RBI Assistant Mains Mock Test, RBI Assistant mains online test 2021
RBI Assistant Mains Mock Test, RBI Assistant mains online test series 2
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
RBI Assistant Mains Mock Test, RBI Assistant mains online test series 2. RBI Exam Online Test 2021, Take CAknowledge RBI Assistant Mains Free Quiz Exam 2021. RBI Assistant Mains Exam Free Online Quiz 2021, RBI Assistant Mains Full Online Mock Test Series 2nd in English. To Analyse your preparation, one should attempt the test series on a regular basis to score more in the examination. Start FREE mock test and get top rank among all applicants whoever are participating in the exam. RBI Assistant Mains Question and Answers in English and Hindi Series 2. Here we are providing RBI Assistant Mains Full Mock Test Paper in English. RBI Assistant Mains Mock Test Series 2nd 2021. Now Test your self for RBI Assistant Mains Exam by using below quiz…
This paper has 50 questions.
Time allowed is 50 minutes.
The RBI Assistant Mains Online Test Series 2nd, RBI Assistant Mains 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
[10 + {4 * ({2/3 + 1/4} * √144/121 + 23) ÷ 12 + 5} – 3] = ?
Correct
[10 + { 4 * (2/3 + 1/4 * √144/121 + 23) ÷ 12 + 5} – 3] = ? [10 + { 4 * (11/12 * 12/11 + 23) ÷ 12 + 5 } – 3] = ?
= [10 + {4 * 24 ÷ 12 + 5} – 3] = ?
= [10 + 13 – 3] = ? = > 20Incorrect
[10 + { 4 * (2/3 + 1/4 * √144/121 + 23) ÷ 12 + 5} – 3] = ? [10 + { 4 * (11/12 * 12/11 + 23) ÷ 12 + 5 } – 3] = ?
= [10 + {4 * 24 ÷ 12 + 5} – 3] = ?
= [10 + 13 – 3] = ? = > 20 
Question 2 of 50
2. Question
Two tests had the same maximum mark. The pass percentages in the first and the second test were 40% and 45% respectively. A candidate scored 216 marks in the second test and failed by 36 marks in that test. Find the pass mark in the first test?
Correct
Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.Incorrect
Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224. 
Question 3 of 50
3. 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 4 of 50
4. Question
2222.2 + 222.22 + 22.222 = ?
Correct
Incorrect

Question 5 of 50
5. Question
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in?
Correct
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.Incorrect
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days. 
Question 6 of 50
6. Question
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
Correct
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
= [3/7 of {21 * (1/3 * 30 ÷ 5) – 7} – 8] = ?
= 3/7 of [{21 * 2 – 7} – 8] = ? => {3/7 * 35 – 8} = ? => 7Incorrect
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
= [3/7 of {21 * (1/3 * 30 ÷ 5) – 7} – 8] = ?
= 3/7 of [{21 * 2 – 7} – 8] = ? => {3/7 * 35 – 8} = ? => 7 
Question 7 of 50
7. Question
What percentage of numbers from 1 to 70 have squares that end in the digit 1?
Correct
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers = 14.
Required percentage = (14/70 * 100) = 20%Incorrect
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers = 14.
Required percentage = (14/70 * 100) = 20% 
Question 8 of 50
8. Question
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = ?
Correct
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = (2)^{2} * (3)^{2} * 1/(2^{3})^{2}
= 4 * 9 * 64 = 2304.Incorrect
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = (2)^{2} * (3)^{2} * 1/(2^{3})^{2}
= 4 * 9 * 64 = 2304. 
Question 9 of 50
9. Question
Six years ago, the ratio of ages of Kunal and Sagar was 6:5. Four years hence, the ratio of their ages will be 11:10. What is Sagar’s age at present?
Correct
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10
10(6x + 10) = 11(5x + 10) => x = 2
Sagar’s present age = (5x + 6) = 16 years.Incorrect
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10
10(6x + 10) = 11(5x + 10) => x = 2
Sagar’s present age = (5x + 6) = 16 years. 
Question 10 of 50
10. Question
Mudit’s age 18 years hence will be thrice his age four years ago. Find Mudit’s present age?
Correct
Let Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years.Incorrect
Let Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years. 
Question 11 of 50
11. Question
The least number of four digits which is divisible by 4, 6, 8 and 10 is?
Correct
LCM = 120
120) 1000 (8
960
——
40
1000 + 120 – 40 = 1080Incorrect
LCM = 120
120) 1000 (8
960
——
40
1000 + 120 – 40 = 1080 
Question 12 of 50
12. Question
The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:10. If P invested the money for 5 months, find for how much time did Q invest the money?
Correct
7*5: 5*x = 7:10
x = 10Incorrect
7*5: 5*x = 7:10
x = 10 
Question 13 of 50
13. Question
There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits?
Correct
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.Incorrect
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36. 
Question 14 of 50
14. Question
A, B, C together started a business. A invested Rs.6000 for 5 months B invested Rs.3600 for 6 months and C Rs.7500 for 3 months. If they get a total profit of Rs.7410. Find the share of A?
Correct
60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000Incorrect
60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000 
Question 15 of 50
15. Question
The amount of water (in ml) that should be added to reduce 9 ml. Lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is?
Correct
4.5 4.5
30% 70%
30% —– 4.5
70% ——? => 10.5 – 4.5 = 6 mlIncorrect
4.5 4.5
30% 70%
30% —– 4.5
70% ——? => 10.5 – 4.5 = 6 ml 
Question 16 of 50
16. Question
A dishonest dealer professes to sell goods at the cost price but uses a false weight and gains 25%. Find his false weight age?
Correct
25 = E/(1000 – E) * 100
1000 – E = 4E
1000 = 5E => E = 200
1000 – 200 = 800Incorrect
25 = E/(1000 – E) * 100
1000 – E = 4E
1000 = 5E => E = 200
1000 – 200 = 800 
Question 17 of 50
17. Question
Find the one which does not belong to that group ?
Correct
Boxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.
Incorrect
Boxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.

Question 18 of 50
18. Question
A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the room?
Correct
Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117Incorrect
Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117 
Question 19 of 50
19. Question
Rs. 1300 is divided into three parts A, B and C. How much A is more than C if their ratio is 1/2:1/3:1/4?
Correct
1/2:1/3:1/4 = 6:4:3
3/13*1300 = 300Incorrect
1/2:1/3:1/4 = 6:4:3
3/13*1300 = 300 
Question 20 of 50
20. 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 21 of 50
21. Question
Find the one which does not belong to that group ?
Correct
11 = 1 1^{3}, 28 = 2 2^{3}, 327 = 3 3^{3}, 416 = 4 4^{2} and 5125 = 5 5^{3}
Except 416, other numbers follow similar pattern.Incorrect
11 = 1 1^{3}, 28 = 2 2^{3}, 327 = 3 3^{3}, 416 = 4 4^{2} and 5125 = 5 5^{3}
Except 416, other numbers follow similar pattern. 
Question 22 of 50
22. Question
A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest is 3% higher, then the invest earned will be 25% more than the interest earned earlier. What is the earlier rate of interest?
Correct
If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) – x/x * 100 = 25
=> x = 12Incorrect
If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) – x/x * 100 = 25
=> x = 12 
Question 23 of 50
23. Question
The area of a triangle will be when a = 1m, b = 2m, c = 3m, a, b, c being lengths of respective sides.
Correct
S = (1 + 2 + 3)/2 = 3
=> No triangle existsIncorrect
S = (1 + 2 + 3)/2 = 3
=> No triangle exists 
Question 24 of 50
24. Question
Find the area of a rhombus whose side is 25 cm and one of the diagonals is 30 cm?
Correct
Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals
bisect at right angles in a rhombus.
BE^{2} + AE^{2} = AB^{2}
25^{2} = 15^{2} + AE^{2} AE = √(625 – 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d_{1}d_{2}
= 1/2 * 40 * 30 = 600 sq.cm.Incorrect
Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals
bisect at right angles in a rhombus.
BE^{2} + AE^{2} = AB^{2}
25^{2} = 15^{2} + AE^{2} AE = √(625 – 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d_{1}d_{2}
= 1/2 * 40 * 30 = 600 sq.cm. 
Question 25 of 50
25. Question
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
Correct
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.Incorrect
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters. 
Question 26 of 50
26. 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 27 of 50
27. Question
Find the one which does not belong to that group ?
Correct
Except Japanian, all others are appropriate usage of citizenship.
Incorrect
Except Japanian, all others are appropriate usage of citizenship.

Question 28 of 50
28. Question
If the perimeter of a rectangular garden is 600 m, its length when its breadth is 100 m is?
Correct
2(l + 100) = 600 => l = 200 m
Incorrect
2(l + 100) = 600 => l = 200 m

Question 29 of 50
29. Question
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.
Correct
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63
Incorrect
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

Question 30 of 50
30. Question
The curved surface of a sphere is 64 π cm^{2}. Find its radius?
Correct
4 πr^{2 }= 64 => r = 4
Incorrect
4 πr^{2 }= 64 => r = 4

Question 31 of 50
31. Question
P is three times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?
Correct
P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.Incorrect
P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days. 
Question 32 of 50
32. Question
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?
Correct
LCM = 1400
1400 – 6 = 1394Incorrect
LCM = 1400
1400 – 6 = 1394 
Question 33 of 50
33. Question
A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is?
Correct
S.I. for 3 years = (12005 – 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 – 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5) = 12%Incorrect
S.I. for 3 years = (12005 – 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 – 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5) = 12% 
Question 34 of 50
34. Question
The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?
Correct
Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.Incorrect
Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm. 
Question 35 of 50
35. Question
Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?
Correct
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? → 28 mIncorrect
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? → 28 m 
Question 36 of 50
36. Question
In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg?
Correct
Incorrect

Question 37 of 50
37. Question
(50 – ?/29)% of 4200 = 3√196
Correct
(50 – ?/29)% of 4200 = 3√196
Incorrect
(50 – ?/29)% of 4200 = 3√196

Question 38 of 50
38. Question
Sachin is younger than Rahul by 4 years. If their ages are in the respective ratio of 7:9, how old is Sachin?
Correct
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.Incorrect
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years. 
Question 39 of 50
39. Question
Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article?
Correct
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000Incorrect
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000 
Question 40 of 50
40. Question
The price of an article has been reduced by 25%. In order to restore the original price the new price must be increased by?
Correct
100
75
——
75 —— 25
100 —— ? => 33 1/3%Incorrect
100
75
——
75 —— 25
100 —— ? => 33 1/3% 
Question 41 of 50
41. Question
Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 9:17. The first number is:
Correct
(3x9):(5x9) = 9:17
x = 12 => 3x = 36Incorrect
(3x9):(5x9) = 9:17
x = 12 => 3x = 36 
Question 42 of 50
42. Question
The sale price of an article including the sales tax is Rs. 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 12%, then the cost price of the article is:
Correct
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500Incorrect
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500 
Question 43 of 50
43. Question
Two persons A and B take a field on rent. A puts on it 21 horses for 3 months and 15 cows for 2 months; B puts 15 cows for 6months and 40 sheep for 7 1/2 months. If one day, 3 horses eat as much as 5 cows and 6 cows as much as 10 sheep, what part of the rent should A pay?
Correct
3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3Incorrect
3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3 
Question 44 of 50
44. Question
What is the difference between the largest and the smallest number written with 7, 7, 0, 7?
Correct
7770
7077
————
693Incorrect
7770
7077
————
693 
Question 45 of 50
45. Question
What distance will be covered by a bus moving at 72 kmph in 30 seconds?
Correct
72 kmph = 72 * 5/18 = 20 mps
D = Speed * time = 20 * 30 = 600 m.Incorrect
72 kmph = 72 * 5/18 = 20 mps
D = Speed * time = 20 * 30 = 600 m. 
Question 46 of 50
46. Question
A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water?
Correct
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8Incorrect
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8 
Question 47 of 50
47. Question
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 2:3 and B:C = 2:5. If the total runs scored by all of them are 75, the runs scored by B are?
Correct
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18Incorrect
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18 
Question 48 of 50
48. Question
Thirty men can do a work in 24 days. In how many days can 20 men can do the work, given that the time spent per day is increased by onethird of the previous time?
Correct
Let the number of hours working per day initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
30 * 24 * x = 20 * d_{2} * (4x)/3 => d_{2} = (30 * 24 * 3)/(24 * 4) = 27 days.Incorrect
Let the number of hours working per day initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
30 * 24 * x = 20 * d_{2} * (4x)/3 => d_{2} = (30 * 24 * 3)/(24 * 4) = 27 days. 
Question 49 of 50
49. Question
What is the difference between the largest number and the least number written with the figures 3, 4, 7, 0, 3?
Correct
74330 Largest
30347 Smallest
————
43983Incorrect
74330 Largest
30347 Smallest
————
43983 
Question 50 of 50
50. Question
A sum of money is to be distributed among A, B, C, D in the proportion of 5:2:4:3. If C gets Rs. 1000 more than D, what is B’s share?
Correct
Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x – 3x = 1000 => x = 1000.
B’s share = Rs. 2x = 2 * 1000 = Rs. 2000.Incorrect
Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x – 3x = 1000 => x = 1000.
B’s share = Rs. 2x = 2 * 1000 = Rs. 2000.