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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Two pipes P and Q can fill a tank in 10 hrs and 12 hrs respectively. If the two pipes are opened at 9 AM in the morning, at what time the pipe P should be closed to fill the tank exactly at 3.00 PM?1). 13:30 PM2). 1:00 PM3). 2:00 PM4). 12:00 PM |
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Answer» According to the given CONDITIONS , P fills 1/10 TH of the tank in 1 hour And Q fills 1/12 th of the tank in 1 hour It takes total 6 HOURS to fill the tank Let both the taps be open for X hours ∴ x × (1/10 + 1/12) + (6 - x) × (1/12) = 1 ∴ 11x/60 + ½ - x/12 = 1 ∴ x = 5 ∴ P should be closed after 5 hours i.e at 2:00 PM |
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| 152. |
A cylindrical water tank is completely filled at 10 am. There are two leakages in the tank. One of them is at bottom of tank and other is at half of the height of tank. Each of them drains out the water at 2 litres per minute. The tank gets emptied 11 am. Find the capacity of tank.1). 100 litres2). 120 litres3). 144 litres4). 160 litres |
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Answer» Let the CAPACITY of TANK be T LITERS. One of the leakages is at bottom of tank and other at half the height of tank. ⇒Till the tank gets half empty, both leakages take out water, and after that only one leakage take out water. Rate of drainage of each leakage = 2 liters per minute Time taken to make tank half empty = (T/2)/2 × 2 = (T/8) minutes. Time taken to make tank empty from half empty = (T/2)/2 = (T/4) minutes. Time between 11 am and 10 am = 1 hour = 60 minutes. ⇒ Total time taken to empty tank = 60 minutes ⇒ (T/8) + (T/4) = 60. ⇒ (3T/8) = 60 ⇒ T = 160 ∴ Capacity of tank is 160 liters |
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| 153. |
A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days, B had to leave and A alone completed the remaining work, The whole work was completed in :1). 10 days2). 8 days3). 12 days4). 15 days |
| Answer» OPTION 3 is the RIGHT ONE | |
| 154. |
Raj and Ram working together do a piece of work in 10 days. Rajalone can do it in 12 days. Ram alone will do the work in1). 20 days2). 40 days3). 50 days4). 60 days |
| Answer» | |
| 155. |
A and B together can complete a piece of work in 72 days, B and C together can complete it in 120 days, and A and C together in 90 days. In what time can A alone complete the work ?1). 80 days2). 100 days3). 120 days4). 150 days |
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| 156. |
A is 50% as efficient as B.C. does half of the work done by A and B together. If C alone does the work in 20 days, then A, B and C together can do the work in1). 52/3 days2). 62/3 days3). 6 days4). 7 days |
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Answer» ⇒ Let the TIME taken by B to complete the work be x days ⇒ B’s one-day work = 1/x ⇒ Time taken by A to complete the work be 2x days ⇒ A’s one-day work = 1/2x ⇒ Work done by C in 1 day = (1/2) × (1/x + 1/2x) ⇒ Work done by C in 1 day = 3/4x ⇒ given C ALONE finish the work in 20 days ⇒ 3/4x = 1/20 ⇒ x = 15 ⇒ Work done by (A + B + C) for 1 day = (1/x + 1/2x + 3/4x) ⇒ Work done by (A + B + C) for 1 day = 1/30 + 1/15 + 1/20 = 9/60 = 3/20 ∴ REQD. time = 6 (2/3) days |
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| 157. |
A man, a woman and a boy can complete a job in 3, 4 and 12 days respectively. How many boys must assist 1 man and 1 woman to complete the job in $\frac{1}{4}$ of a day ?1). 12). 43). 194). 41 |
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| 158. |
1). 92). 83). 104). 7 |
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Answer» Pipes P and Q can FILL a tank in 12 HOURS and 36 hours Q’s one-day work = 1/36 P and Q‘s one-day work = 1/12 + 1/36 = 4/36 = 1/9 Total time taken by both P and Q to fill the tank = 1/(1/9) = 9 hours |
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| 159. |
If the expenditure of gas on burning 6 burners for 6 hours a day for 8 days is Rs. 450, then how many burners can be used for 10 days at 5 hours a day for Rs. 625 ?1). 122). 163). 44). 8 |
| Answer» OPTION 4 is the RIGHT ANSWER | |
| 160. |
A particular job can be completed by a team of 10 men in 12 days. The same Job can be completed by a team of 10 women in 6 days. How many days are needed to complete the Job if the two teams work together?1). 4 days2). 6 days3). 9 days4). 18 days |
| Answer» ANSWER for this QUESTION is 4 DAYS | |
| 161. |
A does half as much work as B in three- fourth of the time. If together they take 18 days to complete a work , how much time shall B take to do it alone ?1). 30 days2). 35 days3). 40 days4). 45 days |
| Answer» OPTION 1 is the RIGHT ANSWER | |
| 162. |
A and B together can do a piece of work in 6 days. If A can alone do the work in 18 days, then the number of days required for B to finish the work is1). 102). 123). 94). 15 |
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| 163. |
If 3 men or 4 women can plough a field in 43 days, how long will 7 men and 5 women take to plough it ?1). 10 days2). 11 days3). 9 days4). 12 days |
| Answer» | |
| 164. |
1). 102). \(6\frac{1}{3}\)3). \(7\frac{1}{3}\)4). 7 |
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Answer» A can complete the WORK in 10 days. ⇒ A’s 1day’s work = 1/10 B can complete the work in 20 days. ⇒ B’s 1 day’s work = 1/20 C can complete the work in 15 days. ⇒ C’s 1day’s work = 1/15 (A + B)’s 1 day’s work = (1/10) + (1/20) = 3/20(A + B implies A and B) (A + C)’s 1 day’s work = (1/10) + (1/15) = 1/6(A + C implies A and C) According to the QUESTION, A is assisted by B and C on ALTERNATE days. ⇒Their 2 day’s work = (3/20) + (1/6) = 19/60 [? (A + B) work on 1st day and (A + C) work on 2nd day] ⇒ Their (3 × 2 =) 6 day’s work = 3 × (19/60) = 57/60 ⇒ Remaining work = 1 - (57/60) = 3/60 = 1/20 On 7th day (A + B) will work. (A + B) do (3/20)th work in 1 day ⇒ Time TAKEN by (A + B) to do remaining work = (1/3) days ∴ The work will get completed in = $(6 + \frac{1}{3} \Rightarrow 6\frac{1}{3})$ days |
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| 165. |
A and B can do a piece of work in 10 days B and C can do it in 12 days, C and A in 15 days. In how many days will C finish it alone ?1). 24 days2). 30 days3). 40 days4). 60 days |
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| 166. |
There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. Find the percentage of work done by tap A itself?1). 72%2). 70%3). 71%4). 65% |
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Answer» Tap A can fill the TANK in 25 hrs. Tap B cab fill the tank in 20 hrs. Tap C can fill the tank in 10 hrs. Total volume of the tank filled by 3 pipes = LCM of (25, 20, 10) = 100 units ⇒ Pipe A can fill 4 units of water in 1 hr. ⇒ Pipe B can fill 5 units of water in 1 hr. ⇒ Pipe C can fill 10 units of water in 1 hr. ⇒ Pipe (A + B + C) can fill 19 units of water in 1 hr. According to question, Pipe (A + B + C) are opened for 1 hr, then tap C is closed. ⇒ 19 units of water are filled in the tank. After 4th hr, tap B is also closed ⇒ for 3 hrs, pipe A and B are open ⇒ (4 + 5) × 3 = 27 units of water are filled. ⇒ 100 - (19 + 27) = 54 units of water are filled by tap A alone. Total units of water filled by tap A is 4 × 1 + 4 × 3 + 54 = 4 + 12 + 54 = 70 units ∴ Percentage of work DONE by tap A = 70% |
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| 167. |
Shan can complete a task in 16 days. When Shan and Jyoti work together on the same task, they receive their wages in the ratio 7 ∶ 4. In how many days will Jyoti complete the task alone?1). 24 days2). 26 days3). 29 days4). 28 days |
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Answer» Let Jyoti will complete the task alone in ‘x’ days Jyoti’s 1 DAY work = 1/x Shan’s 1 day work = 1/16 When they work TOGETHER, RATIO of their WAGES = 7 ? 4 ⇒ 1/16 ? 1/x = 7 ? 4 ⇒ x ? 16 = 7 ? 4 ⇒ x = 7/4 × 16 = 28 ∴ Jyoti will complete the task alone in 28 days |
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| 168. |
1). 33/82). 40/93). 41/104). 50/11 |
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Answer» Given that, A can do a task in 8 days. ∴ 1 day’s WORK by A = 1/8 Also, B can do it in 10 days ∴ 1 day’s work by B = 1/10 1 day’s work by both A and B together = $(\FRAC{1}{8}\; + \;\frac{1}{{10}}\; = \;\frac{{5\; + \;4}}{{40}}\; = \;\frac{9}{{40}})$ To do the whole work, time taken by A and B together = $(\frac{1}{{\frac{9}{{40}}}}\; = \;\frac{{40}}{9})$ |
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| 169. |
TWO workers A and B working together completed a Job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days.To complete the Job alone, A would require1). 5 1/5 days2). 6 1/4 days3). 7 1/2 days4). 8 3/4 days |
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Answer» This question was asked some where in PREVIOUS year PAPERS of ssc, and CORRECT answer was option 2 |
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| 170. |
Two taps P and Q can fill a tank in 24 hours and 18 hours respectively. If the two taps are opened at 11 a.m., then at what time (in p.m.) should the tap P be closed to completely fill the tank at exactly 2 a.m?1). 52). 23). 34). 4 |
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Answer» Let the CAPACITY of the tank = 72 litres Efficiency of pipe P = 72/24 = 3 litres/hour Efficiency of pipe Q = 72/18 = 4 litres/hour According to the question, Total time in which tank has filled = 15 hours Let the pipe P was closed at after n hours According to the question ⇒ 3 × (n) + 4 × 15 = 72 ⇒ 3n + 60 = 72 ⇒ 3n = 12 ⇒ n = 4 So, 4 hours after 11:00 am. that MEANS 3:00 PM ∴ the tap P should be closed at 3 : 00 pm |
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| 171. |
A can do a piece of work in 12 days and B in 15days.They work together for 5 days and then B left The days taken by A to finish the remaining work is1). 32). 53). 104). 121 |
| Answer» OPTION 1 is the RIGHT ANSWER | |
| 172. |
A and B together can complete a piece of work in 18 days, B and C in 24 days and A and C in 36 days. In how many days, will all of them together complete the work ?1). 16 days2). 15 days3). 12 days4). 10 days |
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| 173. |
How many men need to be employed to complete a Job in 5 days, if 15 men can complete $\frac{1}{3}$ of the job in 7 days ?1). 202). 213). 244). 63 |
| Answer» OPTION 4 : 63 is CORRECT | |
| 174. |
If 10 men can do a piece of work in 12 days, then what will be the time taken by 12 men to do the same piece of work?1). 12 days2). 10 days3). 9 days4). 8 days |
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Answer» When 10 MEN WORK for 12 days, then Total work = 10 × 12 = 120 12 men can COMPLETE the work = 120/12 = 10 days |
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| 175. |
A and B can do a piece of work in 45 and 40 days repectively. They began the work together but A left after some days and B finished the remaining work in 23 days. A left after1). 6 days2). 9 days3). 12 days4). 5 days |
| Answer» OPTION 2 : - 9 DAYS | |
| 176. |
Seventy-five men are employed to lay down a railway line in 3 months. Due to certain emergency conditions, the work was to be finished in 18 days. How many more men should be employed to complete the work in the desired time ?1). 3002). 3253). 3504). 375 |
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| 177. |
1).2). $5043). $2524). Data inadequate |
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Answer» Amount of work Peter can do in an HOUR = (1/20) Amount of work Quagmire can do in an hour = (1/5) Let Goldman is able to complete ONE piece of work in 'n' hours, ⇒ Then the amount of work Goldman does in an hour = 1/n ⇒ THUS, total work done by all 3 in an hour = (1/20) + (1/5) + (1/n) ⇒ Total amount of work done by all 3 in a couple of hours (2 hours) = {(1/20) + (1/5) + (1/n)} × 2---- (1) From given data, The entire piece of work is completed in 2 hours. ⇒ {(1/20) + (1/5) + (1/n)} × 2 = 1 ⇒ {(1/20) + (1/5) + (1/n)} = ½ ⇒ (1/n) = ½ – {(1/20) + (1/5)} ⇒ (1/n) = ½ – (5/20) ⇒ (1/n) = ¼ As Goldman does 1/4th of the work in an hour. ⇒ They FINISHED the work in 2 hours Therefore, part of the work done by Goldman = 2 × ¼ = ½ ∴ He gets ½ × 1008 = $504 |
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| 178. |
A contractor undertook a contract for a construction work to be completed in 180 days. He engaged 120 labours. But only 1/6th of the work was done in 1/5th of the scheduled time. The additional number of labours that will be required to complete the work in time, is1). 202). 403). 304). 15 |
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Answer» Total Scheduled TIME = 180 DAYS 1/5th of the scheduled time = 180/5 = 36 days According to the given information, 120 labours can do 1/6th of the total work in 36 days ∴ remaining work = 5/6th of total work $(\FRAC{{{M_1}{D_1}}}{{{W_1}}} = \frac{{{M_2}{D_2}}}{{{W_2}}})$ Where, M = Number of men, D = number of days, W = Amount of work $(\RIGHTARROW \frac{{120 \TIMES 36}}{{\frac{1}{6}}} = \frac{{{M_2}\; \times 144}}{{\frac{5}{6}}})$ ⇒ M2 = 150 ∴ Additional Labours required = 150 – 120 = 30 |
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| 179. |
P and Q can complete a task in 50 and 20 days respectively. In how many days can they complete 70% of the task if they work together?1). 20 days2). 30 days3). 40 days4). 10 days |
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Answer» P’s one - day work = 1/50 Q’s one - day work = 1/20 One - day work of both P and Q = 1/50 + 1/20 = 7/100 Total TIME taken to COMPLETE the work = 100/7 days Time taken to complete 70% of work = (70/100) × (100/7) = 10 days |
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| 180. |
Kamal can do a work in 15 days. Bimal is 50 per cent more efficient than Kama! in doing the work. In how many days will Bimal do that work ?1). 14 days2). 12 days3). 10 days4). $10\frac{1}{2}$ days |
| Answer» HELLO, 10 DAYS is CORRECT | |
| 181. |
1). \(63\frac{1}{3}\) hours2). 75 hours3). 60 hours4). 56 hours |
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Answer» Rahim does (7/10) part of a work in = 21 HOURS ∴ Rahim completes the work in = 21 × (10/7) hours = 30 hours ∴ Part of work Rahim can do in an hour = 1/30 Remaining part of the work = 1 – (7/10) = 3/10 Bablu and Rahim together DONE 3/10th part of the work in 6 hours ∴ Bablu and Rahim can COMPLETE the work = 6 × (10/3) hours = 20 hours ∴ Part of work Bablu and Rahim can do in an hour = 1/20 ∴ Part of work Bablu alone can do in an hour = Part of work Bablu and Rahim can do in an hour – Part of work Rahim can do in an hour = (1/20) – (1/30) = 1/60 ∴ Bablu alone can complete the work in 60 hours |
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| 182. |
A cistern can be filled by two pipes filling separately in 12 and 16 min respectively. Both pipes are opened together for a certain time but being closed, only 7/8 of the full quantity of water flows through the former and only 5/6 through the latter pipe. The obstruction however being suddenly removed the cistern is filled in 3 min from that moment. How long was it before the full flow began?1). 2.5 min2). 4.5 min3). 3.5 min4). 5.5 min |
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Answer» Let the PIPE be opened for x min ⇒ WORK DONE in 1 min by first pipe = 1/12 ⇒ Work done in 1 min by second pipe = 1/16 ⇒ Work done by both in 3 min = 3 × {(1/12) + (1/16)} = 21/48 ⇒ REMAINING work i.e. 27/48 is done by (7/8) first pipe and (5/6) second pipe ⇒ x × [{(7/8) × (1/12)} + {(5/6) × (1/16)} = 27/48 ⇒ (1/8) x = 27 / 48 ∴ x = 4.5 min |
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| 183. |
1). 502). 403). 454). 10 |
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Answer» Let n be the NUMBER of workers and x be the work done by 1 worker in 1 day ∴ According to the 1st condition, ⇒ nx + (n - 1)x + (n - 2)x + … + x = 1 ∴ x × n(n + 1)/2 = 1 ∴ x × n(n + 1) = 2- - - - - (1) Also According to the 2nd given condition, (55n/100) × nx = 1 ∴ x = 20/(11 × n × n)- - - - - (2) On substituting (2) in (1) we get, 20/11n2 × (n2 + n) = 2 (n2 + n)/n2 = 11/10 1 + 1/n = 11/10 ∴ n = 10 ∴ There are 10 workers in the group. |
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| 184. |
If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man.1 woman and 1 boy in1). 21 days2). 24 days3). 26 days4). 33 days |
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| 185. |
5 class A workers and 4 class B workers are assigned a job to paint a circular park. They work for one day and complete painting till half the radius of the circular park. The next day, class B workers go on strike and returns the day after as the management resolves their problems. This creates a delay of 4 hours and 48 min in overall. If their payment is based on their efficiency, what will be the ratio of their pay?1). 16 : 52). 5 : 163). 8 : 54). 5 : 8 |
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Answer» Let a class A worker complete the work in ‘a’ days and a class B worker complete the work in ‘b’ days ⇒ Amount of work done by a class A worker in 1 day = 1/a ⇒ Amount of work done by a class B worker in 1 day = 1/b ⇒ Amount of work done by 5 class A and 4 class B workers in 1 day = 5/a + 4/b = 1/4 [? Half the radius, quarter the AREA of the CIRCLE] ⇒ 20/a + 16/b = 1 → 1 So, they would have completed the work in 4 days if the strike did not happen. But, it takes 4 days 4 HOURS and 48 min due to the strike Class A workers work for 4 days 4 hours 48 min whereas class B workers work for 3 days 4 hours 48 min ⇒ 4 hours 48 min = (4/24) + {48/ (60 × 24)} = 0.2 day ⇒ Total amount of work done = 4.2 × (5/a) + 3.2 × (4/a) ⇒ 1 = 21/a + 12.8/b → 2 Comparing EQUATIONS 1 and 2 ⇒ 21/a + 12.8/b = 20/a + 16/b ⇒ 1/a = 3.2/b ⇒ a/b = 1/3.2 Efficiency ∝ 1/Time taken ∴ Ratio of their efficiencies = 3.2 : 1 = 16 : 5 |
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| 186. |
1). 20 days2). 18 days3). 22 days4). 24 days |
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Answer» A is 4 times as productive as C and B is half as productive as A; ∴ Ratio of efficiencies of A, B and C = 4 : 2 : 1 Since B left 15 DAYS and THUS the work got delayed by 2 days; Suppose the work would have completed in ‘X’ days, if B had not left; ∴ (4 + 2 + 1) × 15 + (4 + 1) × (x – 15 + 2) = (4 + 2 + 1) × x ⇒ 105 + 5X – 65 = 7x ⇒ 2x = 40 ⇒ x = 20 ∴ The work got COMPETED in 22 days (= 20 + 2) |
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| 187. |
A, B and C individually can do a work in 10 days, 12 days and 15 days respectively. If they start working together, then the number of days required to finish the work is1). 16 days2). 8 days3). 4 days4). 2 days |
| Answer» HELLO, 4 DAYS is CORRECT | |
| 188. |
Two men can do a piece of work in x days. But y women can do that in 3 days. Then the ratio of the work done by 1 man and 1 woman is1). 3y:2x2). 2x:3y3). x:y4). 2y:3x |
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| 189. |
1). 9 days2). 8 days3). 8.1 days4). 8.2 days |
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Answer» Time taken by 4 MEN to BUILD a SMALL house = 12 DAYS ⇒ Time taken by one man to build a small house = 12 × 4 = 48 days ⇒ Time taken by 6 men to build the same house = 48/6 = 8 days ∴ Time taken by one man to build the same house = 8 days |
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| 190. |
A alone can do a work in 20 days. B is 25% more efficient than A. A and B started working and worked for 4 days. If C alone completed the remaining job in 22 days. How many days C alone takes to complete the entire job?1). 24 days2). 36 days3). 12 days4). 20 days |
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Answer» A alone can do work in 20 days. B is 25% more efficient than A. ⇒ B can do a work in 16 days. Let the total UNIT of work ( LCM of 20 and 16) = 80 UNITS ⇒ A’s one day work = 4 ⇒ B’s one day work = 5 As A and B left after working for 4 days. ⇒ Work done by A and B in 4 days = 36 work ⇒ Work left = 44 units is done by C in 22 days. ⇒ C’s one day work = 2 units ∴ 80 units of work can be completed by C alone in 80/2 = 40 days |
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| 191. |
A and B can do a Job together in 12 days. A is 2 times as efficient as B. In how many days can B alone complete the work ?1). 18 days2). 9 days3). 36 days4). 12 days |
| Answer» 36 DAYS is the BEST SUITED | |
| 192. |
Some carpenters promised to do a Job in 9 days but 5 of them were absent and remaining men did the Job in 12 days. The original number of carpenters was1). 242). 203). 164). 18 |
| Answer» OPTION 2 is the RIGHT ANSWER | |
| 193. |
A alone can complete a work in 25 days while B can complete the same work in 20 days. They worked together for 5 days and then A left. In how many days the remaining work will be completed?1). 10 days2). 11 days3). 9 days4). 12 days |
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Answer» One DAY’s WORK DONE by A = 1/25 One day’s work done by B = 1/20 Work done by (A + B) TOGETHER in 1 day = 1/25 + 1/20 = 9/100 So, Work done by (A + B) together in 5 days = 5 × 9/100 = 45/100 Remaining work = 1 – 45/100 = 55/100 1 work completed by B = 20 days Hence, 55/100 part of work completed by B = 20 × 55/100 = 11 days |
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| 194. |
1). 502). 753). 854). 90 |
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Answer» Time taken by Rajan and GEETA to complete the WORK = 6 days As we KNOW, work is directly proportional to time, w/t = constant Time taken by Ramesh to complete work = 15 days According to the question, 1/15 + 1/x = 1/6 Where x = time taken by Geeta to complete the same work, ⇒ 1/x = 1/6 – 1/15 ⇒ x = 10 days For finding the EFFICIENCY of Rajan, 1/10 + 1/y = 1/9 Where, y = time taken by Rajan alone to complete the same work, ⇒ 1/y = 1/9 – 1/10 ⇒ 1/y = 1/90 ∴ y = work efficiency of Rajan = 90 days |
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| 195. |
Pipe A takes 36 hours to fill the tank and pipe B takes 12 hours. Pipe C takes 72 hours to empty the tank. If all the three pipes are opened, how much time will the tank take to be filled?1). \(10\frac{2}{7}\) hours2). \(8\frac{5}{7}\) hours3). \(9\frac{2}{7}\) hours4). \(6\frac{5}{7}\) hours |
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Answer» ⇒ Part of the tank pipe A FILLS in one minute $(= \frac{1}{{36}})$ ⇒ Part of the tank pipe B fills in one minute $(= \frac{1}{{12}})$ ⇒ Part of the tank pipe A and B fill in one minute $(= \frac{1}{{36}} + \frac{1}{{12}} = \frac{1}{9})$ ⇒ Part of the tank pipe C EMPTIES in one minute $(= \frac{1}{{72}})$ The RATE of filling is more than the rate of emptying ⇒ NET part of the tank filled in one minute = $(\frac{1}{9} - \frac{1}{{72}} = \frac{7}{{72}})$ ∴ Time taken to fill the tank is $(\frac{{72}}{7} = 10\frac{2}{7})$ hours |
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| 196. |
If A and B together can complete a work in 12 days, B and C together in 15 days and C and A together in 20 days, then B alone can complete the work in1). 30 days2). 25 days3). 24 days4). 20 days |
| Answer» 20 DAYS is the BEST SUITED | |
| 197. |
A can cultivate $\frac{2}{5}$ th of a land in 6 days and B can cultivate $\frac{1}{3}$ rd of the same land in 10 days. Working together A and B can cultivate $\frac{4}{5}$th of the land in :1). 4 days2). 5 days3). 8 days4). 10 days |
| Answer» OPTION 3 is the RIGHT ANSWER | |
| 198. |
10 men can complete a piece of work in 34 days, 16 women can complete the same piece of work in 56 days. 5 men and 18 women work together for 20 days. If only women were to complete the remaining piece of work in 4 days along with a boy who does 10 units of work per day, how many women would be required?1). 532). 683). 584). 54 |
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Answer» The amount of work done by 10 MEN in 34 DAYS = amount of work done by 16 women in 56 days ⇒ 10 ? men ? 34 = 16 ? women ? 56 ⇒ Men/women = 224/85 ⇒ Amount of work done by a man in 1 day = 224 ⇒ Amount of work done by a women in 1 day = 85 ⇒ Total work = 10 ? men ? 34 = 10 ? 224 ? 34 = 76160 ⇒ Work done by 6 men and 18 women in 20 days = (5 ? men + 18 ? women) ? 20 ⇒ (5? 224 + 18 ? 85) ? 20 = 53000 ⇒ Work LEFT = 76160 – 53000 = 23160 Work done by Boy for 4 days = 4 × 10 = 40 Units ⇒ Remaining work = 23160 – 40 = 23120 ∴ Number of women required to finish the remaining work = $(\frac{{23120}}{{4\; \times \;85}}\; = \;68)$ women |
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| 199. |
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped on the second day, four more workers dropped on third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed?1). 282). 243). 254). 30 |
| Answer» OPTION option 3 is the CORRECT ANSWER | |
| 200. |
A piece of work was finished by A, B and C together. A and B together finished 60% of the work and B and C together finished 70% of the work. Who among the three is most efficient?1). A2). B3). C4). A or B |
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Answer» Let the TOTAL AMOUNT of WORK be a ⇒ Work done by A and B together = 0.6a ⇒ Work done by C ALONE = a – 0.6a = 0.4a ⇒ Work done by B and C together = 0.7a ⇒ Work done by B alone = 0.7a – 0.4a = 0.3a ⇒ Work done by A alone = a – 0.7a = 0.3a ∴ the most efficient is C |
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