## 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?

CorrectLet 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.IncorrectLet 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^{7m-5}, solve for m.Correct3

^{4m+1}= 3^{7m-5}equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.Incorrect3

^{4m+1}= 3^{7m-5}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?

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 4 of 50
##### 4. Question

The average of first 10 natural numbers is?

CorrectSum of 10 natural no. = 110/2 = 55

Average = 55/10 = 5.5IncorrectSum 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 one-third of the whole trip by plane and the distance traveled by train is three-fifth of the distance traveled by bus. If he traveled by train, plane and bus, then find the distance traveled by bus?

CorrectTotal 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.IncorrectTotal 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?

CorrectA+B+C = 590

5A = 6B = 8C = x

A:B:C = 1/5:1/6:1/8

= 24:20:15

15/59 * 590 = Rs.150IncorrectA+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?

CorrectTotal 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.IncorrectTotal 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 = ?

Correct3 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 = 13¾Incorrect3 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?

CorrectSpeed = 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.IncorrectSpeed = 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?CorrectI. (a + 6)(a + 5) = 0

=> a = -6, -5

II. (b + 5)(b + 1) = 0

=> b = -5, -1 => a ≤ bIncorrectI. (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:

CorrectL.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.IncorrectL.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?

CorrectLet 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 metersIncorrectLet 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?

Correct12 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.Incorrect12 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?

CorrectSuppose 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.IncorrectSuppose 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 three-fifth is 50. What is the number?

CorrectLet the number be x. Then,

x – 3/5 x = 50 => 2/5 x = 50

x = (50 * 5)/2 = 125.IncorrectLet 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 = ?

Correct64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967

Incorrect64309 – 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?

Correct12 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.Incorrect12 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

Correct4500 * x = 3375

= x = 3375/4500 = 3/4Incorrect4500 * 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?

CorrectT = (121 + 165)/ (80 + 65) * 18/5

T = 7.15IncorrectT = (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 ?

CorrectExcept 132, other numbers are odd numbers.

IncorrectExcept 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?

CorrectWork 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.IncorrectWork 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?

CorrectB = 1/16 – 1/24 = 1/48 => 48 days

IncorrectB = 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?

CorrectD = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = 125/3

T = 500 * 3/125 = 12 secIncorrectD = 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 = ?

Correct421 * 0.9 + 130 * 101 + 10000

= 378.9 + 1313 + 10000 = 23508.9 = 23500Incorrect421 * 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?

Correct1/6 – 1/15 = 1/10 => 10

Incorrect1/6 – 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?

Correct441 = P(21/20)

^{2}

P = 400Incorrect441 = P(21/20)

^{2}

P = 400 - Question 30 of 50
##### 30. Question

Find the one which does not belong to that group ?

Correct48, 75, 84 and 57 are divisible by 3 but not 35.

Incorrect48, 75, 84 and 57 are divisible by 3 but not 35.

- Question 31 of 50
##### 31. Question

³√4900 + 123 = ? / 33.004

Correct³√4913 + 123 = ? / 33

=> ? * 1/33 = 140

=> ? = 140 * 33 = 4620Incorrect³√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%?

Correct15 CP = 18 SP

18 — 3 CP loss

100 — ? => 16 2/3% lossIncorrect15 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

Correct9 x + x = 13 + 27

10 x = 40 => x = 4Incorrect9 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:

CorrectLet 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 = 8:9:24IncorrectLet 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?

CorrectAverage = (312 + 162 + 132 + 142 + 122)/5 = 870/5 = 174

IncorrectAverage = (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?

Correct16 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 = 11/53Incorrect16 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?

Correct22/7 r

^{2}= 616 => r = 14

2 * 22/7 * 14 = 88Incorrect22/7 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?

CorrectThree 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.IncorrectThree 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 ?

CorrectExcept Real, all others are rhyming words.

IncorrectExcept 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?

CorrectGiven 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%IncorrectGiven 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?

CorrectSum = (50 * 100) / (2 * 5) = Rs. 500

Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25

C.I. = (551.25 – 500) = Rs. 51.25.IncorrectSum = (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?

CorrectB’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.IncorrectB’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?

CorrectSpeed = 144 * 5/18 = 40 m/sec

Time taken = 100/40 = 2.5 sec.IncorrectSpeed = 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?

CorrectLet 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.IncorrectLet 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?

CorrectLet 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%.IncorrectLet 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 two-digit numbers, which remain the same when the digits interchange their positions, is:

CorrectAverage = (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.IncorrectAverage = (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?CorrectThe 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 = 0IncorrectThe 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?

CorrectA+B+C = 590

5A = 6B = 8C = x

A:B:C = 1/5:1/6:1/8

= 24:20:15

15/59 * 590 = Rs.150IncorrectA+B+C = 590

5A = 6B = 8C = x

A:B:C = 1/5:1/6:1/8

= 24:20:15

15/59 * 590 = Rs.150