IBPS Clerk Mains Quantitative Aptitude Free Online Mock Test  1
Finish Quiz
0 of 50 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
Information
IBPS Clerk Mains Quantitative Aptitude Online Test. Quantitative Aptitude Questions for IBPS Clerk Mains Free Online Test, IBPS Clerk Free Mock Test Series. IBPS Free Mock Test for Clerk Exam. The below online test on IBPS Clerk Recruitment will help you for preparation of the post of Clerk. IBPS Quantitative Aptitude Online Test in English. IBPS Clerk Quantitative Aptitude Quiz 2021. The IBPS Full online mock test paper is free for all students. IBPS Mock test for Quantitative Aptitude Subject Via Online Mode. Here we are providing Quantitative Aptitude Question for IBPS Clerk Exam in English. IBPS Clerk Mains Exam free Online Mock Test Series 2021. Now Test your self for IBPS Exam by using below quiz…
This paper has 50 questions.
Time allowed is 50 minutes.
The IBPS online Mock Test Exam is Very helpful for all students. Now Scroll down below n click on “Start Quiz” or “Start Test” and Test yourself.
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 50 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 Answered
 Review

Question 1 of 50
1. Question
Study the following graph carefully and answer the given questions. Description of income and expenditure of a company in 7 months of the year 2015 (in Rs. thousands)
Profit = Income – Expenditure , Loss = Expenditure – Income
Profit% = (Profit/Expenditure)*100
Loss% = (Loss/Expenditure )*100What is the average expenditure of the company in the given months(in thousands) (Approximately)?
Correct
Average expenditure of company = 2720/7 = Rs. 389 thousands
Incorrect
Average expenditure of company = 2720/7 = Rs. 389 thousands

Question 2 of 50
2. Question
Study the following graph carefully and answer the given questions. Description of income and expenditure of a company in 7 months of the year 2015 (in Rs. thousands)
Profit = Income – Expenditure , Loss = Expenditure – Income
Profit% = (Profit/Expenditure)*100
Loss% = (Loss/Expenditure )*100What is the respective ratio between the percentage profits earned by the company in the months of February and May?
Correct
Profit percent of company
= February = [(780 – 580)/580]*100 = (200/580) *100
May = [(560320)/320]*100 = 240/320 *100
= 200/580 : 240/32020*32 :58*24 = 40:87
Incorrect
Profit percent of company
= February = [(780 – 580)/580]*100 = (200/580) *100
May = [(560320)/320]*100 = 240/320 *100
= 200/580 : 240/32020*32 :58*24 = 40:87

Question 3 of 50
3. Question
Study the following graph carefully and answer the given questions. Description of income and expenditure of a company in 7 months of the year 2015 (in Rs. thousands)
Profit = Income – Expenditure , Loss = Expenditure – Income
Profit% = (Profit/Expenditure)*100
Loss% = (Loss/Expenditure )*100What is the difference between the total profit earned by the company in the months of February, April and July and that earned in the months of January, March and June?
Correct
Total profit earned by the company in the months of February, April and July
= 200 + 280 + 140 = Rs. 620 thousands
Profit earned in the months of January, March and June = 340 + 180 + 260 = Rs.780 thousands
Difference = 780 – 620 = Rs. 160 thousandsIncorrect
Total profit earned by the company in the months of February, April and July
= 200 + 280 + 140 = Rs. 620 thousands
Profit earned in the months of January, March and June = 340 + 180 + 260 = Rs.780 thousands
Difference = 780 – 620 = Rs. 160 thousands 
Question 4 of 50
4. Question
Study the following graph carefully and answer the given questions.Description of income and expenditure of a company in 7 months of the year 2015 (in Rs. thousands)
Profit = Income – Expenditure , Loss = Expenditure – Income
Profit% = (Profit/Expenditure)*100
Loss% = (Loss/Expenditure )*100By what percent is the profit earned by the company in the month of February less than that earned in the month of January (Approx)?
Correct
Profit of company :
February = 780 – 580 = Rs. 200 thousand
January = 660 – 320 = Rs. 340 thousand
Required % = [(340 – 200)/340]*100 = 41%Incorrect
Profit of company :
February = 780 – 580 = Rs. 200 thousand
January = 660 – 320 = Rs. 340 thousand
Required % = [(340 – 200)/340]*100 = 41% 
Question 5 of 50
5. Question
Study the following graph carefully and answer the given questions. Description of income and expenditure of a company in 7 months of the year 2015 (in Rs. thousands)
Profit = Income – Expenditure , Loss = Expenditure – Income
Profit% = (Profit/Expenditure)*100
Loss% = (Loss/Expenditure )*100How many months, the income of company was more than the average income during the given months?
Correct
Average income of company = 4360/7 = Rs. 623 thousand
Hence, the answer is 3 (Ian., Feb. and Apr.)Incorrect
Average income of company = 4360/7 = Rs. 623 thousand
Hence, the answer is 3 (Ian., Feb. and Apr.) 
Question 6 of 50
6. Question
Study the charts carefully and answer the related questions. The bar graph shows the profit % earned on six different articles . The piechart shows the cost price of six different articles sold by a store in 2015.
The selling price of fan and cooler together is approximately what percent of the selling price of T.V. and radio together?
Correct
CP of fan = (12/100)*72000 = 8640
Profit% = 25%
SP of fan = (8640*125)/100 = 10800
CP of cooler = (18/100)*72000 = 12960
Profit% = 48%
SP of cooler = (12960*148)/100 = 19180.80
Total SP of fan and cooler together = 10800 + 19180.80 = 29980.80
CP of TV = (21/100)*72000 = 15120
Profit% = 42%
SP of TV = (15120*142)/100 = 21470.40
CP of radio = (10/100)*72000 = 7200
Profit % = 36%
SP of radio = (7200*136)/100 = 9792
Total SP of TV and radio = 21470.40 + 9792 = 31262.40
Required % = (29980.80/31262.40)*100 = 95.9% = 96%(approx.)Incorrect
CP of fan = (12/100)*72000 = 8640
Profit% = 25%
SP of fan = (8640*125)/100 = 10800
CP of cooler = (18/100)*72000 = 12960
Profit% = 48%
SP of cooler = (12960*148)/100 = 19180.80
Total SP of fan and cooler together = 10800 + 19180.80 = 29980.80
CP of TV = (21/100)*72000 = 15120
Profit% = 42%
SP of TV = (15120*142)/100 = 21470.40
CP of radio = (10/100)*72000 = 7200
Profit % = 36%
SP of radio = (7200*136)/100 = 9792
Total SP of TV and radio = 21470.40 + 9792 = 31262.40
Required % = (29980.80/31262.40)*100 = 95.9% = 96%(approx.) 
Question 7 of 50
7. Question
Study the charts carefully and answer the related questions. The bar graph shows the profit % earned on six different articles . The piechart shows the cost price of six different articles sold by a store in 2015.
The marked price of TV is what percent of the cost price of cooler ?
Correct
Since ,we don’t have discount of TV .So,we can’t find the marked price.
Incorrect
Since ,we don’t have discount of TV .So,we can’t find the marked price.

Question 8 of 50
8. Question
Study the charts carefully and answer the related questions. The bar graph shows the profit % earned on six different articles . The piechart shows the cost price of six different articles sold by a store in 2015.
What is the difference between the selling price of LED and that of TV?
Correct
CP of LED = (26/100)*72000 = 18720
Profit% = 28%
SP of LED = (18720*128)/100 = 23961.60
SP of TV = 21470.40
Required difference = 23961.60 – 21470.40 = 2491.20Incorrect
CP of LED = (26/100)*72000 = 18720
Profit% = 28%
SP of LED = (18720*128)/100 = 23961.60
SP of TV = 21470.40
Required difference = 23961.60 – 21470.40 = 2491.20 
Question 9 of 50
9. Question
Study the charts carefully and answer the related questions. The bar graph shows the profit % earned on six different articles . The piechart shows the cost price of six different articles sold by a store in 2015.
The selling price of fan is approx. what percent of the selling price of watch ?
Correct
CP of fan = (12/100)*72000 = 8640
Profit% = 25%
SP of fan = (8640*125)/100 = 10800
CP of watch = (13/100)*72000 = 9360
Profit % = 35%
SP of watch = (9360*135)/100 = 12636
Required % = (10800/12636)*100 = 85.47% = 85%Incorrect
CP of fan = (12/100)*72000 = 8640
Profit% = 25%
SP of fan = (8640*125)/100 = 10800
CP of watch = (13/100)*72000 = 9360
Profit % = 35%
SP of watch = (9360*135)/100 = 12636
Required % = (10800/12636)*100 = 85.47% = 85% 
Question 10 of 50
10. Question
Study the charts carefully and answer the related questions. The bar graph shows the profit % earned on six different articles . The piechart shows the cost price of six different articles sold by a store in 2015.
The marked price of radio is approx. what percent of the selling price of LED if 20% discount is available on radio?
Correct
CP of radio = (10/100)*72000 = 7200
Profit% = 36%
SP of radio = (7200*136) /100 = 9792
Discount on radio = 20%
MP of radio = (9792/80)*100 = 12240
CP of LED = (26/100)*72000 = 18720
Profit% = 28%
SP of LED = (18720*128)/100 = 23961.60
Required % = (12240/23961.60)*100 = 51.08% = 51%Incorrect
CP of radio = (10/100)*72000 = 7200
Profit% = 36%
SP of radio = (7200*136) /100 = 9792
Discount on radio = 20%
MP of radio = (9792/80)*100 = 12240
CP of LED = (26/100)*72000 = 18720
Profit% = 28%
SP of LED = (18720*128)/100 = 23961.60
Required % = (12240/23961.60)*100 = 51.08% = 51% 
Question 11 of 50
11. Question
What is the ratio of the cost price of mobile and cost price of laptop?
Correct
ost Price of mobile = 25000 * 100/ (100+25) = 20000
Cost Price of laptop = 20000 * 100/ (100+25) = 16000
Required Ratio = 20000 : 16000 = 5 : 4Incorrect
ost Price of mobile = 25000 * 100/ (100+25) = 20000
Cost Price of laptop = 20000 * 100/ (100+25) = 16000
Required Ratio = 20000 : 16000 = 5 : 4 
Question 12 of 50
12. Question
Market Price of T.V is how much percent more than the cost price of T.V ?
Correct
Amount of profit on T.V = 2000
Cost Price = 2000* 100/10 = 20000
Selling Price = 20000 + 2000 = 22000
Market Price = Selling Price * 100 / (100 – Discount%)
= 20000 * 100 /80 = 27500
Required Percentage = [(27500 20000) /20000] * 100 = 37.5%Incorrect
Amount of profit on T.V = 2000
Cost Price = 2000* 100/10 = 20000
Selling Price = 20000 + 2000 = 22000
Market Price = Selling Price * 100 / (100 – Discount%)
= 20000 * 100 /80 = 27500
Required Percentage = [(27500 20000) /20000] * 100 = 37.5% 
Question 13 of 50
13. Question
What is the ratio between the percentage of profit and percentage of discount of Mobile ?
Correct
Percentage of profit on Mobile = 25%
Percentage of discount on Mobile = [40000 – 25000/40000]*100
= 37.5%
Required Ratio = 25 : 37.5
= 2 : 3Incorrect
Percentage of profit on Mobile = 25%
Percentage of discount on Mobile = [40000 – 25000/40000]*100
= 37.5%
Required Ratio = 25 : 37.5
= 2 : 3 
Question 14 of 50
14. Question
Selling Price of D.V.D is what percent of cost price of A.C ?
Correct
Market Price of D.V.D = 500 * 100/20 = 2500
Selling Price of D.V.D = 2500 – 500 = 2000
Selling Price of A.C = 50000 * 60 / 100 = 30000
Cost Price of A.C = 30000 * 100/80 = 37500
Required Percentage = (2000/37500)*100 = 5.3%Incorrect
Market Price of D.V.D = 500 * 100/20 = 2500
Selling Price of D.V.D = 2500 – 500 = 2000
Selling Price of A.C = 50000 * 60 / 100 = 30000
Cost Price of A.C = 30000 * 100/80 = 37500
Required Percentage = (2000/37500)*100 = 5.3% 
Question 15 of 50
15. Question
What is the amount (in Rupees) of profit /loss of shop by selling all electronic goods ?
Correct
Selling Price of A.C = 50000 * 60/100 = 30000
Loss on A.C = 30000 * 100/80 – 3000 = Rs.7500
Market Price of D.V.D = 500 * 100 /20 = 2500
Selling Price of D.V.D = 2500 – 500 = 2000
Profit on D.V.D =2000 – 1800 = Rs.200
Profit on T.V = Rs. 2000
Profit on Mobile = 25000 – 25000 * 100/25 = Rs. 5000
Profit on Laptop = 20000 – 20000 * 100 / 125 = Rs. 4000
Profit on Radio = 10,000 * 25 /100 = Rs.2500Total Profit = 7500 + 200 + 2000 + 5000 + 4000 + 2500
= Rs. 6200Incorrect
Selling Price of A.C = 50000 * 60/100 = 30000
Loss on A.C = 30000 * 100/80 – 3000 = Rs.7500
Market Price of D.V.D = 500 * 100 /20 = 2500
Selling Price of D.V.D = 2500 – 500 = 2000
Profit on D.V.D =2000 – 1800 = Rs.200
Profit on T.V = Rs. 2000
Profit on Mobile = 25000 – 25000 * 100/25 = Rs. 5000
Profit on Laptop = 20000 – 20000 * 100 / 125 = Rs. 4000
Profit on Radio = 10,000 * 25 /100 = Rs.2500Total Profit = 7500 + 200 + 2000 + 5000 + 4000 + 2500
= Rs. 6200 
Question 16 of 50
16. Question
4003×6621015=?×3 What will come in place of question mark (?) in each of the following questions?
Correct
(4000+3)×6621015=?×3
or, ?= (26419821015) /3=81061
Incorrect
(4000+3)×6621015=?×3
or, ?= (26419821015) /3=81061

Question 17 of 50
17. Question
(6√7+√7) (4√7+8√7) – (19)²=? What will come in place of question mark (?) in each of the following questions?
Correct
?= (6√7+√7) (4√7+8√7) (19)²
=24×7+48×7+4×7+8×7(19)²
=168+336+28+56361=227
Incorrect
?= (6√7+√7) (4√7+8√7) (19)²
=24×7+48×7+4×7+8×7(19)²
=168+336+28+56361=227

Question 18 of 50
18. Question
(5555/50) + (645/25) +3991/26=? What will come in place of question mark (?) in each of the following questions?
Correct
?= (5555/50) + 645/25+3991/26
=111.1+25.8+153.5=290.4
Incorrect
?= (5555/50) + 645/25+3991/26
=111.1+25.8+153.5=290.4

Question 19 of 50
19. Question
√33489×√2601 (83)² = (?)²+ (37)² What will come in place of question mark (?) in each of the following questions?
Correct
√33489×√2601 (83)² = (?)²+ (37)²
or, 183×516889= (?)² +1369
Or, 93336889= (?)²+ 1369 or,
(?)²=24441369=1075
∴ ?= √1075=√25×43=5√43
Incorrect
√33489×√2601 (83)² = (?)²+ (37)²
or, 183×516889= (?)² +1369
Or, 93336889= (?)²+ 1369 or,
(?)²=24441369=1075
∴ ?= √1075=√25×43=5√43

Question 20 of 50
20. Question
5 (27/37)×4(51/52)×11(1/7) +2(3/4) =? What will come in place of question mark (?) in each of the following questions?
Correct
?= (212/37)×(259/32)×(78/7)+ (11/4)
=318+ (11/4) =320.75
Incorrect
?= (212/37)×(259/32)×(78/7)+ (11/4)
=318+ (11/4) =320.75

Question 21 of 50
21. Question
√930.25+√1482.2545% of 180+46.5=√? What will come in place of question mark (?) in each of the following questions?
Correct
√?= √930.25+ √1482.25 (45×180)/180+46.5
=30.5+38.5.45×1.8+46.5=115.581=34.5
∴ ?=34.5×34.5=1190.25
Incorrect
√?= √930.25+ √1482.25 (45×180)/180+46.5
=30.5+38.5.45×1.8+46.5=115.581=34.5
∴ ?=34.5×34.5=1190.25

Question 22 of 50
22. Question
79% of 790 +1/3 of 675/0.5=? What will come in place of question mark (?) in each of the following questions?
Correct
?= [(790×790)/100]+(1/3)×(675/0.5)
=624.1+450=1074.1
Incorrect
?= [(790×790)/100]+(1/3)×(675/0.5)
=624.1+450=1074.1

Question 23 of 50
23. Question
[(1728)^{0.6}/ (144)^{0.3}×(0.0144)^{0.4}]/ (10)^{0.4}= (?)^{2 What will come in place of question mark (?) in each of the following questions?}
Correct
(?)^{2}= [(1728)^{0.6}/ (144)^{0.3}×(0.0144)^{0.4}]/ (10)^{0.4}
= [(12)^{1.8}/ (12)^{0.6}×(0.12)^{0.8}]/ (10)^{0.4}
= [(12)^{1.8}/ (12)^{0.6}×(12)^{0.8}]/ (10)^{0.4}×(10)^{1.6}
= (12)^{1.80.6+0.8}/ (10)^{0.4+1.6 }= (12)^{2}/ 10^{2}
Or, ?= √[12/10]² =12/10 =1.2
Incorrect
(?)^{2}= [(1728)^{0.6}/ (144)^{0.3}×(0.0144)^{0.4}]/ (10)^{0.4}
= [(12)^{1.8}/ (12)^{0.6}×(0.12)^{0.8}]/ (10)^{0.4}
= [(12)^{1.8}/ (12)^{0.6}×(12)^{0.8}]/ (10)^{0.4}×(10)^{1.6}
= (12)^{1.80.6+0.8}/ (10)^{0.4+1.6 }= (12)^{2}/ 10^{2}
Or, ?= √[12/10]² =12/10 =1.2

Question 24 of 50
24. Question
174% of 445+9% of 167+√1521=? What will come in place of question mark (?) in each of the following questions?
Correct
?= [(174×445)/5] + [(9×167)/100] +39
=774.3+15.03+39 =828.33
Incorrect
?= [(174×445)/5] + [(9×167)/100] +39
=774.3+15.03+39 =828.33

Question 25 of 50
25. Question
19(1/7) +26(2/3) 9(1/3) +5(1/7) =? What will come in place of question mark (?) in each of the following questions?
Correct
19(1/7) +26(2/3)9(1/3) +5(1/7) =?
Or, ?= (19+26+59) + [(1/7) + (2/3) – (1/3) + (1/7)]
=41+ (3+147+3)/21 =41(13/21)
Incorrect
19(1/7) +26(2/3)9(1/3) +5(1/7) =?
Or, ?= (19+26+59) + [(1/7) + (2/3) – (1/3) + (1/7)]
=41+ (3+147+3)/21 =41(13/21)

Question 26 of 50
26. Question
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.4×2 + 3x – 27 = 0,
II. 3y2 – 20y + 32 = 0Correct
x =2.25, 3
y = 8/3, 4
Put all values on number line and analyze the relationship
3……..2.25……..8/3……….4
Incorrect
x =2.25, 3
y = 8/3, 4
Put all values on number line and analyze the relationship
3……..2.25……..8/3……….4

Question 27 of 50
27. Question
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} + 19x + 21 = 0,
II. 3y^{2} – 19y – 14 = 0Correct
x = 3, – 1.75
y = 2/3, 7
Put all values on number line and analyze the relationship
3…….. 1.75………2/3……..7Incorrect
x = 3, – 1.75
y = 2/3, 7
Put all values on number line and analyze the relationship
3…….. 1.75………2/3……..7 
Question 28 of 50
28. Question
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. 15y^{2} – 29y + 12 = 0Correct
x = 3/4, 3
y = 3/5, 4/3
Put all values on number line and analyze the relationship
3/4…….3/5…….4/3………3Incorrect
x = 3/4, 3
y = 3/5, 4/3
Put all values on number line and analyze the relationship
3/4…….3/5…….4/3………3 
Question 29 of 50
29. Question
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. 3×2 – 5x – 12 = 0,
II. 3y2 – 8y – 16 = 0Correct
x = 4/3, 3
y = 4/3, 4
Put all values on number line and analyze the relationship
4/3…….. 3……..4Incorrect
x = 4/3, 3
y = 4/3, 4
Put all values on number line and analyze the relationship
4/3…….. 3……..4 
Question 30 of 50
30. Question
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} + 2x – 21 = 0,
II. 3y^{2} – 19y + 28 = 0Correct
x = 3, 7/3
y = 7/3, 4
Put all values on number line and analyze the relationship
3………. 7/3……..4Incorrect
x = 3, 7/3
y = 7/3, 4
Put all values on number line and analyze the relationship
3………. 7/3……..4 
Question 31 of 50
31. Question
62 87 187 420 812 1437 In each of the following questions number series, there is a wrong number that does not follow the pattern of the series. Find the wrong number.
Correct
62 + 5^2 = 62 + 25 = 87
87 + 10^2 = 87 + 100 = 187
187 + 15^2 = 187 + 225 = 412
412 + 20^2 = 412 + 400 = 812
812 + 25^2 = 812 + 625 = 1437Incorrect
62 + 5^2 = 62 + 25 = 87
87 + 10^2 = 87 + 100 = 187
187 + 15^2 = 187 + 225 = 412
412 + 20^2 = 412 + 400 = 812
812 + 25^2 = 812 + 625 = 1437 
Question 32 of 50
32. Question
120 137 170 222 290 375 In each of the following questions number series, there is a wrong number that does not follow the pattern of the series. Find the wrong number.
Correct
120 + 1 * 17 = 137
137 + 2 *17 = 171
171 + 3 * 17 = 222
222 + 4 * 17 = 290
290 + 5 * 17 = 375Incorrect
120 + 1 * 17 = 137
137 + 2 *17 = 171
171 + 3 * 17 = 222
222 + 4 * 17 = 290
290 + 5 * 17 = 375 
Question 33 of 50
33. Question
550 542 537 521 496 460 In each of the following questions number series, there is a wrong number that does not follow the pattern of the series. Find the wrong number.
Correct
550 – 2^2 = 546
546 – 3^2 = 537
537 – 4^2 = 521
521 – 5^2 = 496
496 – 6^2 = 460Incorrect
550 – 2^2 = 546
546 – 3^2 = 537
537 – 4^2 = 521
521 – 5^2 = 496
496 – 6^2 = 460 
Question 34 of 50
34. Question
12 48 168 500 1260 2520 In each of the following questions number series, there is a wrong number that does not follow the pattern of the series. Find the wrong number.
Correct
12 * 4 = 48
48 * 3.5 = 168
168 * 3 = 504
504 * 2.5 = 1260
1260 * 2 = 2520Incorrect
12 * 4 = 48
48 * 3.5 = 168
168 * 3 = 504
504 * 2.5 = 1260
1260 * 2 = 2520 
Question 35 of 50
35. Question
24 536 487 703 670 742 In each of the following questions number series, there is a wrong number that does not follow the pattern of the series. Find the wrong number.
Correct
24 + 8^3 = 536
536 – 7^2 = 487
487 + 6^3 = 703
703 – 5^2 = 678
678 + 4^3 = 742Incorrect
24 + 8^3 = 536
536 – 7^2 = 487
487 + 6^3 = 703
703 – 5^2 = 678
678 + 4^3 = 742 
Question 36 of 50
36. Question
A and B both sold the same kind of Bike. The price of each Bike is Rs. 36,000. A gives a discount of 15% on whole, while B gives a discount of 10% on the first Rs 25,000 and 15% discount on the rest. What is the difference between their selling prices?
Correct
A’s Discount = 15% of 36000=5400
B’s Discount = 10% of 25000 + 15% of 11000 = 2500+ 1650=4150
the difference in selling price is same as difference in discount = 5400 4150 = Rs1250.Incorrect
A’s Discount = 15% of 36000=5400
B’s Discount = 10% of 25000 + 15% of 11000 = 2500+ 1650=4150
the difference in selling price is same as difference in discount = 5400 4150 = Rs1250. 
Question 37 of 50
37. Question
The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:
Correct
Let the no be x and y.
x*y=9375 and x/y=15
xy/(x/y)=9375/15
y2=625
y=25
then x=15y=15*25=375
375+25=400.Incorrect
Let the no be x and y.
x*y=9375 and x/y=15
xy/(x/y)=9375/15
y2=625
y=25
then x=15y=15*25=375
375+25=400. 
Question 38 of 50
38. Question
A bag contains 8 green, 6 white and 10 blue balls. Four balls are drawn at random from the bag. The probability that all of them are Green, is:
Correct
P= n(e)/n(s)
=8C4/24C4
=(8*7*6*5) /(24*23*22*21)
=5/759.Incorrect
P= n(e)/n(s)
=8C4/24C4
=(8*7*6*5) /(24*23*22*21)
=5/759. 
Question 39 of 50
39. Question
P salary is 75% more than Q’s. P got a raise of 40% on his salary while Q got a raise of 25% on his salary. By what percent is P’s salary more than Q’s?
Correct
Let Q’s salary = 100; Q’s salary after rise = 125.
Then P’s salary = 175.
P’s salary after rise = 245.
Difference between P’s and Q’s salary = 245125 = 120; .
Then 120*100/125 = 96%.Incorrect
Let Q’s salary = 100; Q’s salary after rise = 125.
Then P’s salary = 175.
P’s salary after rise = 245.
Difference between P’s and Q’s salary = 245125 = 120; .
Then 120*100/125 = 96%. 
Question 40 of 50
40. Question
A shopkeeper sold 12 watches at a profit of 20% and 8 watches at a profit of 10%. If he had sold all the 20 watches at a profit of 15%, then his profit would have been reduced by Rs. 36. What is the cost price of each watch?
Correct
Let CP of each watches = 10 Rs.
In 1st case
Total cp of 20 watches = 200
Total sp of 20 watches =(120+24)+(80+8) =232
Profit = 232 – 200 = 32
In Second case profit = 15% of 200 = 30
So, 32 – 30 = 36
1 = 18
10 = 180.Incorrect
Let CP of each watches = 10 Rs.
In 1st case
Total cp of 20 watches = 200
Total sp of 20 watches =(120+24)+(80+8) =232
Profit = 232 – 200 = 32
In Second case profit = 15% of 200 = 30
So, 32 – 30 = 36
1 = 18
10 = 180. 
Question 41 of 50
41. Question
Total number of men, women and children working in a factory is 27. They earn Rs. 6000 in a day. If the sum of the wages of all men, all women and all children is in ratio of 18: 10: 12 and if the wages of an individual man, woman and child is in ratio 6: 5: 3, then how much a man earn in a day?
Correct
Ratio of number of men, women and children,
= (18/6): (10/5):(12/3) = 3:2:4
Total (Men +Women +Children) = 27
3X +2X +4X = 27
9X = 27
X = 3
Hence, number of men = 3x = 9
Share of all men = (18*6000)/40 = Rs. 2700 [18+10+12 =40]
Thus, share of each man = 2700/9 = Rs. 300.Incorrect
Ratio of number of men, women and children,
= (18/6): (10/5):(12/3) = 3:2:4
Total (Men +Women +Children) = 27
3X +2X +4X = 27
9X = 27
X = 3
Hence, number of men = 3x = 9
Share of all men = (18*6000)/40 = Rs. 2700 [18+10+12 =40]
Thus, share of each man = 2700/9 = Rs. 300. 
Question 42 of 50
42. Question
Two students appeared at an examination. One of them secured 20 marks more than the other and his marks was 60% of the sum of their marks. The marks obtained by them are:
Correct
Let their marks be (x + 20) and x.
Then,
x+20=60/100(x+20+x)
==> 5(x + 20) = 3(2x + 20)
==>5x+100=6x+60
==>x=40.
So, their marks are 40 and 60.Incorrect
Let their marks be (x + 20) and x.
Then,
x+20=60/100(x+20+x)
==> 5(x + 20) = 3(2x + 20)
==>5x+100=6x+60
==>x=40.
So, their marks are 40 and 60. 
Question 43 of 50
43. Question
Two cyclist start from the same place in opposite direction the first cyclist goes towards north at30km/hr and the second cyclist goes towards south at 25km/hr. Approximately What time will they take to be 62.5km?
Correct
Both 1hr travelled 30+25=55km
Then 55 == 1
62.5? ==>62.5/55=1 7.5/55hrs
7.5/55hrs= 60 mints
7.5? = 8.1mins (approx.)
So time=1hr 8mins.Incorrect
Both 1hr travelled 30+25=55km
Then 55 == 1
62.5? ==>62.5/55=1 7.5/55hrs
7.5/55hrs= 60 mints
7.5? = 8.1mins (approx.)
So time=1hr 8mins. 
Question 44 of 50
44. Question
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
Correct
2(A + B + C)’s 1 day’s work = (1/30+1/24+1/20) =15/120=1/8
A+B+C=1/(2*8)=1/16.
Work done in 10days=10/16=5/8.
Remaining =3/8work.
A’s 1day work=1/16 – 1/24=1/48
Then 48*3/8=18days.Incorrect
2(A + B + C)’s 1 day’s work = (1/30+1/24+1/20) =15/120=1/8
A+B+C=1/(2*8)=1/16.
Work done in 10days=10/16=5/8.
Remaining =3/8work.
A’s 1day work=1/16 – 1/24=1/48
Then 48*3/8=18days. 
Question 45 of 50
45. Question
Starting from my office, I reach the house 25 min late if I walk at 4kmph. Instead, if I walk at 6 kmph, I reach the house 15 min early. How far is my house from my office?
Correct
Distance=[Time diff *(s1*s2)]/(s1s2) s1=4; s2=6
=(40/60 *24)/2
=8km.Incorrect
Distance=[Time diff *(s1*s2)]/(s1s2) s1=4; s2=6
=(40/60 *24)/2
=8km. 
Question 46 of 50
46. Question
A loss of 10% is made by selling an article. Had it been sold for Rs 75 more, there would have been a profit of 5%. The initial loss is what percentage of the profit earned, if the article was sold at a profit of 20%?
Correct
Use shortcut for these type of questions :
CP of article = 75 × 100/ [5 – (10)] = Rs 500 (+5 for 5% profit, 10 for 10% loss)
So loss was = 10/100 * 500 = Rs 50
and new profit = 20/100 * 500 = Rs 100
So required % = 50/100 * 100 = 50%Incorrect
Use shortcut for these type of questions :
CP of article = 75 × 100/ [5 – (10)] = Rs 500 (+5 for 5% profit, 10 for 10% loss)
So loss was = 10/100 * 500 = Rs 50
and new profit = 20/100 * 500 = Rs 100
So required % = 50/100 * 100 = 50% 
Question 47 of 50
47. Question
Vanya inquires about a mobile having same price at two shops. One offers 30% discount and the other offers 25%. She has just sufficient amount of Rs 21,000 to purchase mobile at 30% discount, how much amount she has less than the amount required to purchase mobile at 25% discount?
Correct
Let MP of mobile = Rs x
So at 30% discount, she gets mobile at Rs 70% of x
So 7x/10 = 21000
Solve, x = 30,000
So SP at 25% discount = 75% of 30,000= Rs 22,500
Required difference = 22500 – 21000 = Rs 1500Incorrect
Let MP of mobile = Rs x
So at 30% discount, she gets mobile at Rs 70% of x
So 7x/10 = 21000
Solve, x = 30,000
So SP at 25% discount = 75% of 30,000= Rs 22,500
Required difference = 22500 – 21000 = Rs 1500 
Question 48 of 50
48. Question
Two containers contain mixture of milk and water. Container A contains 25% water and rest milk. Container B contains 36% water and rest milk. How much amount should be mixed from container A to 50 litres of container B so as to get a new mixture having water to milk in ratio 2 : 5?
Correct
In resultant mixture, water is 2/7 * 100 = 200/7%
So by method of allegation:
25%……………………………36%
…………..200/7%
52/2%……………………….25/2%
So ratio is 52/2 : 25/2 = 52 : 25.
So 52/25 = x/50
So x = 104 LIncorrect
In resultant mixture, water is 2/7 * 100 = 200/7%
So by method of allegation:
25%……………………………36%
…………..200/7%
52/2%……………………….25/2%
So ratio is 52/2 : 25/2 = 52 : 25.
So 52/25 = x/50
So x = 104 L 
Question 49 of 50
49. Question
A boat can row in upstream at 8km/hr and in downstream at 18 km/hr. find the current and in still water rate of man and what is the time taken by a man cover the distance of 36km along the stream?
Correct
Rate of current = U.S +D.S /2 = (8+18)/2 =13km/hr
Speed of boat = U.S – D.S /2 = (188)/2 =5km/hr
(In still water)
Time taken by the boat along river= distance / downstream speed
= 36/18 = 2hrs
Incorrect
Rate of current = U.S +D.S /2 = (8+18)/2 =13km/hr
Speed of boat = U.S – D.S /2 = (188)/2 =5km/hr
(In still water)
Time taken by the boat along river= distance / downstream speed
= 36/18 = 2hrs

Question 50 of 50
50. Question
Average age of 20 boys and 40 girls is 12 . if the number of boys reduced half and the number of girls increased half the average remains same. ThenThe total number of age of one boy and one girl?
Correct
20b +40 g = 720
10b+ 60g =840
*2= 20b+120g=1680
20b+40g = 720
80 g= 960
G=12 years; by solving this, boys=12 years
Age of one boy+ one girl =24 years
Incorrect
20b +40 g = 720
10b+ 60g =840
*2= 20b+120g=1680
20b+40g = 720
80 g= 960
G=12 years; by solving this, boys=12 years
Age of one boy+ one girl =24 years