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## Arithmetic: Average Test-3

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Question 1 |

The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle number is

11.5 | |

11.4 | |

11.3 | |

11.0 |

Question 1 Explanation:

Sum of 11 numbers = 11 × 10.9

Sum of first 6 numbers = 6 × 10.5

Sum of last 6 numbers = 6 × 11.4

The middle number

= 6 × 10.5 + 6 × 11.4 – 11 × 10.9

= 63 + 68.4 – 119.9

= 131.4 – 119.9 = 11.5

Sum of first 6 numbers = 6 × 10.5

Sum of last 6 numbers = 6 × 11.4

The middle number

= 6 × 10.5 + 6 × 11.4 – 11 × 10.9

= 63 + 68.4 – 119.9

= 131.4 – 119.9 = 11.5

Question 2 |

The average age of 80 boys in a class is 15 years. The average age of a group of 15 boys in the class is 16 years and the average age of another 25 boys in the class is 14 years. What is the average age of the remaining boys in the class?

15.25 years | |

14 years | |

14.75 years | |

Cannot be determined |

Question 2 Explanation:

Let the required average age be x years

$ \displaystyle \begin{array}{l}\therefore \,\,80\times 15=15\times 16+25\times 14+40\times x\\\Rightarrow \,40x=1200-240-350=610\\\therefore \,\,x=\frac{610}{40}=15.25\,\,years\end{array}$

$ \displaystyle \begin{array}{l}\therefore \,\,80\times 15=15\times 16+25\times 14+40\times x\\\Rightarrow \,40x=1200-240-350=610\\\therefore \,\,x=\frac{610}{40}=15.25\,\,years\end{array}$

Question 3 |

The ages of Melwyn and Louis are in the ratio of 7: 10 respectively. After 6 years the ratio of their ages will be 17: 23. What is the difference in their ages?

8 years | |

4 years | |

12 years | |

10 years |

Question 3 Explanation:

Let the present age of Melwyn be 7k

Age after 6 years=7k+6

Then present age of Louis=10k

age after 6 years=10k+6

$ \displaystyle \begin{array}{l}\frac{7k+6}{10k+6}=\frac{17}{23}\\\Rightarrow 170k+102=161k+138\\\Rightarrow 9k=138-102=36\\\Rightarrow k=\frac{36}{9}=4\\\therefore \,\,\,\operatorname{Re}quired\,\,\,\,difference\\=10k-7k=3k\\=3\times 4=12\,\,\,years\end{array}$

Age after 6 years=7k+6

Then present age of Louis=10k

age after 6 years=10k+6

$ \displaystyle \begin{array}{l}\frac{7k+6}{10k+6}=\frac{17}{23}\\\Rightarrow 170k+102=161k+138\\\Rightarrow 9k=138-102=36\\\Rightarrow k=\frac{36}{9}=4\\\therefore \,\,\,\operatorname{Re}quired\,\,\,\,difference\\=10k-7k=3k\\=3\times 4=12\,\,\,years\end{array}$

Question 4 |

The mean temperature of Monday to Wednesday was 37

^{o}C and of Tuesday to Thursday was 34^{o}C. If the temperature on Thursday was 4/5^{th}that of Monday, then what was the temperature on Thursday?36.5 ^{o}C | |

36°C | |

35.5 ^{o}C | |

34°C |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}M+T+W=3\times 37={{111}^{o}}C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( i \right)\,\\T+W+Th=3\times 34={{102}^{o}}C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( ii \right)\,\\Subtracting\,\,\,\,equation\,\,\,\left( ii \right)\,\,from\,\,\,equation\,\,\left( i \right).\\M-Th={{\left( 111-102 \right)}^{o}}C\\\Rightarrow \frac{5}{4}Th-Th={{9}^{o}}C\\\Rightarrow \frac{Th}{4}={{9}^{o}}C\Rightarrow Th=9\times 4={{36}^{o}}C\end{array}$

Question 5 |

The average weight of 29 students in a class is 48 kg. If the weight of the teacher is included, the average weight rises by 500 g. Find the weight of the teacher.

57 kg | |

60 kg | |

65 kg | |

63 kg |

Question 5 Explanation:

Total weight of 29 students = 29 × 48 = 1392 kg

If teacher weight is included, then total weight

= 30 × 48.5 = 1455 kg

Weight of teacher

= 1455 – 1392

= 63 kg

If teacher weight is included, then total weight

= 30 × 48.5 = 1455 kg

Weight of teacher

= 1455 – 1392

= 63 kg

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