Indian Bank PO Prelims Online Test Series 5, Indian Bank Mock test
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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…
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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?
Correct
LCM = 1400
1400 – 6 = 1394Incorrect
LCM = 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:
Correct
Let the number be x. Then,
3x^{2} – 4x = x + 50
3x^{2} – 5x – 50 = 0
(3x + 10)(x – 5) = 0
x = 5Incorrect
Let 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?
Correct
d = 3d d = d
a√2 = 3d a√2 = d
a = 3d/√2 a = d/√2 => 3: 1Incorrect
d = 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?
Correct
Suppose 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.Incorrect
Suppose 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?
Correct
SP 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/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (6250/3) 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%Incorrect
SP 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/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (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 * ?) = 4
Correct
(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?
Correct
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000Incorrect
10 * 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?
Correct
Speed = 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.Incorrect
Speed = 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?
Correct
Let 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.Incorrect
Let 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?
Correct
504/M = 384/800
(504 * 800) / 384 = M
M = 1050Incorrect
504/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?
Correct
Principal = (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.Incorrect
Principal = (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?
Correct
8820 — 441
100 — ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000Incorrect
8820 — 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?
Correct
Time 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 = 60Incorrect
Time 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?
Correct
LCM = 1260
1260 + 7 = 1267Incorrect
LCM = 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?
Correct
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.Incorrect
Let 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 ?
Correct
27, 36, 72 and 45 are divisible by 9, but not 30.
Incorrect
27, 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?
Correct
Time = (100 * 81) / (450 * 4.5) = 4 years
Incorrect
Time = (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}.
Correct
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840Incorrect
a^{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?
Correct
Here, 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 = ¹²C₇Incorrect
Here, 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?
Correct
x^{2} + 7x – 6x + 42 = 0
x(x + 7) – 6(x + 7) = 0
(x + 7)(x – 6) = 0 => x = 7, 6Incorrect
x^{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?
Correct
Ratio 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.Incorrect
Ratio 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.?
Correct
Amount = [25000 * (1 + 12/100)^{3}]
= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20
C.I. = (35123.20 – 25000) = Rs. 10123.20Incorrect
Amount = [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?
Correct
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.
Incorrect
Percentage 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.?
Correct
Let 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.Incorrect
Let 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?
Correct
If 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.Incorrect
If 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:
Correct
Let 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.Incorrect
Let 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?
Correct
1/3 * π * 3 * 3 * 4 = 12 π
Incorrect
1/3 * 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?
Correct
9M + 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 daysIncorrect
9M + 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?
Correct
Let the sum lent to C be Rs. x. Then,
(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120
7/25 x = (1120 – 700) => x = 1500Incorrect
Let 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:
Correct
10% of x = 20% of y
10x/100 = 20y/100 => x/10 = y/5
x/y = 10/5 = 2/1
x:y = 2:1.Incorrect
10% 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.
Correct
Let 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 = 25Incorrect
Let 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?
Correct
Relative 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.Incorrect
Relative 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 onethird of onefifth of a number, the result is onefifteenth of the number. Find the number?
Correct
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60Incorrect
Let 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?
Correct
1 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.Incorrect
1 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?
Correct
30% ———— 50
45% ———— 25
———————
15% ———— 75
30% ————– ?
150 + 50 = 200 MarksIncorrect
30% ———— 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?
Correct
The 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, 200Incorrect
The 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?
Correct
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270Incorrect
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270 
Question 40 of 50
40. Question
1296 ÷ (24 * 0.75) = ?
Correct
1296 ÷ 18 = ?
=> ? = 72Incorrect
1296 ÷ 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?
Correct
Incorrect

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.
Correct
Let 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 : 16Incorrect
Let 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?
Correct
a^{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.Incorrect
a^{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?
Correct
Let 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.Incorrect
Let 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?
Correct
Principal = [4913 / (1 + 25/(4 * 100))^{3}]
= 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096.Incorrect
Principal = [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 = ?
Correct
Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020Incorrect
Given 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?
Correct
a/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.Incorrect
a/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?
Correct
Ratio 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.Incorrect
Ratio 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?
Correct
1600 — 28 days
1600 — 24
1200 — ?
1600*24 = 1200*x
x = 32 daysIncorrect
1600 — 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?
Correct
D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 secIncorrect
D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec