SEBI Online Mock Test 5, SEBI Online Test in English Series 5
SEBI Online Mock Test 5, SEBI Online Test in English Series 5
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SEBI Online Mock Test 5, SEBI Online Test in English Series 5, SEBI Free Online Test Series 5. SEBI Exam Online Test 2019, SEBI Free Mock Test Exam 2019. SEBI Exam Free Online Quiz 2019, SEBI Full Online Mock Test Series 5th in English. RRB Online Test for All Subjects, SEBI Free Mock Test Series in English. SEBI Free Mock Test Series 5. SEBI English Language Online Test in English Series 5th. 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 5. Here we are providing SEBI Full Mock Test Paper in English. SEBI Mock Test Series 5th 2019. Now Test your self for SEBI Exam by using below quiz…
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Question 1 of 50
1. Question
At present, the ratio between the ages of Arun and Deepak is 4:3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present?
Correct
Let the present ages of Arun and Deepak be 4x and 3x years respectively.
Then, 4x + 6 = 26 => x = 5
Deepak’s age = 3x = 15 years.Incorrect
Let the present ages of Arun and Deepak be 4x and 3x years respectively.
Then, 4x + 6 = 26 => x = 5
Deepak’s age = 3x = 15 years. 
Question 2 of 50
2. Question
If 3^{4m+1} = 3^{7m5}, solve for m.
Correct
3^{4m+1} = 3^{7m5} equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.
Incorrect
3^{4m+1} = 3^{7m5} equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.

Question 3 of 50
3. 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 4 of 50
4. Question
The average of first 10 natural numbers is?
Correct
Sum of 10 natural no. = 110/2 = 55
Average = 55/10 = 5.5Incorrect
Sum of 10 natural no. = 110/2 = 55
Average = 55/10 = 5.5 
Question 5 of 50
5. Question
A man traveled a total distance of 1800 km. He traveled onethird of the whole trip by plane and the distance traveled by train is threefifth of the distance traveled by bus. If he traveled by train, plane and bus, then find the distance traveled by bus?
Correct
Total distance traveled = 1800 km.
Distance traveled by plane = 600 km.
Distance traveled by bus = x
Distance traveled by train = 3x/5
=> x + 3x/5 + 600 = 1800
=> 8x/5 = 1200 => x = 750 km.Incorrect
Total distance traveled = 1800 km.
Distance traveled by plane = 600 km.
Distance traveled by bus = x
Distance traveled by train = 3x/5
=> x + 3x/5 + 600 = 1800
=> 8x/5 = 1200 => x = 750 km. 
Question 6 of 50
6. Question
Rs.590 is divided amongst A, B, C so that 5 times A’s share, six times B’s share and eight times C’s share are all equal. Find C’s share?
Correct
A+B+C = 590
5A = 6B = 8C = x
A:B:C = 1/5:1/6:1/8
= 24:20:15
15/59 * 590 = Rs.150Incorrect
A+B+C = 590
5A = 6B = 8C = x
A:B:C = 1/5:1/6:1/8
= 24:20:15
15/59 * 590 = Rs.150 
Question 7 of 50
7. Question
A salt manufacturing company produced a total of 5000 tonnes of salt in January of a particular year. Starting from February its production increased by 100 tonnes every month over the previous months until the end of the year. Find its average monthly production for that year?
Correct
Total production of salt by the company in that year = 5000 + 5100 + 5200 + …. + 6100 = 66600.
Average monthly production of salt for that year = 66600/12 = 5550.Incorrect
Total production of salt by the company in that year = 5000 + 5100 + 5200 + …. + 6100 = 66600.
Average monthly production of salt for that year = 66600/12 = 5550. 
Question 8 of 50
8. Question
Positive integers indicated by x and y satisfy 3 1/x * y 2/5 = 13 3/4, the fractions being in their lowest terms, then x = ? and y = ?
Correct
3 1/x * y 2/5 = 13¾
=> (3x + 1) / x + (5y + 1) / 2 += 55 / 4
only choice (4) i.e x = 8 and y = 4 satisfies the given equation.
3 (1/8) * 4 (2/5) = (25 / 8 ) * (22 / 5) = 55/4 = 13¾Incorrect
3 1/x * y 2/5 = 13¾
=> (3x + 1) / x + (5y + 1) / 2 += 55 / 4
only choice (4) i.e x = 8 and y = 4 satisfies the given equation.
3 (1/8) * 4 (2/5) = (25 / 8 ) * (22 / 5) = 55/4 = 13¾ 
Question 9 of 50
9. Question
A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Correct
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.Incorrect
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m. 
Question 10 of 50
10. Question
I. a^{2} + 11a + 30 = 0,
II. b^{2} + 6b + 5 = 0 to solve both the equations to find the values of a and b?Correct
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ bIncorrect
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ b 
Question 11 of 50
11. Question
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Correct
L.C.M of 5, 6, 4 and 3 = 60.
On dividing 2496 by 60, the remainder is 37. Number to be added = 60 – 37 = 23.Incorrect
L.C.M of 5, 6, 4 and 3 = 60.
On dividing 2496 by 60, the remainder is 37. Number to be added = 60 – 37 = 23. 
Question 12 of 50
12. Question
How many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
Correct
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
800+x = (65/3) * 60
800+x = 20 * 65 = 1300
x = 1300 – 800 = 500 metersIncorrect
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
800+x = (65/3) * 60
800+x = 20 * 65 = 1300
x = 1300 – 800 = 500 meters 
Question 13 of 50
13. Question
Twelve men can complete a piece of work in 32 days. The same work can be completed by 16 women in 36 days and by 48 boys in 16 days. Find the time taken by one man, one woman and one boy working together to complete the work?
Correct
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.Incorrect
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together. 
Question 14 of 50
14. Question
The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A?
Correct
(A + B) – (B + C) = 12
A – C = 12Incorrect
(A + B) – (B + C) = 12
A – C = 12 
Question 15 of 50
15. Question
Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Correct
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
Therefore 20x + 25(x – 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.Incorrect
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
Therefore 20x + 25(x – 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m. 
Question 16 of 50
16. Question
The difference between a number and its threefifth is 50. What is the number?
Correct
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.Incorrect
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125. 
Question 17 of 50
17. Question
64309 – 8703 + 798 – 437 = ?
Correct
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967
Incorrect
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967

Question 18 of 50
18. Question
If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?
Correct
12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km.Incorrect
12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km. 
Question 19 of 50
19. Question
4500 * ? = 3375
Correct
4500 * x = 3375
= x = 3375/4500 = 3/4Incorrect
4500 * x = 3375
= x = 3375/4500 = 3/4 
Question 20 of 50
20. Question
Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
Correct
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15Incorrect
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15 
Question 21 of 50
21. Question
Find the one which does not belong to that group ?
Correct
Except 132, other numbers are odd numbers.
Incorrect
Except 132, other numbers are odd numbers.

Question 22 of 50
22. Question
In what time will Rs.4000 lent at 3% per annum on simple interest earn as much interest as Rs.5000 will earn in 5 years at 4% per annum on simple interest?
Correct
(4000*3*R)/100 = (5000*5*4)/100
R = 8 1/3Incorrect
(4000*3*R)/100 = (5000*5*4)/100
R = 8 1/3 
Question 23 of 50
23. Question
Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days, six men left and six women joined, then in hoe many more days will the work be completed?
Correct
Work done by a women in one day = 1/2 (work done by a man/day)
One women’s capacity = 1/2(one man’s capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women – 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.Incorrect
Work done by a women in one day = 1/2 (work done by a man/day)
One women’s capacity = 1/2(one man’s capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women – 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days. 
Question 24 of 50
24. Question
A and B can finish a work in 16 days while A alone can do the same work in 24 days. In how many days B alone will complete the work?
Correct
B = 1/16 – 1/24 = 1/48 => 48 days
Incorrect
B = 1/16 – 1/24 = 1/48 => 48 days

Question 25 of 50
25. Question
Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?
Correct
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 secIncorrect
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec 
Question 26 of 50
26. Question
(332% of 2113) / 42 = ?
Correct
(3.32/100 * 2113) * 1/42 = (70.1516) * 1/42 = 167
Incorrect
(3.32/100 * 2113) * 1/42 = (70.1516) * 1/42 = 167

Question 27 of 50
27. Question
421 * 0.9 + 130 * 101 + 10000 = ?
Correct
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500Incorrect
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500 
Question 28 of 50
28. Question
AA and B together can do a work in 6 days. If A alone can do it in 15 days. In how many days can B alone do it?
Correct
1/6 – 1/15 = 1/10 => 10
Incorrect
1/6 – 1/15 = 1/10 => 10

Question 29 of 50
29. Question
Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.441 in 2 years?
Correct
441 = P(21/20)^{2}
P = 400Incorrect
441 = P(21/20)^{2}
P = 400 
Question 30 of 50
30. Question
Find the one which does not belong to that group ?
Correct
48, 75, 84 and 57 are divisible by 3 but not 35.
Incorrect
48, 75, 84 and 57 are divisible by 3 but not 35.

Question 31 of 50
31. Question
³√4900 + 123 = ? / 33.004
Correct
³√4913 + 123 = ? / 33
=> ? * 1/33 = 140
=> ? = 140 * 33 = 4620Incorrect
³√4913 + 123 = ? / 33
=> ? * 1/33 = 140
=> ? = 140 * 33 = 4620 
Question 32 of 50
32. Question
The C.P of 15 books is equal to the S.P of 18 books. Find his gain% or loss%?
Correct
15 CP = 18 SP
18 — 3 CP loss
100 — ? => 16 2/3% lossIncorrect
15 CP = 18 SP
18 — 3 CP loss
100 — ? => 16 2/3% loss 
Question 33 of 50
33. Question
Solve the equation for x : 6x – 27 + 3x = 4 + 9 – x
Correct
9 x + x = 13 + 27
10 x = 40 => x = 4Incorrect
9 x + x = 13 + 27
10 x = 40 => x = 4 
Question 34 of 50
34. Question
If A:B:C = 2:3:4, then A/B : B/C : C/A is equal to:
Correct
Let A = 2x, B = 3x and C = 4x.
Then, A/B = 2x/3x = 2/3, B/C = 3x/4x = 3/4 and C/A = 4x/2x = 2/1
A/B : B/C : C/A = 2/3 : 3/4 : 2/1 = 8:9:24Incorrect
Let A = 2x, B = 3x and C = 4x.
Then, A/B = 2x/3x = 2/3, B/C = 3x/4x = 3/4 and C/A = 4x/2x = 2/1
A/B : B/C : C/A = 2/3 : 3/4 : 2/1 = 8:9:24 
Question 35 of 50
35. Question
Find the average of the series : 312, 162, 132, 142 and 122?
Correct
Average = (312 + 162 + 132 + 142 + 122)/5 = 870/5 = 174
Incorrect
Average = (312 + 162 + 132 + 142 + 122)/5 = 870/5 = 174

Question 36 of 50
36. Question
Tanya’s grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. Eight years ago, what was the ratio of Tanya’s age to that of her grandfather?
Correct
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53Incorrect
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53 
Question 37 of 50
37. Question
2003 * 2004 – 2001 * 2002 = ?
Correct
(2000 + 3)(2000 + 4) – (2000 + 1)(2000 + 2) = ?
Since (2000 * 2000) – (2000 * 2000) is equal to zero. ?
= (8000 + 6000 + 12) – (4000 + 2000 + 2)
=> ? = 14012 – 6002 = 8010Incorrect
(2000 + 3)(2000 + 4) – (2000 + 1)(2000 + 2) = ?
Since (2000 * 2000) – (2000 * 2000) is equal to zero. ?
= (8000 + 6000 + 12) – (4000 + 2000 + 2)
=> ? = 14012 – 6002 = 8010 
Question 38 of 50
38. Question
If the area of circle is 616 sq cm then its circumference?
Correct
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88Incorrect
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88 
Question 39 of 50
39. Question
The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
Correct
Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.
(2x – 2)^{2} + (2x)^{2} + (2x + 2)^{2} = 1460
4x^{2} – 8x + 4 + 4x^{2} + 8x + 4 = 1460
12x^{2} = 1452 => x^{2} = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24.Incorrect
Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.
(2x – 2)^{2} + (2x)^{2} + (2x + 2)^{2} = 1460
4x^{2} – 8x + 4 + 4x^{2} + 8x + 4 = 1460
12x^{2} = 1452 => x^{2} = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24. 
Question 40 of 50
40. Question
Find the one which does not belong to that group ?
Correct
Except Real, all others are rhyming words.
Incorrect
Except Real, all others are rhyming words.

Question 41 of 50
41. Question
The simple interest accrued on an amount Rs.10,000 at the end of two years is same as the compound interest on Rs.8,000 at the end of two years. The rate of interest is same in both the cases. What is the rate of interest?
Correct
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50%Incorrect
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50% 
Question 42 of 50
42. Question
If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same sum at the rate and for the same time?
Correct
Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25
C.I. = (551.25 – 500) = Rs. 51.25.Incorrect
Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25
C.I. = (551.25 – 500) = Rs. 51.25. 
Question 43 of 50
43. Question
A can finish a work in 18 days B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
Correct
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days.Incorrect
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days. 
Question 44 of 50
44. Question
In what time will a train 100 m long cross an electric pole, it its speed be 144 km/hr?
Correct
Speed = 144 * 5/18 = 40 m/sec
Time taken = 100/40 = 2.5 sec.Incorrect
Speed = 144 * 5/18 = 40 m/sec
Time taken = 100/40 = 2.5 sec. 
Question 45 of 50
45. Question
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
Correct
Let the two consecutive positive integers be x and x + 1
x^{2} + (x + 1)^{2} – x(x + 1) = 91
x^{2} + x – 90 = 0
(x + 10)(x – 9) = 0 => x = 10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.Incorrect
Let the two consecutive positive integers be x and x + 1
x^{2} + (x + 1)^{2} – x(x + 1) = 91
x^{2} + x – 90 = 0
(x + 10)(x – 9) = 0 => x = 10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10. 
Question 46 of 50
46. Question
An automobile financier claims to be lending money at S.I., but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes?
Correct
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%.Incorrect
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%. 
Question 47 of 50
47. Question
25 * 25 / 25 + 15 * 40 = ?
Correct
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.
Incorrect
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.

Question 48 of 50
48. Question
The average of the twodigit numbers, which remain the same when the digits interchange their positions, is:
Correct
Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99)/9
= [(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55]/9
= [(4 * 110) + 55]/9 = 495/9 = 55.Incorrect
Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99)/9
= [(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55]/9
= [(4 * 110) + 55]/9 = 495/9 = 55. 
Question 49 of 50
49. Question
Find the quadratic equations whose roots are the reciprocals of the roots of 2x^{2} + 5x + 3 = 0?
Correct
The quadratic equation whose roots are reciprocal of 2x^{2} + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)^{2} + 5(1/x) + 3 = 0
=> 3x^{2} + 5x + 2 = 0Incorrect
The quadratic equation whose roots are reciprocal of 2x^{2} + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)^{2} + 5(1/x) + 3 = 0
=> 3x^{2} + 5x + 2 = 0 
Question 50 of 50
50. Question
Rs.590 is divided amongst A, B, C so that 5 times A’s share, six times B’s share and eight times C’s share are all equal. Find C’s share?
Correct
A+B+C = 590
5A = 6B = 8C = x
A:B:C = 1/5:1/6:1/8
= 24:20:15
15/59 * 590 = Rs.150Incorrect
A+B+C = 590
5A = 6B = 8C = x
A:B:C = 1/5:1/6:1/8
= 24:20:15
15/59 * 590 = Rs.150