We are providing you with Quant Section Mock for the upcoming India Post Payments Bank Scale I Officer Prelim Exam. It contains 35 questions and time limit is 25 minutes.
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Review
Question 1 of 35
1. Question
1 points
Directions (1-5): Complete the series following a certain pattern.
5, 7, 17, 55, 225, ?
Correct
× 1 + 2, × 2 + 3, × 3 + 4, ……
Incorrect
× 1 + 2, × 2 + 3, × 3 + 4, ……
Unattempted
× 1 + 2, × 2 + 3, × 3 + 4, ……
Question 2 of 35
2. Question
1 points
Directions (1-5): Complete the series following a certain pattern.
Directions (1-5): Complete the series following a certain pattern.
33, 16, 15, 21, 40, ?
Correct
× 0.5 – 0.5, × 1 – 1, × 1.5 – 1.5, ×2 – 2……
Incorrect
× 0.5 – 0.5, × 1 – 1, × 1.5 – 1.5, ×2 – 2……
Unattempted
× 0.5 – 0.5, × 1 – 1, × 1.5 – 1.5, ×2 – 2……
Question 6 of 35
6. Question
1 points
Directions (6-10): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 3x^{2} + 19x + 28 = 0,
II. 6y^{2} – y – 2 = 0
Correct
Incorrect
Unattempted
Question 7 of 35
7. Question
1 points
Directions (6-10): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 5x^{2} – 4x – 12 = 0,
II. 3y^{2} – 19y + 20 = 0
Correct
Incorrect
Unattempted
Question 8 of 35
8. Question
1 points
Directions (6-10): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 3x^{2} – 19x + 28 = 0,
II. 4y^{2} – 11y + 6 = 0
Correct
Incorrect
Unattempted
Question 9 of 35
9. Question
1 points
Directions (6-10): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 4x^{2} – 9x – 9 = 0,
II. 4y^{2} + 11y + 6 = 0
Correct
Incorrect
Unattempted
Question 10 of 35
10. Question
1 points
Directions (6-10): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 4x^{2} + 17x + 18 = 0,
II. 3y^{2} + 4y – 4 = 0
Correct
Incorrect
Unattempted
Question 11 of 35
11. Question
1 points
Directions (11-15): Simplify the questions taking approximations (You are not required to get the exact answer.)
22.02% of 449.89 + 34.99% of 240.01 = ?
Correct
Incorrect
Unattempted
Question 12 of 35
12. Question
1 points
Directions (11-15): Simplify the questions taking approximations (You are not required to get the exact answer.)
223.29 + 6234.06 – 2969.89 = ? + 450.09
Correct
Incorrect
Unattempted
Question 13 of 35
13. Question
1 points
Directions (11-15): Simplify the questions taking approximations (You are not required to get the exact answer.)
√7920 – 14.96% of 179.86 = ? + 22.03
Correct
Incorrect
Unattempted
Question 14 of 35
14. Question
1 points
Directions (11-15): Simplify the questions taking approximations (You are not required to get the exact answer.)
x% of 1128.79 + 64.05% of 272.22 = 24^{2} + 84.04
Correct
Incorrect
Unattempted
Question 15 of 35
15. Question
1 points
Directions (11-15): Simplify the questions taking approximations (You are not required to get the exact answer.)
A shopkeeper bought 120 kg of rice at Rs 10 per kg. Of this, he sold 80 kg at 15% profit. Now to gain a total of 20%, at what rate per kg should the shopkeeper sell the remaining rice?
Correct
Let remaining (120-80) = 40 kg sold at x% profit.
By allegation method:
(80 kg)…………………(40 kg)
15% …………………….x%
…………….20%
(x-20)…………..………5
So (x-20)/5 = 80 kg/40 kg
So it gives x = 30%
Now
CP of 40 kg = 40 × 10 = Rs 400
So at 30% profit SP of 40 kg is (130/100) × 400 = Rs 520
This gives SP of 1 kg = 520/40 = Rs 13
Incorrect
Let remaining (120-80) = 40 kg sold at x% profit.
By allegation method:
(80 kg)…………………(40 kg)
15% …………………….x%
…………….20%
(x-20)…………..………5
So (x-20)/5 = 80 kg/40 kg
So it gives x = 30%
Now
CP of 40 kg = 40 × 10 = Rs 400
So at 30% profit SP of 40 kg is (130/100) × 400 = Rs 520
This gives SP of 1 kg = 520/40 = Rs 13
Unattempted
Let remaining (120-80) = 40 kg sold at x% profit.
By allegation method:
(80 kg)…………………(40 kg)
15% …………………….x%
…………….20%
(x-20)…………..………5
So (x-20)/5 = 80 kg/40 kg
So it gives x = 30%
Now
CP of 40 kg = 40 × 10 = Rs 400
So at 30% profit SP of 40 kg is (130/100) × 400 = Rs 520
This gives SP of 1 kg = 520/40 = Rs 13
Question 17 of 35
17. Question
1 points
While filling an application form, weight of one of the candidates is mistakenly written 10% more than the actual weight. How much percent should the weight be decreased if it was found that his actual weight is 60 kg?
Correct
Weight filled in form = (110/100)*60 = 66 kg
So weight should be reduced by = (66-60)/66 × 100 = 100/11%
Incorrect
Weight filled in form = (110/100)*60 = 66 kg
So weight should be reduced by = (66-60)/66 × 100 = 100/11%
Unattempted
Weight filled in form = (110/100)*60 = 66 kg
So weight should be reduced by = (66-60)/66 × 100 = 100/11%
Question 18 of 35
18. Question
1 points
Anil, Ajay and Arun start a business. They invest Rs 20,000, Rs 25,000 and Rs 24,000 respectively. After 8 months Anil and Arun add Rs 2000 each to their investments while Ajay withdrew Rs 1000 from his investment. Find the share of the one who invested the least in the business if their total profit after one year is Rs 1,26,000.
A work can be completed by A, B and C in 20, 15 and 30 days respectively. A and B started the work and work for 3 days after which C replaces A. In how much time the work is completed?
Correct
Work that A and B complete in 1 day = 1/20 + 1/15 = 7/60. So in 3 days they complete 3 × 7/60 = 7/20 of work
Now B and C takes over the work, In 1 day they complete = 1/15 + 1/30 = 1/10 of work
So to complete 7/20 of work, they both require (7/20) × (10/1) = 7/2 days
So total work completed in 3 + 3.5 days = 6.5 days
Incorrect
Work that A and B complete in 1 day = 1/20 + 1/15 = 7/60. So in 3 days they complete 3 × 7/60 = 7/20 of work
Now B and C takes over the work, In 1 day they complete = 1/15 + 1/30 = 1/10 of work
So to complete 7/20 of work, they both require (7/20) × (10/1) = 7/2 days
So total work completed in 3 + 3.5 days = 6.5 days
Unattempted
Work that A and B complete in 1 day = 1/20 + 1/15 = 7/60. So in 3 days they complete 3 × 7/60 = 7/20 of work
Now B and C takes over the work, In 1 day they complete = 1/15 + 1/30 = 1/10 of work
So to complete 7/20 of work, they both require (7/20) × (10/1) = 7/2 days
So total work completed in 3 + 3.5 days = 6.5 days
Question 20 of 35
20. Question
1 points
A box contains 3 blue, 5 yellow and 4 black balls. If two balls are drawn at random, what is the probability that no ball is yellow in color?
Correct
Total balls = 12
Not yellow means both 2 balls of either blue or black color
So prob. = ^{7}C_{2} / ^{12}C_{2} = 7/22
Incorrect
Total balls = 12
Not yellow means both 2 balls of either blue or black color
So prob. = ^{7}C_{2} / ^{12}C_{2} = 7/22
Unattempted
Total balls = 12
Not yellow means both 2 balls of either blue or black color
So prob. = ^{7}C_{2} / ^{12}C_{2} = 7/22
Question 21 of 35
21. Question
1 points
Directions (21 – 25): Study the following pie chart and answer the questions that follow:
The number of I20 cars sold in March is approximately what percent of the number of I20 cars sold in all the given months?
Correct
60/(180+120+60+220+240) * 100
Incorrect
60/(180+120+60+220+240) * 100
Unattempted
60/(180+120+60+220+240) * 100
Question 22 of 35
22. Question
1 points
Directions (21 – 25): Study the following pie chart and answer the questions that follow:
The number of Santro cars sold in April month is what percent less than the number of I20 cars sold in January and May together?
Correct
(420-250)/420 * 100
Incorrect
(420-250)/420 * 100
Unattempted
(420-250)/420 * 100
Question 23 of 35
23. Question
1 points
Directions (21 – 25): Study the following pie chart and answer the questions that follow:
What is the average number of Santro cars sold over the given months?
Correct
(200+150+70+250+180)/5
Incorrect
(200+150+70+250+180)/5
Unattempted
(200+150+70+250+180)/5
Question 24 of 35
24. Question
1 points
Directions (21 – 25): Study the following pie chart and answer the questions that follow:
Sale of both cars in April and May together is approximately what percent greater than the sale of both the cars in remaining months?
Correct
(890 – 780)/780 * 100
Incorrect
(890 – 780)/780 * 100
Unattempted
(890 – 780)/780 * 100
Question 25 of 35
25. Question
1 points
Directions (21 – 25): Study the following pie chart and answer the questions that follow:
Sale of both cars in January is approximately what percent of sale of both cars in April?
Correct
380/470 * 100
Incorrect
380/470 * 100
Unattempted
380/470 * 100
Question 26 of 35
26. Question
1 points
A person lent out certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and 2 years is 47 : 15. The rate of interest per annum is
Correct
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 15/47
Solve, r = 13 1/3
Incorrect
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 15/47
Solve, r = 13 1/3
Unattempted
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 15/47
Solve, r = 13 1/3
Question 27 of 35
27. Question
1 points
An article is marked at Rs 25,000. It was bought by the trader at successive discounts of 20% and 5%. After this he spent Rs 1,000 on its transportation and finally sold the article for Rs 25,000. What is his profit% in the whole transaction?
Correct
Successive discounts of 20% and 5% makes overall discount of (-20) + (-5) + (-20)(-5)/100 = -25 +1 = -24%
So he bought the article for [(100-24)/100] * 25000 = 19,000
Spent 1000 on repairs, so total CP = 1000 + 19000 = 20,000
SP = 25,000
So profit% = (5000/20000) * 100
Incorrect
Successive discounts of 20% and 5% makes overall discount of (-20) + (-5) + (-20)(-5)/100 = -25 +1 = -24%
So he bought the article for [(100-24)/100] * 25000 = 19,000
Spent 1000 on repairs, so total CP = 1000 + 19000 = 20,000
SP = 25,000
So profit% = (5000/20000) * 100
Unattempted
Successive discounts of 20% and 5% makes overall discount of (-20) + (-5) + (-20)(-5)/100 = -25 +1 = -24%
So he bought the article for [(100-24)/100] * 25000 = 19,000
Spent 1000 on repairs, so total CP = 1000 + 19000 = 20,000
SP = 25,000
So profit% = (5000/20000) * 100
Question 28 of 35
28. Question
1 points
The length and the breadth of a rectangular card are increased by 1 m and 2 m respectively and due to this the area of the card increased by 66 sq. m. But if the length and breadth are decreased by 2 m and 1 m respectively, area is decreased by 51 sq. m. Find the perimeter of the original card.
Correct
Let original length = l, breadth = b, so area = lb
When l and b increased:
(l+1)(b+2) = lb + 66
Solve, 2l + b = 64
When both decreased:
(l-2)(b-1) = lb – 51
Solve, l + 2b = 53
Now solve both equations, l = 25, b = 14
Perimeter = 2(25+14)
Incorrect
Let original length = l, breadth = b, so area = lb
When l and b increased:
(l+1)(b+2) = lb + 66
Solve, 2l + b = 64
When both decreased:
(l-2)(b-1) = lb – 51
Solve, l + 2b = 53
Now solve both equations, l = 25, b = 14
Perimeter = 2(25+14)
Unattempted
Let original length = l, breadth = b, so area = lb
When l and b increased:
(l+1)(b+2) = lb + 66
Solve, 2l + b = 64
When both decreased:
(l-2)(b-1) = lb – 51
Solve, l + 2b = 53
Now solve both equations, l = 25, b = 14
Perimeter = 2(25+14)
Question 29 of 35
29. Question
1 points
A person withdraws certain amount from bank. The person decides to spend 45% of the amount withdrawn. He spends 25% on his food items and 5% on repair of furniture and the remaining Rs 16,380 he gave to his wife. What is the amount withdrawn from bank?
Correct
Let amount withdrawn = x
Person decides to spend = (45/100)*x
He spends 25% + 5% = 30%, i.e. left with 70% of (45/100)*x
So (70/100) *(45/100) * x = 16380
Solve, x = 52000
Incorrect
Let amount withdrawn = x
Person decides to spend = (45/100)*x
He spends 25% + 5% = 30%, i.e. left with 70% of (45/100)*x
So (70/100) *(45/100) * x = 16380
Solve, x = 52000
Unattempted
Let amount withdrawn = x
Person decides to spend = (45/100)*x
He spends 25% + 5% = 30%, i.e. left with 70% of (45/100)*x
So (70/100) *(45/100) * x = 16380
Solve, x = 52000
Question 30 of 35
30. Question
1 points
A man can row 14 km/hr in still water. When the river is running at 6 km/hr, it takes him 7 hours to row to a place and come back. How far is the place?
Correct
B is speed of boat in still water, R is speed of stream
Time is total time taken for upstream and downstream
Distance = time * [B^2 – R^2] / 2*B
=7 * [14^2 – 6^2] / 2*14
Incorrect
B is speed of boat in still water, R is speed of stream
Time is total time taken for upstream and downstream
Distance = time * [B^2 – R^2] / 2*B
=7 * [14^2 – 6^2] / 2*14
Unattempted
B is speed of boat in still water, R is speed of stream
Time is total time taken for upstream and downstream
Distance = time * [B^2 – R^2] / 2*B
=7 * [14^2 – 6^2] / 2*14
Question 31 of 35
31. Question
1 points
Directions (31-35): Study the following table carefully. Some values are missing. Complete that based on given information in each question to answer the question.
The table shows the number of students in 5 classes with total 2130 students in these 5 classes. The table also shows the percentage of students who like 4 different subjects – Math, English, Science and Computer with each student in particular class liking 1 subject.
What is the total number of students in classes II and V who like Science?
Correct
In V, science % = 100 – (20+22+20) = 38%
Required Ans = (30/100)*380 + (38/100)*350
Incorrect
In V, science % = 100 – (20+22+20) = 38%
Required Ans = (30/100)*380 + (38/100)*350
Unattempted
In V, science % = 100 – (20+22+20) = 38%
Required Ans = (30/100)*380 + (38/100)*350
Question 32 of 35
32. Question
1 points
Directions (31-35): Study the following table carefully. Some values are missing. Complete that based on given information in each question to answer the question.
The table shows the number of students in 5 classes with total 2130 students in these 5 classes. The table also shows the percentage of students who like 4 different subjects – Math, English, Science and Computer with each student in particular class liking 1 subject.
If in class I, students who like Computer are 40% more than students who like Math, then what is the number of students who like Science in class I?
Correct
Let % of students who like Math in I is x%. So
[28/100 * 450 – x/100 * 450]/[ x/100 * 450] * 100 = 40
Which is [(28-x)/x] * 100 = 40
Solve, x = 20
So % of students who like Science is 100 – (20+18+28) = 34%
So ans = 34/100 * 450
Incorrect
Let % of students who like Math in I is x%. So
[28/100 * 450 – x/100 * 450]/[ x/100 * 450] * 100 = 40
Which is [(28-x)/x] * 100 = 40
Solve, x = 20
So % of students who like Science is 100 – (20+18+28) = 34%
So ans = 34/100 * 450
Unattempted
Let % of students who like Math in I is x%. So
[28/100 * 450 – x/100 * 450]/[ x/100 * 450] * 100 = 40
Which is [(28-x)/x] * 100 = 40
Solve, x = 20
So % of students who like Science is 100 – (20+18+28) = 34%
So ans = 34/100 * 450
Question 33 of 35
33. Question
1 points
Directions (31-35): Study the following table carefully. Some values are missing. Complete that based on given information in each question to answer the question.
The table shows the number of students in 5 classes with total 2130 students in these 5 classes. The table also shows the percentage of students who like 4 different subjects – Math, English, Science and Computer with each student in particular class liking 1 subject.
If number of students in class III is 10% less than number of students in class IV, then what is the difference between number of students who like Computer in these 2 class and who like English in these 2 classes?
Correct
Total students in III and IV = 2130 – (450+380+350) = 950
So if in class IV, students is x, then in III – 90/100 * x
So x + 90x/100 = 950
Solve, x = 500, so in IV = 500, and in III = 90/100 * 500 = 450
So required answer = [32/100 * 450 + 35/100 * 500] – [20/100 * 450 + 25/100 * 500]
Or = 12/100 * 450 + 10/100 * 500
Incorrect
Total students in III and IV = 2130 – (450+380+350) = 950
So if in class IV, students is x, then in III – 90/100 * x
So x + 90x/100 = 950
Solve, x = 500, so in IV = 500, and in III = 90/100 * 500 = 450
So required answer = [32/100 * 450 + 35/100 * 500] – [20/100 * 450 + 25/100 * 500]
Or = 12/100 * 450 + 10/100 * 500
Unattempted
Total students in III and IV = 2130 – (450+380+350) = 950
So if in class IV, students is x, then in III – 90/100 * x
So x + 90x/100 = 950
Solve, x = 500, so in IV = 500, and in III = 90/100 * 500 = 450
So required answer = [32/100 * 450 + 35/100 * 500] – [20/100 * 450 + 25/100 * 500]
Or = 12/100 * 450 + 10/100 * 500
Question 34 of 35
34. Question
1 points
Directions (31-35): Study the following table carefully. Some values are missing. Complete that based on given information in each question to answer the question.
The table shows the number of students in 5 classes with total 2130 students in these 5 classes. The table also shows the percentage of students who like 4 different subjects – Math, English, Science and Computer with each student in particular class liking 1 subject.
If a same criterion as taken in question 23 is taken, in which class total students who like English and Computer is more?
Correct
I – (18+28)/100 * 450 = 207
II – 55/100 * 380 = 209
III – (20+32)/100 * 450 = 234
IV – (25+35)/100 * 500 = 300
V – (22+20)/100 * 350 = 147
Incorrect
I – (18+28)/100 * 450 = 207
II – 55/100 * 380 = 209
III – (20+32)/100 * 450 = 234
IV – (25+35)/100 * 500 = 300
V – (22+20)/100 * 350 = 147
Unattempted
I – (18+28)/100 * 450 = 207
II – 55/100 * 380 = 209
III – (20+32)/100 * 450 = 234
IV – (25+35)/100 * 500 = 300
V – (22+20)/100 * 350 = 147
Question 35 of 35
35. Question
1 points
Directions (31-35): Study the following table carefully. Some values are missing. Complete that based on given information in each question to answer the question.
The table shows the number of students in 5 classes with total 2130 students in these 5 classes. The table also shows the percentage of students who like 4 different subjects – Math, English, Science and Computer with each student in particular class liking 1 subject.
If in class II, number of students who like Computer is 19 more than the number of students who like Science in class V, then what is the number of students who like English in class II?
Correct
% of students who like Science in class V = 100 – (20+22+20) = 38%
Number of students who like Science in class V = 38/100 * 350 = 133
So number of students who like Computer in class II is 133+19 = 152
So number of students who like English in class II = 380 – [152 + ((15+30)/100 * 380)] = 57
Incorrect
% of students who like Science in class V = 100 – (20+22+20) = 38%
Number of students who like Science in class V = 38/100 * 350 = 133
So number of students who like Computer in class II is 133+19 = 152
So number of students who like English in class II = 380 – [152 + ((15+30)/100 * 380)] = 57
Unattempted
% of students who like Science in class V = 100 – (20+22+20) = 38%
Number of students who like Science in class V = 38/100 * 350 = 133
So number of students who like Computer in class II is 133+19 = 152
So number of students who like English in class II = 380 – [152 + ((15+30)/100 * 380)] = 57
Are wah?
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