The table shows the speed of the stream in different days in the week and some data are missing.

Day | Speed of stream |
---|---|

Monday | 8 |

Tuesday | – |

Wednesday | 12 |

Thursday | – |

Friday | 4 |

**1) If the time taken by boat to travel upstream on Friday is equal to the time taken by it to travel downstream on Monday and the speed of boat in still water on Monday is 15 kmph then find the speed of boat in still water on Friday?**

- a) 14.5kmph

b) 15kmph

c) 16.5kmph

d) 12kmph

e) None of these

**Answer**: Option (a)

**Explanation**:

Day | Upstream distance | Downstream distance | Speed of stream |
---|---|---|---|

Monday | 576 | 552 | 8 |

Tuesday | 324 | 432 | – |

Wednesday | 360 | 672 | 12 |

Thursday | 288 | 408 | – |

Friday | 252 | 336 | – |

(252)/(b – 4) = (552)/(15 + 8)

2(b – 4) = 21

b = 14.5kmph

**2) If the speed of boat in still water on Tuesday was 10kmph and the speed of boat in still water on Wednesday was 80% of Tuesday and time taken to travel downstream on Tuesday is 18 hrs more than the time taken by it to travel upstream on Wednesday, then find the speed of stream on Tuesday?**

- a) 2kmph

b) 6kmph

c) 4kmph

d) 8kmph

e) None of these

**Answer**: Option (a)

**Explanation**: Speed of boat on Tuesday = 10kmph

Speed of boat on Wednesday = 10 * 80/100 = 8kmph

Downstream distance on Tuesday = 18/100 * 2400 = 432km

Upstream distance on Wednesday = 20/100 * 1800 = 360

Speed of stream on Wednesday = 12kmph

(432/(10 + w)) – (360/(8 + 12)) = 18

432/10 + w = 36

10 + w = 12

W = 2kmph

**3) On Thursday the speed of stream is 33(1/3)% of the speed of the stream on Wednesday. If the time taken to cover upstream distance on Thursday is 20% more than the downstream distance, what was the speed of boat
on that day?**

- a) 15(3/7) kmph

b) 18(3/7) kmph

c) 12(3/7) kmph

d) 11(3/7) kmph

e) None of these

**Answer**: Option (a)

**Explanation**: Speed of the stream on Thursday = 12 * 100/300 = 4kmph

120/100 (Downstream journey time) = upstream journey time

120/100 (408/(b+4)) = 288/(b-4)

6/5(102/(b + 4)) = 72/(b – 4)

17b – 68 = 10b + 40

7b = 108

b = 108/7 = 15 3/7

**4) The speed of boat on Monday is 12 kmph more than the speed of stream on that day and time for upstream journey on Friday is same as time for upstream journey on Monday. What was the downstream journey tim (approximate) on Friday?**

- a) 12 hours

b) 16 hours

c) 25 hours

d) 24 hours

e) 18 hours

**Answer**: Option (c)

**Explanation**: Speed of stream on Monday = 8kmph

Speed of boat on Monday = 8 + 12 = 20kmph

Upstream journey on Monday = 576/12 = 48 hours

Stream speed on Friday = 4kmph

Upstream journey time on Friday = 48 hours

252/b-4 = 48

b = 9.25 kmph

Downstream journey time on Friday = 336/(4 + 9.25)

= 25 hours (Approximately)

**5) On Thursday ratio of speed of boat in still water in going upstream to downstream is 5:3 and also difference in speed of boat in still water in going upstream and downstream is 4 kmph. If the time taken by boat to cover upstream and downstream is same on Thursday, find the speed of stream?**

- a) 3(11/29) kmph

b) 2(11/29) kmph

c) 4(12/19) kmph

d) 5(13/19) kmph

e) None of these

**Answer**: Option (a)

**Explanation**: Difference 5x – 3x = 4

2x = 4

X = 2

Speed of boat in still water in upstream = 5(2) = 10kmph

Speed of boat in still water in downstream = 3(2) = 6kmph

Distance of downstream on Thursday = 408

Distance of upstream on Thursday = 288

408/(6 + w) = 288/10-w

W = 3(11/29) kmph

**Directions (6-10)**: Study the following information carefully and answer the given questions:

There are three Cities City A, City B and City C. Certain Number of Projects are allocated to these cities Under Scheme P, Scheme Q and Scheme R in 2018.

**Scheme P**: The Number of Projects in City B is 2/3rd of the Number of projects in city B under Scheme Q. The Number of Projects in City C is 4/5th of the Number of projects in city C under Scheme R. The Total Number of Projects in city A and City C is equal under Scheme P.

**Scheme Q**: The Total Number of Projects under Scheme Q is 375. The Number of Projects in City B is 75less than that of number of projects in City A and C together. The Number of Projects in City A is 5/6th of the number of projects in City A under Scheme R.

**Scheme R**: The Number of Projects under Scheme R is 4/5th of the number of Projects under Scheme Q. The Number of Projects in City A is equal to the number of Projects in City B under Scheme Q. The Number of Projects in city C under Scheme Q and Scheme R is Equal.

City/Schemes | Scheme P | Scheme Q | Scheme R |
---|---|---|---|

City A | 80 | 125 | 150 |

City B | 100 | 150 | 50 |

City C | 80 | 100 | 100 |

Total | 260 | 375 | 300 |

**Scheme Q**: The Total Number of Projects under Scheme Q = 375

The Number of Projects in City B is 75 less than that of number of projects in City A and C together.

Consider the number of projects in City A and C is X, then X + (X – 75) = 375, on solving we get, X = 225, then number of projects in City B = 150 projects.

**Scheme P**: The Number of Projects in City B is 2/3rd of the Number of projects in city B under Scheme Q.

**Scheme P**: Number of Projects in City B = 2/3rd * 150 = 100 projects.

City/Schemes | Scheme P | Scheme Q | Scheme R |
---|---|---|---|

City A | |||

City B | 100 | 150 | |

City C | |||

Total | 375 |

**Scheme Q**: The Number of Projects in City A is 5/6th of the number of projects in City A under Scheme R.

City/Schemes | Scheme P | Scheme Q | Scheme R |
---|---|---|---|

City A | \((\frac{5}{6})^th\) | ||

City B | 100 | 150 | |

City C | |||

Total | 375 |

**Scheme R**: The Number of Projects under Scheme R is 4/5th of the number of Projects under Scheme Q.

Therefore, the total number of projects in Scheme R is 300.

The Number of Projects in City A is equal to the number of Projects in City B under Scheme Q.

Therefore number of projects in city A under scheme R is 150 projects.

City/Schemes | Scheme P | Scheme Q | Scheme R |
---|---|---|---|

City A | 125 | 150 | |

City B | 100 | 150 | |

City C | 100 | ||

Total | 375 | 300 |

**Scheme R**: The Number of Projects in city C under Scheme Q and Scheme R is Equal. So there are 100 projects in city C under scheme R. And also for city B under scheme R should have 50 projects.

**Scheme P**: The Number of Projects in City C is 4/5th of the Number of projects in city C under Scheme R. Number of Projects in City C= 4/5 * 100 = 80

The Total Number of Projects in city A and City C is equal under Scheme P.

Therefore number of projects in City A is 80 projects. So the final table becomes

City/Schemes | Scheme P | Scheme Q | Scheme R |
---|---|---|---|

City A | 80 | 125 | 150 |

City B | 100 | 150 | 50 |

City C | 80 | 100 | 100 |

Total | 260 | 375 | 300 |

- a) City C under Scheme P and City A under scheme R together

b) City B under Scheme P and City C under scheme P together

c) City A under Scheme Q and City B under scheme R together

d) City C under Scheme Q and City A under scheme R together

e) None of those given as options

**Answer**: Option (d)

**Explanation**: In 2019, The projects allocated for city B under Scheme Q is increases by 30%

= 150 * 130/100

= 195

The project allocated for city C under scheme R is decreases by 45% = 100 * 55/100 = 55

Thus the total number of projects in 2019 for city B under scheme Q and City C under Scheme R = 250

From the given options 250 projects obtained only from option D.

**7) What is the average number of projects under Scheme Q and Scheme R for City C and City B together?**

- a) 75

b) 80

c) 100

d) 150

e) None of those given as option

**Answer**: Option (c)

**Explanation**: The average number of projects under Scheme Q and Scheme R for City C and City B together = (150 + 100 + 50 + 100)/4 = 400/4 = 100

**8) If the cost of each projects is 15 crore for scheme P, and 19Crore for Scheme Q, and 17 crore for Scheme R, then how much fund allocated for City C under these three schemes (In Million)?**

- a) 48000

b) 3800

c) 4800

d) 54000

e) None of those given as option

**Answer**: Option (a)

**Explanation**: Fund allocated for City C under these three schemes = (15 * 80) + (100 * 19) + (100 * 17) = 4800 crores = 48000 Millions

**9) What is the total number of projects under Scheme P and Scheme Q together?**

- a) 560

b) 675

c) 720

d) None of those given as option

e) 635

**Answer**: Option (e)

**Explanation**: The total number of projects under Scheme P and Scheme Q together = 260 + 375 = 635 Projects

**10) In City X, The number of projects allocated under scheme P is 20% more than that of number of projects allocated for city C under scheme P, and the number of projects allocated under scheme Q is 40% less than that of number of projects allocated for city B under scheme Q and the total number of projects in city X is 300, then how many projects are allocated for City X under scheme R?**

- a) 154

b) 100

c) 94

d) 114

e) None of those given as option

**Answer**: Option (d)

**Explanation**: The number of projects allocated under scheme P is 20% more than that of number of projects allocated for city C

under scheme P = 80 * 120/100 = 96

The number of projects allocated under scheme Q is 40% less than that of number of projects allocated for city B

under scheme Q = 150 * 60/100 = 90

The total number of projects in city X is 300

Then number of projects allocated for City X under scheme R = 300 – (96 + 90) = 114 projects

**Directions (11-15)**: Study the following information carefully and answer the given questions:

The line graph shows that total number of manufacturing parts produced by five different companies in the year 2016 and 2017.

The table shows that the percentage of Defective parts and percentage of non- defective parts that are not passed

the quality test in the year 2016.

The table shows that the percentage of Defective parts and percentage of non- defective parts that are not passed the quality test in the year 2016.

COMPANIES | % of Defective parts |
% of non- defective parts that are not passed the quality test |
---|---|---|

A | 15% | 50% |

B | 30% | 25% |

C | 25% | 10% |

D | 20% | 25% |

E | 28% | 50% |

**11) If both the years the percentage of defective parts from company D is equal, and in 2017 only 10% of nondefective parts are not passed the quality test for company D, then what is the total number of non-defective parts that are passed the quality test in both years from company D?**

- a) 1656

b) 306

c) 1556

d) 1666

e) None of those given as option

**Answer**: Option (c)

**Explanation**:

Total number of manufacturing parts from company D in 2017 = 1360

From the given question there 20% parts that are defective in 2017 from company D = 1360 *20 /100 = 272

Non defective parts = 1360 – 272 = 1088

10% of non-defective parts are not passed the quality test for company D = 10% of 1088 = 108

Non-defective parts that are passed the quality test for company D = 1088 – 108 = 980

The total number of non-defective parts that are passed the quality test in both years from company D = 576 + 980 = 1556 parts

- a) 3: 7

b) 53: 52

c) 4 : 9

d) 19 : 17

e) Cannot be determined

**Answer**: Option (e)

**Explanation**: There is no such information to find out the defective and non-defective parts from Company B and Company C in 2017. So we are not able to find out the Non-defective parts that are passed by quality test from company B and Company C in 2017. So cannot be determined is the answer.

**13) In 2017, the percentage of defective parts from company A, B, and E are 20%, 25% and 15% respectively, then the total number of Non-defective parts that are passed from quality test from company C and D in 2016 is approximately what percentage more or less than that of the total number of Non-defective parts from A, B and E together in 2017?**

- a) 38% More

b) 50% less

c) 28 % More

d) 40% less

e) None of those given as option

**Answer**: Option (d)

**Explanation**:

COMPANIES | Total number of Manufacturing Parts |
% of Defective parts |
Defective parts |
Non- Defective parts |
---|---|---|---|---|

A | 960 | 20% | 192 | 768 |

B | 1280 | 25% | 320 | 960 |

E | 1000 | 15% | 150 | 850 |

The total number of Non-defective parts that are passed from quality test from company C and D in 2016 = 972 + 576

= 1548

Total number of Non-defective parts from A, B and E together in 2017 = 768 + 960 + 850 = 2578

Required percentage = [(2578 – 1548) /2578] * 100 = 39.99% = 40% less

**14) What is the average number of non-defective parts that are not passed the quality test from all the company except D in 2016?**

- a) 254

b) 248

c) 220

d) None of those given as option

e) 252

**Answer**: Option (e)

**Explanation**:

The average numbers of non-defective parts that are not passed the quality test from all the company except D in 2016 = (272 + 196 + 108 + 432)/4 = 252

- a) 664

b) 786

c) 544

d) 684

e) None of those given as option

**Answer**: Option (a)

**Explanation**:

The difference between the non- defective parts that are passed the quality test from Company B and C together in 2016 = 588 + 972 = 1560

The non- defective parts that are not passed the quality test from Company A, D and E together in 2016 = 272 + 192 + 432 = 896

Difference = 1560 – 896 = 664

The following table shows the number of classes taken by each tutors in different days and the total amount given to the professor per class for the certain course also given

Tutors | Number of classes taken on Monday, Tuesday and Wednesday by each |
Number of classes taken on Thursday and Friday by each |
Salary per class (In Rs.) |
---|---|---|---|

A | 2 | 0 | 5000 |

B | 3 | – | 8000 |

C | 1 | 3 | 6000 |

D | 2 | 2 | 4000 |

Note:

Saturday and Sunday are holidays

“-“ is missing value, we have to find the value according to the question.

**16) Find the ratio of the number of lectures taken by A to that of the number of lectures taken by D in a week?**

- a) 3 : 5

b) 4 : 7

c) 5 : 9

d) 11 : 13

e) None of these

**Answer**: Option (a)

**Explanation**: The number of lectures taken by A in a week

⇒ (2*3) + (0*2) = 6

The number of lectures taken by D in a week

⇒ (2*3) + (2*2) = 6 + 4 = 10

Required ratio = 6 : 10 = 3 : 5

**17) Find the earnings made by C if he teaches for 6 weeks?**

- a) Rs. 306000

b) Rs. 282000

c) Rs. 348000

d) Rs. 324000

e) None of these

**Answer**: Option (d)

**Explanation**: The number of lectures taken by C in a week

⇒ (1*3) + (3*2) = 3 + 6 = 9

The number of lectures taken by C in 6 weeks

⇒ 9*6 = 54

The earnings made by C if he teaches for 6 weeks

⇒ 54*6000 = Rs. 324000

**18) Find the difference between the earnings made by C for 3 weeks to that of the earnings made by D for 2 weeks?**

- a) Rs. 74000

b) Rs. 66000

c) Rs. 82000

d) Rs. 70000

e) None of these

**Answer**: Option (c)

**Explanation**: The number of lectures taken by C in a week

⇒ (1*3) + (3*2) = 3 + 6 = 9

The earnings made by C for 3 weeks

⇒ 9*3*6000 = Rs. 162000

The number of lectures taken by D in a week

⇒ (2*3) + (2*2) = 6 + 4 = 10

The earnings made by D for 2 weeks

⇒ 10*2*4000 = Rs. 80000

Required difference = 162000 – 80000 = Rs. 82000

**19) If B takes 2 classes each on Thursday and Friday, then how much he can earn in a week?**

- a) Rs. 116000

b) Rs. 104000

c) Rs. 95000

d) Rs. 88000

e) None of these

**Answer**: Option (b)

**Explanation**: B takes 2 classes each on Thursday and Friday

The number of classes taken by B in a week

= (3*3) + (2*2) = 9 + 4 = 13

B’s earning in a week

⇒ 13*8000 = Rs. 104000

**20) If the amount of Rs. 3.12 lakhs was given to the tutor B for 3 weeks, then find the number of class/ classes taken by the tutor B in Thursday and Friday each?**

- a) 2

b) 4

c) 3

d) 1

e) None of these

**Answer**: Option (a)

**Explanation**: Total number of classes taken by the professor B in 3 weeks

⇒ 312000/8000 = 39 classes

According to the question,

(9*3) + 3 * (Tuesday + Thursday) = 39

3 * (Tuesday + Thursday) = 39 – 27

3*(Tuesday + Thursday) = 12

Tuesday + Thursday = 4 classes

The each 2 classes taken by the tutor B in Thursday and Friday

**Directions (21-25)**: Bar chart given below shows different discount rates are given for different products of different shops, for some products discount rate is missing which you have to find out according to data given in different questions if they are necessary. Answer the following questions with the help given Bar chart.

Selling price is same for a particular product (excluding cooking oil and sugar) for all shops. (MP= market price, CP=Cost price, SP= selling price)

- a) Rs3450

b) Rs Rs3600

c) Rs4270

d) Rs3300

e) None of these

**Answer**: Option (b)

**Explanation**: (SP/75 × 100 + SP/90 × 100 + SP/80 100)/3 = 3990

Solving this we will get SP=2160

Then MP of soap by shop B = 2160/90 × 100 = Rs. 3600

**22) Difference between MP of Olive oil of Shop A and shop B is Rs 504 then find MP of Olive oil for Shop C?**

- a) Rs5814

b) Rs5678

c) Rs4678

d) Rs6234

e) None of these

**Answer**: Option (a)

**Explanation**: SP/85 × 100 – SP/95 × 100 = 504

SP = (504 × 17 × 19)/40

= 4069.8

MP by shop C= (SP/70) × 100 = Rs. 5814

**23) If MP of cooking oil is same for all shops and Average SP of cooking oil for Shop A and shop B is Rs 3728 and average SP of cooking oil for shop B and shop C is Rs 3368, then find SP of cooking oil by shop C?**

- a) Rs2256

b) Rs2860

c) Rs1890

d) Rs2450

e) Rs2160

**Answer**: Option (e)

**Explanation**: (MP × 84)/100 + (MP × X)/100 = 3728 —(1)

(MP × X)/100 + (MP × 72)/100 = 3368 —(2)

Subtracting equation 1 from equation 2 we get

(MP × 12)/100 = 360

Thus MP = Rs 3000

Then SP of shop is 72% of MP which is Rs 2160

**24) If difference between MP and SP for rice in shop B is Rs 741 find average MP of rice of shop A and shop C?**

- a) Rs6420

b) Rs5360

c) Rs5440

d) Rs6640

e) None of these

**Answer**: Option (c)

**Explanation**: If discount is Rs 741 in shop B then SP of rice is = (741/15) × 85 = 4199

MP of rice by shop A = (4199/65) × 100 = 6460

MP of rice by shop C = (4199/95) × 100 = 4420

Average of MP of these two shops is = Rs 5440

**25) If market price is equal for all shops for sugar. Ratio of discount for sugar of shop A and B is 1/3, difference between SP for sugar of shop A and C is Rs 780, if difference SP of shop A is 680 more than shop B, then find SP of sugar by shop C?**

- a) Rs2428

b) Rs2256

c) Rs2786

d) Rs2280

e) None of these

**Answer**: Option (d)

**Explanation**: Ratio of discount for sugar by shop B is 30%

According to given question discount by shop A will be 10%

Thus we have mp × 90/100 – mp × 70/100 = 680

After solving this we have MP= Rs 3400

And difference between SP of shop A and shop C is 780

(i.e) 3400 × 90/100 – sp of shop C = 780

SP by shop C is = 3060 – 780 = Rs2280

Total number who attended the workshop = Number of Literates + Number of illiterates

Day | No. of literates (males+ Females) |
Overall ratio (illiterate : literates) (out of those who attended) |
Number of males (literates + illiterates) (out of those who attended) |
---|---|---|---|

Monday | 420 | 5:6 | 250 |

Tuesday | 350 | 3:5 | 240 |

Wednesday | 320 | 5:4 | 320 |

Thursday | 300 | 6:5 | 300 |

Friday | 420 | 2:3 | 320 |

**1) The total number of people (literates + illiterates) who attended the workshop on Monday was what % more than those who attended on Friday?**

- a) 12%

b) 10%

c) 15%

d) 18%

e) None of these

**Answer**: Option (b)

**Explanation**:

Total number of people who attended the workshop on Monday

= 420 * (11/6) = 770

Total number of people who attended the workshop on Friday

= 420 * (5/3) = 700

Required % more = [(770-700)*100]/700 = 10 %

**2) On Friday, if 192 illiterate males attended the workshop, what was the number of literate females who attended the workshop on that day?**

- a) 292

b) 300

c) 275

d) 280

e) None of these

**Answer**: Option (a)

**Explanation**:

Number of Literate males on Friday = 320 – 192 = 128

Literate female on Friday = 420 – 128 = 292

**3) On Saturday, if the number of illiterates (males + females) increased by 40 % and that of literates (males + females) reduced by 20 %, as compared to Tuesday, what was the difference between the number of literates and illiterates who attended the workshop on Saturday?**

- a) 12

b) 14

c) 15

d) 18

e) None of these

**Answer**: Option (b)

**Explanation**:

Number of illiterate (males + female) on Tuesday = 350 * (3/5) =210

Now, Required difference = (210 * 140/100) – (350 * 80/100)

= 294 – 280 = 14

**4) What is the average number of illiterates (males + females) who attended the workshop on Monday, Wednesday and Thursday?**

- a) 390

b) 400

c) 300

d) 370

e) None of these

**Answer**: Option (d)

**Explanation**:

Required average = {(420 * 5/6 + 320 * 5/4 + 300 * 6/5)}/3

= (350 + 400 + 360)/3 = 370

**5) What is the ratio of the total number of males (Literates + Illiterates) who attended the workshop on Tuesday and Friday together to that of females (literates and Illiterates) who attended the workshop on the same days together?**

- a) 5:9

b) 9:7

c) 7:9

d) 1:3

e) None of these

**Answer**: Option (c)

**Explanation**:

Number of Illiterate (male + female) on Tuesday = 350 * 3/5 = 210

Number of illiterate (males + females) on Friday = 320 * (5/4) = 400

Number of females (literate + Illiterate) on Tuesday = (350 + 210 – 240) = 320

Similarly on Friday = (320 + 400 – 320) = 400

Required ratio = (240 + 320): (320 + 400) = 7: 9

**Directions (6-10)**: Study the information carefully to answer the following questions.

Data regarding number of employees working in various departments in Company A and B in the year 2018. Both Companies have six departments namely Production, HR, Finance, R&D, Marketing and Accounts. The total number of employees in company A is 9000. In Company A, number of employees in production, HR and finance together is 60 % of the total number of employees. The number of employees in R&D, Marketing and Accounts were 1300, 1440 and 860 respectively. The number of employees in Production department was 25 % more than that of finance department. In company B, the number of employees in Marketing was 900 and they constituted 12 % of the total

number of employees. Also the number of employees in Marketing was 40 % less than that of HR department. The number of employees in production from company B was 10 % less than the same department from Company A. The Number of employees in accounts is 500. Number of employees in finance and R&D department is same. Total Number of employees in finance and R&D together were double the total Number of employees in Marketing and accounts together.

Number of employees in R&D (in A) = 1300

Number of employees in marketing (in A) = 1440

Number of employees in accounts (in A) = 860

The number of employees in Marketing was 20 % less than that of HR department.

So, 1440 = (100 – 20) % of HR department

Number of employees in HR (in A) = 1440 * (100/80) = 1800

We have, Production + HR + Finance = 5400

Production + 1800 + Finance = 5400

Production + Finance = 3600 ……………….. (1)

Let the number of employees in finance be X.

Given that the number of employees in the Production department was 25 % more than that of the finance department.

So, the number of employees in Production in A = (125/100) X

Now equation (1) becomes,

(125/100)X + X = 3600

X = 1600.

Number of employees in finance (in A) = 1600

Number of employees in production (in A) = (125/100) * 1600 = 2000

Number of employees in marketing (in B)= 900

12% of total Number of employees in B = 900

Total Number of employees in B= (100/12) * 900 = 7500

Given that, the number of employees in Marketing was 40 % less than that of HR department

So, 900 = (100 – 40)% of Number of employees in HR (in B)

Number of employees in HR (in B) = 1500

Given that, the number of employees in production from company B was 10 % less than the same department from

Company A.

So, Number of employees in Production (in B) = (100 – 10) % 0f 2000 = (90/100) * 2000 = 1800

Number of employees in accounts (in B) = 500

Also given that, the total Number of employees in finance and R&D together were double the total Number of

employees in Marketing and accounts together

So, (Finance + R&D (in B) = 2 * (Marketing + accounts)(in B)

(Finance + Finance) (in B) = 2*(900 + 500) since, finance = R&D

2 * Finance = 2 * 1400

Number of employees in Finance (in B)= 1400

Number of employees in R&D (in B)= 1400

Subject | Company A (9000) | Company B ( 7500) |
---|---|---|

R&D | 1300 | 1400 |

Marketing | 1440 | 900 |

Accounts | 860 | 500 |

Production | 2000 | 1800 |

HR | 1800 | 1500 |

Finance | 1600 | 1400 |

- a) 700

b) 200

c) 400

d) 600

e) 900

**Answer**: Option (e)

**Explanation**:

Total Number of employees in Marketing and accounts together (in A) = 1440 + 860 = 2300

Total Number of employees in Marketing and accounts together (in B) = 900 + 500 = 1400

Required difference = 2300 – 1400 = 900

**7) 3/4th of the number of R&D employees in Company A was female. If the number of female R&D employees in Company A is less than that of Company B by 175, what is the number of male R&D employee in Company B?**

- a) 600

b) 400

c) 500

d) 100

e) 800

**Answer**: Option (a)

**Explanation**:

The number of female R&D employee in Company A = (3/4 * 1300) = 975

The number of female R&D employee in company B = 975 – 175 = 800

The number of male R&D employees in company B = 1400 – 800 = 600

**8) What is the respective ratio between the total number of employees in finance and production together in Company A and that in the same courses together in company B?**

- a) 1:9

b) 7:3

c) 4:9

d) 9:8

e) 3:2

**Answer**: Option (d)

**Explanation**:

Total number of employees in finance and production together in A = 1600 + 2000 = 3600

Total number of employees in finance and production together in B = 1400 + 1800 = 3200

Required ratio = 3600 : 3200 = 9:8

**9) Number of HR employees in Company B is what percent less than that in Company A?**

- a) 10/4%

b) 50/3%

c) 26/7%

d) 12/6%

e) 43/6%

**Answer**: Option (b)

**Explanation**:

Number of HR employees in Company B = 1500

Number of HR employees in Company A = 1800

Required percentage = {(1800 – 1500) * 100}/1800 = 50/3 %

**10) Total number of employees in Company A, is what percent to that of in Company B?**

- a) 120%

b) 160%

c) 216%

d) 567%

e) 230%

**Answer**: Option (b)

**Explanation**:

Total number of employees in Company A = 9000

Total number of employees in Company B = 7500

Required percentage = {9000 * 100}/7500 = 120%

**Directions (11-15)**: Study the information carefully to answer the following questions.

In the following table there are five colleges in which total student and percentage of arts students and the ratio of civil and mechanical engineering students are given. Calculate the missing data if necessary:

College | Total number of students |
Percentage of arts students |
Ratio of civil to mechanical engineering students |
---|---|---|---|

A | 1500 | 35% | – |

B | – | 40% | – |

C | – | – | 7:3 |

D | 2100 | – | 3:2 |

E | 1750 | – | – |

**11) If the ratio of boys and girls in college A for civil engineering students are 4:1 and the civil engineering students are 50% more than the mechanical engineering students. Then find the difference of boys and girls in civil department?**

- a) 120

b) 430

c) 351

d) 320

e) 270

**Answer**: Option (c)

**Explanation**:

Total number of engineering student in college A = (100 – 35) % of 1500

= (65/100) * 1500 = 975

Given that the civil engineering students are 50% more than the mechanical engineering students.

Let x be the number of mechanical engineering students in college A.

Then number of civil engineering students in college A = x + 50% of x = 150% of x

So, x + (150/100)x = 975

(250/100)x = 975

x = 390

So, the no of mechanical engineering students = 390

The number of civil engineering students = 585

Given that the ratio of boys and girls in college A for civil engineering students = 4:1

So, the difference of boys and girls in mechanical students = (3/5) * 585 = 351

**12) If the total engineering student in college E is 525 and students in civil department are 12 ½% less than the students in mechanical department and the engineering student in college D is 735. Then find ratio of civil engineering student in college D and E?**

- a) 445:247

b) 441:243

c) 453:247

d) 441:247

e) 441:249

**Answer**: Option (d)

**Explanation**:

Given that, the total engineering student in college E = 525

Let the number of mechanical engineering students in college E be X.

Then the number of civil engineering student in college E = X- 12 ½% of X = 87 ½% of X

Therefore, X + 87 ½% of X = 525

187 ½% of X = 525

X = 280

So, the number of mechanical engineering students in college E = 280

And the number of civil engineering students in college E = 525 – 280 = 245

The number of engineering student in college D = 735

The number of civil engineering in college D= (3/5)*735 = 441

Required ratio = 441:247

**13) If arts student in college A is 375 less than arts student in college B. Then the total student in college D is what percent more or less than the total students in college B?**

- a) 5 1/7%

b) 8 1/7%

c) 2 1/7%

d) 7 1/7%

e) 1 1/7%

**Answer**: Option (d)

**Explanation**:

Number of arts student in college A = (35/100) * 1500 = 525

Number of arts student in college B = 525 + 375 = 900

Let the total number of students in college B be X.

We know that, there are 40% of students in college B are arts.

So, (40/100) * X = 900

X=2250

(i.e) Total number of students in college B = 2250

Required percent = [(2250 – 2100)/2100] * 100 = 7 1/7%

**14) If total student in college C is 1380 and total arts student in college C is equal to the total students in engineering. And the ratio of boys and girls in college C in arts is 5:1. If 20% of boys are transferred to college E, then find the total students in college E?**

- a) 1546

b) 1456

c) 1585

d) 1865

e) 1687

**Answer**: Option (d)

**Explanation**:

Given that, total student in college C = 1380 and total arts student in college C = total engineering students in college C.

So, total arts student + total engineering student = 1380

Total arts student + total arts student = 1380

Therefore, total arts student = 690

Number of arts boys in college C = (5/6) * 690 = 575

If 20% of boys are transferred to college E, then the total students in college E = 1750 + (20/100) * 575 = 1865

**15) Suppose there is another college X in which arts students are 2/5th of arts student in college A and engineering student in college X is 40% of total students of college D then what is the total students in X?**

- a) 1050

b) 1205

c) 1640

d) 1550

e) 4520

**Answer**: Option (a)

**Explanation**:

Total students in X = [(2/5) * (35/100) * 1500] + [(40/100) * 2100]

= 210 + 840

= 1050

**Directions (16-20)**: Given below is table which shows the ratio of efficiency of both Anand and Abinav on different days and total time taken by Anand and Abinav to complete the work 1 if they complete whole work with the efficiency of different days.

Days | Efficiency of Anand and Abinav |
Time taken by both to complete work hours |
---|---|---|

Jan 1 | 3:2 | 3 |

Jan 2 | 3:2 | 4 |

Jan 3 | 7:9 | 6 |

Jan 4 | 8:9 | 5 |

Jan 5 | 5:4 | 8 |

There is also the line graph which shows the time taken by Abinav to complete work 2 if it completes whole work with efficiency of different days.

- a) 115/2 hours

b) 111/13 hours

c) 108/19 hours

d) 110/19 hours

e) 110/13 hours

**Answer**: Option (c)

**Explanation**:

Let Anand and Abinav can do 3𝑥 and 2𝑥 unit of work 1 in one hour respectively.

So, total work 1 done by both = (3𝑥 + 2𝑥) * 4

= 20𝑥

Anand alone will complete work 1 = 20𝑥/3𝑥 = 20/3 hours

Abinav alone will complete work 1 = 20x/2x = 10 hours

Ratio of efficiency of Anand and Ajay = 5: 3

Ratio of time taken by Anand and Ajay = 3: 5

Ajay alone will complete work 1 = 20/(3×3) × 5 hours

= 100/9 hours

Let total time taken in completing work 1 is 𝑦

So, 2/(20/3) + (𝑦−2)/10 + (𝑦−2)/(100/9) = 1

(𝑦−2)/10 + 9(𝑦−2)/100 = 7/10

10𝑦 − 20 + 9𝑦 − 18 = 70

𝑦 = 108/19 hours

**17) If a part of work 2 completed by 4 women in 5 hours equals to the part of work 2 done by Abinav on Jan 3 in 7 hours and ratio of efficiency of a women and a children to complete work 2 is 5 : 3 then in what time work 2 will be completed by 3 children.**

- a) 120/9 hours

b) 200/9 hours

c) 100/11hours

d) 210/11 hours

e) 150/21 hours

**Answer**: Option (b)

**Explanation**:

Part of work 2 done by Abinav on Jan 3 in 7 hours = 7/14 = 1/2

This part of work done by 4 women in 5 hours

So whole work will be completed by 4 women in = 10 hours

One women will complete it in = 40 hours 3 children will complete it in = 40 × 5/3 × 3 = 200/9 hours

**18) 𝒙 can complete a work in (𝒏–𝒎) hours while 𝒚 can complete the same work in (𝒏+𝒎) hours where 𝒎 is the time taken by Anand to complete work 2 on Jan 2 and 𝒏 is time taken by Anand to complete work 2 on Jan 5. Find the time in which x and y together can complete the work**

- a) 3/2 hours

b) 7/4 hours

c) 7/5 hours

d) 8/3 hours

e) 9/5 hours

**Answer**: Option (b)

**Explanation**:

Ratio of efficiency Anand and Abinav on Jan 2 = 3: 2

Let Anand and Abinav does 3𝑥 and 2𝑥 work in one hour

And Abinav completes work 2 in 9 hours

So, total work = 9 × 2𝑥 = 18𝑥

Anand will complete work 2 in 18𝑥/3𝑥

= 6 hours

So, 𝑚 = 6

Similarly 𝑛 = 10 × 4𝑥/5𝑥 = 8

Total x and y will complete the work in = (8 −6)(8 + 6)/(8 − 6) + (8 + 6)

**19) Anand and Abinav started to complete work 1, alternatively starting from Anand on first hour on Jan 1. Then time taken by Anand and Abinav in completing 80% of work 1, alternatively on Jan 1 is what percent more or less than time taken by Anand and Abinav together to complete work 2 together on Jan 5.**

- a) 3%

b) 5%

c) 8%

d) 15%

e) 6%

**Answer**: Option (b)

**Explanation**:

Let Anand and Abinav can do 3𝑥 and 2𝑥 work in one hour on Jan 1

Then 80% of total work 1 = 4/5(3𝑥 + 2𝑥) × 3 = 12𝑥

In 4 hours 10𝑥 work 1 is completed working alternatively and remaining 2𝑥 is complete by Anand on 5th hour

So total time =(4 + 2𝑥/3𝑥)hours = 14/3 hours

Ratio of efficiency on Jan 5 is 5: 4

Ratio of time taken to complete work will be 4: 5

But Abinav completes work 2 in 10 hours on Jan 5

So, Anand will complete work 2 in 8 hours on Jan 5

Together they will complete work 2 in = 8 × 10/18 = 40/9hours

Required percentage=(14/3 − 40/9)/(40/9) × 100 = ((42 − 40)/9)/(40/9) × 100 = 2/40 × 100 = 5%

**20) If Abinav with another person Ajay works on work 2 on Jan 5 for 2 hours than 80% of work 2 is completed then, time taken by Ajay alone to finish work 2 is what percent to time taken by Abinav to finish work 1 with efficiency of Jan 5**

a) 500/27%

b) 400/13%

c) 300/17%

d) 400/21%

e) 500/21%

**Answer**: Option (a)

**Explanation**:

Let Ajay complete work 2 in 𝑥 hours

According to question, 2/10 + 2/𝑥 = 4/5

2/𝑥 = 4/5 − 1/5

2/𝑥 = 3/5

𝑥 = 10/3

Time taken by Abinav to finish work 1 on Jan 5 = (5 + 4) * 8/4

= 18 hours

Required percentage = 10/(3 × 18) × 100 = 500/27%

**Directions (21–25)**: Study the following information carefully and answer the questions give below: (2 marks)

The following table shows the ratio of time taken by pipes to fill the tank.

A:F | 3:4 |

B:G | 3:2 |

C:H | 5:6 |

D:I | 4:5 |

E:J | 9:10 |

**21) Pipe A and Pipe B opened simultaneously for 4 minutes, then closed and then pipe F and pipe Q are opened for 2 minutes, then closed. Find the time taken by pipe G to fill the remaining part of the tank.**

- a) 7 minutes

b) 169/36 minutes

c) 5 minutes

d) 178/39 minutes

e) None of these

**Answer**: Option (b)

**Explanation**:

Part of the tank filled by pipe A in one minute = 1/12

Part of the tank filled by pipe B in one minute = 1/15

Time taken by pipe F to fill the tank = 4/3 x 12 = 16 minutes

Part of the tank filled by pipe F in one minute = 1/16

Time taken by pipe G to fill the tank = 2/3 x 15 = 10 minutes

Part of the tank filled by pipe G in one minute = 1/10

Part of the tank emptied by pipe Q in one minute = 1/18

Let required time = t minutes

According to the question

4/15 + 4/16 + 2/16 – 2/18 + t/10 = 1

⇒ t/10 = 1 – 4/15 – ¼ – 1/8 + 1/9

⇒ t/10 = (360 – 96 – 90 – 45 + 40)/360

⇒ t/10 = 169/360

⇒ t = 169/360 x 10

⇒ t = 169/36 minutes

**22) Efficiency of pipe K is twice the efficiency of pipe A and efficiency of pipe L is 1.5 times the efficiency of pipe S. Pipe C and pipe K are opened simultaneously for 3 minutes and then closed. Find the time taken by pipe L and pipe R together to empty the filled part of the tank.**

- a) 4 minutes

b) 47/12 minutes

c) 39/10 minutes

d) 5 minutes

e) None of these

**Answer**: Option (c)

**Explanation**:

Part of the tank filled by pipe A in one minute = 1/12

Part of the tank filled by pipe K in one minute = 2/12 = 1/6

Part of the tank filled by pipe C in one minute = 1/10

Part of the tank emptied by pipe S in one minute = 1/24

Part of the tank emptied by pipe L in one minute = 1.5/24 = 1/16

Part of the tank emptied by pipe R in one minute = 1/16

Part of the tank filled by pipe C and pipe K in 3 minutes = 3/10 + 3/16

= (24 + 15)/80

= 39/80

Let the required time taken = t minutes

t/16 + t/16 = 39/80

⇒ 2t/16 = 39/80

⇒ t = 39/80 x 16/2

⇒ t = 39/10 minutes

**23) Time taken by pipe M to fill the tank is 20% more than the time taken by pipe I to fill the tank and efficiency of pipe N is twice the efficiency of pipe J. Time taken by pipe M and pipe N to fill the tank is what percent of the time taken by pipe D and pipe E together to fill the tank.**

- a) 67.67%

b) 74.44%

c) 98.48%

d) 81.14%

e) 83.33%

**Answer**: Option (c)

**Explanation**:

Time taken by pipe I to fill the tank = 5/4 x 8 = 10 minutes

Part of the tank filled by pipe I in one minute = 1/10

Time taken by pipe M to fill the tank = 10 x 120/100 = 12 minutes

Part of the tank filled by pipe M in one minute = 1/12

Time taken by pipe J to fill the tank = 10/9 x 18 = 20 minutes

Part of the tank filled by pipe J in one minute = 1/20

Part of the tank filled by pipe N in one minute = 2/20 = 1/10

Part of the tank filled by pipe D in one minute = 1/8

Part of the tank filled by pipe E in one minute = 1/18

Let the time taken by pipe M and pipe N to fill the tank = t minutes

And the time taken by pipe D and pipe E to fill the tank = k minutes

t/12 + t/10 = 1

⇒ (5t + 6t)/60 = 1

⇒ 11t/60 = 1

⇒ t = 60/11 minutes

And

k/8 + k/18 = 1

(9k + 4k)/72 = 1

⇒ k = 72/13 minutes

Required percentage = (60/11)/(72/13) x 100

= 60/11 x 13/72 x 100

= 98.48%

**24) Find the respective ratio of time taken by pipe B, pipe G and pipe P together to fill the tank and time taken by pipe E, pipe H and pipe S together to fill the tank.**

- a) 4:5

b) 5:6

c) 6:7

d) 3:4

e) None of these

**Answer**: Option (b)

**Explanation**:

Part of the tank filled by pipe B in one minute = 1/15

Time taken by pipe G to fill the tank = 2/3 x 15 = 10 minutes

Part of the tank filled by pipe G in one minute = 1/10

Part of the tank emptied by pipe P in one minute = 1/20

Part of the tank filled by pipe E in one minute = 1/18

Time taken by pipe H to fill the tank = 6/5 x 10 = 12 minutes

Part of the tank filled by pipe H in one minute = 1/12

Part of the tank emptied by pipe S in one minute = 1/24

Let the time taken by pipe B, pipe g and pipe P together to fill the tank = t minutes

And the time taken by pipe E, pipe H and pipe S together to fill the tank = k minutes

t/15 + t/10 – t/20 = 1

⇒ (4t + 6t – 3t)/60 = 1

⇒ 7t/60 = 1

⇒ t = 60/7 minutes

And

k/18 + k/12 – k/24 = 1

⇒ (4k + 6k – 3k)/72 = 1

⇒ 7k/72 = 1

⇒ k = 72/7 minutes

Required ratio = 60/7: 72/7 = 5:6

**25) Pipe A, pipe C and pipe E are opened simultaneously for 4 minutes then closed and pipe P and pipe S are opened for 2 minutes then closed. Find the time taken by pipe G and pipe J to fill the remaining part of the tank.**

- a) 27/13 days

b) 41/27 days

c) 31/11 days

d) 51/29 days

e) None of these

**Answer**: Option (b)

**Explanation**:

Part of the tank filled by pipe A in one minute = 1/12

Part of the tank filled by pipe C in one minute = 1/10

Part of the tank filled by pipe E in one minute = 1/18

Part of the tank emptied by pipe P in one minute = 1/20

Part of the tank emptied by pipe S in one minute = 1/24

Time taken by pipe G to fill the tank = 2/3 x 15 = 10 minutes

Part of the tank filled by pipe G in one minute = 1/10

Time taken by pipe J to fill the tank = 10/9 x 18 = 20 minutes

Part of the tank filled by pipe J in one minute = 1/20

Let the required time taken = t minutes

4/12 + 4/10 + 4/18 – 2/20 – 2/24 + t/10 + t/20 = 1

⇒ 1/3 + 2/5 + 2/9 – 1/10 – 1/12 + (2t + t)/20 = 1

⇒ (60 + 72 + 40 – 18 – 15)/180 + 3t/20 = 1

⇒ 139/180 + 3t/20 = 1

⇒ 3t/20 = 1 – 139/180

⇒ 3t/20 = (180 – 139)/180

⇒ 3t/20 = 41/180

⇒ t = 41/180 x 20/3

⇒ t = 41/27 days