**Quantitative Aptitude Quiz Set-10**

#### I. Missing Data in the table :

**Directions for questions 1 to 3 :**A team of 5 players Arpit, Bimal, Chatur, Dinu and Elan participated in a ‘Freaket’ tournament and played four matches (1 to 4). The following table gives partial information about their individual scores and the total runs scored by the team in each match.

Each column has two values missing. These are the runs scored by the two lowest scorers in that match.None of the two missing values is more than 10% of the total runs scored in that match.

Runs scored by player | Match-1 | Match-2 | Match-3 | Match-4 | |

Runs scored by | Arpit | ? | 100 | ? | 53 |

Bimal | 88 | 65 | ? | 52 | |

Chatur | ? | ? | 110 | ? | |

Dinu | 72 | 75 | 20 | 56 | |

Elan | 60 | ? | 78 | ? | |

Total | 270 | 300 | 240 | 200 |

01.What is the maximum possible percentage contribution of Arpit in the total runs scored in the four matches?

A. 19.7%

B. 19.9%

C. 20.1%

D. 20.2%

02.If the absolute difference between the total runs scored by Arpit and Chatur in the four matches is minimum possible then what is the absolute difference between total runs scored by Bimal and Elan in the four matches?

A. 32

B. 37

C. 27

D. Cannot be determined

03.The players are ranked 1 to 5 on the basis of the total runs scored by them in the four matches, with the highest scorer getting Rank 1. If it is known that no two players scored the same number of total runs, how many players are there whose rank can be exactly determined?

A. 0

B. 1

C. 3

D. 5

II.Directions: Study the following information carefully and answer the questions that follow:

Percentage distribution of teachers who teach six different subjects

**Total number of teachers = 1800**

**04. If two-ninths of the teachers who teach Physics are females, then the number of male Physics teachers is approximately what per cent of the total number of teachers who teach Chemistry?**

A. 43%

B. 65%

C. 70%

D. 57%

**05. What is the total numbers of teachers who teach Chemistry, English and Biology?**

A. 1116

B. 1456

C. 1180

D. 1000

**06. What is difference between the total number of employees teacher who teach English and Physics together and total number of teacher who teach Mathematics and Biology together?**

A. 500

B. 650

C. 342

D. 400

**07. What is the ratio of the teachers who teach mathematics to the number of teachers who teach Hindi?**

A. 7:13

B. 26:7

C. 14:7

D. 13:8

**08. If the percentage of Mathematics teachers is increased by 50% and percentage of Hindi teachers is decreased by 25% , what will be the total number of Mathematics and Hindi teachers together?**

A. 400

B. 430

C. 450

D. 459

**ANSWERS**

**I.**

**01.**

**Option A**

**Explanation:**

Maximum possible runs scored by Arpit in Match-1 = 27

Maximum possible runs scored by Arpit in Match-3 = 19

Maximum possible percentage contribution:

(27+100+19+53)/(270+300+240+200)x100% = 199/1010×100%

= 19.7%

= 19.7%

**02.**

**Option B**

**Explanation:**

Maximum possible total runs scored by Chatur in the four matches = 27 + 30 + 110 + 20 = 187.

In such a case minimum possible total runs scored by Arpit in the four matches = 23 + 100 + 13 + 53 = 189.

Difference = 189 – 187 = 2 (minimum possible) Subsequently total runs scored by Bimal in the four matches = 88 + 65 + 19 + 52 = 224.

Also, total runs scored by Elan in the four matches = 60 + 30 + 78 + 19 = 187

Absolute difference = 224 – 187 = 37

**03.**Option C

**Explanation:**

Individual ranges for total score:

Arpit-> 189-199

Bimal-> 218-224

Chatur-> 182-187

Dinu-> 223

Elan-> 187-188

Least total will be of Chatur (Rank 5)

2nd least will be Elan (Rank 4)

Rank 3 must be of Arpit.

It is not possible to determine the exact ranks of Bimal and Dinu.

**II.**

**04. Option D**

**Explanation :**

If | 2 | th of physics are female teachers, then |

9 |

Male teachers = 1 – | 2 | = | 7 | th of physics |

9 | 9 |

Let's take x% of male physics teacher equal to Chemistry teachers, then

⇒ x% of Chemistry teacher = | 7 | th of physics |

9 |

⇒ x% of (23% of 1800) = | 7 | th of (17% of 1800) |

9 |

⇒ x% of 23 = | 7 | th of 17 |

9 |

⇒ x = | 7 | × | 17 | × 100 = 57.4% ≈ 57% |

9 | 23 |

**05. Option A**

**Explanation :**

(Chemistry + English + Biology)% of Total number of teachers

⇒ (23 + 27 + 12)% of 1800

⇒ 62% of 1800

⇒ (23 + 27 + 12)% of 1800

⇒ 62% of 1800

⇒ | 62 | × 1800 = 1116. |

100 |

**06. Option C****Explanation :**

Req. difference = [(English + Physics) – (Mathematics + Biology)]% of 1800

⇒ [(27 + 17) – (13 + 12)]% of 1800

⇒ (44 – 25)% of 1800

⇒ [(27 + 17) – (13 + 12)]% of 1800

⇒ (44 – 25)% of 1800

⇒ 19% of 1800 = | 19 | × 1800 = 342. |

100 |

**07. Option D****Explanation :**

No. of teachers who teaches Mathematics : No. of teachers who teaches Hindi

13% of 1800 : 8% of 1800

13 : 8.

**08. Option D**

**Explanation :**

Percentage of Mathematics teacher = 13%

After increased by 50% it becomes, 13 + 6.5 = 19.5%

Similarly,

Percentage of Hindi teacher = 8%

After increased by 25% it becomes, 8 – 2 = 6%

Now, percentage of total teachers becomes, 19.5 + 6 = 25.5%

Therefore, 25.5% of 1800

After increased by 50% it becomes, 13 + 6.5 = 19.5%

Similarly,

Percentage of Hindi teacher = 8%

After increased by 25% it becomes, 8 – 2 = 6%

Now, percentage of total teachers becomes, 19.5 + 6 = 25.5%

Therefore, 25.5% of 1800

⇒ | 25.5 | × 1800 = 25.5 × 18 = 459. |

100 |