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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 14: refer to the graphs given below and answer the questions that follow:
Note: Productivity of wheat in India = Total wheat production / Area under wheat production.
Figure 1: Production and Export
The population of India in 2003 2004 was:
Correct
Solution: For the period of 200304 quantity of wheat (in million kg) which was not exported was 645 215 = 430 million kg. Hence total population = Wheat not exported/ Per capita consumption = 430/0.544 = 790 million
Incorrect
Solution: For the period of 200304 quantity of wheat (in million kg) which was not exported was 645 215 = 430 million kg. Hence total population = Wheat not exported/ Per capita consumption = 430/0.544 = 790 million

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 14: refer to the graphs given below and answer the questions that follow:
Note: Productivity of wheat in India = Total wheat production / Area under wheat production.
Figure 1: Production and Export
Say the area under production was reduced by 10% in 200405 as compared to that of 200304 then the rate of increase/decrease in the productivity of wheat in 200405 was:
Correct
Solution: Say the area under the productivity in the year 20032004 is ‘z’ hectare. Productivity in 2002004 = 645/z, Productivity in 20042005 = 660/0.9z = 733/z (approx). Hence, rate of increase in productivity in 20042005 = ((733/z) – (645/z))/(645/z)*100 = 88/645*100 = 13.6%
Incorrect
Solution: Say the area under the productivity in the year 20032004 is ‘z’ hectare. Productivity in 2002004 = 645/z, Productivity in 20042005 = 660/0.9z = 733/z (approx). Hence, rate of increase in productivity in 20042005 = ((733/z) – (645/z))/(645/z)*100 = 88/645*100 = 13.6%

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 14: refer to the graphs given below and answer the questions that follow:
Note: Productivity of wheat in India = Total wheat production / Area under wheat production.
Figure 1: Production and Export
The ratio of the average wheat that is exported to that of the wheat produced from 20002001 to 20042005 is:
Correct
Solution: The average wheat exported is (207+189+209+215+220)/5 = 208 million kg. The average wheat production is (421+561+581+645+660)/5 = 574.8 million kg. Hence the required ratio = 208/574.8 = 0.36
Incorrect
Solution: The average wheat exported is (207+189+209+215+220)/5 = 208 million kg. The average wheat production is (421+561+581+645+660)/5 = 574.8 million kg. Hence the required ratio = 208/574.8 = 0.36

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 14: refer to the graphs given below and answer the questions that follow:
Note: Productivity of wheat in India = Total wheat production / Area under wheat production.
Figure 1: Production and Export
The average of per capita availability of wheat in the given period was:
Correct
Solution: Average of per capita availability = (487 +464 + 510 + 544 + 566)/ 5 = 514.2g
Incorrect
Solution: Average of per capita availability = (487 +464 + 510 + 544 + 566)/ 5 = 514.2g

Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for Question 56: The questions consist of two statements numbered “I and II” given below it. You have to decide whether the data provided in the statement are sufficient to answer the questions?
An aquarium shaped cuboid provides 60 cm2 of water surface area per fish. How many fishes are there in aquarium?
Statement I: The dimensions of an aquarium are 50, 60, 70 cms.
Statement II: The aquarium is filled to a depth of 45 cms.Correct
Solution: Both the statements do not give us any information about the number of fishes, had there been given total volume occupied by fishes and volume per fish, or about density then it would be of any help. Hence we cannot find the no. of fishes by both the statements.
Incorrect
Solution: Both the statements do not give us any information about the number of fishes, had there been given total volume occupied by fishes and volume per fish, or about density then it would be of any help. Hence we cannot find the no. of fishes by both the statements.

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections for Question 56: The questions consist of two statements numbered “I and II” given below it. You have to decide whether the data provided in the statement are sufficient to answer the questions?
the dimensions of a room are 20, 30, 40 feet. What is the floor area of the room?
Statement I: Length of the room > 30
Statement II: Total area of four walls is 3600 square feet.Correct
Solution: Statement II is alone sufficient to answer the question as we get the area of walls must be 30*40*2 + 20* 30*2 = 3600 (by hit and trial method). Hence, we would get the floor area as 20*40= 800 square feet. Hence statement II is sufficient.
Incorrect
Solution: Statement II is alone sufficient to answer the question as we get the area of walls must be 30*40*2 + 20* 30*2 = 3600 (by hit and trial method). Hence, we would get the floor area as 20*40= 800 square feet. Hence statement II is sufficient.

Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 710: Read the following and answer the following questions:
A set of 10 pipes (set X) can fill 70% of a tank in 7 minutes. Another set of 5 pipes (Set Y) fills 3/8 of the tank in 3 minutes, a third set of 8 pipes (set Z) can empty 5/10 of the tank in 10 minutes.
How many minutes will it take to fill the tank if all the 23 pipes are opened at the same time?
Correct
Solution: First set of 10 pipes operate at 10%/min (filling) i.e. Filling done by 1 pipe = 1%/ minute. Second set of five pipes operate at 12.5%/ min (filling) i.e. Filling done by 1 pipe = 2.5%/ min. Similarly, Set Z (Emptying) = 5 %/ minute, emptying per pipe =0.625% per minute. If all the 23 pipes are opened the per minute rate will be: 10 + 12.5 – 5 = 17.5 %. Therefore time taken = 100/17.5 = 40/7= 5 5/7 minutes
Incorrect
Solution: First set of 10 pipes operate at 10%/min (filling) i.e. Filling done by 1 pipe = 1%/ minute. Second set of five pipes operate at 12.5%/ min (filling) i.e. Filling done by 1 pipe = 2.5%/ min. Similarly, Set Z (Emptying) = 5 %/ minute, emptying per pipe =0.625% per minute. If all the 23 pipes are opened the per minute rate will be: 10 + 12.5 – 5 = 17.5 %. Therefore time taken = 100/17.5 = 40/7= 5 5/7 minutes

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 710: Read the following and answer the following questions:
A set of 10 pipes (set X) can fill 70% of a tank in 7 minutes. Another set of 5 pipes (Set Y) fills 3/8 of the tank in 3 minutes, a third set of 8 pipes (set Z) can empty 5/10 of the tank in 10 minutes.
If 4 pipes are closed in set Z, and all others remain open, how long will it take to fill the tank?
Correct
Solution: If 4 of the taps of set Z are closed, the net work done by Set Z would be 2.5% while the work done by Sets X and Y would remain 10% and 12.5% respectively. Thus, the total work per minute would be 20% and hence the tank would take 5 minutes to fill up.
Incorrect
Solution: If 4 of the taps of set Z are closed, the net work done by Set Z would be 2.5% while the work done by Sets X and Y would remain 10% and 12.5% respectively. Thus, the total work per minute would be 20% and hence the tank would take 5 minutes to fill up.

Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 710: Read the following and answer the following questions:
A set of 10 pipes (set X) can fill 70% of a tank in 7 minutes. Another set of 5 pipes (Set Y) fills 3/8 of the tank in 3 minutes, a third set of 8 pipes (set Z) can empty 5/10 of the tank in 10 minutes.
If the tank is half full and set X and Set Y are closed, how many minutes will it take for set Z to empty the tank if alternate taps of set Z are closed.
Correct
Solution: Again if we close 4 taps of set Z, the rate of emptying would be 2.5% per minute. A half filled tank would contain 50% f the capacity and hence would take 50/2.5 = 20 minutes to empty.
Incorrect
Solution: Again if we close 4 taps of set Z, the rate of emptying would be 2.5% per minute. A half filled tank would contain 50% f the capacity and hence would take 50/2.5 = 20 minutes to empty.

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 710: Read the following and answer the following questions:
A set of 10 pipes (set X) can fill 70% of a tank in 7 minutes. Another set of 5 pipes (Set Y) fills 3/8 of the tank in 3 minutes, a third set of 8 pipes (set Z) can empty 5/10 of the tank in 10 minutes.
If one pipe is added for set X and set Y and set Z’s capacity is increased by 20% on its original value and all the taps are opened at 2.58 p.m. then at what time does the tank get filled? (If it is initially empty.)
Correct
Solution: The rate per minute with the given changes (in percent terms) would be: Set X = 11%, set Y= 15% and Set Z= 6% Hence, net rate = 11+156 = 20% per minute and it would take 5 minutes for the tank to fill. If all the pipes are opened at 2.58 p.m., the tank would get filled at 3:03.
Incorrect
Solution: The rate per minute with the given changes (in percent terms) would be: Set X = 11%, set Y= 15% and Set Z= 6% Hence, net rate = 11+156 = 20% per minute and it would take 5 minutes for the tank to fill. If all the pipes are opened at 2.58 p.m., the tank would get filled at 3:03.
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