>> Aptitude >> Unitary Method

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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

1). If cost of 24 oranges is Rs. 72, then find out the cost of 120 oranges.
Because, Cost of 24 oranges = Rs. 72 Therefore, Cost of 1 orange = Rs. \( \Large \frac{72}{24} \) Therefore, Cost of 120 oranges = \( \Large \frac{72}{24} \times 120 = 3 \times 120 \) = Rs.360 | ||||

2). If cost of 15 eggs is Rs.75, then find out the cost of 4 dozens eggs.
Because, Cost of 15 eggs = Rs.75 | ||||

3). A worker makes a toy in every 2 h. If he works for 80 h, then how many toys will he make?
Let number of toys be x. More hours, More toys (Direct proportion) 360 : 60 :: 192 : x Therefore, x = \( \Large \frac{60 \times 192 }{360} \) = 32 bananas | ||||

4). 12 men can do a piece of work in 24 days. How many days are needed to complete the work, if 8 men are engaged in the same work?
12 men can do the work in 24 days. | ||||

5). If 45 m of a uniform rod weighs 171 kg, then what will be the weight of 12 m of the same rod?
Because, Weight of 45 m rod = 171 kg Therefore, Weight of 1 m rod = \( \Large \frac{171}{45} \) kg Therefore, Weight of 12 mm rod = \( \Large \frac{171}{45} \times 12 \) = 45.6 kg | ||||

6). 22 men can complete a job in 16 days. In how many days, will 32 men complete that job?
Because, 22 men do the work in 16 days. Therefore, 1 man will do the work in \( \Large 16 \times 22 \) days. Therefore, 32 men will do the job in = \( \Large \frac{16 \times 22 }{32} \) days Therefore, Required number of days = \( \Large \frac{16 \times 22 }{32} \) = 11 days | ||||

7). If 10 spiders can catch 10 flies in 10 min, then how many flies can 200 spiders catch in 200 min?
Let the required number of files be More spiders, More flies (Direct proportion) More time, More flies (Direct Proportion) Spiders 10 : 200 ) Time 10 : 200 ) :: 10 : x Therefore, \( \Large \left(10 \times 10 \times x\right) = \left(200 \times 200 \times 10\right) \) Therefore, x = \( \Large \frac{200 \times 200 \times 10}{10 \times 10} \) = 4000 files | ||||

8). 2000 soldiers in a fort had enough food for 20 days. But some soldiers were transferred to another fort and the food lasted for 25 days. How many soldiers were transferred?
Let the number of soldiers transferred be x. Now, the food would last for 25 days for \( \Large \left(2000 - x\right) \) soldiers. Less men, More days (Indirect proportion) 25 : 20 :: 2000 : 22000-x = 2000 - x = \( \Large \frac{2000 \times 20}{25} \) = 2000 - x = 1600 Therefore, x = 2000 - 1600 = 400 | ||||

9). If in a hostel, food is available for 45 days for 50 students. For how many days will this food be sufficient for 75 students?
Because, For 50 students, food is sufficient for 45 days. Therefore, For 1 student, food is sufficient for \( \Large 45 \times 50 \) days. Therefore, For 75 students, food is sufficient for \( \Large \frac{45 \times 50}{75} \), i.e..for 30 days. | ||||

10). In a garrison, there was food for 1000 soldiers for one month. After 10 days 1000 more soldiers joined the garrison. How long would the soldiers be able to carry on with the remaining food?
Let us assume that each soldier eats one unit of food per day. Thus, total units of food at the beginning will be \( \Large 1000 \times 30 \) = 30000. After 10 days 1000 soldiers would have eaten \( \Large 1000 \times 10 \) = 10000 units of food. Thus, food left after 10 days equals 20000 units. Now, there are total of 2000 soldiers who eat one unit of food every day. So, the number of days that 20000 units of food will serve 2000 soldiers is \( \Large \frac{20000}{2000} \) = 10. |