SEBI Online Test in English Series 2, SEBI Online Mock Test 2
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SEBI Online Test in English Series 2, SEBI Online Mock Test 2, SEBI Free Online Test Series 2, SEBI Free Mock Test Exam 2019. SEBI Exam Free Online Quiz 2019, SEBI Full Online Mock Test Series 2nd in English. RRB Online Test for All Subjects, SEBI Free Mock Test Series in English. SEBI Free Mock Test Series 2. SEBI English Language Online Test in English Series 2nd. SEBI Quantitative Aptitude Quiz 2019, SEBI Reasoning Ability Free Online Test. Take SEBI Online Quiz. The SEBI Full online mock test paper is free for all students. SEBI Question and Answers in English and Hindi Series 2. Here we are providing SEBI Full Mock Test Paper in English. SEBI Mock Test Series 2nd 2019. Now Test your self for SEBI Exam by using below quiz…
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Question 1 of 50
1. Question
Two men and three women working 7 hours a day finish a work in 5 days. Four men and four women working 3 hours a day complete the work in 7 days. The number of days in which only 7 men working 4 hours a day will finish the work is?
Correct
2M + 3W —– 35 h
4M + 4W —— 21 h
7M ——? d
70M + 105W = 84M +84M
21W = 14M => 2M = 3W
4 * 35 = 7 * x => x = 20 hours
20/4 = 5 daysIncorrect
2M + 3W —– 35 h
4M + 4W —— 21 h
7M ——? d
70M + 105W = 84M +84M
21W = 14M => 2M = 3W
4 * 35 = 7 * x => x = 20 hours
20/4 = 5 days 
Question 2 of 50
2. Question
In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes?
Correct
Let the total number of valid votes be x.
70% of x = 70/100 * x = 7x/10
Number of votes secured by the other candidate = x – 7x/100 = 3x/10
Given, 7x/10 – 3x/10 = 172 => 4x/10 = 172
=> 4x = 1720 => x = 430.Incorrect
Let the total number of valid votes be x.
70% of x = 70/100 * x = 7x/10
Number of votes secured by the other candidate = x – 7x/100 = 3x/10
Given, 7x/10 – 3x/10 = 172 => 4x/10 = 172
=> 4x = 1720 => x = 430. 
Question 3 of 50
3. Question
The least square number which divides 8, 12 and 18 is?
Correct
LCM = 72
72 * 2 = 144Incorrect
LCM = 72
72 * 2 = 144 
Question 4 of 50
4. Question
The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
Correct
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.Incorrect
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years. 
Question 5 of 50
5. Question
In a three digit number, the hundred digit is 2 more than the tens digit and the units digit is 2 less than the tens digit. If the sum of the digits is 18, find the number?
Correct
Let the three digit numbers be 100a + 10b + c
a = b + 2
c = b – 2
a + b + c = 3b = 18 => b = 6
So a = 8 and b = 4
Hence the three digit number is: 864Incorrect
Let the three digit numbers be 100a + 10b + c
a = b + 2
c = b – 2
a + b + c = 3b = 18 => b = 6
So a = 8 and b = 4
Hence the three digit number is: 864 
Question 6 of 50
6. Question
If the cost price of 12 pens is equal to the selling price of 8 pens, the gain percent is:
Correct
Let C.P. of each pen be Re. 1.
Then, C.P. of 8 pens = Rs. 8; S.P. of 8 pens = Rs. 12.
Gain % = 4/8 * 100 = 50%Incorrect
Let C.P. of each pen be Re. 1.
Then, C.P. of 8 pens = Rs. 8; S.P. of 8 pens = Rs. 12.
Gain % = 4/8 * 100 = 50% 
Question 7 of 50
7. Question
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
Correct
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10.Incorrect
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10. 
Question 8 of 50
8. Question
Two cylinders are of the same height. Their radii are in the ratio 1: 3. If the volume of the first cylinder is 40 cc. Find the volume of the second cylinder?
Correct
r_{1} = x r_{2} = 3x
Π * x^{2} * h = 40
Π9 x^{2}h = 40 * 9 = 360Incorrect
r_{1} = x r_{2} = 3x
Π * x^{2} * h = 40
Π9 x^{2}h = 40 * 9 = 360 
Question 9 of 50
9. Question
Rs.800 amounts to Rs.920 in 3 years at simple interest. If the interest is increased by 3%, it would amount to how much?
Correct
(800*3*3)/100 = 72
920 + 72 = 992Incorrect
(800*3*3)/100 = 72
920 + 72 = 992 
Question 10 of 50
10. Question
The monthly incomes of A and B are in the ratio 5 : 2. B’s monthly income is 12% more than C’s monthly income. If C’s monthly income is Rs. 15000, then find the annual income of A?
Correct
B’s monthly income = 15000 * 112/100 = Rs. 16800
B’s monthly income = 2 parts —> Rs. 16800
A’s monthly income = 5 parts = 5/2 * 16800 = Rs. 42000
A’s annual income = Rs. 42000 * 12 = Rs. 504000Incorrect
B’s monthly income = 15000 * 112/100 = Rs. 16800
B’s monthly income = 2 parts —> Rs. 16800
A’s monthly income = 5 parts = 5/2 * 16800 = Rs. 42000
A’s annual income = Rs. 42000 * 12 = Rs. 504000 
Question 11 of 50
11. Question
Find the one which does not belong to that group ?
Correct
All are blood relation except Daughterinlaw.
Incorrect
All are blood relation except Daughterinlaw.

Question 12 of 50
12. Question
33/4 of 4500 + ? = 118500
Correct
? = 118500 – 33/4 * 4500 = 8.1375 * 10^{4}
Incorrect
? = 118500 – 33/4 * 4500 = 8.1375 * 10^{4}

Question 13 of 50
13. Question
The food in a camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?
Correct
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days.Incorrect
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days. 
Question 14 of 50
14. Question
A pupil’s marks were wrongly entered as 83 instead of 63. Due to the average marks for the class got increased by half. The number of pupils in the class is:
Correct
Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2
x/2 = (83 – 63) => x/2 = 20 => x = 40.Incorrect
Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2
x/2 = (83 – 63) => x/2 = 20 => x = 40. 
Question 15 of 50
15. Question
A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 700 m long?
Correct
L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1900 m
T = D/S
T = 1900/10
T = 190 SecIncorrect
L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1900 m
T = D/S
T = 1900/10
T = 190 Sec 
Question 16 of 50
16. Question
Three years ago the average age of a family of six members was 19 years. A boy have been born, the average age of the family is the same today. What is the age of the boy?
Correct
6 * 22 = 132
7 * 19 = 133
————–
1Incorrect
6 * 22 = 132
7 * 19 = 133
————–
1 
Question 17 of 50
17. Question
The cost price of 13 articles is equal to the selling price of 11 articles. Find the profit percent?
Correct
13 CP = 11 SP
11 — 2 CP
100 — ? =>18 2/11%Incorrect
13 CP = 11 SP
11 — 2 CP
100 — ? =>18 2/11% 
Question 18 of 50
18. Question
In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case?
Correct
Assume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)
= 1200(30)(3) kg
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg
As the same provisions are available
=> 1200(30)(3) = (1200 + x)(25)(2.5)
x = [1200(30)(3)]/[(25)(2.5)] – 1200
x = 528Incorrect
Assume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)
= 1200(30)(3) kg
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg
As the same provisions are available
=> 1200(30)(3) = (1200 + x)(25)(2.5)
x = [1200(30)(3)]/[(25)(2.5)] – 1200
x = 528 
Question 19 of 50
19. Question
The sum of three consecutive integers is 102. Find the lowest of the three?
Correct
Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33.Incorrect
Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33. 
Question 20 of 50
20. Question
Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Correct
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}
Incorrect
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}

Question 21 of 50
21. Question
Two numbers are 30% and 37% are less than a third number .How much percent is the second number less than the first?
Correct
I II III
70 63 100
70 ——– 7
100 —— ? => 10%Incorrect
I II III
70 63 100
70 ——– 7
100 —— ? => 10% 
Question 22 of 50
22. Question
Find the greatest number which is such that when 697, 909 and 1227 are divided by it, the remainders are all the same?
Correct
Incorrect

Question 23 of 50
23. Question
If one third of 3/4 of a number is 21. Then find the number?
Correct
x * 1/3 * 3/4 =21 => x = 84
Incorrect
x * 1/3 * 3/4 =21 => x = 84

Question 24 of 50
24. Question
Two numbers have a H.C.F of 16 and a product of two numbers is 2560. Find the L.C.M of the two numbers?
Correct
L.C.M of two numbers is given by
(Product of the two numbers) / (H.C.F of the two numbers) = 2560/16 = 160.Incorrect
L.C.M of two numbers is given by
(Product of the two numbers) / (H.C.F of the two numbers) = 2560/16 = 160. 
Question 25 of 50
25. Question
A shopkeeper purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs. 6.50 per kg. What will be his gain percent?
Correct
C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 %Incorrect
C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 % 
Question 26 of 50
26. Question
A room is 15m long, 4m broad and 3m height. Find the cost of white washing its four walls at 50p per m^{2 }?
Correct
2 *3(15 + 4) = 114
114 * 1/2 = Rs.57Incorrect
2 *3(15 + 4) = 114
114 * 1/2 = Rs.57 
Question 27 of 50
27. Question
Two trains of equal are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
Correct
Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 – 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.Incorrect
Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 – 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50. 
Question 28 of 50
28. Question
A man misses a bus by 40 minutes if he travels at 30 kmph. If he travels at 40 kmph, then also he misses the bus by 10 minutes. What is the minimum speed required to catch the bus on time?
Correct
Let the distance to be travelled to catch the bus be x km
x/30 – x/40 = 30/60 => (4x – 3x)/120 = 1/2 => x = 60 km
By traavelling 30 kmph time taken = 60/30 = 2 hours
By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph.Incorrect
Let the distance to be travelled to catch the bus be x km
x/30 – x/40 = 30/60 => (4x – 3x)/120 = 1/2 => x = 60 km
By traavelling 30 kmph time taken = 60/30 = 2 hours
By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph. 
Question 29 of 50
29. Question
The sum and the product of two numbers are 25 and 144 respectively, the difference of the number is?
Correct
x + y = 25
xy = 144
(x y)^{2} = (x + y)^{2} – 4xy
(x y)^{2 }= 625 – 576 => (x – y) = 7Incorrect
x + y = 25
xy = 144
(x y)^{2} = (x + y)^{2} – 4xy
(x y)^{2 }= 625 – 576 => (x – y) = 7 
Question 30 of 50
30. Question
Ramu bought an old car for Rs. 42000. He spent Rs. 13000 on repairs and sold it for Rs. 64900. What is his profit percent?
Correct
Total CP = Rs. 42000 + Rs. 13000 = Rs. 55000 and SP = Rs. 64900
Profit(%) = (64900 – 55000)/55000 * 100 = 18%Incorrect
Total CP = Rs. 42000 + Rs. 13000 = Rs. 55000 and SP = Rs. 64900
Profit(%) = (64900 – 55000)/55000 * 100 = 18% 
Question 31 of 50
31. Question
A man purchases 8 pens for Rs.9 and sells 9 pens for Rs.8, how much profit or loss does he make?
Correct
81 — 17
100 —– ? è 20.98%lossIncorrect
81 — 17
100 —– ? è 20.98%loss 
Question 32 of 50
32. Question
A and B’s salaries together amount to Rs. 2,000. A spends 95% of his salary and B spends 85% of his. If now their savings are the same, what is A’s salary?
Correct
(5/100) A = (15/100) B
A = 3B
A + B = 2000
4B = 2000 => B = 500
A = 1500Incorrect
(5/100) A = (15/100) B
A = 3B
A + B = 2000
4B = 2000 => B = 500
A = 1500 
Question 33 of 50
33. Question
Find the one which does not belong to that group ?
Correct
All except Architect are manufactures.
Incorrect
All except Architect are manufactures.

Question 34 of 50
34. Question
Find the one which does not belong to that group ?
Correct
Cat, Dog, Tiger and Lion are carnivores, while Elephant is a herbivore.
Incorrect
Cat, Dog, Tiger and Lion are carnivores, while Elephant is a herbivore.

Question 35 of 50
35. Question
A can do a piece of work in 20 days. B in 15 days A and C in 12 days. In how many days can A finish the work if he is assisted by B on one day and C on the next, alternately?
Correct
A + B = 1/20 + 1/15 = 7/60
A + C = 1/20 + 1/12 = 8/60
7/60 + 8/60 = 15/60 = 1/4
4 days * 2 = 8 daysIncorrect
A + B = 1/20 + 1/15 = 7/60
A + C = 1/20 + 1/12 = 8/60
7/60 + 8/60 = 15/60 = 1/4
4 days * 2 = 8 days 
Question 36 of 50
36. Question
In a twodigit number, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
Correct
Let the ten’s digit be x. Then, unit’s digit = x + 2. Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2
(11x + 2)(2x + 2) = 144
2×2 + 26x – 140 = 0
(x – 2)(11x + 35) = 0
x = 2
Hence, required number = 11x + 2 = 24.Incorrect
Let the ten’s digit be x. Then, unit’s digit = x + 2. Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2
(11x + 2)(2x + 2) = 144
2×2 + 26x – 140 = 0
(x – 2)(11x + 35) = 0
x = 2
Hence, required number = 11x + 2 = 24. 
Question 37 of 50
37. Question
60% of a number is added to 120, the result is the same number. Find the number?
Correct
(60/100) * X + 120 = X
2X = 600
X = 300Incorrect
(60/100) * X + 120 = X
2X = 600
X = 300 
Question 38 of 50
38. Question
The area of a parallelogram is 128sq m and its altitude is twice the corresponding base. Then the length of the base is?
Correct
2x * x = 128 => x= 8
Incorrect
2x * x = 128 => x= 8

Question 39 of 50
39. Question
The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?
Correct
The word MEADOWS has 7 letters of which 3 are vowels.
VVV
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 144.Incorrect
The word MEADOWS has 7 letters of which 3 are vowels.
VVV
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 144. 
Question 40 of 50
40. Question
The length of each side of an equilateral triangle having an area of 4√3 cm^{2} is?
Correct
√3/4a^{2 }= 4√3 → a = 4
Incorrect
√3/4a^{2 }= 4√3 → a = 4

Question 41 of 50
41. Question
The average marks in mathematics scored by the pupils of a school at the public examination were 39. If four of these pupils who actually scored 5, 12, 15 and 19 marks at the examination had not been sent up, the average marks for the school would have been 44. Find the number of pupils sent up for examination from the school?
Correct
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25Incorrect
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25 
Question 42 of 50
42. Question
A sum of money becomes triple itself in 5 years at simple interest. How many years will it become six times at the same rate?
Correct
100 — 200 — 5
200 — 5
100 — 2 1/2
——————
600 — 12 ½ yearsIncorrect
100 — 200 — 5
200 — 5
100 — 2 1/2
——————
600 — 12 ½ years 
Question 43 of 50
43. Question
A batsman in his 17^{th} innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17^{th}innings?
Correct
16x + 85 = 17(x + 3)
x = 34 + 3 = 37Incorrect
16x + 85 = 17(x + 3)
x = 34 + 3 = 37 
Question 44 of 50
44. Question
A man gets a simple interest of Rs.500 on a certain principal at the rate of 5% p.a in two years. Find the compound interest the man will get on twice the principal in two years at the same rate.
Correct
Let the principal be Rs.P
S.I at 5% p.a in 8 years on Rs.P = Rs.500
(P)(8)(5)/100 = 500
P = 1250
C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years
=2500{ [1 + 5/100]^{2} – 1} = 2500{ 21^{2} – 20^{2} /20^{2}}
= 2500/400(441 – 400)
= 25/4(41) = 1025/4 = Rs.256.25Incorrect
Let the principal be Rs.P
S.I at 5% p.a in 8 years on Rs.P = Rs.500
(P)(8)(5)/100 = 500
P = 1250
C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years
=2500{ [1 + 5/100]^{2} – 1} = 2500{ 21^{2} – 20^{2} /20^{2}}
= 2500/400(441 – 400)
= 25/4(41) = 1025/4 = Rs.256.25 
Question 45 of 50
45. Question
1: 3 = 1 2/3: x. The value of x is?
Correct
x * 1 = 3 * 5/3
x = 5Incorrect
x * 1 = 3 * 5/3
x = 5 
Question 46 of 50
46. Question
Three pipes of same capacity can fill a tank in 8 hours. If there are only two pipes of same capacity, the tank can be filled in.
Correct
The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours.Incorrect
The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours. 
Question 47 of 50
47. Question
A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?
Correct
D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec = 3 minsIncorrect
D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec = 3 mins 
Question 48 of 50
48. Question
The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Correct
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.Incorrect
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9. 
Question 49 of 50
49. Question
Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. In how much time will the trains cross each other?
Correct
Relative speed = (54 + 72)* 5/18 = 7 * 5 = 35 mps.
The time required = d/s = (100 + 100 + 200)/35
= 400/35 = 80/7 sec.Incorrect
Relative speed = (54 + 72)* 5/18 = 7 * 5 = 35 mps.
The time required = d/s = (100 + 100 + 200)/35
= 400/35 = 80/7 sec. 
Question 50 of 50
50. Question
Find the one which does not belong to that group ?
Correct
37, 47, 67 and 17 are prime numbers but not 27.
Incorrect
37, 47, 67 and 17 are prime numbers but not 27.