November 14, 2017
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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 ...
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
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.
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?
The number of non-negative real roots of 2x – x – 1 = 0 equals
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
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m + 1
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 - 2m + 1
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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