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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Refer to the table and answer the given questions.
NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.
Rahul purchased a mobile and sold it for a loss (loss % given in the table). From that money, he purchased another article and sold it for a gain of (Profit % given in the table). What is the overall gain or loss?
Correct
Solution: CP = 8400, Therefore SP after 5% loss = 8400 * 95/100 =7980 CP for another article = 7980, Therefore SP after 5% profit = 7980 * 105/100 = 8379 Difference = 8400 – 8379 =Rs 21 loss.
Incorrect
Solution: CP = 8400, Therefore SP after 5% loss = 8400 * 95/100 =7980 CP for another article = 7980, Therefore SP after 5% profit = 7980 * 105/100 = 8379 Difference = 8400 – 8379 =Rs 21 loss.

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Refer to the table and answer the given questions.
NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.
A, B and C invests rupees 8000, 12000 and 10000 respectively in a business. At the end of the year the balance sheet shows a loss of initial investment. Find the share of loss of B.
Correct
Solution: Total loss after one year = 30000*40/100 = 12000, Ratio of shares of A, B and C =4:6:5, Therefore share of B = (6/15)*12000 = 4800
Incorrect
Solution: Total loss after one year = 30000*40/100 = 12000, Ratio of shares of A, B and C =4:6:5, Therefore share of B = (6/15)*12000 = 4800

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Refer to the table and answer the given questions.
NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.
A Shop Keeper sells two bags for Rs. 500 each. On one, he gets % Profit (as per the table) and on the other he gets % loss (as per the table). His profit or loss in the entire transaction was?
Correct
Solution: Net effect = 1414 (14*14/100) = – 49/25%
Incorrect
Solution: Net effect = 1414 (14*14/100) = – 49/25%

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Refer to the table and answer the given questions.
NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.
The price of TV was marked up by % (as per the table). It was sold at a discount of 20% on the marked price. What was the profit percent of the cost price?
Correct
Solution: Without putting the actual values this sum can be solved as: MP= CP + 40% of CP, Let CP = 100, MP = 140, Discount is 20% on MP, therefore SP = 140 20% 0f 140 = 140 28 = 112, therefore profit% = (SP CP/CP)/100 = (112100)/100 =12%
Incorrect
Solution: Without putting the actual values this sum can be solved as: MP= CP + 40% of CP, Let CP = 100, MP = 140, Discount is 20% on MP, therefore SP = 140 20% 0f 140 = 140 28 = 112, therefore profit% = (SP CP/CP)/100 = (112100)/100 =12%

Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Refer to the table and answer the given questions.
NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.
Preethi sold a machine to Shalini at a profit of % (as per table). Shalini sold this machine to Arun at a loss of % (as per table). If Preethi paid Rs.5000 for this machine, then find the cost price of machine for Arun?
Correct
Solution: Profit % = 30% Loss % = 20%, CP for Shalini = 5000 + 1500 = 6500. CP for Arun = 6500= 1300= Rs.5200 or simply CP for Arun= 5000 * 130/100 * 80/100 = Rs. 5200
Incorrect
Solution: Profit % = 30% Loss % = 20%, CP for Shalini = 5000 + 1500 = 6500. CP for Arun = 6500= 1300= Rs.5200 or simply CP for Arun= 5000 * 130/100 * 80/100 = Rs. 5200

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeAjay borrows Rs. 1500 from two moneylenders. He pays interest at the rate of 12% per annum for one loan and at the rate of 14% per annum for the other. The total interest he pays for the entire year is Rs. 186. How much does he borrow at the rate of 12%?
Correct
Solution: The average rate of interest he pays is 186*100/1500 = 12.4 %. The average rate of interest being 12.4%, it means that the ratio in which the two amounts would be distributed would be (1412.4)/ (12.412) = 1.6 : 0.4 = 4:1, Thus borrowing at 12% would be Rs. 1200
Incorrect
Solution: The average rate of interest he pays is 186*100/1500 = 12.4 %. The average rate of interest being 12.4%, it means that the ratio in which the two amounts would be distributed would be (1412.4)/ (12.412) = 1.6 : 0.4 = 4:1, Thus borrowing at 12% would be Rs. 1200

Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe odds against an event is 5:3 and the odds in favour of another independent event is 7 : 5. Find the probability that at least one of the two events occur.
Correct
Solution: P(E1) = 3/8, P(E2) =7/12. Event definition is E1 occurs and E2 does not occur or E1 occurs and E2 occurs or E1 does not occur and E2 occurs. i.e. (3/8)*(5/12)+ (3/8)*(7/12)+(5/8)*(7/12)= 71/96
Incorrect
Solution: P(E1) = 3/8, P(E2) =7/12. Event definition is E1 occurs and E2 does not occur or E1 occurs and E2 occurs or E1 does not occur and E2 occurs. i.e. (3/8)*(5/12)+ (3/8)*(7/12)+(5/8)*(7/12)= 71/96

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections for Question 810 : 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?
What is the distance between city P and city Q?
Statement I: B reached P one hour earlier than A to Q.
Statement II: The difference between speeds of A and B is 30 kmph.
Statement III: Two persons A and B started simultaneously from P to Q, with their speeds in the ratio 4 : 5.Correct
Solution: Speeds of A and B is in ratio 4:5, hence time taken will be in ratio 5:4, since difference between their time taken is 1, hence time taken is 5 hr and 4 hr respectively. Now diff in speed for A and B is 30 kmph hence speed is 120 kmph and 150 respectively. Total distance covered = distance between P and Q = distance covered by A = 5*120 = 600 km. Hence, all the statements are required.
Incorrect
Solution: Speeds of A and B is in ratio 4:5, hence time taken will be in ratio 5:4, since difference between their time taken is 1, hence time taken is 5 hr and 4 hr respectively. Now diff in speed for A and B is 30 kmph hence speed is 120 kmph and 150 respectively. Total distance covered = distance between P and Q = distance covered by A = 5*120 = 600 km. Hence, all the statements are required.

Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections for Question 810 : 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?
What is the area of the rectangle?
Statement I: The perimeter of the rectangle is 188 cm.
Statement II: The length of diagonal of the rectangle is 74 cm
Statement III: The ratio of length to breadth of the rectangle is 35 : 12.Correct
Solution: Since perimeter=188 cm, L+B = 94 cm—(I), L2 +B2 = 74*74 –(II), L:B = 35:12. Hence using any 2, the values of x and y can be found out.
Incorrect
Solution: Since perimeter=188 cm, L+B = 94 cm—(I), L2 +B2 = 74*74 –(II), L:B = 35:12. Hence using any 2, the values of x and y can be found out.

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections for Question 810 : 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?
What is the two digit number?
Statement I: The number obtained by interchanging the digits of the number is greater than the original number by 10.
Statement II: Difference between the two digits of the number is 2.
Statement III: Sum of the two digits of the number is 15.Correct
Solution: Let the number be 10x + y.
From I: (10y + x) – (10x + y) = 10
9x9y = 10 ——(1)
From II: y – x = 2 ——(2)
From III: x + y = 15 —–(3). It seems that the equation can be solved by 1 and 2 but both cancels hence answer cannot be determined. Hence III and either I or IIIncorrect
Solution: Let the number be 10x + y.
From I: (10y + x) – (10x + y) = 10
9x9y = 10 ——(1)
From II: y – x = 2 ——(2)
From III: x + y = 15 —–(3). It seems that the equation can be solved by 1 and 2 but both cancels hence answer cannot be determined. Hence III and either I or II
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