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## Arithmetic: Average Test-4

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Question 1 |

The average score of boys in an examination in a school is 71 and that of the girls is 73. The average score of the school is 71.8. The ratio of the number of boys to that of the girls that appeared in the examination is

1: 2 | |

3: 2 | |

2: 3 | |

4: 2 |

Question 1 Explanation:

Let the total number of boys and girls be B and G respectively.

Total score of boys = 71B

Total score of girls = 73G

Total score of the class = 71.8 (B+G)

71B + 73G = 71.8 (B+G)

$ \displaystyle \Rightarrow \,0.8B=1.2G\Rightarrow \frac{B}{G}=\frac{1.2}{0.8}=\frac{3}{2}$

Total score of boys = 71B

Total score of girls = 73G

Total score of the class = 71.8 (B+G)

71B + 73G = 71.8 (B+G)

$ \displaystyle \Rightarrow \,0.8B=1.2G\Rightarrow \frac{B}{G}=\frac{1.2}{0.8}=\frac{3}{2}$

Question 2 |

The average monthly expenditure of a family was Rs.2200 during the first 3 months; Rs.2250 during the next 4 months and Rs.3120 during the last 5 months of a year. If the total savings during the year were Rs.1260, then the average monthly income was

Rs.2605 | |

Rs.2805 | |

Rs.2705 | |

Rs.2905 |

Question 2 Explanation:

Total expenditure for the year

= [2200 × 3 + 2250 × 4 + 3120 × 5]

= 6600 + 9000 + 15600

= Rs.31200

Total saving = Rs.1260

Total income = expenses + savings = 31200 + 1260 = Rs.32460

Average income = 32460/12 = Rs. 2705

= [2200 × 3 + 2250 × 4 + 3120 × 5]

= 6600 + 9000 + 15600

= Rs.31200

Total saving = Rs.1260

Total income = expenses + savings = 31200 + 1260 = Rs.32460

Average income = 32460/12 = Rs. 2705

Question 3 |

The mean of 1, 7, 5, 3, 4 and 10 is m. The mean of 10, 12, 3, 5, 1, 7 and p is m + 1 and their median is q. Then, q is equal to

4 | |

5 | |

10 | |

12 |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}Given,\,\,\,\,\,m=\frac{1+7+5+3+4+10}{6}\\\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m=5\\and\,\,\,\,\left( m+1 \right)\,\,\,=\frac{10+12+3+5+1+7+p}{7}\\\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{38+p}{7}=m+1\\\therefore \,\,p=42-38=4\\\therefore \,\,Number\,\,\,are\,\,1,\,3,\,4,\,5,\,7,\,10,\,12.\\\therefore \,\,Median\,\,\,q=5\end{array}$

Question 4 |

The difference between the present ages of Arun and Deepak is 14 years. Seven years ago the ratio of their ages was 5:7 respectively. What is Deepak's present age?

49 years | |

42 years | |

63 years | |

none of these |

Question 4 Explanation:

Seven years ago, let Arun’s and Deepak’s age be 5x and 7x years respectively.

Therefore Arun’s present age = (5x + 7) years

Deepak present age = (7x + 7) years

According to the question,

7x + 7 – 5x – 7 = 14

2x = 14

x= 14/2=7

Therefore Deepak’s present age = 7x + 7

= 7 × 7 +7 = 56 years

Therefore Arun’s present age = (5x + 7) years

Deepak present age = (7x + 7) years

According to the question,

7x + 7 – 5x – 7 = 14

2x = 14

x= 14/2=7

Therefore Deepak’s present age = 7x + 7

= 7 × 7 +7 = 56 years

Question 5 |

Four numbers are written in a row. The average of first two numbers is 7, the average of the middle two terms is 2.3 and the average of the last two numbers is 8.4. The average of first number and the last number is

5.9 | |

10.7 | |

13.1 | |

Cannot be determined |

Question 5 Explanation:

Let the numbers be a, b, c, and d respectively

$ \displaystyle \begin{array}{l}a+b=14\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.............\left( i \right)\\b+c=4.6\,\,\,\,\,\,\,\,\,\,\,\,\,..............\left( ii \right)\\c+d=16.8\,\,\,\,\,\,\,\,................\left( iii \right)\\By\,\,\,\,equations\,\,\,\,\left( i \right)-\left( ii \right)+\left( iii \right)\\a+d=14-4.6+16.8=26.2\\\therefore \,\,\frac{a+d}{2}=\frac{26.2}{2}=13.1\end{array}$

$ \displaystyle \begin{array}{l}a+b=14\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.............\left( i \right)\\b+c=4.6\,\,\,\,\,\,\,\,\,\,\,\,\,..............\left( ii \right)\\c+d=16.8\,\,\,\,\,\,\,\,................\left( iii \right)\\By\,\,\,\,equations\,\,\,\,\left( i \right)-\left( ii \right)+\left( iii \right)\\a+d=14-4.6+16.8=26.2\\\therefore \,\,\frac{a+d}{2}=\frac{26.2}{2}=13.1\end{array}$

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