RRB SSE Online Test Series 2  RRB SSE Mock Test Series 2020
RRB SSE Online Test Series 2  RRB Mock Test Series2, Online Exam
Finish Quiz
0 of 50 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
Information
RRB SSE Online Test Series 2  RRB Mock Test Series 2, Online Preparation: RRB Senior Section Engineer, Junior Engineer, NonTechnical 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 2021, 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 2021. Now Test your self for RRB SSE Exam by using below quiz…
This paper has 50 questions.
Time allowed is 50 minutes.
The RRB SSE Online Test Series 1st, RRB SSE Free Online Test Exam is Very helpful for all students. Now Scroll down below n click on “Start Quiz” or “Start Test” and Test yourself.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 50 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 Answered
 Review

Question 1 of 50
1. Question
The G.C.D of 1.08, 0.36 and 0.9 is
Correct
Given 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 = 0.18Incorrect
Given 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?
Correct
Milk = 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.Incorrect
Milk = 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?
Correct
Let 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.Incorrect
Let 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:
Correct
LCM = 10080
10080) 99999 (8
90720
———
9279
99999 – 9279 = 90720Incorrect
LCM = 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?
Correct
100*(80/100)*(90/100)*(95/100) = 68.4
100 – 68.4 = 31.6Incorrect
100*(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.
Correct
2 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.Incorrect
2 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?
Correct
1: 2/5 = 5: 2
2/7 * 70 = 20Incorrect
1: 2/5 = 5: 2
2/7 * 70 = 20 
Question 8 of 50
8. Question
Twofifth of onethird of threeseventh of a number is 15. What is 40% of that number?
Correct
Let 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) = 105Incorrect
Let 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?
Correct
Let 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.Incorrect
Let 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
Correct
Incorrect

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?
Correct
Incorrect

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?
Correct
Let 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 = 20Incorrect
Let 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?
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 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?Correct
I. 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 > yIncorrect
I. 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:
Correct
Let 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.Incorrect
Let 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:
Correct
Let 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} = 290/289Incorrect
Let 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?
Correct
1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6^{th} = 294 + 312 – 550 = 56Incorrect
1 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?
Correct
16 * 25 = 20 * x => x = 20 days
Incorrect
16 * 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?
Correct
Net 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.Incorrect
Net 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?
Correct
624/6 = 104 kmph
Incorrect
624/6 = 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?
Correct
In 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.Incorrect
In 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?
Correct
Incorrect

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?
Correct
Incorrect

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?
Correct
Let the average after the 16th inning be P.
So, the average after the 15th inning will be (P3) Hence, 15(P30) + 64 = 16P => P = 19.Incorrect
Let the average after the 16th inning be P.
So, the average after the 15th inning will be (P3) Hence, 15(P30) + 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?
Correct
WC = 2:1
2x + x = 1/14 => x = 1/42
2x = 1/21
A can do the work in 21 days.Incorrect
WC = 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 = ?
Correct
By applying BODMAS rule.
51/4 * (19/17 / 49/6) / 19/3
= 51/4 * ( 19/17 * 6/49) * 3/19 = 27/98Incorrect
By 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+168} = 17^{22}
Incorrect
? = 17^{14+168} = 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?
Correct
Let 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.Incorrect
Let 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%?
Correct
Let 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.Incorrect
Let 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?
Correct
Work done by the tank in 1 hour = (1/2 – 1/3) = 1/14 Leak will empty the tank in 14 hrs.
Incorrect
Work 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?
Correct
Let 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%.Incorrect
Let 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?
Correct
Sum 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.Incorrect
Sum 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?
Correct
Let 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. 16.50Incorrect
Let 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?
Correct
Distance = Relative Speed * Time
= (2 + 3) * 4 = 20 km
[ They are travelling in the opposite direction, relative speed = sum of the speeds].Incorrect
Distance = 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?
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 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?
Correct
26^{2} – 10^{2} = 24^{2}
d_{1 }= 20 d_{2 }= 48
1/2 * 20 * 48 = 480Incorrect
26^{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?
Correct
LCM = 48 + 3 = 51
Incorrect
LCM = 48 + 3 = 51

Question 39 of 50
39. Question
Find the one which does not belong to that group ?
Correct
Except 110, other numbers are divisible by 4.
Incorrect
Except 110, other numbers are divisible by 4.

Question 40 of 50
40. Question
Find the one which does not belong to that group ?
Correct
36 = 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.Incorrect
36 = 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?
Correct
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20Incorrect
Number = (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?
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 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?
Correct
Let 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.Incorrect
Let 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 ?
Correct
30 = 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.Incorrect
30 = 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?Correct
I. (a – 5)(a – 4) = 0
=> a = 5, 4
II. (2b + 3)(b – 4) = 0
=> b = 4, 3/2 => a ≥ bIncorrect
I. (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?
Correct
Time 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.Incorrect
Time 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?
Correct
Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144Incorrect
Let 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?
Correct
4/18 + (x 7)/27 + x/36 = 1
x = 16 daysIncorrect
4/18 + (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?
Correct
By 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.Incorrect
By 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?
Correct
Let 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.877.40Incorrect
Let 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