# RBI Assistant Mains Mock Test, RBI Assistant mains online test series 2

## 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 2019, Take CAknowledge RBI Assistant Mains Free Quiz Exam 2019. RBI Assistant Mains Exam Free Online Quiz 2019, 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** 2019. 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?

CorrectLet 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.IncorrectLet 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?

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 4 of 50
##### 4. Question

2222.2 + 222.22 + 22.222 = ?

CorrectIncorrect - 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?

CorrectClearly, 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%IncorrectClearly, 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?

CorrectLet 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.IncorrectLet 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?

CorrectLet Mudit’s present age be ‘m’ years.

m + 18 = 3(m – 4)

=> 2m = 30 => m = 15 years.IncorrectLet 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?

CorrectLCM = 120

120) 1000 (8

960

——-

40

1000 + 120 – 40 = 1080IncorrectLCM = 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?

Correct7*5: 5*x = 7:10

x = 10Incorrect7*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?

CorrectLet 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.IncorrectLet 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?

Correct60*5:36*6:75*3

100: 72: 75

100/247 * 7410 = 3000Incorrect60*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?

Correct4.5 4.5

30% 70%

30% —– 4.5

70% ——? => 10.5 – 4.5 = 6 mlIncorrect4.5 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?

Correct25 = E/(1000 – E) * 100

1000 – E = 4E

1000 = 5E => E = 200

1000 – 200 = 800Incorrect25 = 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 ?

CorrectBoxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.

IncorrectBoxing, 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?

CorrectLength = 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 = 117IncorrectLength = 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?

Correct1/2:1/3:1/4 = 6:4:3

3/13*1300 = 300Incorrect1/2:1/3:1/4 = 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 ?

Correct11 = 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.Incorrect11 = 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?

CorrectIf 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 = 12IncorrectIf 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.

CorrectS = (1 + 2 + 3)/2 = 3

=> No triangle existsIncorrectS = (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?

CorrectConsider 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.IncorrectConsider 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?

CorrectB 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.IncorrectB 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

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 27 of 50
##### 27. Question

Find the one which does not belong to that group ?

CorrectExcept Japanian, all others are appropriate usage of citizenship.

IncorrectExcept 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?

Correct2(l + 100) = 600 => l = 200 m

Incorrect2(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.

CorrectRequired probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

IncorrectRequired 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?Correct4 πr

^{2 }= 64 => r = 4Incorrect4 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?

CorrectP = 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.IncorrectP = 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?

CorrectLCM = 1400

1400 – 6 = 1394IncorrectLCM = 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?

CorrectS.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%IncorrectS.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?

CorrectArea of the square = s * s = 5(125 * 64)

=> s = 25 * 8 = 200 cm

Perimeter of the square = 4 * 200 = 800 cm.IncorrectArea 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?

CorrectSpeed Ratio = 1:7/6 = 6:7

Time Ratio = 7:6

1 ——– 7

4 ——— ? → 28 mIncorrectSpeed 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?

CorrectIncorrect - 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?

CorrectLet 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.IncorrectLet 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?

CorrectLet 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 = 15000IncorrectLet 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?

Correct100

75

——-

75 ——- 25

100 —— ? => 33 1/3%Incorrect100

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(3x-9):(5x-9) = 9:17

x = 12 => 3x = 36Incorrect(3x-9):(5x-9) = 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:

Correct110% of S.P. = 616

S.P. = (616 * 100)/110 = Rs. 560

C.P = (110 * 560)/112 = Rs. 500Incorrect110% 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?

Correct3h = 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 = 1/3Incorrect3h = 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?

Correct7770

7077

————-

693Incorrect7770

7077

————-

693 - Question 45 of 50
##### 45. Question

What distance will be covered by a bus moving at 72 kmph in 30 seconds?

Correct72 kmph = 72 * 5/18 = 20 mps

D = Speed * time = 20 * 30 = 600 m.Incorrect72 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?

Correct30 — 3 DS = 10

? —- 1

18 —- 3 US = 6

? —- 1 M = ?

M = (10 + 6)/2 = 8Incorrect30 — 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?

CorrectA:B = 2:3

B:C = 2:5

A:B:C = 4:6:15

6/25 * 75 = 18IncorrectA: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 one-third of the previous time?

CorrectLet 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.IncorrectLet 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?

Correct74330 Largest

30347 Smallest

————-

43983Incorrect74330 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?

CorrectLet 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.IncorrectLet 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.