# Indian Bank PO Prelims Online Test Series 5th, Indian Bank Mock test 2019

## Indian Bank PO Prelims Online Test Series 5, Indian Bank Mock test

#### 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

Indian Bank PO Prelims Online Test Series 5th, Indian Bank Mock test Series 5. Indian Bank PO Pre Free Mock Test Exam 2019. CAknowledge provide free Indian Bank PO Pre Exam Online Quiz 2019. Indian Bank every year invites application from the young and bright candidates for the post of JMG I as Probationary officers. Indian Bank conducts online recruitment to admission to postgraduate diploma in banking & finance course offered through Manipal global education services 2019. Here we are providing** Indian Bank PO Pre. Full Mock Test Paper in English. Indian Bank PO Pre. **Mock Test **Series 5th** 2019. Now Test your self for Indian Bank PO Pre. Exam by using below quiz…

This paper has** 50 questions**.

Time allowed is **50 minutes**.

The **Indian Bank PO Pre. Online Test Series 5th, Indian Bank PO Pre. Free Online 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)

Average score | |

Your score |

#### 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

The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?

CorrectLCM = 1400

1400 – 6 = 1394IncorrectLCM = 1400

1400 – 6 = 1394 - Question 2 of 50
##### 2. Question

Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is:

CorrectLet the number be x. Then,

3x^{2}– 4x = x + 50

3x^{2}– 5x – 50 = 0

(3x + 10)(x – 5) = 0

x = 5IncorrectLet the number be x. Then,

3x^{2}– 4x = x + 50

3x^{2}– 5x – 50 = 0

(3x + 10)(x – 5) = 0

x = 5 - Question 3 of 50
##### 3. Question

What is the ratio between perimeters of two squares one having 3 times the diagonal then the other?

Correctd = 3d d = d

a√2 = 3d a√2 = d

a = 3d/√2 a = d/√2 => 3: 1Incorrectd = 3d d = d

a√2 = 3d a√2 = d

a = 3d/√2 a = d/√2 => 3: 1 - Question 4 of 50
##### 4. Question

A take twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in?

CorrectSuppose A, B and C take x, x/2 and x/3 respectively to finish the work.

Then, (1/x + 2/x + 3/x) = 1/2

6/x = 1/2 => x = 12

So, B takes 6 hours to finish the work.IncorrectSuppose A, B and C take x, x/2 and x/3 respectively to finish the work.

Then, (1/x + 2/x + 3/x) = 1/2

6/x = 1/2 => x = 12

So, B takes 6 hours to finish the work. - Question 5 of 50
##### 5. Question

A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?

CorrectSP of first article = 1000

Profit = 20%

CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3

SP of second article = 1000

Loss = 20%

CP = (SP)*[100/(100-L)] = 5000/4 = 1250

Total SP = 2000

Total CP = 2500/3 + 1250 = 6250/3

CP is more than SP, he makes a loss.

Loss = CP-SP = (6250/3)- 2000 = 250/3

Loss Percent = [(250/3)/(6250/3)]*100 =

0.04 * 100 = 4%IncorrectSP of first article = 1000

Profit = 20%

CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3

SP of second article = 1000

Loss = 20%

CP = (SP)*[100/(100-L)] = 5000/4 = 1250

Total SP = 2000

Total CP = 2500/3 + 1250 = 6250/3

CP is more than SP, he makes a loss.

Loss = CP-SP = (6250/3)- 2000 = 250/3

Loss Percent = [(250/3)/(6250/3)]*100 =

0.04 * 100 = 4% - Question 6 of 50
##### 6. Question

(18

^{2}– 9^{2}* 3) / (675 * ?) = 4Correct(18

^{2}– 9^{2}* 3) / (675 * ?) = 4

=> 27 * ? = 324 – 243

=> 27 * ? = 27(12 – 9) => ? = 3Incorrect(18

^{2}– 9^{2}* 3) / (675 * ?) = 4

=> 27 * ? = 324 – 243

=> 27 * ? = 27(12 – 9) => ? = 3 - Question 7 of 50
##### 7. Question

Calculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?

Correct10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x

10 * 20 * 90 = 15 * 2 * x => x = 6000Incorrect10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x

10 * 20 * 90 = 15 * 2 * x => x = 6000 - Question 8 of 50
##### 8. Question

A 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec. What is the length of the platform?

CorrectSpeed = 300/18 = 50/3 m/sec.

Let the length of the platform be x meters.

Then, (x + 300)/39 = 50/3

3x + 900 = 1950 => x = 350 m.IncorrectSpeed = 300/18 = 50/3 m/sec.

Let the length of the platform be x meters.

Then, (x + 300)/39 = 50/3

3x + 900 = 1950 => x = 350 m. - Question 9 of 50
##### 9. Question

In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can?

CorrectLet the capacity of the can be T litres.

Quantity of milk in the mixture before adding milk = 4/9 (T – 8)

After adding milk, quantity of milk in the mixture = 6/11 T.

6T/11 – 8 = 4/9(T – 8)

10T = 792 – 352 => T = 44.IncorrectLet the capacity of the can be T litres.

Quantity of milk in the mixture before adding milk = 4/9 (T – 8)

After adding milk, quantity of milk in the mixture = 6/11 T.

6T/11 – 8 = 4/9(T – 8)

10T = 792 – 352 => T = 44. - Question 10 of 50
##### 10. Question

In an exam, a candidate secured 504 marks our of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 384 marks. What is the value of M?

Correct504/M = 384/800

(504 * 800) / 384 = M

M = 1050Incorrect504/M = 384/800

(504 * 800) / 384 = M

M = 1050 - Question 11 of 50
##### 11. Question

A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% C.I. The sum borrowed was?

CorrectPrincipal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)

= [882/(1 + 5/100) + 882/(1 + 5/100)^{2}]

= (882 * 20)/21 + (882 * 400)/441 = Rs. 1640.IncorrectPrincipal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)

= [882/(1 + 5/100) + 882/(1 + 5/100)^{2}]

= (882 * 20)/21 + (882 * 400)/441 = Rs. 1640. - Question 12 of 50
##### 12. Question

What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?

Correct8820 —- 441

100 —- ? => 5%

x *105/100 * 105/100 = 8820

x*1.1025=8820

x=8820/1.1025 => 8000Incorrect8820 —- 441

100 —- ? => 5%

x *105/100 * 105/100 = 8820

x*1.1025=8820

x=8820/1.1025 => 8000 - Question 13 of 50
##### 13. Question

The wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?

CorrectTime for one revolution = 60/15 = 4

60/ 20 = 3

60/48 = 5/4

LCM of 4, 3, 5/4

LCM of Numerators/HCF of Denominators =

60/1 = 60IncorrectTime for one revolution = 60/15 = 4

60/ 20 = 3

60/48 = 5/4

LCM of 4, 3, 5/4

LCM of Numerators/HCF of Denominators =

60/1 = 60 - Question 14 of 50
##### 14. Question

The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?

CorrectLCM = 1260

1260 + 7 = 1267IncorrectLCM = 1260

1260 + 7 = 1267 - Question 15 of 50
##### 15. Question

A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?

CorrectLet the sum be Rs. x.

(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840

=> 12x/100 = 840 => x = 7000.IncorrectLet the sum be Rs. x.

(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840

=> 12x/100 = 840 => x = 7000. - Question 16 of 50
##### 16. Question

Find the one which does not belong to that group ?

Correct27, 36, 72 and 45 are divisible by 9, but not 30.

Incorrect27, 36, 72 and 45 are divisible by 9, but not 30.

- Question 17 of 50
##### 17. Question

How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

CorrectTime = (100 * 81) / (450 * 4.5) = 4 years

IncorrectTime = (100 * 81) / (450 * 4.5) = 4 years

- Question 18 of 50
##### 18. Question

Find the length of the wire required to go 15 times round a square field containing 69696 m

^{2}.Correcta

^{2}= 69696 => a = 264

4a = 1056

1056 * 15 = 15840Incorrecta

^{2}= 69696 => a = 264

4a = 1056

1056 * 15 = 15840 - Question 19 of 50
##### 19. Question

A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?

CorrectHere, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅

= ¹²C₅ = ¹²C₇IncorrectHere, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅

= ¹²C₅ = ¹²C₇ - Question 20 of 50
##### 20. Question

Find the roots of quadratic equation: x

^{2}+ x – 42 = 0?Correctx

^{2}+ 7x – 6x + 42 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7)(x – 6) = 0 => x = -7, 6Incorrectx

^{2}+ 7x – 6x + 42 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7)(x – 6) = 0 => x = -7, 6 - Question 21 of 50
##### 21. Question

In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?

CorrectRatio of speeds = 3:4

Ratio of times taken = 4:3

Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.

Then, 4x – 3x = 30/60 => x = 1/2

Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.IncorrectRatio of speeds = 3:4

Ratio of times taken = 4:3

Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.

Then, 4x – 3x = 30/60 => x = 1/2

Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs. - Question 22 of 50
##### 22. Question

A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?

Correct(x – 5)/15 + x/20 = 1

x = 11 3/7 daysIncorrect(x – 5)/15 + x/20 = 1

x = 11 3/7 days - Question 23 of 50
##### 23. Question

What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12% p.a.?

CorrectAmount = [25000 * (1 + 12/100)

^{3}]

= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20

C.I. = (35123.20 – 25000) = Rs. 10123.20IncorrectAmount = [25000 * (1 + 12/100)

^{3}]

= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20

C.I. = (35123.20 – 25000) = Rs. 10123.20 - Question 24 of 50
##### 24. Question

In a mixture of milk and water, the proportion of milk by weight was 80%. If, in a 180 gm mixture, 36 gms of pure milk is added, what would be the percentage of milk in the mixture formed?

CorrectPercentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.

IncorrectPercentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.

- Question 25 of 50
##### 25. Question

Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?

CorrectLet the distance traveled be x km.

Then, x/10 – x/15 = 2

3x – 2x = 60 => x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.

So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph.IncorrectLet the distance traveled be x km.

Then, x/10 – x/15 = 2

3x – 2x = 60 => x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.

So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph. - Question 26 of 50
##### 26. Question

A dishonest dealer professes to sell his goods at Cost Price but still gets 20% profit by using a false weight. What weight does he substitute for a kilogram?

CorrectIf the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.

If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.

How many grams he has to give instead of one kilogram(1000 gm).

120 gm —— 100 gm

1000 gm —— ?

(1000 * 100)/120 = 2500/3 = 833 1/3 grams.IncorrectIf the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.

If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.

How many grams he has to give instead of one kilogram(1000 gm).

120 gm —— 100 gm

1000 gm —— ?

(1000 * 100)/120 = 2500/3 = 833 1/3 grams. - Question 27 of 50
##### 27. Question

A father said to his son, “I was as old as you are at present at the time of your birth.” If the father’s age is 38 years now, the son’s age five years back was:

CorrectLet the son’s present age be x years.

Then, (38 – x) = x

2x = 38 => x = 19

Son’s age 5 years back = (19 – 5) = 14 years.IncorrectLet the son’s present age be x years.

Then, (38 – x) = x

2x = 38 => x = 19

Son’s age 5 years back = (19 – 5) = 14 years. - Question 28 of 50
##### 28. Question

The radius of the base of cone is 3 cm and height is 4 cm. Find the volume of the cone?

Correct1/3 * π * 3 * 3 * 4 = 12 π

Incorrect1/3 * π * 3 * 3 * 4 = 12 π

- Question 29 of 50
##### 29. Question

9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days?

Correct9M + 12B —– 12 days

12M + 12B ——- 10 days

10M + 10B ——-?

108M + 144B = 120M +120B

24B = 12M => 1M = 2B

18B + 12B = 30B —- 12 days

20B + 10B = 30B —–? => 12 daysIncorrect9M + 12B —– 12 days

12M + 12B ——- 10 days

10M + 10B ——-?

108M + 144B = 120M +120B

24B = 12M => 1M = 2B

18B + 12B = 30B —- 12 days

20B + 10B = 30B —–? => 12 days - Question 30 of 50
##### 30. Question

A lends Rs. 2500 to B and a certain to C at the same time at 7% p.a. simple interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is?

CorrectLet the sum lent to C be Rs. x. Then,

(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120

7/25 x = (1120 – 700) => x = 1500IncorrectLet the sum lent to C be Rs. x. Then,

(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120

7/25 x = (1120 – 700) => x = 1500 - Question 31 of 50
##### 31. Question

If 10% of x = 20% of y, then x:y is equal to:

Correct10% of x = 20% of y

10x/100 = 20y/100 => x/10 = y/5

x/y = 10/5 = 2/1

x:y = 2:1.Incorrect10% of x = 20% of y

10x/100 = 20y/100 => x/10 = y/5

x/y = 10/5 = 2/1

x:y = 2:1. - Question 32 of 50
##### 32. Question

The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 40 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.

CorrectLet the number of students who wrote the exam be x.

Total marks of students = 80 x.

Total marks of (x – 5) students = 90(x – 5)

80x – (5 * 40) = 90(x – 5)

250 = 10x => x = 25IncorrectLet the number of students who wrote the exam be x.

Total marks of students = 80 x.

Total marks of (x – 5) students = 90(x – 5)

80x – (5 * 40) = 90(x – 5)

250 = 10x => x = 25 - Question 33 of 50
##### 33. Question

A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?

CorrectRelative speed = 120 + 80 = 200 km/hr.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230.IncorrectRelative speed = 120 + 80 = 200 km/hr.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230. - Question 34 of 50
##### 34. Question

The first three terms of a proportion are 3, 9 and 12. The fourth term is?

Correct(9*12)/3 = 36

Incorrect(9*12)/3 = 36

- Question 35 of 50
##### 35. Question

When 2 is added to half of one-third of one-fifth of a number, the result is one-fifteenth of the number. Find the number?

CorrectLet the number be

2 + 1/2[1/3(a/5)] = a/15

=> 2 = a/30 => a = 60IncorrectLet the number be

2 + 1/2[1/3(a/5)] = a/15

=> 2 = a/30 => a = 60 - Question 36 of 50
##### 36. Question

10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

Correct1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days.Incorrect1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days. - Question 37 of 50
##### 37. Question

A candidate who gets 30% of the marks fails by 50 marks. But another candidate who gets 45% marks gets 25 marks more than necessary for passing. Find the number of marks for passing?

Correct30% ———— 50

45% ———— 25

———————-

15% ————- 75

30% ————– ?

150 + 50 = 200 MarksIncorrect30% ———— 50

45% ———— 25

———————-

15% ————- 75

30% ————– ?

150 + 50 = 200 Marks - Question 38 of 50
##### 38. Question

The incomes of two persons A and B are in the ratio 3:4. If each saves Rs.100 per month, the ratio of their expenditures is 1:2 . Find their incomes?

CorrectThe incomes of A and B be 3P and 4P.

Expenditures = Income – Savings

(3P – 100) and (4P – 100)

The ratio of their expenditure = 1:2

(3P – 100):(4P – 100) = 1:2

2P = 100 => P = 50

Their incomes = 150, 200IncorrectThe incomes of A and B be 3P and 4P.

Expenditures = Income – Savings

(3P – 100) and (4P – 100)

The ratio of their expenditure = 1:2

(3P – 100):(4P – 100) = 1:2

2P = 100 => P = 50

Their incomes = 150, 200 - Question 39 of 50
##### 39. Question

A, B and C rents a pasture for Rs.870. A put in 12 horses for 8 months, B 16 horses for 9 months and 18 horses for 6 months. How much should C pay?

Correct12*8 :16*9 = 18*6

8: 12: 9

9/29 * 870 = 270Incorrect12*8 :16*9 = 18*6

8: 12: 9

9/29 * 870 = 270 - Question 40 of 50
##### 40. Question

1296 ÷ (24 * 0.75) = ?

Correct1296 ÷ 18 = ?

=> ? = 72Incorrect1296 ÷ 18 = ?

=> ? = 72 - Question 41 of 50
##### 41. Question

On dividing number by 5, we get 3 as remainder. What will be the remainder when the square of this number is divided by 5?

CorrectIncorrect - Question 42 of 50
##### 42. Question

The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square.

CorrectLet the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.

a2 = 4096 = 2^{12}

a = (2^{12})^{1/2}= 2^{6}= 64

L = 2a and b = a – 24

b : l = a – 24 : 2a = 40 : 128 = 5 : 16IncorrectLet the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.

a2 = 4096 = 2^{12}

a = (2^{12})^{1/2}= 2^{6}= 64

L = 2a and b = a – 24

b : l = a – 24 : 2a = 40 : 128 = 5 : 16 - Question 43 of 50
##### 43. Question

If a and b are the roots of the equation x

^{2}– 9x + 20 = 0, find the value of a^{2}+ b^{2}+ ab?Correcta

^{2}+ b^{2}+ ab = a^{2}+ b^{2}+ 2ab – ab

i.e., (a + b)^{2}– ab

from x^{2}– 9x + 20 = 0, we have

a + b = 9 and ab = 20. Hence the value of required expression (9)^{2}– 20 = 61.Incorrecta

^{2}+ b^{2}+ ab = a^{2}+ b^{2}+ 2ab – ab

i.e., (a + b)^{2}– ab

from x^{2}– 9x + 20 = 0, we have

a + b = 9 and ab = 20. Hence the value of required expression (9)^{2}– 20 = 61. - Question 44 of 50
##### 44. Question

When 73

^{2}is subtracted form the square of a number, the answer that is obtained is 5280. What is the number?CorrectLet the number be x

x^{2}– 73^{2}= 5280

x^{2}= 5280 + (70+3)^{2}= 5280 + 4900 + 420 + 9 = 10609

= 10000 + 2(100)(3) + 3^{2}= (100 + 3)^{2}

x = 100 + 3 = 103.IncorrectLet the number be x

x^{2}– 73^{2}= 5280

x^{2}= 5280 + (70+3)^{2}= 5280 + 4900 + 420 + 9 = 10609

= 10000 + 2(100)(3) + 3^{2}= (100 + 3)^{2}

x = 100 + 3 = 103. - Question 45 of 50
##### 45. Question

The principal that amounts to Rs. 4913 in 3 years at 6 1/4 % per annum C.I. compounded annually, is?

CorrectPrincipal = [4913 / (1 + 25/(4 * 100))

^{3}]

= 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096.IncorrectPrincipal = [4913 / (1 + 25/(4 * 100))

^{3}]

= 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096. - Question 46 of 50
##### 46. Question

9 3/4 + 7 2/17 – 9 1/15 = ?

CorrectGiven sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)

= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)

= 7 + (765 + 120 – 68)/1020 = 7 817/1020IncorrectGiven sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)

= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)

= 7 + (765 + 120 – 68)/1020 = 7 817/1020 - Question 47 of 50
##### 47. Question

Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x

^{2}+ 8x + 4 = 0?Correcta/b + b/a = (a

^{2}+ b^{2})/ab = (a^{2}+ b^{2}+ a + b)/ab

= [(a + b)^{2}– 2ab]/ab

a + b = -8/1 = -8

ab = 4/1 = 4

Hence a/b + b/a = [(-8)^{2}– 2(4)]/4 = 56/4 = 14.Incorrecta/b + b/a = (a

^{2}+ b^{2})/ab = (a^{2}+ b^{2}+ a + b)/ab

= [(a + b)^{2}– 2ab]/ab

a + b = -8/1 = -8

ab = 4/1 = 4

Hence a/b + b/a = [(-8)^{2}– 2(4)]/4 = 56/4 = 14. - Question 48 of 50
##### 48. Question

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?

CorrectRatio of times taken by A and B = 100:130 = 10:13

Suppose B takes x days to do the work.

x = (23 * 13)/10 = 299/10

A’s 1 day work = 1/23; B’s 1 day work = 10/299

(A + B)’s 1 day work = (1/23 + 10/299) = 1/13

A and B together can complete the job in 13 days.IncorrectRatio of times taken by A and B = 100:130 = 10:13

Suppose B takes x days to do the work.

x = (23 * 13)/10 = 299/10

A’s 1 day work = 1/23; B’s 1 day work = 10/299

(A + B)’s 1 day work = (1/23 + 10/299) = 1/13

A and B together can complete the job in 13 days. - Question 49 of 50
##### 49. Question

1600 men have provisions for 28 days in the temple. If after 4 days, 400 men leave the temple, how long will the food last now?

Correct1600 —- 28 days

1600 —- 24

1200 —- ?

1600*24 = 1200*x

x = 32 daysIncorrect1600 —- 28 days

1600 —- 24

1200 —- ?

1600*24 = 1200*x

x = 32 days - Question 50 of 50
##### 50. Question

How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?

CorrectD = 110 + 170 = 280 m

S = 60 * 5/18 = 50/3

T = 280 * 3/50 = 16.8 secIncorrectD = 110 + 170 = 280 m

S = 60 * 5/18 = 50/3

T = 280 * 3/50 = 16.8 sec