November 14, 2017

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## Mathematics Test-21

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Question 1 |

**Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centers at P and R respectively, and S is the centre of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to ...**

A | 28 |

B | 20 |

C | 40 |

D | 36 |

Question 2 |

**The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f…. is**

A | v |

B | x |

C | u |

D | w |

Question 2 Explanation:

The number of terms of the series forms the sum of first n natural numbers i.e. n(n + 1)/2. Thus the first 23 letters will account for the first (23 x 24)/2 = 276 terms of the series. The 288th term will be the 24th letter viz. x.

Question 3 |

**Let p and q be the roots of the quadratic equation x2 – (? - 2) x - ? - 1 = 0. What is the minimum possible value of p2 + q2?**

A | 4 |

B | 3 |

C | 5 |

D | 0 |

Question 4 |

**The number of non-negative real roots of 2x – x – 1 = 0 equals**

A | 3 |

B | 0 |

C | 2 |

D | 1 |

Question 5 |

**Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?**note: read a2 as

**a square 2**

A | The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1 |

B | The minimum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m + 1 |

C | The maximum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1 |

D |
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1 |

Question 5 Explanation:

(a + b + c + d)2 = (4m + 1)2
Thus, a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd) = 16m2 + 8m + 1
a2 + b2 + c2 + d2 will have the minimum value if (ab + ac + ad + bc + bd + cd) is the maximum.
This is possible if a = b = c = d = (m + 0.25) ……….since a + b + c + d = 4m + 1
In that case 2((ab + ac + ad + bc + bd + cd) = 12(m + 0.25)2 = 12m2 + 6m + 0.75
Thus, the minimum value of a2 + b2 + c2 + d2
= (16m2 + 8m + 1) – 2(ab + ac + ad + bc + bd + cd) = (16m2 + 8m + 1) – (12m2 + 6m + 0.75)
= 4m2 + 2m + 0.25
Since it is an integer, the actual minimum value = 4m2 + 2m + 1

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