Important for :

1

D

Sum of the heights of the first six students = 170 × 6 = 1020 cmSum of the heights of the last eight students = 175 × 8 = 1400 cm

Sum of the heights of the total 16 students = 180 × 16 = 2880

Sum of the height of the left 2 students = 2880 – 1020 – 1400 = 460

Average height of the left 2 students = | 460 | = 230 cm |

2 |

Hence, option D is correct.

2

3

B

Man days for which food is available = 200 × 50 = 10000Available food is enough for 1 student for 10000 days

Food used by 200 students in 10 days = 200 × 10 man days of food = 2000

Man days of food left = 10000 – 2000 = 8000 man days of food

Total number of students now = 200 + 50 = 250

Remaining food can be used for 250 students for

= | 8000 | days = 32 days |

250 |

Hence, option B is correct.

4

D

Total weight of friends P, Q, R, S and T = (x + 6) x 5 = 5 (x + 6) kg

So, total weight of P, Q and S = 5 (x + 6) – 2 (x – 6) = (3x + 42) = 3 (x + 14)

Weight of U = (x + 6 – 5) × 6 – 5 (x + 6) = 6 (x + 1) – 5 (x + 6) = (x – 24) kg

According to the question,

[3 (x + 14) + (x – 24)] = 94.5 × 4

4x + 18 = 378

4x = 360; x = 90

Hence, option D is correct.

5

E

Let the average weight of all five = 100kSo, weight of A = 90k, B = 112k and C = 94k

Let the weight of D = d and that of E = e

90k + 112k + 94k + d + e | = 100k |

5 |

d + e = 204k

d : e = 6 : 11 → d = | 6 | × 204 = 72k → e = 132k |

17 |

Difference = 132k – 72k = 60k

60k = 75

So, k = | 75 | = 1.25 |

60 |

Average weight of all the five persons = 100 × 1.25 = 125kg

Hence ,option E is correct.

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