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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.
| 101. |
A contract is to be completed in 50 days and 105 men were set to work, each working 8 hour a day. After 25 days, 2/5th of the work is finished. How many additional men should be employed so that the work may be completed on time, each man now working 9 hour a day?1). 342). 363). 354). 37 |
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Answer» Answer: C) 35 35 is correct answer. According to the formula, |
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| 102. |
P, Q and R contract to do a work for Rs. 1650, P can do a work in 8 days, Q can do a work in 12 days and R can do a work in 15 days, then find the share of R?1). Rs. 4002). Rs. 4203). Rs. 8004). Rs. 350 |
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Answer» <P>Let the total work be 120 units(LCM of 8, 12 and 15) ⇒ ONE DAY work of P = 15 units ⇒ One day work of Q = 10 units ⇒ One day work of R = 8 units P, Q and R contract to do a work for Rs. 1650 ∴ Share of R = 8/33 × 1650 = Rs. 400 |
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| 103. |
1). 28 days2). 30 days3). 15 days4). 10 days |
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Answer» Suppose Shashi takes K days to do the work. ∴ Sachin takes $(\LEFT( {2 \times \frac{3}{4}K} \RIGHT) = \frac{3}{2}K)$ days to do it From question, ⇒ (Shashi + Sachin)’s ONE DAY’s work $(= \frac{1}{{18}})$ ∴ $(\frac{1}{K} + \frac{2}{{3K}} = \frac{1}{{18}})$ Or K = 30 So shashi can do the same work in 30 days. |
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| 104. |
A contractor undertook to complete a project in 90 days and employed 60 men on it. After 60 days, he found that $\frac{3}{4}$ of the work has already been completed. How many men can he discharge so that the project may be completed exactly on time ?1). 402). 203). 304). 15 |
| Answer» 20 SEEMS CORRECT. | |
| 105. |
While working together on a task, Amit and Sumit received Rs. 600 and Rs. 400 respectively. If Amit can do the task alone in 12 days, then how much time will Sumit take to do the task alone?1). 8 days2). 9 days3). 15 days4). 18 days |
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Answer» Let X be the number of days taken by SUMIT to finish the work alone. Given: Amit takes 12 days to complete the work alone. Salary of Amit = Rs.600 Salary of Sumit = Rs.400 Work done by Amit in one day = 1/12 Work done by Sumit in one day = 1/x Ratio of their wages = Ratio of their efficiencies 600 : 400 = (1/12) : (1/x) 3/2 = x/12 x = (3/2)(12) x = 18 days So, Sumit takes 18 days to complete the work alone. The correct OPTION is 4). 18 days |
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| 106. |
3 boys and 4 girls can do a piece of work in 4.5 days. Also, 9 boys and 6 girls can do the same work in 2 days. In how many days will 6 boys and 8 girls finish the same work?1). 13/2 days2). 9/4 days3). 7/2 days4). 11/4 days |
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Answer» LET 1 BOY’s 1 DAY work be ‘x’ and 1 GIRL’s 1 day work be ‘y’. Hence, we have, ⇒ 3x + 4y = 1/(4.5) ⇒ 3x + 4y = 2/9…(1) Also, ⇒ 9x + 6y = 1/2…(2) MULTIPLYING eq. (1) by 3 and subtracting from (2), we get, ⇒ 9x + 6y – 9x – 12y = 1/2– 2/3 ⇒ – 6y = –1/6 ⇒ y = 1/36 Substituting in (1), ⇒ x = 1/27 ⇒ 6 boys and 8 girls 1 day work = 6(1/27) + 8(1/36) = 2/9 + 2/9 = 4/9 ∴ 6 boys and 8 girls will finish the same piece of work in 9/4 days. |
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| 107. |
A can do a piece of work in 6 days. B is 25 % more efficient than A. How long would B alone take to finish this work ?1). $4\frac{4}{5}$ days2). $3\frac{1}{3}$ days3). $5\frac{1}{4}$ days4). $2\frac{2}{3}$ days |
| Answer» | |
| 108. |
Jyothl can do $\frac{3}{4}$ of a Job in 12 days. Mala is twice as efficient as Jyothi. In how many days will Mala finish the job ?1). 6 days2). 8 days3). 12 days4). 16 days |
| Answer» | |
| 109. |
Quantity 2: If A is twice as efficient as B and they both can do a piece of work in 12 days, then what percentage of the work will B do alone in 20 days?1). Quantity 1 > Quantity 22). Quantity 1 < Quantity 23). Quantity 1 ≥ Quantity 24). Quantity 1 ≤ Quantity 2 |
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Answer» Solving for QUANTITY 1: A’s 1 day WORK = 1/15 B’s 1 day work = ½ × 1/15 = 1/30 (A + B) ’s 1 day work = 1/15 + 1/30 = 1/10 In 8 days, work done = 8/10 ⇒ Quantity 1 = (8/10) × 100 = 80% Solving for Quantity 2: Let B’s 1 day work = 1/x A’s 1 day work = 2/x (A + B) ’s 1 day work = 1/x + 2/x = 3/x ⇒ 3/x = 12 ⇒ x = 36 Hence, B’s 1 day work = 1/36 In 20 days, work done by B = 20 × 1/36 = 5/9 ⇒ Quantity 2 = (5/9) × 100 = 55.55% ∴ Quantity 1 > Quantity 2 |
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| 110. |
16 men are able to complete a piece of work in 12 days working 14 hours a day. How long will 28 men, working 12 hours a day. take to complete the work ?1). 10 days2). 7 days3). 8 days4). 6 days |
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| 111. |
Peeyush has done 1/3rd of a job in 30 days, Sanjiv completes the rest of the job in 60 days. In how many days can they together do the job?1). 15 days2). 45 days3). 30 days4). 10 days |
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Answer» ⇒ TIME taken by Peeyush to complete the work = 30 × 3 = 90 days RATE of doing work of Peeyush = 1/90 ⇒ Time taken by Sanjiv to complete the work = 60 × (3/2) = 90 days Rate of doing work of Sanjiv = 1/90 If both of them worked together ⇒ Then rate of doing work when both worked together = (1/90) + (1/90) = 2/90 = 1/45 ∴ time taken to complete the work if they both worked together is 45 days. |
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| 112. |
A and B undertook to do a piece of work for Rs. 48,000. A could do it alone in 5 days and B could do it alone in 8 days. With the help of C and D, they finished the work in 3 days. If the work done by C was twice that of the work done by D, then the difference between the shares of A and C was:1). Rs. 129602). Rs. 340803). Rs. 206404). Rs. 28000 |
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Answer» Work done by A in 1 day = 1/5 Work done by B in 1 day = 1/8 Let the work done by D in 1 day be 1/x ⇒ Work done by C in 1 day = 2/x(GIVEN) ⇒ Work done by all in 1 day$(= \frac{1}{5}\; + \;\frac{1}{8}\; + \;\frac{1}{x}\; + \;\frac{2}{x})$ Now, when all of them work, they finish the work in 3 days. ⇒ Work done by all in 1 day = 1/3 $(\begin{array}{l} \frac{1}{5}\; + \;\frac{1}{8}\; + \;\frac{1}{x}\; + \;\frac{2}{x}\; = \;\frac{1}{3}\\ \frac{3}{x}\; = \;\frac{1}{3} - \frac{1}{5} - \frac{1}{8}\; = \;\frac{{40 - 24 - 15}}{{120}}\; = \;\frac{1}{{120}} \END{array})$ ⇒ x = 120 × 3 = 360 Time taken by C = 180 days ⇒ A’s total payment = 48000 × (3/5) = 28800 ⇒ C’s total payment = 48000 × (3/180) = 800 ∴ Difference = 28800 - 800 = 28000 |
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| 113. |
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do the work in1). 15 days2). 20 days3). 25 days4). 30 days |
| Answer» HELLO, 25 DAYS is CORRECT | |
| 114. |
A garrison of 450 men had provision for 42 days, when given at the rate of 550 gms per head. At the end of 10 days a reinforcement arrives and it was found that the provisions will last 15 days more, when given at the rate of 825 gms per head.What is the strength of the reinforcement?1). 702). 1903). 2304). 640 |
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Answer» Total food = 450 × 42 × .550 = 10395 kg After 10 days; food left = 10395 - 10 × 450 × 0.550 = 7920 kg Let the REINFORCEMENTS has X soldiers Food consumed in 15 days = (450 + x) × 15 × .825 (450 + x) × 15 × .825 = 7920 ⇒ x = 190 |
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| 115. |
Ganga and Saraawati, working separately can mow a field in 8 and 12 hours respectively. If they work in stretches of one hour alternately. Ganga beginning at 9 a.m., when will the moving be completed?1). 6 p.m.2). 6.30 p.m.3). 5 p.m.4). 5.30 p.m. |
| Answer» | |
| 116. |
Raj and Rahul’s work efficiency is in the ratio 3∶ 2. If Raj can do a work in 12 days, in how many days will Rahul do the work alone?1). 8 days2). 9 days3). 15 days4). 18 days |
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Answer» LET Rahul take ‘x’ days to do the WORK alone Raj’s 1 DAY work = 1/12 Rahul’s 1 day work = 1/x Ratio of their work efficiencies = (1/12)? (1/x) = 3? 2 ⇒ x/12 = 3/2 ⇒ x = (12 × 3)/2 = 18 days ∴ Rahul will do the work alone in 18 days |
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| 117. |
1). 24402). 11203). 16804). 560 |
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| 118. |
Two pipes X and Y can fill an empty tank in 16 hours and 20 hours respectively. Pipe Z alone can empty the completely filled tank in 25 hours. Firstly both pipes X and Y are opened and after 6 hours pipe Z is also opened. What will be the total time (in hours) taken to completely fill the tank?1). 80/72). 67/73). 28/34). 304/29 |
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Answer» LET the tank be filled in ‘a’ hrs. Part filled by X in 1 HR. = 1/16 ⇒ Part filled by X in a hrs. = a/16 Part filled by Y in 1 hr. = 1/20 ⇒ Part filled by Y in a hrs. = a/20 Part emptied by Z in 1 hr. = 1/25 ⇒ Part emptied by Z in (a – 6) hrs. = (a – 6)/25 ⇒ a/16 + a/20 – (a – 6)/25 = 1 ⇒ a(1/16 + 1/20 – 1/25) = 1 – 6/25 ⇒ a(29/400) = 19/25 ⇒ a = 400/29 × 19/25 ⇒ a = 304/29 ∴ The tank will be filled in 304/29 hrs. |
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| 119. |
P can complete $\frac{1}{4}$ of a work in 10 days, Q can complete 40% of the same work in 15 days, R, completes $\frac{1}{3}$ of the work in 13 days and S, $\frac{1}{6}$ of the work in 7 days. Who will be able to complete the work first ?1). P2). Q3). R4). S |
| Answer» Q : - is CORRECT HENCE OPTION 2 | |
| 120. |
x can copy 80 pages in 20 hours, x and y together can copy 135 pages in 27 hours. Then y can copy 20 pages in1). 20 hours2). 3 hours3). 24 hours4). 12 hours |
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| 121. |
A labourer was appointed by a contractor on the condition that he would be paid Rs. 75 for each day of his work but would be fined at the rate of Rs.15 per day for his absence, apart from losing his wages. After 20 days, the contractor paid the labourer Rs. 1140. The number of days the labourer abstained from work1). 32). 53). 44). 2 |
| Answer» CORRECT ANSWER is: OPTION 3 | |
| 122. |
Working 5 hours a day, A can complete a work in 8 days and working 6 hours a day, B can complete the same work in 10 days. Working 8 hours a day, they can jointly complete the work in1). 3 days2). 4 days3). 4.5 days4). 5.4 days |
| Answer» 3 DAYS : - OPTION 1 | |
| 123. |
A can complete $\frac{1}{3}$ of a work in 5 days and B, $\frac{2}{5}$ of the work in 10 days. In how many days both A and B together can complete the work ?1). 10 days2). $9\frac{3}{8}$ days3). $8\frac{4}{5}$ days4). $7\frac{1}{2}$ days |
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Answer» Its very SIMPLE QUESTION $9\frac{3}{8}$ days is the correct answer. |
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| 124. |
Three persons undertake to complete a piece of work for Rs. 1,200. The first person complete the work in 8 days, second person in 12 days and third person in 16 days.They complete the work with the help of a fourth person in 3 days. What does the fourth person get?1). Rs. 1802). Rs. 2003). Rs. 2254). Rs. 250 |
| Answer» RS. 225 is the BEST SUITED | |
| 125. |
Three taps A, B and C can fill a tank in 20, 24 and 30 hours respectively. How long (in hours) would the three taps take to fill the tank if all of them are opened together?1). 82). 123). 94). 6 |
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Answer» A fill the tank in one DAY = 1/20 B fill the tank in one day = 1/24 C fill the tank in one day = 1/30 Fill the tank (A + B + C) in one day = (1/20) + (1/24) + (1/30) = 15/120 = 1/8 |
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| 126. |
A and B together can do a work in 8 days, B and C together in 6 days while C and A together in 10 days, if they all work together,the work will be completed in :1). $3\frac{3}{4}$ days2). $3\frac{3}{7}$ days3). $5\frac{5}{47}$ days4). $4\frac{4}{9}$ days |
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| 127. |
If 12 men working 8 hours a day complete the work in 10 days, how long would 16 men working$7\frac{1}{2}$ hours a day take to complete the same work ?1). 72). 63). 104). 8 |
| Answer» CORRECT ANSWER is: 8 | |
| 128. |
Working efficiencies of P and Q for completing a piece of work are in the ratio 3:4. The number of days to be taken by them to complete the work will be in the ratio1). 3:22). 2:33). 3:44). 4:3 |
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| 129. |
8 children and 12 men complete a certain piece of work in 9 days. Each child takes twice the time taken by a man to flnish the work. In how many days will 12 men finish the same work ?1). 9 days2). 13 days3). 12 days4). 15 days |
| Answer» OPTION option 3 is the CORRECT ANSWER | |
| 130. |
A can finish a work in 18 days and B can do the same work in half the time taken by A. Then working together what part of the same work they can finish in a day ?1). $\frac{1}{6}$2). $\frac{2}{5}$3). $\frac{1}{9}$4). $\frac{2}{7}$ |
| Answer» HELLO, $\FRAC{1}{6}$ is CORRECT | |
| 131. |
10 women can do a piece of work in 6 days, 6 men can do same work in 5 days and 8 children can do it in 10 days. What is the ratio of the efficiency of a woman, a man and a child respectively?1). 4 : 6 : 32). 4 : 5 : 33). 2 : 4 : 34). 4 : 8 : 3 |
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Answer» Work done by 1 WOMEN in 1 day is 1/10 × 1/6 = 1/60 Work done by 1 men in 1 day is 1/6 × 1/5 = 1/30 Work done by 1 CHILDREN in 1 day is 1/8 × 1/10 = 1/80 ⇒ RATIO of their efficiencies is 1/60 : 1/30 : 1/80 ∴ their efficiencies are in RATION 4 : 8 : 3 |
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| 132. |
A takes three times as long as B and C together to do a Job. B takes four times as long as A and C together to do the work. If all the three, working together can complete the Job in 24 days, then the number of days. A alone will take to finish the Job is1). 1002). 963). 954). 90 |
| Answer» ANSWER for this QUESTION is 96 | |
| 133. |
Three persons A, B and C can do a piece of work 20, 15 and 30 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?1). 18 days2). 10 days3). 15 days4). 12 days |
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Answer» The work DONE by A, B and C in 1 day respectively = $(\frac{1}{{20}},\frac{1}{{15}},\frac{1}{{30}})$ The work done by A in 2 days = $(\frac{1}{{20}}\;)$× 2 = $(\frac{1}{{10}})$ The work done in 1 day by A, B and C together = $(\frac{1}{{20}}\;)$+ $(\;\frac{1}{{15}}\;)$ + $(\;\frac{1}{{30\;}})$ = $(\frac{9}{{60}})$ The total work done in span of 3 days = $(\frac{1}{{10}})$+$(\frac{9}{{60}})$ = $(\frac{1}{4})$ ¼ The of the total work is done in 3 days. So time to COMPLETE the work = 12 days. |
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| 134. |
Pipe A, B and C can fill an empty tank alone in 6 hours, 8 hours and 12 hours respectively. A tank is initially 1/6 filled. Pipes A and B are simultaneously opened in the tank and closed after 2 hours. Now, if pipe C is opened in the tank, in how much time will it be filled?1). 2 hrs.2). 3 hrs.3). 4 hrs.4). 6 hrs. |
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Answer» Part filled by A in 1 hr. = 1/6 Part filled by B in 1 hr. = 1/8 Part filled by C in 1 hr. = 1/12 Now, pipes A and B are opened for 2 hrs. ⇒ Part filled by A and B together in 2 hrs. = 2(1/6 + 1/8) = 2 × 7/24 = 7/12 ? TANK is initially 1/6 filled, Remaining part of tank to be filled = 1 - 1/6 - 7/12 = 1 - 3/4 = 1/4 ∴ Pipe C will fill the remaining part in = (1/4)/(1/12) = 3 hrs |
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| 135. |
If 12 carpenters working 6 hours a day can make 460 chairs in 240 days, then the number of chairs made by 18 carpenters in 36 days each working 8 hours a day is1). 922). 1323). 1384). 126 |
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| 136. |
P can do a work in 10 days. Q can do the same work in 15 days. If they work together for 5 days, how much of the work will they complete?1). 1/22). 2/33). 1/34). 5/6 |
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Answer» WORK done by P in 1 day = 1/10 Work done by Q in 1 day = 1/15 Work done by P and Q TOGETHER in 1 day = 1/10 + 1/15 = 1/6 ∴ work done by P and Q together in 5 DAYS = 1/6 × 5 = 5/6 |
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| 137. |
In a company, employees were shortlisted for a project. Previously, they added x employees in a team for a particular project with a target to finish it in 30 days. But after 20 days, they felt the task can take longer duration if they would not increase number of members for the project, so they added 500 more members to the team. Then the project took only 5 more days for completion. Find the value of x?1). 10002). 7503). 5004). 250 |
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Answer» Initially NUMBER of employee = x After 20 days number of employee = x + 500 According to Question, 30 × x = [(20 × x) + {(x + 500) × 5}] 30x = 20X + 5x + 2500 30x – 25X = 2500 5x = 2500 x = 500 |
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| 138. |
Five women can do a work in thirty six days. If the ratio between the capacity of a man and a woman is 3 : 1, then find how may days it will take 5 men to complete the same work?1). 12 days2). 15 days3). 18 days4). 108 days |
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Answer» Time taken by 5 WOMEN to do a work = 36 DAYS Time taken by 1 women to do the same work = 36 × 5 = 180 days According to the GIVEN information, Man/Woman = 3 : 1 Man = (1/3) × Woman ⇒ Time taken by 1 men to complete the same work = 1/3 × 180 ∴ Time taken by 5 men to complete the same work = (1/5) × 1/3 × 180 = 12 days |
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| 139. |
1). 7 days2). 2 days3). 5 days4). 11 days |
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Answer» AMOUNT receive at the end of the month = Rs. 10,450 Number of days ATTENDED = 10450/550 = 19 days Number of Sundays in month of March = 5 ∴ Number of leave = 31 - 19 - 5 = 7 days = 1 week |
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| 140. |
To complete a job, Abhay alone takes thrice as long as Bittu and Chanakya together. Bittu would take five times as long as Abhay and Chanakya together. All three together complete the work in 10 days. How many days are required by Chanakya to finish the work alone?1). 120/7 days2). 129/7 days3). 106/7 days4). 85/7 days |
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Answer» Let, Abhay, Bittu and Chanakya can COMPLETE the work alone in x, y and z days respectively. Abhay alone takes thrice as LONG as Bittu and Chanakya together 1/x = 1/3 (1/y + 1/z) 3/x = 1/y + 1/z----(1) Bittu alone takes 5 times as long as Abhay and Chanakya together 1/y = 1/5 (1/x + 1/z) 5/y = 1/x + 1/z----(2) All three together complete the work in 10 days 1/x + 1/y + 1/z----(3) (3) – (1) 4/x = 1/10 ⇒ x = 40 (3) – (2) 6/y = 1/10 ⇒ y = 60 Putting this value in (2) 5/60 = 1/40 + 1/z 1/z = 1/12 – 1/40 = 7/120 Z = 120/7 days |
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| 141. |
A tap can fill a tank in 10 hours. After filling half the tank, four more similar taps are opened. What is the total time taken to fill the tank completely?1). 5 hours 45 minutes2). 6 hours3). 5 hours 48 minutes4). 6 hours 45 minutes |
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Answer» Given, a tap can fill a TANK in 10 hours. ∴ In 1 hour, the tap fills 1/10th PART of the tank. To fill half the tank, time required = 5 hours. Now, part of tank EMPTY = ½ Given, 4 more similar taps are opened. ∴ In 1 hour, 5 taps fill part of tank = 5 × 1/10 = ½ ∴ to fill the half filled tank, the taps require 1 hr. Total time taken = 5 hr + 1 hr = 6 hr |
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| 142. |
Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time the tank will be filled?1). 7 hrs2). 8 hrs3). 7 hrs. 30 min.4). 8 hrs. 30 min. |
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Answer» Time taken by A , B and C to fill/empty the tank completely = 10 , 12 and 20 hours respectively Hence part of tank filled by A, B and EMPTIED by C in 1 hour = 1/10, 1/12 and 1/20 respectively Part of tank filled by all 3 pipes in 1 min $(= \FRAC{1}{{10}} + \frac{1}{{12}}\; - \frac{1}{{20}})$ $(= \frac{6}{{60}}\; + \frac{5}{{60}}-\frac{3}{{60}})$ $(= \frac{8}{{60}})$ = 2/15 Hence it will take 15/2 hours IE 7 hours and 30 mins to completely fill the tank |
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| 143. |
A pump can fill a tank with water in 3 hours. Because of a leak it took \(3\frac{1}{5}hours\) to fill the tank. The leak can drain all the water in the tank in1). 54 hours2). 36 hours3). 42 hours4). 48 hours |
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Answer» The pump can FILL the tank in 3 HOURS ∴ Part of tank filled by the pump in 1 hour = 1/3 Let, the leak can DRAIN all the WATER in ‘x’ hours ∴ Part of tank leaked by the drain in 1 hour = 1/x Given, Because of a leak pump took $(3\frac{1}{5}hours = \frac{{16}}{5}hours)$ to fill the tank $(\THEREFORE \frac{1}{3} \times \frac{{16}}{5} - \frac{1}{x} \times \frac{{16}}{5} = 1)$ ⇒ 16/15 - 16/5x = 1 ⇒ 16/5x = 16/15 – 1 = 1/15 ⇒ 5x = 16 × 15 ⇒ x = 48 hours |
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| 144. |
Sunil completes a work in 4 days, whereas Dinesh completes the work in 6 days. Ramesh works $1\frac{1}{1}$ times as fast as Sunil. The three together can complete the work in1). $1\frac{5}{12}$ days2). $1\frac{5}{7}$ days3). $1\frac{3}{8}$ days4). $1\frac{5}{19}$ days |
| Answer» HELLO, $1\frac{5}{19}$ DAYS is CORRECT | |
| 145. |
50 men can construct a road in 60 days. They started the work together. But at the end of every 15th day, 10 men left the work. What part of work was left incomplete at the end of allotted time?1). 1/42). 3/103). 7/104). 17/20 |
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Answer» ? 50 MEN can construct the road in60 days ∴ Part of road constructed by 1 man in 1 DAY $(= \frac{1}{{50 \times 60}} = \frac{1}{{3000}})$ Now, for the first 15 days, all 50 men worked. part of work done in first 15 days $(= 15 \times 50 \times \frac{1}{{3000}} = \frac{1}{4})$ After 15 days, 10 men left the work. ∴For next 15 days 40 men worked, part of work done in next 15 days $(= 15 \times 40 \times \frac{1}{{3000}} = \frac{1}{5})$ Again, 10 men left the work after these 15 days, (i.e. after 30 days in total). ∴ For next 15 days 30 men worked, part of work done in next 15 days $(= 15 \times 30 \times \frac{1}{{3000}} = \frac{3}{{20}})$ Now, after 45 days, 10 more men left the work. ∴ For the LAST 15 days, 20 men worked. part of work done in last 15 days $(= 15 \times 20 \times \frac{1}{{3000}} = \frac{1}{{10}})$ ∴ Part of work finished in all 60 days $(= \frac{1}{4} + \frac{1}{5} + \frac{3}{{20}} + \frac{1}{{10}} = \frac{{14}}{{20}} = \frac{7}{{10}})$ ⇒ Part of work left = 1 – (7/10) = 3/10 Hence 3/10thpart of work REMAINED incomplete. |
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| 146. |
X and Y can complete a work in 5 days. Y and Z can complete the same work in 6 days. Z and X can complete the same work in 15/2 days. In how many days will X, Y and Z together complete the work?1). 22). 43). 34). 2/5 |
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Answer» Time taken by X + Y to COMPLETE a work = 5 DAYS Time taken by Y + Z to complete a work = 6 days Time taken by Z + X to complete a work = 15/2 days Total work = LCM of (5, 6 and 15/2) = 30 units Efficiency of X + Y together = 30/5 = 6 units/day Efficiency of Y + Z together = 30/6 = 5 units/day Efficiency of Z + X together = 30/(15/2) = 4 units/day Efficiency of X + Y + Z together = 2(X + Y + Z) = 6 + 5 + 4 = 15 units/day X + Y + Z = 15/2 = 7.5 units/day ∴ Total time taken by X + Y + Z together to complete the work = 30/7.5 = 4 days |
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| 147. |
L alone can do a work in 100 hours. M alone can do the same in 150 hours. L, M and N together can do the same work in 30 hours. In how many hours can N alone do the work?1). 652). 603). 404). 90 |
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Answer» Given, M’s 1 hour’s work = 1/150 Let N can complete work in n days. N’s 1 hour’s work = 1/n Given, (L + M + N)’s 1 hour’s work = 1/30 ⇒ 1/30 = (1/100) + (1/150) + (1/n) ⇒ 1/30 = 1/60 + 1/n ⇒ 1/n = 1/30 - 1/60 ⇒ 1/n = 1/60 ∴ N can complete work in 60 days. |
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| 148. |
A and B can do a piece of work, in 12 days, B and C in 8 days and C and A in 6 days. How long would B take to do the same work alone ?1). 24 days2). 32 days3). 40 days4). 48 days |
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Answer» This question was ASKED some where in PREVIOUS YEAR papers of ssc, and correct answer was OPTION 4 |
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| 149. |
A tank has a leak which would empty the completely filled cistern in 20 hours. If the tank is full of water and a tap is opened which admits 2 liters of water per minute in the tank, the leak now takes 30 hours to empty the tank. How many liters of water does the tank hold?1). 24002). 45003). 12004). 7200 |
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Answer» Given data- Let the total water holding capacity of cistern be x litres A LEAK which would empty the completely FILLED cistern in 20 hours ⇒ Rate at which leak is taken PLACE= x/20 liter/hour→ 1 Water is admitted by a pump at the rate of 2 liters/ minute ⇒ Rate at which water is been admitted in cistern = 2× 60 ⇒ Rate at which water is been admitted in cistern= 120 liters/ hour→2 When both process of leakage and water admission takes place simultaneously Resultant rate of water leakage$( = \;\FRAC{x}{{20}} - 120)$→ 3 It takes 30 hours to empty the tanks with both actions takes place simultaneously ∴ after 30 hours of time water level in cistern will b 0 ⇒ Original water level- total water leakage after 30 hours = 0 $(\begin{ARRAY}{l} \Rightarrow \;x - \;30 \times \;\left( {\frac{x}{{20}} - 120} \right) = \;0\\ \Rightarrow \;x - \frac{{30\left( {{\rm{x}} - 2400} \right)}}{{20}} = 0 \end{array})$ ⇒ 2x = 3 (x – 2400) ⇒ 2x = 3x – 7200 ⇒ x = 7200 liters |
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If Kavita is 70% more efficient than Savita and 50% less efficient than Babita, Savita is what percent less efficient than Babita?1). 29%2). 71%3). 96%4). 120% |
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Answer» Kavita’s EFFICIENCY = (100 + 70)% of Savita’s efficiency = 1.7 × Savita’s efficiency Also, Kavita’s efficiency = (100 – 50)% of Babita’s efficiency = 0.5 × Babita’s efficiency Hence, ⇒ 1.7 × Savita’s efficiency = 0.5 × Babita’s efficiency ⇒ Savita’s efficiency = (0.5/1.7) × Babita’s efficiency ⇒ Savita’s efficiency = 0.29 × Babita’s efficiency ⇒ Savita’s efficiency = (1 - 0.71) × Babita’s efficiency ⇒ Savita’s efficiency = (100 – 71)% × Babita’s efficiency ∴ Savita is 71% LESS efficient than Babita. |
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