>> Aptitude >> Unitary Method

PrimeFaces.cw("DataGrid","widget_frm_base_j_idt45",{id:"frm_base:j_idt45"});

Contents:

- Aptitude
- Approximation
- Average
- Boat and Stream
- Compound interest
- Discount
- Linear Equations
- Mensuration
- Mixture and Allegation
- Number series
- Number System
- Partnership
- Percentage
- Permutation and combination
- Pipes and Cisterns
- Probability
- Problem on ages
- Profit and Loss
- Ratio and Proportions
- Simple and compound interest
- Time and Distance
- Time and work
- Trains
- Unitary Method
- Word problems
- Work and Wages

21). If 30 men working 9 h per day can reap a field in 16 days, in how many days, will 36 men reap the field working 8 h per day?
Let the required number of days be x. | ||||

22). In a race, Ravi covers 5 km in 20 min. How much distance will he cover in 100 min?
Because, Distance covered in 20 min = 5 km Therefore, Distance covered in 1 min = \( \Large \frac{5}{20} \) km Therefore, Distance covered in 100 min = \( \Large \frac{5}{20} \times 100 = 5 \times 5 \) = 25 km | ||||

23). A garrison of 1000 men had provisions for 48 days. However, a reinforcement of 600 men arrived. How long will now food last for?
Because, For 1000 men, provision lasts for 48 days. Therefore, For 1 man, provision lasts for \( \Large \left(48 \times 1000\right) \) days. For \( \Large \left(1000 + 600\right) \) men, provision will last for \( \Large \frac{48 \times 1000}{1600} \) days Therefore, Required number of days = \( \Large \frac{48 \times 1000}{1600} = 3 \times 10 \) = 30 days | ||||

24). A tree is 12 m tall and casts an 8 m long shadow. At the same time, a flag pole casts a 100 m long shadow. How long is the flag pole?
Because, 8 m shadow means original height = 12 m. Therefore, 1 m shadow means original height = \( \Large \frac{12}{8} \) m Therefore, 100 m shadow means original height = \( \Large \frac{12}{8 \times 100m}=\frac{6}{4} \times 100 \) = \( \Large 6 \times 25 \) = 150 m | ||||

25). If 12 persons working 16 h per day earn Rs.33600 per week, then how much will 18 persons earn working 12 h per day?
Let the required earnings be Rs.x. More persons, More earnings (Direct proportion) Less hours per day, Less earnings (Direct proportion) Persons 36 : 30 :: 33600 : x Hours per day 8 : 9 Therefore, \( \Large \left(12 \times 16 \times x\right)=\left(18 \times 12 \times 33600\right) \) Therefore, \( \Large x = \frac{18 \times 12 \times 33600}{12 \times 16} = 18 \times 2100 \) = Rs.37800. | ||||

26). A man and a boy working together can complete a work in 24 days. If for the last 6 days, the man alone does the work, then it is completed in 26 days. How long will the boy take to complete the work alone?
Let man's 1 day's work = \( \Large \frac{1}{m} \) and boy's 1 day's work = \( \Large \frac{1}{n} \) 1 day's work man and boy = \( \Large \frac{1}{24} \) Man's 6 day's work = \( \Large \frac{6}{m} \) Now, for 20 days, both man and boy do the work and for last 6 days, only man does the work. According to the question, \( \Large \frac{1}{m}+\frac{1}{n}=\frac{1}{24} \) => \( \Large 20 \left(\frac{1}{m}+\frac{1}{n}\right)+\frac{6}{m}=1 \) => \( \Large \left(20 \times \frac{1}{24}\right)+\frac{6}{m}=1 \) \( \Large \frac{6}{m}= \left(1-\frac{20}{24}\right)=\frac{4}{24}=\frac{1}{6} \) \( \Large \frac{1}{m}=\frac{1}{36} \) Now from eq. (i) \( \Large \frac{1}{m}+\frac{1}{n}=\frac{1}{24} \) \( \Large \frac{1}{36}+\frac{1}{n}=\frac{1}{24} \) => \( \Large \frac{1}{n} = \left(\frac{1}{24 }- \frac{1}{36}\right)=\frac{1}{72} \) Hence, the boy alone can do the work in 72 days. | ||||

27). 25 men can reap a field in 20 days. When should 15 men leave the work, if the whole field is to be reaped in \( \Large 37\frac{1}{2} \) days after they leave the work?
Let 15 men work for m days. Work done in 1 days = \( \Large \frac{m}{20} \) Remaining work = \( \Large \left(1 - \frac{m}{20}\right) \) 25 men's 1 days work = \( \Large \frac{1}{20} \) 1 man's 1 days work = \( \Large \frac{1}{20} \times \frac{1}{25}=\frac{1}{500} \) 10 men's 1 days work= \( \Large \frac{1}{500} \times 10 = \frac{1}{50} \) 10 men's \( \Large \frac{75}{2} \) days work =\( \Large \frac{1}{50} \times \frac{75}{2}=\frac{75}{100}=\frac{3}{4} \) Therefore, \( \Large \left(1 - \frac{m}{20}\right)=\frac{3}{4} => \frac{m}{20}=\frac{1}{4} \) => \( \Large m = \frac{1}{4} \times 20 \) Clearly, 15 men leave after 5 days. | ||||

28). If 12 engines consume 30 metric tonnes of coal when each is running 18h per day, how much coal will be required for 16 engines, each running 24 h per day, it being given that 6 engines of former type consume as much as 8 engines of latter type?
Let the required quantity of Coal consumed be x. More engines, More coal consumption (Direct proportion) More hours, More coal consumption (Direct proportion) Less rate of consumption, Less coal consumption (Direct proportion) Engines 12 : 16 ) ) :: 30 : x Working hours 18 : 24 ) Rate of consumption \( \frac{1}{6} \) : \( \frac{1}{8} \) ) Therefore, => \( \Large m = \frac{1}{4} \times 20 \) \( \Large 12 \times 18 \times \frac{1}{6} \times x = 16 \times 24 \times \frac{1}{8} \times 30 \) => 36x = 1440 Therefore, x = \( \frac{1440}{36} \) = 40 Hence, quantity of coal consumed will be 40 tonnes. | ||||

29). 8 men can complete a work in 12 days, 4 women can complete it in 48 days and 10 children can complete the same work in 24 days. In how many days can 10 men, 4 women and 10 children complete the same work?
1 man can finish the work in \( \Large 8 \times 12 \) = 96 days 1 woman can finish the work in \( \Large 4 \times 48 \) = 1 92 days 1 child can finish the work in \( \Large 10 \times 24 \) = 240 days 1 man's 1 day's work = \( \Large \frac{1}{96} \) 1 woman's 1 day's work = \( \Large \frac{1}{192} \) 1 child's 1 day's work = \( \Large \frac{1}{240} \) (10 men + 4 women + 10 children)'s 1 day's work = \( \Large \left(\frac{10}{96}+\frac{4}{192}+\frac{10}{240}\right) \) = \( \Large \left(\frac{5}{48}+\frac{1}{48}+\frac{1}{24}\right) \) =\( \Large \left(\frac{5+1+2}{48}\right) \) = \( \Large \frac{8}{48} = \frac{1}{6} \) Hence, they will finish the work in 6 days. | ||||

30). 20 men complete one-third of a work in 20 days. How many more men should be employed to finish the rest of the work in 25 more days?
Work done = \( \Large \frac{1}{3} \) Remaining work = \( \Large \left(1-\frac{1}{3}\right)= \left(\frac{3-1}{3}\right)=\frac{2}{3} \) Let the number of more men to be employed be x. More work, More men (Direct proportion) More days, Less men (Indirect proportion) Work \( \Large \frac{1}{3} \) : \( \Large \frac{2}{3} \) :: 20 : (20+x) Days 25 : 20 Therefore, \( \Large \frac{1}{3} \times 25 \times \left(20+x\right) = \frac{2}{3} \times 20 \times 20 \) => \( \Large \left(20+x\right) = \frac{800}{25} = 32 \) Therefore, x = 32 - 20 = 12 |