# IBPS Clerk Mains Quantitative Aptitude Online Test, Mock Test Series

## IBPS Clerk Mains Quantitative Aptitude Free Online Mock Test - 1

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**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 2019. 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 2019. Now Test your self for IBPS Exam by using below quiz…

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- 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 )*100**What is the average expenditure of the company in the given months(in thousands) (Approximately)?**CorrectAverage expenditure of company = 2720/7 = Rs. 389 thousands

IncorrectAverage 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 )*100**What is the respective ratio between the percentage profits earned by the company in the months of February and May?**CorrectProfit percent of company

= February = [(780 – 580)/580]*100 = (200/580) *100

May = [(560-320)/320]*100 = 240/320 *100

= 200/580 : 240/32020*32 :58*24 = 40:87

IncorrectProfit percent of company

= February = [(780 – 580)/580]*100 = (200/580) *100

May = [(560-320)/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 )*100**What 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?**CorrectTotal 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 thousandsIncorrectTotal 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% = (Profit/Expenditure)*100

Loss% = (Loss/Expenditure )*100**By what percent is the profit earned by the company in the month of February less than that earned in the month of January (Approx)?**

CorrectProfit of company :

February = 780 – 580 = Rs. 200 thousand

January = 660 – 320 = Rs. 340 thousand

Required % = [(340 – 200)/340]*100 = 41%IncorrectProfit 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

Profit% = (Profit/Expenditure)*100

Loss% = (Loss/Expenditure )*100**How many months, the income of company was more than the average income during the given months?**CorrectAverage income of company = 4360/7 = Rs. 623 thousand

Hence, the answer is 3 (Ian., Feb. and Apr.)IncorrectAverage 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 pie-chart 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?**CorrectCP 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.)IncorrectCP 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 pie-chart 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 ?**CorrectSince ,we don’t have discount of TV .So,we can’t find the marked price.

IncorrectSince ,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 pie-chart 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?**CorrectCP 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 = 2491.20IncorrectCP 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

**The selling price of fan is approx. what percent of the selling price of watch ?**CorrectCP 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%IncorrectCP 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

**The marked price of radio is approx. what percent of the selling price of LED if 20% discount is available on radio?**CorrectCP 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%IncorrectCP 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?**Correctost Price of mobile = 25000 * 100/ (100+25) = 20000

Cost Price of laptop = 20000 * 100/ (100+25) = 16000

Required Ratio = 20000 : 16000 = 5 : 4Incorrectost 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 ?**CorrectAmount 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%IncorrectAmount 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 ?**CorrectPercentage of profit on Mobile = 25%

Percentage of discount on Mobile = [40000 – 25000/40000]*100

= 37.5%

Required Ratio = 25 : 37.5

= 2 : 3IncorrectPercentage 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 ?**CorrectMarket 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%IncorrectMarket 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 ?**CorrectSelling 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. 6200IncorrectSelling 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

4003×66-21015=?×3 What will come in place of question mark (?) in each of the following questions?

Correct(4000+3)×66-21015=?×3

or, ?= (264198-21015) /3=81061

Incorrect(4000+3)×66-21015=?×3

or, ?= (264198-21015) /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+56-361=227

Incorrect?= (6√7+√7) (4√7+8√7)- (19)²

=24×7+48×7+4×7+8×7-(19)²

=168+336+28+56-361=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, 183×51-6889= (?)² +1369

Or, 9333-6889= (?)²+ 1369 or,

(?)²=2444-1369=1075

∴ ?= √1075=√25×43=5√43

Incorrect√33489×√2601- (83)² = (?)²+ (37)²

or, 183×51-6889= (?)² +1369

Or, 9333-6889= (?)²+ 1369 or,

(?)²=2444-1369=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+√1482.25-45% 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=115.5-81=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=115.5-81=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)

^{1.8-0.6+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)

^{1.8-0.6+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?**Correct19(1/7) +26(2/3)-9(1/3) +5(1/7) =?

Or, ?= (19+26+5-9) + [(1/7) + (2/3) – (1/3) + (1/7)]

=41+ (3+14-7+3)/21 =41(13/21)

Incorrect19(1/7) +26(2/3)-9(1/3) +5(1/7) =?

Or, ?= (19+26+5-9) + [(1/7) + (2/3) – (1/3) + (1/7)]

=41+ (3+14-7+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 = 0Correctx =2.25, -3

y = 8/3, 4

Put all values on number line and analyze the relationship

-3……..2.25……..8/3……….4

Incorrectx =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 = 0Correctx = -3, – 1.75

y = -2/3, 7

Put all values on number line and analyze the relationship

-3…….. -1.75………-2/3……..7Incorrectx = -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 = 0Correctx = -3/4, 3

y = 3/5, 4/3

Put all values on number line and analyze the relationship

-3/4…….3/5…….4/3………3Incorrectx = -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 = 0Correctx = -4/3, 3

y = -4/3, 4

Put all values on number line and analyze the relationship

-4/3…….. 3……..4Incorrectx = -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 = 0Correctx = -3, 7/3

y = 7/3, 4

Put all values on number line and analyze the relationship

-3………. 7/3……..4Incorrectx = -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.**Correct62 + 5^2 = 62 + 25 = 87

87 + 10^2 = 87 + 100 = 187

187 + 15^2 = 187 + 225 = 412

412 + 20^2 = 412 + 400 = 812812 + 25^2 = 812 + 625 = 1437

Incorrect62 + 5^2 = 62 + 25 = 87

87 + 10^2 = 87 + 100 = 187

187 + 15^2 = 187 + 225 = 412

412 + 20^2 = 412 + 400 = 812812 + 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.**

Correct120 + 1 * 17 = 137

137 + 2 *17 = 171

171 + 3 * 17 = 222

222 + 4 * 17 = 290

290 + 5 * 17 = 375Incorrect120 + 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.**Correct550 – 2^2 = 546

546 – 3^2 = 537

537 – 4^2 = 521

521 – 5^2 = 496

496 – 6^2 = 460Incorrect550 – 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.**Correct12 * 4 = 48

48 * 3.5 = 168

168 * 3 = 504

504 * 2.5 = 1260

1260 * 2 = 2520Incorrect12 * 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.**Correct24 + 8^3 = 536

536 – 7^2 = 487

487 + 6^3 = 703

703 – 5^2 = 678

678 + 4^3 = 742Incorrect24 + 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?**CorrectA’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.IncorrectA’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:

CorrectLet 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.IncorrectLet 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:

CorrectP= n(e)/n(s)

=8C4/24C4

=(8*7*6*5) /(24*23*22*21)

=5/759.IncorrectP= 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?

CorrectLet 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 = 245-125 = 120; .

Then 120*100/125 = 96%.IncorrectLet 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 = 245-125 = 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?

CorrectLet 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.IncorrectLet 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?

CorrectRatio 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.IncorrectRatio 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:

CorrectLet 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.IncorrectLet 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?

CorrectBoth 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.IncorrectBoth 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?

Correct2(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.Incorrect2(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?

CorrectDistance=[Time diff *(s1*s2)]/(s1-s2) s1=4; s2=6

=(40/60 *24)/2

=8km.IncorrectDistance=[Time diff *(s1*s2)]/(s1-s2) 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%?

CorrectUse 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%IncorrectUse 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?

CorrectLet 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 1500IncorrectLet 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?

CorrectIn 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 LIncorrectIn 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?

CorrectRate of current = U.S +D.S /2 = (8+18)/2 =13km/hr

Speed of boat = U.S – D.S /2 = (18-8)/2 =5km/hr

(In still water)

Time taken by the boat along river= distance / downstream speed

= 36/18 = 2hrs

IncorrectRate of current = U.S +D.S /2 = (8+18)/2 =13km/hr

Speed of boat = U.S – D.S /2 = (18-8)/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?

Correct20b +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

Incorrect20b +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