Hello Aspirants,
Welcome to Online Quants Test in AffairsCloud.com. We are starting Cumulus Course for IBPS PO 2017 Exam and we are creating sample questions in Quants section, type of question will be asked in IBPS PO Main!!!
Click Here to View Cumulus Course : IBPS PO 2017 Main
IBPS PO Main 2017: Reasoning Test– 7 PM Every Day
IBPS PO Main 2017: Quants Test– 8 PM Every Day
IBPS PO Main 2017: English Test– 9 PM Every Day
IBPS PO Main 2017: GA Test– 10 PM Every Day
Help: Share Our IBPS PO 2017 Course page to your Friends & FB Groups
_______________________
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
All the Best
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Quantitative Aptitude 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (15): Study the following charts and answer the following questions:
Between 1995 and 2000, in which year has the average employment per factory shown an increment compared to previous year, but decreased in the next year?
Correct
Solution: In 1999, average employment per factory is grater than in 1998 and less than in 2000
Incorrect
Solution: In 1999, average employment per factory is grater than in 1998 and less than in 2000

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (15): Study the following charts and answer the following questions:
If in 1996, 20100 factories had 659 employments on an average, the remaining factories had an average employment of:
Correct
Solution: In 1996, for 20100 factories, employees = 13,245,900; average=606. Total employees = 18,361,350. So average for remaining factories = 509
Incorrect
Solution: In 1996, for 20100 factories, employees = 13,245,900; average=606. Total employees = 18,361,350. So average for remaining factories = 509

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (15): Study the following charts and answer the following questions:
The total employment in 1999 is how many times to that in 1996?
Correct
Solution: Total employment in 1999 = 36240*736, Total employment in 1996 + 609*30150. Ratio= 36240*736/(30150*609) = 1.45
Incorrect
Solution: Total employment in 1999 = 36240*736, Total employment in 1996 + 609*30150. Ratio= 36240*736/(30150*609) = 1.45

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (15): Study the following charts and answer the following questions:
In which of the following fiveyear period, the number of factories has shown maximum increase?
Correct
Solution: From the figure it can be seen that the maximum increase in the number of factories is in 19952000.
Incorrect
Solution: From the figure it can be seen that the maximum increase in the number of factories is in 19952000.

Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (15): Study the following charts and answer the following questions:
Among the given period, in how many years has the trend of the average employment per factory not changed with respect to the previous year’s change?
Correct
Solution: We do not have year wise data in between 1975 to 1980 & similarly between 19801985, etc. Hence, this can’t be determined from the given data.
Incorrect
Solution: We do not have year wise data in between 1975 to 1980 & similarly between 19801985, etc. Hence, this can’t be determined from the given data.

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (67): Study the following charts and answer the questions that follow:
A, B and C enter into a partnership. They invest money 3 times in a year (12 months) in equal intervals. They start by investing money in ratio 4 : 2 : 3 for the first interval. For second interval, they invest money in ratio 3 : 4 : 3. And for the third interval, A invests double his previous investment, and B and C invest Rs 200 more than their respective previous investments. At the end of second interval, C’s total investment was Rs 200 less than that of A that time.
If the ratio of total profit to B’s share in profit after a year is 34 : 11, find the total investment made by C?
Correct
Solution: 3 equal intervals in 12 months = 4 months each
A invests – 4x, 3y and 6y, B invests – 2x, 4y and (4y+200), C invests – 3x, 3y and (3y+200), Now given that 3x + 3y = 4x + 3y – 200,Solve, x = 200
Ratio of profits of A : B : C is4x*4 + 3y*4 + 6y*4 : 2x*4 + 4y*4 + (4y+200)*4 : 3x*4 + 3y*4 + (3y+200)*4 = (4x + 9y) : (2x + 8y + 200) : (3x + 6y + 200)
Put, x = 200, i.e. (200+9y) : (600+8y) : (800+6y)
Now: (600+8y)/(2200+23y) = 11/34
Solve, y = 200
So total investment of C = 3x +3y + (3y+200) = Rs 2000Incorrect
Solution: 3 equal intervals in 12 months = 4 months each
A invests – 4x, 3y and 6y, B invests – 2x, 4y and (4y+200), C invests – 3x, 3y and (3y+200), Now given that 3x + 3y = 4x + 3y – 200,Solve, x = 200
Ratio of profits of A : B : C is4x*4 + 3y*4 + 6y*4 : 2x*4 + 4y*4 + (4y+200)*4 : 3x*4 + 3y*4 + (3y+200)*4 = (4x + 9y) : (2x + 8y + 200) : (3x + 6y + 200)
Put, x = 200, i.e. (200+9y) : (600+8y) : (800+6y)
Now: (600+8y)/(2200+23y) = 11/34
Solve, y = 200
So total investment of C = 3x +3y + (3y+200) = Rs 2000 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirection for questions (67): Study the following charts and answer the questions that follow:
A, B and C enter into a partnership. They invest money 3 times in a year (12 months) in equal intervals. They start by investing money in ratio 4 : 2 : 3 for the first interval. For second interval, they invest money in ratio 3 : 4 : 3. And for the third interval, A invests double his previous investment, and B and C invest Rs 200 more than their respective previous investments. At the end of second interval, C’s total investment was Rs 200 less than that of A that time.
After 8 months, had all invested double the amount than their respective previous investment what would be the total profit at the end of year? Given that difference between the shares of profit of B and C from total profit is Rs 1300 and total of shares of profits of A and C is Rs 16250
Correct
Solution:
Ratio of profits of A : B : C is
4x*4 + 3y*4 + 6y*4 : 2x*4 + 4y*4 + 8y*4 : 3x*4 + 3y*4 + 6y*4, x = 200
So ratio becomes
(800+9y) : (400+12y) : (600+9y)
Now, (400+12y6009y)/(1800+30y) * total profit = 1300
and (800+9y+600+9y)/(1800+30y) * total profit = 16250
Divide both equations and solve, y = 200. So now ratio becomes
(800+9y) : (400+12y) : (600+9y) [put y = 200]
13 : 14 : 12and 2/39 * total profit = 1300
Solve, total profit = Rs 25350Incorrect
Solution:
Ratio of profits of A : B : C is
4x*4 + 3y*4 + 6y*4 : 2x*4 + 4y*4 + 8y*4 : 3x*4 + 3y*4 + 6y*4, x = 200
So ratio becomes
(800+9y) : (400+12y) : (600+9y)
Now, (400+12y6009y)/(1800+30y) * total profit = 1300
and (800+9y+600+9y)/(1800+30y) * total profit = 16250
Divide both equations and solve, y = 200. So now ratio becomes
(800+9y) : (400+12y) : (600+9y) [put y = 200]
13 : 14 : 12and 2/39 * total profit = 1300
Solve, total profit = Rs 25350 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 810: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
Quantity II: Two trains, each 100100 metre long are moving in opposite directions. They cross each other in 88 seconds. If one is moving twice as fast the other, the speed of the faster train isCorrect
Solution: Quantity I:Let length and speed of the train be x metre and y kmph, i.e x/8.4=(y−4.5)×5/18 ⋯(1) and x/8.5=(y−5.4)×5/18 ⋯(2). Dividing (1) by (2) gives
8.5/8.4 = y−4.5/y−5.4 ⇒8.4y−8.4×4.5 =8.5y−8.5×5.4 ⇒0.1y=8.5×5.4−8.4×4.5 ⇒0.1y=45.9−37.8 = 8.1⇒y =81 km/hr
Quantity II: Total distance covered = (100+100)= 200 metre, Time =8 seconds
Let speed of slower train =v m/s, Then the speed of the faster train =2v m/s (since one is moving twice as fast the other), Relative speed =(v+2v)=3v m/s
3v=200/8 ⇒v= 25/3. Speed of the faster train =2v=50/3 m/s =2v=50/3 m/s
=50/3×18/5 km/hr= 60 km/hrIncorrect
Solution: Quantity I:Let length and speed of the train be x metre and y kmph, i.e x/8.4=(y−4.5)×5/18 ⋯(1) and x/8.5=(y−5.4)×5/18 ⋯(2). Dividing (1) by (2) gives
8.5/8.4 = y−4.5/y−5.4 ⇒8.4y−8.4×4.5 =8.5y−8.5×5.4 ⇒0.1y=8.5×5.4−8.4×4.5 ⇒0.1y=45.9−37.8 = 8.1⇒y =81 km/hr
Quantity II: Total distance covered = (100+100)= 200 metre, Time =8 seconds
Let speed of slower train =v m/s, Then the speed of the faster train =2v m/s (since one is moving twice as fast the other), Relative speed =(v+2v)=3v m/s
3v=200/8 ⇒v= 25/3. Speed of the faster train =2v=50/3 m/s =2v=50/3 m/s
=50/3×18/5 km/hr= 60 km/hr 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 810: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine did the cask originally hold?
Quantity II: A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?Correct
Solution: Quantity I: Let initial quantity of wine =x litre. After a total of 4 operations, quantity of wine =x(1−y/x)^ n = x(1−8/x)^4. Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65
⇒x(1−8/x)^4/x= 16/81⇒(1−8/x)^4= (2/3)^4 ⇒(1−8/x)= 2/3 ⇒((x−8)/x)= 2/3 ⇒3x−24= 2x ⇒x=24 litres.
Quantity II: milk contained by the container now
=40(1−4/40)^3= 40(1−1/10)^3= 40×9/10×9/10×9/10=4×9×9×9/100=29.16 litreIncorrect
Solution: Quantity I: Let initial quantity of wine =x litre. After a total of 4 operations, quantity of wine =x(1−y/x)^ n = x(1−8/x)^4. Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65
⇒x(1−8/x)^4/x= 16/81⇒(1−8/x)^4= (2/3)^4 ⇒(1−8/x)= 2/3 ⇒((x−8)/x)= 2/3 ⇒3x−24= 2x ⇒x=24 litres.
Quantity II: milk contained by the container now
=40(1−4/40)^3= 40(1−1/10)^3= 40×9/10×9/10×9/10=4×9×9×9/100=29.16 litre 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 810: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
Quantity II: The area of a parallelogram is 72 cm2 and its altitude is twice the corresponding base. What is the length of the base?Correct
Solution: Let breadth =x cm, Then, length =2x cm
Area =x×2x=2x^2 sq.cm.. Therefore, New length =(2x−5) cm, New breadth =(x+5)cm
New area =(2x−5)(x+5) sq.cm. Given that, new area = initial area +75 sq.cm.
⇒(2x−5)(x+5)=2x^2+75 ⇒2x^2+10x−5x−25=2x^2+75 ⇒5x−25=75 ⇒5x=75+25=100 ⇒x=100/5=20 cm , Length =2x=2×20=40=2x=2×20= 40 cm.
Quantity II: Area of a parallelogram, A =bh, where b is the base and h is the height of the parallelogram. Let base = x cm.
Then, height =2x cm (∵ altitude is twice the base). Since, Area =x×2x=2x^2. Since,
Area is given as 72 cm2 ⇒2x^2=72 ⇒x^2=36 ⇒x= 6 cmIncorrect
Solution: Let breadth =x cm, Then, length =2x cm
Area =x×2x=2x^2 sq.cm.. Therefore, New length =(2x−5) cm, New breadth =(x+5)cm
New area =(2x−5)(x+5) sq.cm. Given that, new area = initial area +75 sq.cm.
⇒(2x−5)(x+5)=2x^2+75 ⇒2x^2+10x−5x−25=2x^2+75 ⇒5x−25=75 ⇒5x=75+25=100 ⇒x=100/5=20 cm , Length =2x=2×20=40=2x=2×20= 40 cm.
Quantity II: Area of a parallelogram, A =bh, where b is the base and h is the height of the parallelogram. Let base = x cm.
Then, height =2x cm (∵ altitude is twice the base). Since, Area =x×2x=2x^2. Since,
Area is given as 72 cm2 ⇒2x^2=72 ⇒x^2=36 ⇒x= 6 cm
_______________________
 Click view Questions button to view Explanation
 Note: We are providing unique questions for you to practice well, have a try !!
 Ask your doubt in comment section, AC Mod’s ll clear your doubts in caring way.
 Share your Marks in Comment Sections