# RRB SSE Online Test Series 2 | RRB SSE Mock Test Series 2020

## RRB SSE Online Test Series 2 | RRB Mock Test Series2, Online Exam

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**RRB SSE Online Test Series 2 | RRB Mock Test Series 2, Online Preparation: **RRB Senior Section Engineer, Junior Engineer, Non-Technical Exam Free Online Test Series. Free *RRB Online Test* for various competitive examination, entrance examination and campus interview. check out Online mock test series for RRB Senior Section Engineer. RRB Online Test for All Subjects, RRB SSE Free Mock Test Series in English. RRB SSE Free Mock Test **Series 2.** RRB SSE English Language Online Test in English **Series 2nd**. RRB SSE Quantitative Aptitude Quiz 2020, RRB SSE Reasoning Ability Free Online Test. Take RRB SSE Online Quiz. The RRB SSE Full online mock test paper is free for all students. RRB SSE Question and Answers in English and Hindi **Series 2**. Here we are providing** RRB SSE Full Mock Test Paper in English. RRB SSE **Mock Test **Series 2nd** 2020. Now Test your self for RRB SSE Exam by using below quiz…

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- Question 1 of 50
##### 1. Question

The G.C.D of 1.08, 0.36 and 0.9 is

CorrectGiven numbers are 1.08, 0.36 and 0.90.

H.C.F of 108, 36 and 90 is 18.

H.C.F of a given numbers = 0.18IncorrectGiven numbers are 1.08, 0.36 and 0.90.

H.C.F of 108, 36 and 90 is 18.

H.C.F of a given numbers = 0.18 - Question 2 of 50
##### 2. Question

125 liters of a mixture of milk and water contains in the ratio 3:2. How much water should now be added so that the ratio of milk and water becomes 3:4?

CorrectMilk = 3/5 * 125 = 75 liters

Water = 50 liters

75 : (50+P) = 3:4

150 + 3P = 400 => P = 50

50 liters of water are to be added for the ratio become 3:4.IncorrectMilk = 3/5 * 125 = 75 liters

Water = 50 liters

75 : (50+P) = 3:4

150 + 3P = 400 => P = 50

50 liters of water are to be added for the ratio become 3:4. - Question 3 of 50
##### 3. Question

The ratio between the present ages of P and Q is 5:7 respectively. If the difference between Q’s present age and P’s age after 6 years is 2. What is the total of P’s and Q’s present ages?

CorrectLet the present ages of P and Q be 5x and 7x years respectively.

Then, 7x – (5x + 6) = 2

2x = 8 => x = 4

Required sum = 5x + 7x = 12x = 48 years.IncorrectLet the present ages of P and Q be 5x and 7x years respectively.

Then, 7x – (5x + 6) = 2

2x = 8 => x = 4

Required sum = 5x + 7x = 12x = 48 years. - Question 4 of 50
##### 4. Question

The greatest number of five digits which is divisible by 32, 36, 40, 42 and 48 is:

CorrectLCM = 10080

10080) 99999 (8

90720

———

9279

99999 – 9279 = 90720IncorrectLCM = 10080

10080) 99999 (8

90720

———

9279

99999 – 9279 = 90720 - Question 5 of 50
##### 5. Question

A single discount equivalent to the discount series of 20%, 10% and 5% is?

Correct100*(80/100)*(90/100)*(95/100) = 68.4

100 – 68.4 = 31.6Incorrect100*(80/100)*(90/100)*(95/100) = 68.4

100 – 68.4 = 31.6 - Question 6 of 50
##### 6. Question

Find the lowest common multiple of 24, 36 and 40.

Correct2 24 – 36 – 40

——————–

2 12 – 18 – 20

——————–

2 6 – 9 – 10

——————-

3 3 – 9 – 5

——————–

1 – 3 – 5

L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360.Incorrect2 24 – 36 – 40

——————–

2 12 – 18 – 20

——————–

2 6 – 9 – 10

——————-

3 3 – 9 – 5

——————–

1 – 3 – 5

L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360. - Question 7 of 50
##### 7. Question

A 70 cm long wire is to be cut into two pieces so that one piece will be 2/5th of the other, how many centimeters will the shorter piece be?

Correct1: 2/5 = 5: 2

2/7 * 70 = 20Incorrect1: 2/5 = 5: 2

2/7 * 70 = 20 - Question 8 of 50
##### 8. Question

Two-fifth of one-third of three-seventh of a number is 15. What is 40% of that number?

CorrectLet the number be x. Then,

2/5 of 1/3 of 3/7 of

x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2

40% of 525/2 = (40/100 * 525/2) = 105IncorrectLet the number be x. Then,

2/5 of 1/3 of 3/7 of

x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2

40% of 525/2 = (40/100 * 525/2) = 105 - Question 9 of 50
##### 9. Question

The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p?

CorrectLet the five consecutive odd numbers of set p be 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

= 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10n + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

The largest number of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.IncorrectLet the five consecutive odd numbers of set p be 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

= 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10n + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

The largest number of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670. - Question 10 of 50
##### 10. Question

CorrectIncorrect - Question 11 of 50
##### 11. Question

What is the least number which when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20 respectively?

CorrectIncorrect - Question 12 of 50
##### 12. Question

The cost of 16 pens and 8 pencils is Rs.352 and the cost of 4 pens and 4 pencils is Rs.96. Find the cost of each pen?

CorrectLet the cost of each pen and pencil be ‘p’ and ‘q’ respectively.

16p + 8q = 352 — (1)

4p + 4q = 96

8p + 8q = 192 — (2)

(1) – (2) => 8p = 160

=> p = 20IncorrectLet the cost of each pen and pencil be ‘p’ and ‘q’ respectively.

16p + 8q = 352 — (1)

4p + 4q = 96

8p + 8q = 192 — (2)

(1) – (2) => 8p = 160

=> p = 20 - Question 13 of 50
##### 13. 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 14 of 50
##### 14. Question

I. x

^{2}+ 11x + 30 = 0,

II. y^{2}+ 15y + 56 = 0 to solve both the equations to find the values of x and y?CorrectI. x

^{2}+ 6x + 5x + 30 = 0

=>(x + 6)(x + 5) = 0 => x = -6, -5

II. y^{2}+ 8y + 7y + 56 = 0

=>(y + 8)(y + 7) = 0 => y = -8, -7

=> x > yIncorrectI. x

^{2}+ 6x + 5x + 30 = 0

=>(x + 6)(x + 5) = 0 => x = -6, -5

II. y^{2}+ 8y + 7y + 56 = 0

=>(y + 8)(y + 7) = 0 => y = -8, -7

=> x > y - Question 15 of 50
##### 15. Question

The L.C.M of two numbers is 48. The numbers are in the ratio 2:3. The sum of numbers is:

CorrectLet the numbers be 2x and 3x.

Then, their L.C.M = 6x. So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.IncorrectLet the numbers be 2x and 3x.

Then, their L.C.M = 6x. So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40. - Question 16 of 50
##### 16. Question

The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is:

CorrectLet the numbers be a and b. Then,

ab = 17, a = 1 and b = 17So, 1/a^{2}+ 1/b^{2}= (a^{2}+ b^{2})/a^{2}b^{2}= (1^{2}+ 17^{2})/(1 * 17)^{2}= 290/289IncorrectLet the numbers be a and b. Then,

ab = 17, a = 1 and b = 17So, 1/a^{2}+ 1/b^{2}= (a^{2}+ b^{2})/a^{2}b^{2}= (1^{2}+ 17^{2})/(1 * 17)^{2}= 290/289 - Question 17 of 50
##### 17. Question

The average of 11 results is 50, if the average of first six results is 49 and that of the last six is 52. Find the sixth result?

Correct1 to 11 = 11 * 50 = 550

1 to 6 = 6 * 49 = 294

6 to 11 = 6 * 52 = 312

6^{th}= 294 + 312 – 550 = 56Incorrect1 to 11 = 11 * 50 = 550

1 to 6 = 6 * 49 = 294

6 to 11 = 6 * 52 = 312

6^{th}= 294 + 312 – 550 = 56 - Question 18 of 50
##### 18. Question

16 men can complete a piece of work in 25 days. In how many days can 20 men complete that piece of work?

Correct16 * 25 = 20 * x => x = 20 days

Incorrect16 * 25 = 20 * x => x = 20 days

- Question 19 of 50
##### 19. Question

Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in?

CorrectNet part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60

The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.IncorrectNet part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60

The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs. - Question 20 of 50
##### 20. Question

A car covers a distance of 624 km in 6 ½ hours. Find its speed?

Correct624/6 = 104 kmph

Incorrect624/6 = 104 kmph

- Question 21 of 50
##### 21. Question

Excluding stoppages, the average speed of a bus is 60 km/hr and including stoppages, the average speed of the bus is 40 km/hr. For how many minutes does the bus stop per hour?

CorrectIn 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 – 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min.IncorrectIn 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 – 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min. - Question 22 of 50
##### 22. Question

Rs.525 among A, B and C such that B may get 2/3 of A and C together get. Find the share of C?

CorrectIncorrect - Question 23 of 50
##### 23. Question

A man walked 20 m to cross a rectangular field diagonally. If the length of the field is 16 cm. Find the breadth of the field is?

CorrectIncorrect - Question 24 of 50
##### 24. Question

A batsman makes a score of 64 runs in the 16th innings and thus increased his average by 3. Find his average after the 16th inning?

CorrectLet the average after the 16th inning be P.

So, the average after the 15th inning will be (P-3) Hence, 15(P-30) + 64 = 16P => P = 19.IncorrectLet the average after the 16th inning be P.

So, the average after the 15th inning will be (P-3) Hence, 15(P-30) + 64 = 16P => P = 19. - Question 25 of 50
##### 25. Question

A is twice as good a work man as B and together they finish the work in 14 days. In how many days A alone can finish the work?

CorrectWC = 2:1

2x + x = 1/14 => x = 1/42

2x = 1/21

A can do the work in 21 days.IncorrectWC = 2:1

2x + x = 1/14 => x = 1/42

2x = 1/21

A can do the work in 21 days. - Question 26 of 50
##### 26. Question

12 3/4 * (1 2/17 / 8 1/6) / 5 4/3 = ?

CorrectBy applying BODMAS rule.

51/4 * (19/17 / 49/6) / 19/3

= 51/4 * ( 19/17 * 6/49) * 3/19 = 27/98IncorrectBy applying BODMAS rule.

51/4 * (19/17 / 49/6) / 19/3

= 51/4 * ( 19/17 * 6/49) * 3/19 = 27/98 - Question 27 of 50
##### 27. Question

(17

^{14}* 17^{16}) / 17^{-8}= ?Correct? = 17

^{14+16-8}= 17^{22}Incorrect? = 17

^{14+16-8}= 17^{22} - Question 28 of 50
##### 28. Question

The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?

CorrectLet the prices of X, Y and Z be 8k, 5k and 3k respectively.

After increase

Price of X = 8k * 125/100 = 10k

Price of Y = 5k * 120/100 = 6k

Price of Z = 3k * (133 1/3)/100 = 4k

Required ratio = 10k : 6k : 4k = 5 : 3 : 2.IncorrectLet the prices of X, Y and Z be 8k, 5k and 3k respectively.

After increase

Price of X = 8k * 125/100 = 10k

Price of Y = 5k * 120/100 = 6k

Price of Z = 3k * (133 1/3)/100 = 4k

Required ratio = 10k : 6k : 4k = 5 : 3 : 2. - Question 29 of 50
##### 29. Question

Jacob brought a scooter for a certain sum of money. He spent 10% of the cost on repairs and sold the scooter for a profit of Rs. 1100. How much did he spend on repairs if he made a profit of 20%?

CorrectLet the C.P. be Rs. x. Then, 20% of x = 1100

20/100 * x = 1100 => x = 5500

C.P. = Rs. 5500, expenditure on repairs = 10%

Actual price = Rs. (100 * 5500)/110 = Rs. 5000

Expenditures on repairs = (5500 – 5000) = Rs. 500.IncorrectLet the C.P. be Rs. x. Then, 20% of x = 1100

20/100 * x = 1100 => x = 5500

C.P. = Rs. 5500, expenditure on repairs = 10%

Actual price = Rs. (100 * 5500)/110 = Rs. 5000

Expenditures on repairs = (5500 – 5000) = Rs. 500. - Question 30 of 50
##### 30. Question

A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 1/3 hours to fill the tank. The leak can drain all the water of the tank in?

CorrectWork done by the tank in 1 hour = (1/2 – 1/3) = 1/14 Leak will empty the tank in 14 hrs.

IncorrectWork done by the tank in 1 hour = (1/2 – 1/3) = 1/14 Leak will empty the tank in 14 hrs.

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

If the price of an article went up by 20%, then by what percent should it be brought down to bring it back to its original price?

CorrectLet the price of the article be Rs. 100.

20% of 100 = 20.

New price = 100 + 20 = Rs. 120

Required percentage = (120 – 100)/120 * 100

= 20/120 * 100 = 50/3 = 16 2/3%.IncorrectLet the price of the article be Rs. 100.

20% of 100 = 20.

New price = 100 + 20 = Rs. 120

Required percentage = (120 – 100)/120 * 100

= 20/120 * 100 = 50/3 = 16 2/3%. - Question 32 of 50
##### 32. Question

The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?

CorrectSum of the 10 numbers = 230

If each number is increased by 4, the total increase =

4 * 10 = 40

The new sum = 230 + 40 = 270 The new average = 270/10 = 27.IncorrectSum of the 10 numbers = 230

If each number is increased by 4, the total increase =

4 * 10 = 40

The new sum = 230 + 40 = 270 The new average = 270/10 = 27. - Question 33 of 50
##### 33. Question

Two varieties of wheat – A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made?

CorrectLet the quantities of A and B mixed be 3x kg and 7x kg.

Cost of 3x kg of A = 9(3x) = Rs. 27x

Cost of 7x kg of B = 15(7x) = Rs. 105x

Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x

Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66

Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50IncorrectLet the quantities of A and B mixed be 3x kg and 7x kg.

Cost of 3x kg of A = 9(3x) = Rs. 27x

Cost of 7x kg of B = 15(7x) = Rs. 105x

Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x

Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66

Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50 - Question 34 of 50
##### 34. Question

Roja and Pooja start moving in the opposite directions from a pole. They are moving at the speeds of 2 km/hr and 3 km/hr respectively. After 4 hours what will be the distance between them?

CorrectDistance = Relative Speed * Time

= (2 + 3) * 4 = 20 km

[ They are travelling in the opposite direction, relative speed = sum of the speeds].IncorrectDistance = Relative Speed * Time

= (2 + 3) * 4 = 20 km

[ They are travelling in the opposite direction, relative speed = sum of the speeds]. - Question 35 of 50
##### 35. Question

[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?

Correct[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?

=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?

=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?

=[8 + {32} ÷ 4] = ? = 16Incorrect[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?

=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?

=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?

=[8 + {32} ÷ 4] = ? = 16 - Question 36 of 50
##### 36. 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 37 of 50
##### 37. Question

The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is?

Correct26

^{2}– 10^{2}= 24^{2}

d_{1 }= 20 d_{2 }= 48

1/2 * 20 * 48 = 480Incorrect26

^{2}– 10^{2}= 24^{2}

d_{1 }= 20 d_{2 }= 48

1/2 * 20 * 48 = 480 - Question 38 of 50
##### 38. Question

The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?

CorrectLCM = 48 + 3 = 51

IncorrectLCM = 48 + 3 = 51

- Question 39 of 50
##### 39. Question

Find the one which does not belong to that group ?

CorrectExcept 110, other numbers are divisible by 4.

IncorrectExcept 110, other numbers are divisible by 4.

- Question 40 of 50
##### 40. Question

Find the one which does not belong to that group ?

Correct36 = 6

^{2}, 49 = 7^{2}, 64 = 8^{2}, 81 = 9^{2}and 100 = 10^{2}.

36, 64, 81 and 100 are squares of composite numbers, but not 49.Incorrect36 = 6

^{2}, 49 = 7^{2}, 64 = 8^{2}, 81 = 9^{2}and 100 = 10^{2}.

36, 64, 81 and 100 are squares of composite numbers, but not 49. - Question 41 of 50
##### 41. Question

In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient?

CorrectNumber = (12 * 35) = 420

Correct quotient = 420/21 = 20IncorrectNumber = (12 * 35) = 420

Correct quotient = 420/21 = 20 - Question 42 of 50
##### 42. 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?

Correctr

_{1}= x r_{2}= 3x

Π * x^{2}* h = 40

Π9 x^{2}h = 40 * 9 = 360Incorrectr

_{1}= x r_{2}= 3x

Π * x^{2}* h = 40

Π9 x^{2}h = 40 * 9 = 360 - Question 43 of 50
##### 43. Question

A certain sum becomes four times itself at simple interest in eight years. In how many years does it become ten times itself?

CorrectLet the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.

R = (100 * 3x)/(x * 8) = 300/8 %

If the sum becomes ten times itself, then interest is 9x.

The required time period = (100 * 9x)/(x * 300/8) = (100 * 9x * 8)/(x * 300) = 24 years.IncorrectLet the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.

R = (100 * 3x)/(x * 8) = 300/8 %

If the sum becomes ten times itself, then interest is 9x.

The required time period = (100 * 9x)/(x * 300/8) = (100 * 9x * 8)/(x * 300) = 24 years. - Question 44 of 50
##### 44. Question

Find the one which does not belong to that group ?

Correct30 = 3

^{3}+ 3, 630 = 5^{4}+ 5, 10 = 2^{3}+ 2, 520 = 8^{3}+ 8 and 130 = 5^{3}+ 5.

30, 10, 130 and 520 can be expressed as n^{3}+ n but not 630.Incorrect30 = 3

^{3}+ 3, 630 = 5^{4}+ 5, 10 = 2^{3}+ 2, 520 = 8^{3}+ 8 and 130 = 5^{3}+ 5.

30, 10, 130 and 520 can be expressed as n^{3}+ n but not 630. - Question 45 of 50
##### 45. Question

I. a

^{2}– 9a + 20 = 0,

II. 2b^{2}– 5b – 12 = 0 to solve both the equations to find the values of a and b?CorrectI. (a – 5)(a – 4) = 0

=> a = 5, 4

II. (2b + 3)(b – 4) = 0

=> b = 4, -3/2 => a ≥ bIncorrectI. (a – 5)(a – 4) = 0

=> a = 5, 4

II. (2b + 3)(b – 4) = 0

=> b = 4, -3/2 => a ≥ b - Question 46 of 50
##### 46. Question

An express traveled at an average speed of 100 km/hr, stopping for 3 min after every 75 kn. How long did it take to reach its destination 600 km from the starting point?

CorrectTime taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 – 1 = 7

Total time of stoppages = 3 * 7 = 21 min

Hence, total time taken = 6 hrs 21 min.IncorrectTime taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 – 1 = 7

Total time of stoppages = 3 * 7 = 21 min

Hence, total time taken = 6 hrs 21 min. - Question 47 of 50
##### 47. Question

A bag contains equal number of Rs.5, Rs.2 and Re.1 coins. If the total amount in the bag is Rs.1152, find the number of coins of each kind?

CorrectLet the number of coins of each kind be x.

=> 5x + 2x + 1x = 1152

=> 8x = 1152 => x = 144IncorrectLet the number of coins of each kind be x.

=> 5x + 2x + 1x = 1152

=> 8x = 1152 => x = 144 - Question 48 of 50
##### 48. Question

Anita, Indu and Geeta can do a piece of work in 18 days, 27 days and 36 days respectively. They start working together. After working for 4 days. Anita goes away and Indu leaves 7 days before the work is finished. Only Geeta remains at work from beginning to end. In how many days was the whole work done?

Correct4/18 + (x -7)/27 + x/36 = 1

x = 16 daysIncorrect4/18 + (x -7)/27 + x/36 = 1

x = 16 days - Question 49 of 50
##### 49. Question

In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?

CorrectBy the time A covers 1000 m, B covers (1000 – 50) = 950 m.

By the time B covers 1000 m, C covers (1000 – 100) = 900 m.

So, the ratio of speeds of A and C =

1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.

So in 1000 m race A beats C by 1000 – 855 = 145 m.IncorrectBy the time A covers 1000 m, B covers (1000 – 50) = 950 m.

By the time B covers 1000 m, C covers (1000 – 100) = 900 m.

So, the ratio of speeds of A and C =

1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.

So in 1000 m race A beats C by 1000 – 855 = 145 m. - Question 50 of 50
##### 50. Question

The cost of 10 kg of apples is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour?

CorrectLet the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.

10a = 24r and 6 * 20.50 = 2r

a = 12/5 r and r = 61.5

a = 147.6

Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5

= 590.4 + 184.5 + 102.5 = Rs.877.40IncorrectLet the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.

10a = 24r and 6 * 20.50 = 2r

a = 12/5 r and r = 61.5

a = 147.6

Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5

= 590.4 + 184.5 + 102.5 = Rs.877.40