frac{1}{(t+6)}=\\frac{2-1}{2t}$
\n$2t+7=\\frac{(t+1)\\cdot(t+6)}{2t}$
\n$3t^2-7t+6=0\u00a0\\longrightarrow\\ t=\\frac{2}{3} $or $t=-3$
\nTherefore total time = $\\frac{2}{3}$hours = 40mins<\/p>\n
\n$\\frac{1}{(t+1)}+\\frac{1}{(t+6)}+\\frac{1}{2t}=\\frac{1}{t}$
\nFrom the options, if time $= 40$ min, that is, $t = \\frac{2}{3}$
\nLHS =\u00a0$\\frac{3}{5} + \\frac{3}{20} + \\frac{3}{4} = \\frac{(12+3+15)}{20} = \\frac{30}{20} = \\frac{3}{2}$
\nRHS = $\\frac{1}{t}=\\frac{3}{2}$
\nThe equation is satisfied only in case of option C
\nHence, C is correct<\/p>\n
\nAssume that a total of\u00a01 min is elapsed for taking the patient into and out of the ambulance?<\/p>\n
\n[CAT 2002]<\/p>\n
\nTherefore, total work to be done = 60 units.
\nFrom 11 AM to 5 PM, work done = 6*6 = 36 units.
\nWork remaining = 60 – 36 = 24 units.
\nWork done in the next 3 hours = 7 units + 8 units + 9 units = 24 units.
\nTherefore, the work gets done by 8 PM.<\/p>\n
\n[CAT 2002]<\/p>\n
\nTherefore, work done by each of the small pumps in one hour = 2 units.
\nLet the total work to be done in filling the tank be 9 units.
\nTherefore, time taken by the big pump if it operates alone = 9\/3 = 3 hours.
\nIf all the pumps operate together, the work done in one hour = 3 + 2*3 = 9 units.
\nTogether, all of them can fill the tank in 1 hour.
\nRequired ratio = 1\/3<\/p>\n
\nAnd C’s efficiency is putting A and B together i.e. = 20 hours $( (\\frac{1}{60} + \\frac{1}{30})^{-1})$
\nNow if all three work together, then it will be completed in x (say) days.<\/p>\n
\nIf he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence.
\nIf he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
\nWhen the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM.
\nSo, Neera must have walked for 55 minutes from 5PM.<\/p>\n