The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero ?

A) 0 | B) -1 |

C) 1 | D) none of these |

Explanation:

Average of 20 numbers = 0.

Sum of 20 numbers = (0 * 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

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Subject: Average - Quantitative Aptitude - Arithmetic Ability

A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ?

A) 4991 | B) 5467 |

C) 5987 | D) 6453 |

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs. 4991.

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Subject: Average - Quantitative Aptitude - Arithmetic AbilityA batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning?

Let the average after 7th inning = x

Then average after 16th inning = x - 3

16(x-3)+87 = 17x

x = 87 - 48 = 39

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Subject: Average - Quantitative Aptitude - Arithmetic Ability

The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers :

A) 4 | B) 8 |

C) 12 | D) 16 |

Explanation:

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57 = 8

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Subject: Average - Quantitative Aptitude - Arithmetic Ability

After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ?

A) 12 | B) 13 |

C) 14 | D) 15 |

Explanation:

i) Let the ages of the five members at present be a, b, c, d & e years.

And the age of the new member be f years.

ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1)

iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years

So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2)

==> a + b + c + d + e = 5x + 15

==> a + b + c + d = 5x + 15 - e ------ (3)

iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above,

The new average is: (5x + 15 - e + f)/5

Equating this to the average age of x years, 3 yrs, ago as in (2) above,

(5x + 15 - e + f)/5 = x

==> (5x + 15 - e + f) = 5x

Solving e - f = 15 years.

Thus the difference of ages between replaced and new member = 15 years.

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Subject: Average - Quantitative Aptitude - Arithmetic Ability