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.

A) 7 and 8 | B) 6 and 7 |

C) 8 and 7 | D) 6 and 8 |

A) 11 | B) 11.111 |

C) 10.50 | D) 10.85 |

A) 3, 32 | B) 12, 8 |

C) 1, 96 | D) 24, 4 |

A) 301 : 60 | B) 7 : 3 |

C) 39 : 11 | D) 60 : 207 |